Toaz - Info Tewari Organic Chemistry Vi PR

ORGANIC CHEMISTRY
A Modern Approach
Volume-I
ABOUT THE AUTHOR
Nimai Tewari is a retired associate Professor, Department of Chemistry,

Katwa College (affiliated to The University of Burdwan), West Bengal.
A PhD in Organic Chemistry from Calcutta University, he has taught
the subject for a period of more than three decades. He has published
various research papers in national and international journals.
Apart from Organic Chemistry—A Modern Approach, Dr Tewari has
authored three more books on Organic Chemistry for undergraduate
and postgraduate students. His research interest includes Organic
Synthesis and Heterocyclic Chemistry.
ORGANIC CHEMISTRY
A Modern Approach
Volume-I
Nimai Tewari
Associate Professor (Retired)
Department of Chemistry
Katwa College
(affiliated to The University of Burdwan)
West Bengal
McGraw Hill Education (India) Private Limited

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Dedicated to
my daughter Aindrila
and
my son-in-law Ritam Mukherjee
whose forbearance, constant
encouragement and inspiration
made this work possible
CONTENTS
Preface xv
1. Structure, Bonding and Properties of Organic Molecules 1.1–1.313

1.1 Hybridization, Bond Lengths, Bond Strengths or Bond Dissociation
Enthalpies, Bond Angles and VSEPR Theory 1.2
1.1.1 Hybridization 1.2
1.1.2 Bond Length 1.9
1.1.3 Bond Dissociation Enthalpy or Bond Dissociation Energy 1.9
1.1.4 Bond Angle 1.10
1.1.5 VESPR Theory and Molecular Geometry 1.11
Solved Problems 1.13
Study Problems 1.21
1.2 Electronegativity and Bond Polarity 1.22
Study Problems 1.42
1.3 Molecular Formula as a Clue to Structure: Double Bond Equivalent (DBE) or
Index of Hydrogen Deficiency (IHD) 1.44
Study Problems 1.46
1.4 Acids and Bases 1.46
1.4.1 Brönsted-Lowry Theory of Acids and Bases 1.46
1.4.2 Lewis Acid-Base Theory 1.52
Study Problems 1.57
1.5 Inductive and Electrometric Effects 1.60
1.5.1 Inductive Effect 1.60
1.5.2 Field Effect 1.67
1.5.3 Electromeric Effect 1.67
Study Problems 1.73
viii Contents
1.6 Resonance and Resonance Effect or Mesomeric Effect 1.75

1.6.1 Resonance Energy 1.77
1.6.2 Rules for Writing Meaningful Resonance Structures 1.78
1.6.3 Relative Contribution of Resonance Structures towards Resonance
Hybrid 1.78
1.6.4 Resonance or Mesomeric Effect 1.80
1.6.5 Isovalent and Heterovalent Resonance 1.81
1.6.6 Effect of Resonance on the Properties of Molecules 1.81
Study Problems 1.112
1.7 Hyperconjugation 1.117
1.7.1 Sacrificial and Isovalent Hyperconjugation 1.119
1.7.2 Effect of Hyperconjugation on the Physical and Chemical Properties of
Molecules and on the Stabilities of Intermediates 1.119
1.8 Steric Effect 1.125
1.8.1 Properties of Molecules Influenced by Steric Effect 1.126
1.8.2 Proton Sponges 1.132
1.8.3 Face Strain or F-Strain 1.133
1.8.4 Steric Acceleration and Steric Retardation 1.134
1.8.5 Bredt’s Rule 1.135
1.9 Intermolecular Forces 1.147
1.9.1 Dipole–Dipole Interactions 1.147
1.9.2 van der Waals Forces 1.149
1.9.3 Effect of Intermolecular Forces on Different Properties of
Compounds 1.150
1.10 Reactive Intermediates 1.170
1.10.1 Nonclassical Carbocation 1.177
1.10.2 Carbonium Ion and Carbenium Ion or Carbocation 1.178
1.11 Tautomerism 1.218
1.11.1 Mechanism of Keto-enol Tautomerism 1.221
1.11.2 Difference between Resonance and Tautomerism 1.222
1.11.3 Position of the Tautomeric Equilibrium 1.223
1.11.4 Ring-chain Tautomerism 1.228
1.11.5 Valence Tautomerism 1.228
Contents ix
1.12 Aromaticity 1.240

1.12.1 Criteria for Aromaticity 1.240
1.12.2 Antiaromatic Compounds 1.242
1.12.3 Nonaromatic Compounds 1.242
1.12.4 Classification of Compounds as Aromatic, Antiaromatic and Nonaromatic
by Comparing their Stabilities with that of the Corresponding Open Chain
Compounds 1.243
1.12.5 Modern Definition of Aromaticity 1.243
1.12.6 Molecular Orbital Energy Diagram of Some Ions and Molecules 1.244
1.12.7 Use of Inscribed Polygon Method to Determine the Relative Energies
of p Molecular Orbitals for Cyclic Planar and Completely Conjugated
Compounds and to Classify Them as Aromatic and Antiaromatic 1.245
1.12.8 Classification of Some Molecules and Ions as Aromatic, Antiaromatic and
Nonaromatic 1.247
1.12.9 Homoaromatic Compounds 1.251
1.12.10 Some Chemical and Physical Consequences of Aromaticity 1.251
1.13 Thermodynamics, Energy Diagrams and Kinetics of Organic Reactions 1.276
1.13.1 Thermodynamics 1.276
1.13.2 Energy Diagram 1.278
1.13.3 Kinetics 1.281
1.13.4 Catalysis 1.283
1.13.5 Hammond Postulate 1.284
1.13.6 Kinetic Control versus Thermodynamic Control of a Chemical Reaction 1.286
1.14 Methods of Determining Mechanisms of Reactions 1.298
1.14.1 Kinetic Isotope Effects 1.305
2. Principles of Stereochemistry 2.1–2.230

Introduction 2.2
2.1 Projection Formulas of Stereoisomers 2.3
2.1.1 Flying-Wedge Projection Formula 2.3
2.1.2 Fischer Projection Formula 2.4
2.1.3 Sawhorse Projection Formula 2.10
2.1.4 Newman Projection Formula 2.11
2.1.5 Interconversion of Projection Formulas 2.12
Study Problems 2.25
2.2 Symmetry Elements 2.29
2.2.1 Simple Axis of Symmetry or Rotational Axis of Symmetry (Cn) 2.29
2.2.2 Plane of Symmetry (s) 2.33
x Contents
2.2.3 Centre of Symmetry (i) 2.36

2.2.4 Alternating Axis of Symmetry (Sn) 2.37
2.2.5 Symmetric, Asymmetric and Dissymmetric Molecules 2.38
Study Problems 2.56
2.3 Isomerism: Constitutional Isomers and Stereoisomers 2.58
2.3.1 Constitutional Isomers 2.58
2.3.2 Stereoisomers 2.59
2.3.3 Enantiomers and Diastereoisomers 2.59
Study Problems 2.79
2.4 Molecular Chirality 2.84
2.4.1 Chiral and Achiral Molecules 2.84
2.4.2 Source of Chirality 2.85
2.4.3 Stereocentre or Stereogenic Centre 2.88
2.4.4 Meso Compounds 2.89
2.5 Configuration and Configurational Nomenclature 2.108
2.5.1 D, L-System of Configurational Designation 2.109
2.5.2 Specification of Configuration: The R, S-System 2.111
2.5.3 Erythro and Threo Nomenclature of Compounds with Two Adjacent Chiral
Centres 2.124
2.5.4 The E-Z System of Designating Alkene Diastereoisomers 2.125
2.5.5 R, S and E, Z Assignment in the Same Molecule (Geometric
Enantiomerism) 2.127
2.5.6 Syn-anti Nomenclature for Aldols 2.128
2.5.7 Number of Stereoisomers for Compounds with Chiral Centres 2.129
2.5.8 Chirotopic and Achirotopic Atom in a Molecule 2.131
2.5.9 Prostereoisomerism and Topicity 2.133
2.6 Optical Activity of Chiral Compounds 2.182
2.6.1 Plane-polarized Light 2.182
2.6.2 Optical Activity 2.182
2.6.3 Polarimeter 2.183
2.6.4 Specific Rotation 2.184
2.6.5 The Necessary and Sufficient Condition (the ultimate criterion) for Optical
Activity 2.186
2.6.6 Racemic Modification 2.187
2.6.7 Enantiomeric Excess (EE) or Optical Purity (EE) 2.187
2.6.8 Racemization 2.188
2.6.9 Resolution of Racemic Modification 2.192
Contents xi
2.7 Conformation of Acyclic Organic Molecules 2.211

2.7.1 Conformations of Ethane (CH3—CH3) 2.213
2.7.2 Conformations of Propane (CH3CH2CH3) 2.214
2.7.3 Conformations of Butane (CH3CH2CH2CH3) 2.215
2.7.4 Conformations of Chloroethane (CH3CH2Cl) 2.218
2.7.5 Conformations of 1,2-dichloroethane (ClCH2CH2Cl) 2.218
2.7.6 Conformations of Some Typical Acyclic Molecules 2.219
2.7.7 Invertomerism 2.220
3. Nucleophilic Substitution Reactions at Saturated Carbon Atom 3.1–3.180

Introduction 3.2
3.1 The SN2 reaction 3.3
3.1.1 Example of SN2 Reaction 3.3
3.1.2 Kinetics of SN2 Reaction 3.3
3.1.3 Mechanism of SN2 Reaction 3.3
3.1.4 Stereochemistry of SN2 Reaction 3.5
3.1.5 Evidence in Favour of SN2 Mechanism 3.6
3.1.6 Factors Influencing SN2 Reaction Rate or SN2 Reactivity 3.7
Study Problems 3.74
3.2 The SN1 Reaction 3.86
3.2.1 Example of SN1 Reaction 3.86
3.2.2 Kinetics of SN1 Reaction 3.86
3.2.3 Mechanism of SN1 Reaction 3.86
3.2.4 Stereochemistry of SN1 Reaction 3.87
3.2.5 Evidence in Favour of SN1 Mechanism 3.90
3.2.6 Factors Influencing SN1 Reaction Rate or SN1 Reactivity 3.91
3.2.7 Carbocation Rearrangements in SN1 Reactions 3.105
3.2.8 Comparison of the SN2 and SN1 Reactions 3.108
3.2.9 Summary of Reactivity of Alkyl Halide in Necleophilic Substitution
Reactions 3.108
3.2.10 Factors Favouring SN1 and SN2 Reactions 3.108
3.2.11 SNi and SNi¢ Mechanisms 3.109
3.2.12 SN1¢ Mechanism 3.112
3.2.13 Isotope Effects and Salt Effects (Methods used to Distinguish between SN1
and SN2 Type Reactions) 3.113
3.3 Neighbouring Group Participation (NGP) 3.146
3.3.1 Definition 3.146
3.3.2 Mechanism of NGP 3.146
3.3.3 Example of NGP 3.146
3.3.4 Anchimeric Assistance 3.147
xii Contents
3.3.5 Evidence for Participation by a Neighbouring Group 3.148

3.3.6 Various Cases of Neighbouring Group Participation 3.148
4. Elimination Reactions 4.1–4.128

Introduction 4.2
4.1 The E2 Reaction 4.4
4.1.1 Example of E2 Reaction 4.4
4.1.2 Kinetics of E2 Reaction 4.4
4.1.3 Mechanism of E2 Reaction 4.4
4.1.4 Stereochemistry of E2 Reaction 4.5
4.1.5 Evidence in Favour of the E2 Mechanism 4.32
4.1.6 Factors Influencing E2 Reaction Rate or E2 Reactivity 4.33
4.1.7 Factors that Govern the Proportions of E2 and SN2 Reactions 4.34
4.1.8 Regioselectivity in b-elimination Reactions (Orientation of p Bond in the
Product Alkene) 4.37
4.1.9 Hofmann Exhaustive Methylation or Hofmann Degradation 4.43
4.1.10 Fragmentations 4.45
4.1.11 Summary of the E2 Reaction 4.47
Study Problems 4.74
4.2 The E1 Reaction 4.86
4.2.1 Example of E1 Reaction 4.86
4.2.2 Kinetics of E1 Reaction 4.87
4.2.3 Mechanism of E1 Reaction 4.87
4.2.4 Stereochemistry of E1 Reaction 4.88
4.2.5 Evidence in Favour of the E1 Mechanism 4.90
4.2.6 Factors Influencing E1 Reaction Rate or E1 Reactivity 4.90
4.2.7 Regioselectivity of E1 Reactions 4.90
4.2.8 Rearrangement of the Carbocation Intermediate Involved in an E1
Reaction 4.93
4.2.9 Acid-Catalyzed Dehydration of Alcohols 4.95
4.2.10 Dehydration using POCl3 and Pyridine 4.99
4.2.11 Factors Influencing the Extent of E1 and E2 Reactions 4.100
4.2.12 Factors that Govern the Proportions of E1 and SN1 Reactions 4.101
4.3 The E1CB Reaction 4.118
4.3.1 Example of E1cB Reaction 4.118
4.3.2 Kinetics of E1cB Reaction 4.118
4.3.3 Mechanism of E1cB Reaction 4.118
4.3.4 The Nature of the Substrate 4.119
Contents xiii
4.3.5 To Distinguish between E1cB and E2 Mechanisms 4.119

4.4 a- or 1,1-Elimination 4.126
4.4.1 Example of a- or 1,1-Elimination Reaction 4.126
4.4.2 Kinetics of a- or 1,1-Elimination Reaction 4.126
4.4.3 Mechanism of a- or 1,1-Elimination Reaction 4.126
4.4.4 Structure of the Substrate Involved in a-Elimination 4.127
Index I.1–I.6
PREFACE
In the course of teaching Organic Chemistry to undergraduate students, I have been

constantly feeling the need of a concise volume that deals with their important topics on
the subject matter. Students have often expressed their difficulty caused by the absence of
such a compact book. My present effort is to meet this long-felt need and the book has been
designed primarily for the students who have taken a basic course in Organic Chemistry
at the undergraduate level.
The book covers some important topics on Organic Chemistry in four chapters. It starts
with a chapter on Structure, Bonding and Properties of Organic Molecules, highlighting
concepts like hybridization, electronegativity and bond polarity, acids and bases, inductive
effect, resonance, steric effect, intermolecular forces, etc.
Stereochemistry is an essential part of the organic chemistry courses. Chapter 2 deals with
the Principles of Stereochemistry. Chapter 3 covers Nucleophilic Substitution Reactions
at Saturated Carbon Atom and Chapter 4 covers the Elimination Reactions.
By following a modern methodology of learning, the book presents a large number of
reactions with discussions supported with mechanistic explanations and diagrams,
wherever needed. Organic chemistry is best learned by solving problems. Each article of
every chapter concludes with a number of Solved as well as Study Problems to provide an
opportunity to the students for self-evaluation.
I believe that this book will be of great utility for the students who have taken a basic
course of Organic Chemistry in B.Sc. (Chemistry) Hons., besides being equally effective
for advance students of Chemistry because of the in-depth discussion in a reader-
friendly language. The book will also be useful for the students preparing for competitive
examinations like NET, SLET, etc.
Acknowledgements
I offer my sincere gratitude to Mr. Kaushik Bellani, MD, McGraw Hill Education (India)
Pvt. Ltd. and Mrs. Vibha Mahajan, Director, Science & Engineering Portfolio for successful
xvi Preface
publication of this book. I also wish to thank Mr. Sumen Sen, Mr. Amit Chatterjee and
Mr. P L Pandita for taking keen interest in publishing this book. I am grateful to all of
them.
I also owe a debt of gratitude to my colleagues for constructive suggestions and to my
students who encouraged me constantly. I appreciate the interest and enthusiasm shown
by my wife Mrs. Dali Tewari and my daughter Andrila Tewari (Mukherjee) during the
long period of preparation of the manuscript.
Valuable suggestions from the readers for the improvement of the book will be most
welcome.
Nimai Tewari
1
CHAPTER
STRUCTURE, BONDING
AND PROPERTIES OF
ORGANIC MOLECULES
Chapter Outline
1.1 Hybridization, Bond Lengths, Bond 1.6.5 Isovalent and Heterovalent
Strengths or Bond Dissociation Resonance
Enthalpies, Bond Angles and VSEPR 1.6.6 Effect of Resonance on the
Theory Properties of Molecules
1.1.1 Hybridization 1.7 Hyperconjugation
1.1.2 Bond Length 1.7.1 Sacrificial and Isovalent
1.1.3 Bond Dissociation Enthalpy or Hyperconjugation
Bond Dissociation Energy 1.7.2 Effect of Hyperconjugation on the
1.1.4 Bond Angle Physical and Chemical Properties
1.1.5 VESPR Theory and Molecular of Molecules and on the Stabilities
Geometry of Intermediates
1.2 Electronegativity and Bond Polarity 1.8 Steric Effect
1.3 Molecular Formula as a clue to 1.8.1 Properties of Molecules Influenced
structure: Double Bond Equivalent by Steric Effect
(DBE) or Index of Hydrogen 1.8.2 Proton Sponges
Deficiency (IHD) 1.8.3 Face Strain or F-Strain
1.4 Acids and Bases 1.8.4 Steric Acceleration and Steric
1.4.1 Brönsted-Lowry Theory of Acids Retardation
and Bases 1.8.5 Bredt’s Rule
1.4.2 Lewis Acid-Base Theory 1.9 Intermolecular Forces
1.5 Inductive and Electrometric Effects 1.9.1 Dipole–Dipole Interactions
1.5.1 Inductive Effect 1.9.2 van der Waals Forces
1.5.2 Field Effect 1.9.3 Effect of Intermolecular Forces on
1.5.3 Electromeric Effect Different Properties of Compounds
1.6 Resonance and Resonance Effect or 1.10 Reactive Intermediates
Mesomeric Effect 1.10.1 Nonclassical Carbocation
1.6.1 Resonance Energy 1.10.2 Carbonium Ion and Carbenium Ion
1.6.2 Rules for Writing Meaningful or Carbocation
Resonance Structures 1.11 Tautomerism
1.6.3 Relative Contribution of Resonance 1.11.1 Mechanism of Keto-enol
Structures towards Resonance Tautomerism
Hybrid 1.11.2 Difference between Resonance and
1.6.4 Resonance or Mesomeric Effect Tautomerism
1.2 Organic Chemistry—A Modern Approach
1.11.3 Position of the Tautomeric Compounds and to Classify Them

Equilibrium as Aromatic and Antiaromatic
1.11.4 Ring-chain Tautomerism 1.12.8 Classification of Some Molecules
1.11.5 Valence Tautomerism and Ions as Aromatic, Antiaromatic
1.12 Aromaticity and Nonaromatic
1.12.1 Criteria for Aromaticity 1.12.9 Homoaromatic Compounds
1.12.2 Antiaromatic Compounds 1.12.10 Some Chemical and Physical
1.12.3 Nonaromatic Compounds Consequences of Aromaticity
1.12.4 Classification of Compounds 1.13 Thermodynamics, Energy Diagrams
as Aromatic, Antiaromatic and and Kinetics of Organic Reactions
Nonaromatic by Comparing 1.13.1 Thermodynamics
their Stabilities with that of 1.13.2 Energy Diagram
the Corresponding Open Chain 1.13.3 Kinetics
Compounds 1.13.4 Catalysis
1.12.5 Modern Definition of Aromaticity 1.13.5 Hammond Postulate
1.12.6 Molecular Orbital Energy Diagram 1.13.6 Kinetic Control versus
of Some Ions and Molecules Thermodynamic Control of a
1.12.7 Use of Inscribed Polygon Method to Chemical Reaction
Determine the Relative Energies 1.14 Methods of Determining Mechanisms
of p Molecular Orbitals for Cyclic of Reactions
Planar and Completely Conjugated 1.14.1 Kinetic Isotope Effects
1.1 HYBRIDIZATION, BOND LENGTHS, BOND STRENGTHS OR BOND

DISSOCIATION ENTHALPIES, BOND ANGLES AND VSEPR THEORY
1.1.1 Hybridization
The process of intermixing of atomic orbitals of the same atom having slightly different
energies so as to redistribute their energies and give new orbitals of equal energies and
identical shapes and sizes is called hybridization.
The number of hybrid orbitals is equal to the number of pure atomic orbitals reshuffled.
The hybrid orbitals are more effective in forming stable bonds as compared to the pure
atomic orbitals and this is because they can undergo more effective overlapping. The
relative overlap of orbitals decreases in the following order: sp > sp2 > sp3 >> p. The hybrid
orbitals are oriented in space in some preferred directions to have a stable arrangement in
which there is minimum repulsion among themselves. Therefore, the type of hybridization
governs the geometrical shapes of the molecules.
Hybrid orbitals Hybridization Geometry Bond angle
2 s + p = sp Linear 180°
3 s + p + p = sp2 Trigonal planar 120°
4 s + p + p + p = sp3 Tetrahedral 109.5°
Structure, Bonding and Proper es of Organic Molecules 1.3
In fact, the number of groups surrounding a particular atom determines its geometry. A
group is either an atom or a lone pair of electrons. Any atom surrounding by two, three
and four groups are linear, trigonal planar and tetrahedral, respectively, and they have
bond angles of 180°, 120° and 109.5°, respectively.
The hybridization of a C, O or N atom can be determined by the number of p bonds it
forms. If it forms no p bond, one p bond and two p bonds, it is sp3, sp2 and sp hybridized,
respectively. All single bonds are s bonds. A double bond consists of one s bond and one p
bond. A triple bond consists of one s bond and two p bonds.
“Hybrid orbital number” method can be used to determine the hybridized state of an atom
within a molecule.
Hybrid orbital number = (number of s bonds) + (number of unshared pair of electrons)
If the hybrid orbital number is 2, the atom is sp hybridized, if it is 3, the atom is sp2
hybridized and if it is 4, the atom is sp3 hybridized.
The electronic configuration of carbon atom in its ground state is 1s2 2s2 2px1 2py1 2pz°,
i.e., one odd electron is present in each of 2px and 2py orbitals of carbon atom. The number
of odd electrons present in the valence shell of an atom generally gives the measure of
covalency of that atom. So, the valency of carbon should be two. However, the valency of
carbon in almost all organic compounds is 4, except a few extremely unstable compounds,
where its valency is 2, like methylene (:CH2), dichloromethylene (:CCl2), etc.
During chemical reaction, the two electrons present in 2s orbital become unpaired by
absorbing energy and one of them is promoted to 2pz orbital. This is the excited state of
carbon atom and the electronic configuration of carbon atom in this state is 1s2 2s1 2px1
2py1 2pz1. Thus, in the excited state, four odd electrons are present in the outermost shell of
carbon atom. The presence of these four unpaired electrons accounts for the tetravalency
of carbon.
sp3-Hybridization: When one s and three p orbitals of the valence shell of a carbon atom
merge together to form four new equivalent orbitals having the same energy and shape,
it results in tetrahedral or sp3 –hybridization. The resulting orbitals are called sp3 hybrid
orbitals
The four sp3 hybrid orbitals each containing one electron are directed towards the four
corners of a regular tetrahedron, making an angle of 109°28′ with one another and the
atom lies at the centre of the tetrahedron. The hybrid orbitals are oriented in such a fashion
in space that there occurs minimum repulsion between them. The formation of sp3 hybrid
orbitals by the combination of s, px, py and pz atomic orbitals may be shown as follows:
Formation of methane (CH4) molecule: During the formation of methane molecule,

one 2s orbital and three 2p orbitals of excited carbon atom undergo hybridization to form
four equivalent sp3 hybrid orbitals. The hybrid orbitals are directed towards the four
corners of a regular tetrahedron. Each hybrid orbitals containing an unpaired electron
overlaps with the 1s orbital of a hydrogen atom resulting in the formation of four C—H s
bonds. Thus, methane molecule possesses a highly stable tetrahedral geometry with each
H—C—H bond angle equal to 109°28′.
It is to be noted that if the four atoms linked covalently to the carbon atom are not the
same, the geometry of the molecule would still be tetrahedral but it may not be regular
in shape, e.g., methyl bromide (CH3Br), bromoform (CHBr3), etc. In these cases, the bond
angles differ slightly from the normal value of 109°28′.
sp2-Hybridization: When one s orbital and two p orbitals of the valence shell of a carbon
atom merge together and redistribute their energies to form three equivalent new
orbitals of equal energy and identical shape, the type of hybridization occurs is called
sp2-hybridization. The new orbitals formed as a result of this hybridization are called sp2
hybrid orbitals.
All three hybrid orbitals each containing one electron lie in one plane and make an angle
of 120° with each other, i.e., they are directed towards the three corners of an equilateral
triangle with the carbon atom in the centre of the triangle. The unhybridized 2pz orbital
(containing one electron) remains perpendicular to the plane of the triangle with its two
lobes above and below that plane. Therefore, a molecule in which the central atom is
sp2-hybridized possesses triangular planar shape and the hybridization is called planar
trigonal hybridization. The formation of sp2 hybrid orbitals by combination of s, px and py
atomic orbitals is shown below.
Formation of ethylene (C2H4) molecule: During the formation of ethylene molecule,

each of the two carbon atoms undergoes sp2-hybridization, leaving the 2pz orbitals
unhybridized. The three sp2 hybrid orbitals of each carbon atom are planar and oriented
at an angle of 120° to each other. The unhybridized 2pz orbitals are perpendicular to
the plane of sp2 hybrid orbitals. One sp2 orbital of one carbon overlaps axially with one
sp2 orbital of the other carbon to form a C—C s bond. The remaining two sp2 orbitals of
each carbon overlap with the half-filled 1s orbitals of two hydrogen atoms resulting in the
formation of a total of three C—H s bonds. The unhybridized 2pz orbitals of one carbon
overlap with that of the other carbon in a sideways fashion to from a p bond between the
two carbon atoms. The p bond consists of two equal electron clouds distributed above and
below the plane of carbon and hydrogen atoms. Since all the six atoms in the molecule
lie in one plane, ethylene is a planar molecule. Each C—C—H or H—C—H bond angle is
nearly equal to 120°.
sp-Hybridization: When one s and one p orbital of the valence shell of a carbon atom merge
together and redistribute their energies to form two equivalent hybrid orbitals of equal
energy and identical shape, it results in sp-hybridization or diagonal hybridization.
Formation of acetylene (C2H2) molecule: In acetylene molecule, the two carbon atoms
are sp-hybridized. There are two unhybridized orbitals (2py and 2pz) on each C atom. Two
sp hybrid orbitals are linear and directed at an angle of 180°. The unhybridized p orbitals
are perpendicular to the sp hybrid orbitals and also perpendicular to each other. One sp
hybrid orbital of one carbon overlaps axially with the similar orbital of the other carbon
to form a C—C s bond. The remaining hybrid orbital of each C atom overlaps with half-
filled 1s orbital of H atom to form a total of two C—H s bond. Thus, acetylene molecule
is linear. The unhybridized py orbitals of two carbons and the unhybridized pz orbitals of
two carbons overlap sideways separately to form two different p bonds. Electron clouds
of one p bond lie above and below the internuclear axis representing the s bond while the
electron clouds of the other p bond lie in front and backside of the internuclear axis. These
two sets of p electron cloud merge into one another to form a cylindrical cloud of electrons
around the internuclear axis surrounding the C—C s bond. Each C—C—H bond angle is
equal to 180°.
p
2px p 2px
2py 2py
s s s p
s s s
H sp C sp sp C sp H H C C H
1s 1s
180°
H C ∫∫ C H
Linear acetylene molecule
1.1.2 Bond Length

Bond length is defined as the equilibrium distance between the centres of the nuclei of
two bonded atoms in a covalent molecule. Bond length depends on the factors such as
(i) size of atoms, (ii) multiplicity of bonds, (iii) s-character of the orbitals and (iv) resonance,
hyperconjugation, etc. Bond length increases with increase in the size of atoms but it
decreases with increase in bond multiplicity. Thus, a triple bond is shorter than a double
bond which in turn is shorter than a single bond. An s orbital is closer to the nucleus
than a p orbital. So, electrons in the s orbital is more tightly held by the nucleus than
the electrons in the p orbital. For this reason, with increase in s-character of the hybrid
orbital, the attractive force on the electron(s) increases and so, the size of the hybrid orbital
decreases. As a consequence, the length of the bond obtained by overlapping of the hybrid
orbitals with the s orbitals of hydrogen, for example, decreases. The s-character of sp3, sp2
and sp hybrid orbitals are 25 percent, 33.33 percent and 50 percent respectively. Thus, the
lengths of the C—H bonds involving C atoms with different hybridization follow the order:
Csp3 —H (1.093 Å) > Csp2 —H (1.078 Å) > Csp — H (1.057 Å).
1.1.3 Bond Dissociation Enthalpy or Bond Dissociation Energy

The bond dissociation enthalpy or bond dissociation energy which is a measure of bond
strength may be defined as the amount of energy required to break one mole of a particular
type of bond between two atoms in the gaseous state so as to produce neutral gaseous
atoms or free radicals. Bond dissociation energies are usually abbreviated by the symbol
DH° and are usually expressed in kJ mol–1 or kcal mol–1. The greater the bond dissociation
energy, stronger the bond. The factors affecting bond dissociation energy are: (i) size of
the bonded atom, (ii) multiplicity of bonds, (iii) s-character of the hybrid orbital involved
in bond formation, etc.
Larger the size of the bonded atoms, greater the bond length and lesser is the bond
dissociation energy. The bond dissociation energy increases with increase in bond
multiplicity. A C ∫∫ C bond is stronger than a C == C bond which in turn is stronger than
a C – C bond. The bond dissociation energy also increases with increase in the s-character
of the hybrid orbital and this is because with increase in s-character of the hybrid orbital,
the electron density in the region of overlap increases.
Bond type Bond dissociation energy
kJ mol–1 (kcal mol–1)
C(sp3) — C(sp3) 346.3 (82.76)
C(sp3) — C(sp2) 357.6 (85.48)
C(sp3) — C(sp) 382.5 (91.42)
C(sp2) — C(sp2) 383.2 (91.58)
C(sp2) — C(sp) 403.7 (96.48)
C(sp) — C(sp) 433.5 (103.6)
Bond Dissociation Energies (kcal mol–1) of Some Chemical Bonds

Bond DH° Bond DH° Bond DH°
H—H 104 F—F 38 Ph—H 110
D—D 106 Cl—Cl 58 CH2==CH CH2—H 88
C—C 83.1 Br—Br 46 Ph CH2—H 85
N—N 38.4 I—I 36 CH3—F 107
O—O 33.2 CH3—H 104 CH3—Cl 84
N—F 136 C2H5—H 98 CH—Br 70
H — Cl 103 (CH3)2 CH—H 95 CH3—I 56
H — Br 88 CH2==CH—H 108 CH2==CH—Cl 82
H—I 71
Another measure of bond strength is Bond Energy. This is actually the average bond
dissociation energy, denoted by E. For example, cleavage of a C—H bond in methane
requires different bond dissociation energies in successive steps as indicated below and
energy of the C—H bond is determined by an average.
CH Æ .CH + H. DH ∞ = 104 Kcal mol –1
4 3 1
.CH Æ CH2 + H. DH 2∞ = 106 Kcal mol -1
3
CH2 Æ .CH + H. DH ∞ = 106 Kcal mol -1

3
.CH Æ .C . + H. DH ∞ = 81 Kcal mol -1
4
CH4 Æ .C. + 4H. Total: 397 Kcal mol -1
Therefore, EC—H = 397/4 = 99.25 kcal mol–1.

However, bond dissociation energy (DH°) is more convenient than bond energy (E) for our
purpose.
1.1.4 Bond Angle

The bond angle is the angle between two bonds around the central atom in a molecule.
For example, H—C—H bond angle in methane (CH4) is 109.5° and H—O—H bond angle
in water is 104.5°. Bond angle is affected by the factors such as (i) type of hybridization,
(ii) number of lone pair of electrons on the central atom, (iii) electronegativity of the
central atom and (iv) electronegativily of the atom bonded to the central atom. The greater
the amount of s-character of the orbital used by carbon to form the bond, the larger the
bond angle. For example, sp-hybridized carbons have bond angles of 180°, sp2-hybridized
carbons have bond angles of 120° and sp3-hydridized carbon have bond angles of 109.5°.
1.1.5 VESPR Theory and Molecular Geometry

We can predict the arrangement of atom in molecules and ions on the basis of a relatively
simple idea called the valence shell electron pair repulsion (VSEPR) theory.
We can apply this theory on the basis of the following considerations:
1. In a molecule or ion, the central atom is covalently bonded to two or more atoms or
groups.
2. Covalent bonds contain shared pair of electrons which are often called bond pairs
or bonding pairs. The unshared electrons of the central atom are called nonbonding
pairs or unshared pairs or lone pairs.
3. Since electron pairs repel each other, the electron pairs of the valence shell tend to
stay as far apart as possible to avoid electronic repulsion.
4. If the central atom is surrounded by bond pairs as well as lone pairs of electrons,
the repulsions among themselves are different. As a result, the molecule possesses
an irregular or distorted geometry. The repulsive interactions of various electron
pairs decrease in the order: lone pair–lone pair (lp – lp)> lone pair–bond pair (lp –
bp) > bond pair–bond pair (bp – bp).
5. The geometry of a molecule is to be settled by considering all of the electron pairs,
bonding and nonbonding. However, the shape of the molecule is to be described
by referring to the positions of the atoms and not by the position of the electron
pairs.
Let us consider the following examples.
Methane (CH4) molecule: In CH4 molecule, the total number of electrons surrounding
the central carbon atom = 4 valence electrons of C atom + 4 electrons of four singly-bonded
H atom = 8 electron or 4 electrons pairs = 4 s bond pairs. The four bond pairs experience
minimum repulsion if they are tetrahedrally oriented, i.e., if all the H—C—H bond angles
are of 109.5°. Hence, the shape of CH4 molecule is tetrahedral.
H
109.5°
C
H
H
H
Tetrahedral
methane molecule
Ammonia (NH3) molecule: In NH3 molecule, the total number of electrons in the
valence shell of the central N atom = 5 valence electrons of N atom + 3 electrons of three
singly-bonded H atom = 8 electron or 4 electron pairs = 3 s bond pairs + 1 lone pair. All
four electron pairs experience minimum repulsion if they occupy the four corners of a
tetrahedron. As lone pair–bond pair repulsion is greater than the bond pair–bond pair
repulsion, the H—N—H bond angle is slightly deviated from the normal tetrahedral angle
(109.5°) and is reduced to 107°, i.e., the tetrahedron is somewhat distorted. Excluding the
lone pair, the shape of the molecule is trigonal pyramidal.
Water (H2O) molecule: In water molecule, the total number of electrons surrounding the
central O atom = 6 valence electron of oxygen atom + 2 electrons of two singly-bonded H
atoms = 8 electrons or 4 electron pairs = 2 s bond pairs + 2 lone pairs.
In order to minimise the extent of mutual repulsion, these four electron pairs are oriented
towards the four corners of a tetrahedron. However, the tetrahedron is somewhat distorted
due to the strong repulsive forces exerted by the lone pairs on each bond pair of electrons.
In fact, the H—O—H bond angle is reduced to 104.5° from the normal tetrahedral angle of
109.5°. Excluding the lone pairs, the shape of the molecule is angular or V-shaped.
H O or O
104.5° H 104.5° H
H
Angular or V-shaped
water molecule
Boron tifluoride (BF3) molecule: In boron trifluoride molecule, the total number of
electrons in the valence shell of the central boron atom = 3 valence electrons of B atom
+ 3 electrons of three singly-bonded F atom = 6 electrons or 3 electrons pairs = 3 s bond
pairs. The three bond pairs experience minimum propulsion if they remain at 120° angle
with respect to each other. Therefore, the geometrical shape of BF3 molecule is trigonal
planar.
F
120° 120°
B
F 120° F
Trigonal planar
boron trifluoride molecule
Acetylene (HCCH) molecule: The number of electrons surrounding each carbon atom
of acetylene molecule = 4 valence electron of carbon + 3 electrons of one triply-bonded C
atom + 1 electron of one singly-bonded H atom = 8 electrons = 4 electron pairs = 2 s bond
pairs + 2p bond pairs. In order to minimise the repulsive forces between the bond pairs, the
shape of acetylene molecule is linear. The effect of electrons involved in the formation of a
p bond is not generally considered in determining the geometrical shape of a molecule.
180° 180°
H——C∫∫C——H
Linear acetylene molecule
1. Give the state of hybridization of the central atom of each of the following
species and predict their shapes.
! @ @
(a) H2O (b) BF3 (c) C H 3 (d) C H 3 (e) :N H 2

(f) H 2S: (g) NH 3 (h) BF4* (i) H3O≈ (j) HCN
≈
(k) CCl4 (l) CO2 (m) BeH2 (n) NH 4
Solution
(a) (b) (c)
F H +
B F C H
F H
sp2; trigonal planar sp2; trigonal planar
(d) (e) (f)

–
–
C N H S H
H
H H
H H sp3; bent or angular
sp3; bent or angular
sp3; trigonal pyramidal
(g) (h) (i)

F
+
H N –
O H
B
H H F H H
sp3; trigonal pyramidal F F sp3; trigonal pyramidal
sp3; tetrahedral
(j) (k) (l)

H—C∫∫N Cl O==C==O
sp; linear
sp; linear
C
Cl
Cl Cl
sp3; tetrahedral
(m) (n)
H—Be—H H
sp; linear
+
N
H
H H
sp3; tetrahedral
2. Draw the structure of a hydrocarbon which contains:

(a) Three sp3-hybridized carbon atoms;
(b) One sp3 and two sp2-hybridized carbon atoms;
(c) One sp and two sp2-hybridized carbon atoms;
(d) Two sp3 and two sp-hybridized carbon atoms.
Solution
(a) (b) (c)
(d)
3. Which orbitals are used to form each bond in methylamine, CH 3 NH 2?

Solution
Each C—H bond is formed by the overlap of an sp3 orbital of carbon with the s orbital of
hydrogen ( Csp3 —H1s). The C—N bond is formed by the overlap of an sp3 orbital of carbon
with sp3 orbital of nitrogen ( Csp3 — N sp3 ) and each N—H bond is formed by the overlap of
an sp3 orbital of nitrogen with the s orbital of hydrogen ( N sp3 —H1s).
: O:
||
4. Answer the following question for the acetaldehyde (CH 3 — C — H)
molecule:
(a) Determine the hybridization of oxygen and the two carbon atoms.
(b) Which orbitals are involved in forming the carbon–oxygen double
bond?
(c) Mention the type of orbital in which the lone pairs reside.
Solution
(a)
(b) The s bond is formed by the end-on overlap of an sp2 orbital of carbon with an sp2
orbital of oxygen and the p bond is formed by the side-by-side overlap of the 2p
orbital of carbon with the 2p orbital of oxygen.
(c) The two sp2 hybrid orbitals are occupied by the two lone pairs of oxygen.
5. Mention the state of hybridization of the starred (*) carbon atoms in each
of the following compounds:
* *
(a) (b) CH 3 C N (c) HC ∫∫ C — CHO
* * –
(d) CH2 == C == CH2 (e) (f)
*
Solution
(a) Three groups around the starred carbon: sp2-hybridized; (b) two groups around the
starred carbon: sp-hybridized; (c) three groups around the starred carbon: sp2-hybridized;
(d) two groups around the starred carbon: sp-hybridized; (e) three groups around the
starred carbon: sp2-hybridized; (f) four groups around the starred carbon; sp3-hybridized.
6. How many s and p bonds are present in each of the following molecules?
(a) CH3—C ∫∫ C—CH == CH—CH3
(b) –CH2CH3
(c) CH3 CH == C == CH CH2 CH3
Solution
(a) s bond = 13; p bond = 3;
(b) s bond = 18; p bond = 3;
(c) s bond 15; p bond 2
7. Designate the state of hybridization of all the atoms of the following
molecule:
Solution The hybridized atoms within the molecule is designated as a, b and c according
to their hybridization status: a = sp3; b = sp2 and c = sp.
8. Draw the orbital picture for each of the following molecules: (a) ethylene
(b) acetylene (c) ketene (CH2 == C == O) (d) acraldehyde (e) acrylonitrile
(f) but-1, 2, 3-triene.
Solution
(a) Ethylene:
(b) Acetylene:
(c) Ketene:
(d) Acraldehyde:
(e) Acrylonitrile:
(f) But-1, 2, 3-triene:
9. Which atoms in each of the following molecules always remain in the

same plane and why?
(a) CH3CH==CH CH3 (b) C6H5C ≡≡ CCH3
(c) CH3CH==C==C==CH2 (d) CH2==CH—C ≡≡ CH
2
Solution The two sp -hybridized carbon atoms and the atom directly attached to them
always remain in the same plane. This is applicable also when the two sp2-hybridzed carbon
atoms are linked through an even number of sp-hybridized carbon atoms. Therefore, in
(a), all atoms excluding the six methyl hydrogens remain in the same plane. In (b), all
atoms excluding the three methyl hydrogens remain in the same plane. In (c), all atoms
excluding the three methyl hydrogens remain in the same plane and in (d), all atoms
remain in the same plane. This is also because the two sp-hybridized carbon atoms and
the atoms directly attached to them remain in the same line.
10. Explain why a p bond is weaker than a s bond.
Solution A p bond is weaker than a s bond because the end-on overlap that forms s bonds
is better than the side-to-side overlap that forms p bonds. This is also because the electron
density in a p bond is farther from the two nuclei as compared to that in a s bond.
11. In CH3CH3, the C—H bond is shorter and stronger than the C—C bond –
explain.
Solution The s orbital of hydrogen is closer to the nucleus than the sp3 orbital of carbon
having less s character. So, the carbon and hydrogen nuclei are closer together in sp3-s
overlap than the carbon nuclei in sp3-sp3 overlap. Because of this, the C—H bond in
ethane (CH3—CH3) is shorter and stronger than the C—C bond. Again, as the percentage
of s character of the overlapping orbitals increases, the electron density in the region
of overlap increases and as a result, the strength of the bond increases. Since there is
greater electron density in the region of sp3-s overlap than in the region of sp3-sp3 overlap,
therefore, the C—H bond is stronger than the C—C bond.
12. Arrange the indicated bonds in each of the following compounds in order
of increasing bond strength and increasing bond length:
bond 1
HC∫∫ C—CH==CH—CH2—CH3
(a) ≠ ≠ ≠ (b) H3C NH—CH2CH3
bond 1 bond 2 bond 3
N CH2—C∫∫ N
bond 2 ≠
bond 3
Solution Greater the bond multiplicity, shorter the bond and greater the bond strength.
Therefore, in compound (a), bond length increases in the order: bond 1 < bond 2 < bond
3 and bond strength increase in the order: bond 3 < bond 2 < bond 1. In compound (b),
the bond length increases in the order: bond 3 < bond 2 < bond 1 and the bond strength
increases in the order: bond 1 < bond 2 < bond 3.
13. Which of the indicated bonds in each pair of compounds is shorter and
why?
(a)
(b)
O
==
(c) CH3—C—H and H—CH2OH

≠ ≠
Solution As the s-character of the overlapping orbitals increases, the nuclei involved in
bond formation becomes closer and the bond becomes shorter. Therefore, (a) the Csp—H
bond in the second compound is shorter than C 2 —H bond in the first compound;
sp
(b) the N sp2 —H bond in the second compound is shorter than the N sp3 —H bond in
the first compound, and (c) the Csp2 —H bond in the first compound is shorter than the
C 3 —H bond in the second compound.
sp
14. Determine the state of hybridization of the indicated atom in each of the
following species:
(a) (b) (c)
– –
–
CH2 NH2
== ≠ ≠
O O2N O2N
(d) (e) (f)
–
O O O O
Ø
== ==
==
==
C6H5—C
Ø– H—C—NH2
O
≠
Solution In any system of the type X = Y – Z:, Z is sp2-hybridized and the unshared
electron pair on it occupies a p orbital to delocalize the electron pair and make the system
conjugated. Therefore, the indicated atoms in (a), (b), (c), (d) and (f) are sp2-hybridized.
However, due to violation of Bredt’s rule (introduction of a double bond is not possible at
the bridgehead position in bridged bicyclic compounds with small rings) delocalization of
the unshared electron pair is not possible in the compound (e) and therefore, the indicated
carbon atom is sp3-hybridized.
15. “Bromination of methane is less exothermic than that of chlorination” —

explain with DH° calculation.
[Bond dissociation energies for C—H = 104 kcal/mol; Br—Br = 46 kcal/mol; H—Br = 87.5
kcal/mol; Cl—Cl = 58 kcal/mol; H—Cl = 103 kcal/mol; C—Cl = 83.5 kcal/mol and C—Br =
70 kcal/mol]
hn
Solution CH 4 + Br – Br ææÆ CH 3 - Br + HBr
DH° (bromination) = [(–70) + (–87.5)] – [(–104) + (–46)]
= (–157.5) – (–150) = –7.5 kcal/mol
hn
CH 4 + Cl – Cl ææÆ CH 3 - Cl + HCl
DH° (chlorination) = [(–83.5) + (103)] – [(–104) + (–58)]

= (–186.5) – (–162) = – 24.5 kcal/mol
Thus, bromination of methane is less exothermic than chlorination of methane.
16. The H—X bond in hydrogen halides becomes longer and weaker as the
atomic mass of the halogen increases—explain.
Solution A p orbital of halogen overlaps with the s orbital of hydrogen to form the
hydrogen–halogen (H—X) bond in hydrogen halides. For bond formation, fluorine uses
the p orbital that belongs to the second shell of electrons while chlorine uses the p orbital
that belongs to the third shell of electrons. Since the average distance from the nucleus is
greater for an electron in the third shell than for an electron in the second shell, therefore,
the average electron density is less in a 3p orbital than in a 2p orbital. Consequently, the
electron density in the region where the s and p orbitals overlap decreases as the size of
the halogen increases. Thus, the hydrogen–halogen bond becomes longer and weaker as
the atomic mass of the halogen increases.
1. A carbon-carbon bond formed by sp2–sp2 overlap is stronger than the one formed
by sp3–sp3 overlap. Explain.
2. Which atom in the ammonium ion (NH≈4 ) has the least electron density and why?
3. What is the state of hybridization of each of the C, O and N atom in the following
compound?
H3 C C∫∫ C—CH—CH2—O—CH3
C
==
O NH2
4. Predict the geometry around each of the indicated atoms:

O
||
(a) B
≠
H3 (b) H 3N Æ B
≠
H3 (c) CH3 —C—OCH3
≠
! !
(d) H 3 O
!
≠
(e) C (CH 3 )3 (f) N H 4
≠ ≠
(g) H C N (h) (CH 3 )2 N H (i) Et 2 N

@
≠ ≠ ≠
@
(j) CH 3 C H 2CH 3 (k) CH 2 = CH– C H 2
≠ ≠
5. Predict the mentioned bond angles:

(a) C—O—C bond angle in CH3COCH3 (b) F—B—F bond angle in BF4@
(c) C—C—N bond angle in C2H5CN (d) H—N—H bond angle in NH ≈4
!
(e) C—O—H bond angle in CH3CH2OH (f) C—N—C bond angle in (CH 3 )2 N H 2
6. Which of the indicated bonds is shorter and why?
(a) (b) CH 3 CH = CH – C ∫ C - CH 3
≠ ≠
NH2
(c) Cl (d) ≠ (e) CH 3 CH = CH – CH = CH - CH 3
9
≠ ≠ OH ≠ ≠
≠
O
||¨
(f) CH 3 — O — CH 2 — C — CH 3 (g) N—CH2CH2N==CHCH2CH3
≠
≠ ≠
7. Give the state of hybridization of the indicated atoms:

≈ ≈
(a) (CH 3 )2 N H 2 (b) (CH 3 )2 O H (c) B F4@
≠ ≠ ≠
@
(d) CH 3 N O2 (e) H3 N Æ BF3 (f) CH2 == CH — CH2
≠ ≠ ≠
(g) CH3CH == CH — NH2 (h) CH2 == C == CH2

≠ ≠
1.2 ELECTRONEGATIVITY AND BOND POLARITY

Electronegativity is the tendency of an atom to pull the bonding electrons towards itself.
It increases across a row of the periodic table (excluding the noble gases) and it decreases
down a column of the periodic table. The electronegativity values of some common elements
are given in the following table.
A bond with the electrons shared equally between the two atoms, i.e., a bond in which
each electron spends as much time in the vicinity of one atom as in the other, is called
a nonpolar covalent bond. For example, the H—H covalent bond, the Br—Br covalent
bond and the C—C covalent bond in ethane etc. are nonpolar covalent bonds. When two
atoms of different electronegativities form a covalent bond, the electrons are not shared
equally between them. The atom with greater electronegativity draws the electron pair
closer to it. As a result of this unequal distribution of the bonding electrons, the bond
acquires a slight positive charge (indicated by the symbol d+) on the end that has the
less electronegative atom and a slight negative charge (indicated by the symbol d–) on
the other end that has the more electronegative atom, i.e., a polarity developes within a
bond. Such a bond is called a polar covalent bond. An example of a polar covalent bond is
the one in hydrogen fluoride (H—F). The fluorine atom, with its greater electronegativity,
pulls the bonding electrons towards itself. As a consequence, the hydrogen atom becomes
somewhat electron deficient and acquires a partial positive charge (d+) while the fluorine
atom becomes somewhat electron rich and acquires a partial negative charge (d–). So,
the hydrogen fluoride molecule is a dipole (Hd+— Fd–). The direction of bond polarity is
indicated by an arrow, the head of which indicates the negative end of the bond. A short
perpendicular line is drawn near the tail of the arrow to indicate the positive end of the
bond.
The extent of polarity in a covalent bond is expressed quantitatively by the physical

property known as ‘dipole moment’ m, which is, in fact, the product of the magnitude of
charge e on any polar end and the distance d between the centres of positive and negative
charges, i.e., m = e × d.
Dipole moment is expressed in Debye unit (D). The charge e and the distance d are of the
order 10–10 esu and 10–8 cm, respectively. Therefore, m is of the order of 10–10 × 10–8 = 10–18
esu.cm (in C.G.S. system). This value of dipole moment is equal to one Debye unit, i.e.,
1D = 10–18 esu. cm. The SI unit of dipole moment is coulomb-meter (C.m.) and 1 C.m. =
2.9962 × 1029 D.
The value of m for a nonpolar molecule is zero. For example, m = 0D for the molecules like
H2, N2, O2, etc. For polar molecules, m has a definite value, e.g., for HF molecule m = 1.91
D. Polarities of molecules increase with increase in the value of m.
Dipole moment is a vector quantity. The dipole moment of a molecule is the result of the
vector sum of the individual bond moments present in the molecule. When a molecule
is formed by two atoms of different electronegativities (i.e., by two different atoms), the
molecule must possess a dipole moment because there is no question to cancellation of the
moment of this only polar bond. However, when a molecule contains two or more polar
bonds, i.e., when a molecule contains more than two atoms, the molecule may or may not
possess dipole moment and in that case the polarity of the molecule depends on the shape
of the molecule. A symmetrical polyatomic molecule possesses no dipole moment because
the individual dipole moments of the bonds cancel each other. It is to be noted that the
symmetry in calculating dipole moment is not the molecular symmetry. H2O molecule, for
example, has a symmetrical structure because it has two planes of symmetry (su planes).
But, it possesses dipole moment (1.85 D) because of its angular structure .
The value of dipole moment of a molecule is equal to the product of the charge e and the
distance, d, between the positive and negative charge centres, i.e., m = e × d, it can be well
understood by the following examples:
(i) Although an H—F (0.92 Å) molecule is smaller than an H—Cl (1.27 Å molecule, it
has a dipole moment larger than HCl (1.75 D compared to 1.03 D). Since fluorine
(electronegativity: 4.0) is more electronegative than chlorine (electronegativity:
3.0), therefore, the magnitude of e in H—F is much higher than that in H—Cl and
for this reason although d in HF is smaller than in HCl (because of smaller size of
F compared to Cl), the product e × d, i.e., m, is larger.
(ii) Chloromethane (CH3Cl) has a larger dipole moment (1.87 D) than fluoromethane
CH3F (1.81 D), even though fluorine is more electronegative than chlorine. Due to
higher electronegativity of fluorine than chlorine, the separation of charge in the
C—X bond, i.e., the magnitude of e in CH3F, is somewhat higher than that in CH3Cl.
However, the C—F bond is shorter than the C—Cl bond (1.42 Å compared to 1.77
Å) and because of this, the value of the product, i.e., m, is larger for chloromethane
than for fluoromethane.
That the polarity of a molecule depends upon the shape of the molecule can
be well understood by the following example. NH3 possesses considerable dipole
moment and although the N—F bonds are more polar than N—H bonds, NF3 (0.26 D) has
a much smaller dipole moment than NH3 (1.46 D). Both NH3 and NF3 with sp3 hybridized
N atom are pyramidal in shape. The unshared electron pair on nitrogen occupying an sp3
orbital contributes a large dipole moment (because an unshared pair has no other atom
attached to it to partially neutralize its negative charge) in the direction opposite to the
triangular base of the pyramid. In NH3, the net moment resulting from three N—H bond
moments adds to the moment contributed by the unshared pair because they act in the
same direction and for this reason, it has considerable dipole moment. In NF3, on the
other hand, the vectorial sum of these N—F bond moments acts in the direction opposite
to that of the moment caused by the unshared pair. Since these moments are of about the
same size, therefore, NF3 has a much smaller dipole moment than NH3.
Resonance often plays an important role in determining the polarities of

molecules. For example:
(i) The dipole moment of ethylchloride, CH3 CH2Cl (2.05 D), is larger than that of vinyl
chloride, CH2 == CH—Cl (1.44 D). Chlorine is more electronegative than carbon
and so, it attracts the C—Cl bonding electrons more towards itself. As a result,
polarity develops in the C—Cl bond in ethyl chloride and the compound shows
considerable dipole moments. In vinyl chloride, on the other hand, the unshared
electron pair on chlorine becomes involved in resonance interaction with the
p-orbital of the double bond. This resonance interaction, therefore, tends to oppose
the usual displacement of electrons towards chlorine. Also, the sp2 hybridized
carbon being more electronegative than the sp3 hybridized carbon is less willing
to release electrons to chlorine. So, although there is still a net displacement of
electrons towards chlorine, it is less than in ethyl chloride. Hence, ethyl chloride is
more polar than vinyl chloride, i.e., the dipole moment of ethyl chloride is greater
than that of vinyl chloride.
(ii) The dipole moment of chlorobenzene is larger than that of fluorobenzene. In

chlorobenzene, the Cl atom withdraws electrons from the ring by its –I effect
and donates electrons to the ring by its +R effect. Because of large size and high
electronegativtiy of chlorine as compared to that of carbon, the resonance electron
donation is much weaker than inductive electron withdrawal. So, the moment
caused by the –I effect is much stronger than the moment caused by the +R effect
and because of this, chlorobenzene possesses a net dipole moment which is relatively
high. In fluorobenzene, on the other hand, the +R effect fluorine even being more
electronegative than chlorine is much more important because of effective p orbital
overlap between the orbitals of carbon and fluorine which are of comparable size.
So, the moment due to inductive electron withdrawal is predominantly balanced
by the moment due to resonance electron donation and the result is that the dipole
moment of fluorobenzene is relatively low.
Aromaticity often plays a role in determining the polarity of a molecule. For

example:
(i) Azulene (a bicyclic hydrocarbon) has a much higher dipole moment. Azulene
contains 10 p-electrons. Redistribution of these electrons between the two rings
generates an aromatic system which consists of an aromatic cycloheptatrienyl
cation and an aromatic cyclopentadienyl anion. Since this aromatic dipolar form is
quite stable and more contributing, azulene has a much higher dipole moment.
(ii) Pyrrole and furan are polar molecules, but the dipole moment of furan is smaller
than and opposite in direction from that of pyrrole. In pyrrole, the unshared
electron pair on nitrogen is highly delocalized with the p-electrons of the ring
for maintaining aromaticily, i.e., to form a delocalized cyclic system of (4n + 2)p
electrons, where n = 1. Since the moment caused by the electron delocalization is
much greater than that caused by the –I effect of nitrogen atom, pyrrole has a net
dipole momemt of 1.81 D and the dipole points towards the ring. In furan, on the
other hand, the unshared electron pair on oxygen is not well delocalized into the
ring due to greater electronegativity of oxygen. Hence, the moment due to electron
delocalization is small and in fact, it is somewhat smaller than that caused by the
–I effect of oxygen. For this reason, the net dipole moment of furan is relatively
small (0.70 D) and the dipole points towards oxygen.
Formation of charge-transfer complex is often responsible for exhibiting dipole

moment by some molecules. 1,3,5-Trinitrobenzene, for example, shows significant
dipole moment in benzene but does not display dipole moment in carbon tetrachloride.
1,3,5-Trinitrobenzene being symmetrical possesses no net dipole moment, i.e., it is a
nonpolar molecule. When it dissolves in benzene, a charge-transfer bonding involving
some kind of donor–acceptor interaction (benzene is the electron donor and the electron
deficient 1,3,5-trinitrobenzene is the electron acceptor) is formed. Due to charge separation
in the resulting complex, the compound shows significant dipole moment in benzene. Since
carbon tetrachloride cannot act as a donor molecule, the compound does not display dipole
moment in carbon tetrachloride.
Conformation plays an important role in determining the polarities of some

molecules. For example:
(i) 1,2-Dichloroethane has a very little dipole moments, where as 1,2-ethanediol
has considersable dipole moment. 1,2-Dichloroethane exists in anti and gauche
staggered form. Since the two C—Cl dipoles in the anti form are antiparallel, its
dipole moment is assumed to be zero. However, the value of dipole moment of the
gauche form is approximately 3.2 D (calculated with reference to C2H5Cl). Because
of dipolar and steric repulsion, the gauche form is less stable than the anti form
by 1.2 kcal/mol and in the vapour phase the compound exists in 88% anti and 12%
gauche form. Because of such distribution of the conformers, the overall dipole
moment of 1,2-dichloroethane is relatively low (1.2 D).
In the gauche conformer of 1,2-ethanediol (ethylene glycol), the two —OH groups
become involved in the formation of intramolecular hydrogen bond. However, in the
anti conformer, no intramolecular hydrogen bond is formed because the two —OH
groups are oppositely placed. The steric and polar repulsion of the hydroxyl groups
in the gauche form is more than outweighed by the energy gained by the formation
of hydrogen bond (5 kcal/mol). Because of this, the compound, particularly in the
gas phase, exists almost exclusively in the gauche form having a finite dipole
moment. For such conformational distribution, 1,2-ethanediol is found to possess
considerable dipole moment.
(ii) The optically inactive (meso) form of 1,2-dichloro-1,2-di-p-tolylethane has a dipole

moment lower than that of the active (d or l) form. The meso form of the compound
exists in the following three conformations:
In the conformer I, bulky tolyl groups are anti to each other and also the two
Cl atoms are anti to each other. So there is no steric interaction caused by the
tolyl groups and repulsive interaction caused by the C—Cl dipoles. Therefore, this
conformer is relatively more stable and it has practically no dipole moment. The
conformers II and III are polar because the two Cl atoms are gauche to each other.
However, these are very unstable because the bulky p-tolyl groups which are
gauch to each other are involved in steric interaction and there occurs repulsive
interaction between the two C—Cl dipoles. Therefore, the most favoured conformer
of meso-1,2-dichloro-1,2-di-p-tolylethane is I and as a consequence, the overall

dipole moment of the meso compound is relatively low.
The optically active form of this compound exists in the following three
conformations:
The conformer IV in which C—Cl dipoles are antiparallel has no appreciable dipole
moment. However, this conformer is unstable because the bulky p-tolyl groups
which are gauche to each other are involved in steric interaction. The conformer
VI has an appreciable dipole moment but it is unstable due to the same steric
interaction and also due to repulsive interaction between the two C—Cl dipoles. In
the conformer V, the two p-tolyl groups are anti to each other. So, it is relatively
stable and the preferred conformer. Since it has an appreciable dipole moment, the
overall dipole moment of active 1,2-dichloro-1,2-ditoylethane is relatively large.
(iii) The dipole moment of cis-1,2-dichlorocyclohexane is larger than that of its trans-
isomer. Each of the conformational isomers of cis-1,2-dichlorcyclohexane has one
axial and one equatorial chlorine atom. As both the forms are equally stable,
50% of the molecules exist in one and 50% in the other form. Since Cl atoms are
gauche to each other in both the forms, they have appreciable dipole moment and
consequently, the overall dipole moment of the compound is relatively large. The
trans-isomer, on the other hand, exists in diequatorial and diaxial forms. The
diequatorial form in which the two Cl atom are gauche to each other has appreciable
dipole moment. Although favoured sterically, it is disfavoured by dipolar repulsion.
The diaxial form with a very small dipole moment, although disfavowred sterically,
is free from dipolar repulsion and is relatively stable. This form, therefore, exists
predominantly and so, the overall dipole moment of this isomer is much lower than
that of the cis-isomer.
(iv) The dipole moment of hexane-3,4-dione is very small, whereas that of cyclohex-
3,5-dien-1,2-dione is very large. Hexane-3,4-dione may exist in two conformations
such as s-cis (cisoid) and s-trans (transoid). In the s-trans conformer, the two C == O
dipoles are antiparallel. Since the C == O bond moment cancels each other, therefore,
this conformer has no net dipole moment. Again, as the negatively polarized
oxygen atoms are as far as possible from each other, this conformer is free from any
repulsive interaction between the two oxygen atoms. In the s-cis conformer, on the
other hand, the C == O dipoles remain on the same side of the double bond make
an angle of 60° with each other. Consequently, a large net moment of two strong
C == O bond moments operates. Again, as the two C == O dipoles are close enough
to repeal each other, this conformer is relatively less stable. For this reason, the
compound exists almost exclusively, in the more stable s-trans conformation and
the overall dipole moment of the compound is very small.
The planar and rigid cyclohex-3,5-dien-1,2-dione molecule exists exclusively in the
highly polar s-cis conformation in which the two C == O bond moments make an
angle of 60° with each other. It cannot exists in the s-trans conformation because
it is very much unstable due to severe angle strain. For this reason, the dipole
moment of the compound is quite large.
Induced moment often plays an important role in determining the dipole

moments of molecules. The expected order of dipole moment of various fluormethanes is
CH2F2 > CH3F ª CHF3 > CF4 . However, the experimental order is: CF3F > CH2F2 > CHF3 > CF4.
This anomaly can be explained in terms of opposing induced moment. In the symmetrical
tetrahedral molecule of CF4, the resultant of three C — F bond moments cancels the fourth
C — F bond moment and because of this, the molecule possesses no net dipole moment.
In CH3F, the moment due to the C—F bond is not cancelled and the molecule possesses
a net dipole moment which is the resultant of C—F and C—H bond moments. The dipole
moment of CH2F2 in which the two C—F bonds make an angle of nearly 109°5 is expected
to be larger than that of CH3F because the resultant of two C—F bond moments must be
greater than one C—F bond moment. Again, the dipole moment of CH3F is expected to
be equal to that of CHF3 because the moment of a —CF3 group is equal to that of a C—F
bond and the moment of a —CH3 group is equal to that of a C—H bond. Thus, the expected
order of dipole moment is CH2F2 Ò CH3F ª CH3F Ò CF4. However, this order does not agree
with the experimental dipole moment values. This can be explained by considering the
moment acting in the opposite direction induced by each C—F dipole in the other. Since
there is only one C—F bond in CH3F, the opposing induced moment is absent in it. So it
has the largest dipole moment. In CH2F2, there are two C—F bond. Thus, opposing induced
moment operates in this case and that partly cancels the resultant moment of two C—F
bonds. Because of this, its dipole moment is smaller than CH3F. In CHF3, there are three
C—F bonds and so, the magnitude of the opposing induced moment is relatively large in
this case. In fact, this reduces the resultant of three C—F bond moments considerably and
because of this its dipole moment is smaller than that of CH2F2 and much smaller than
that of CH3F. Hence, the order of decreasing dipole moments of these four fluoromethanes
in CH3F > CH2F2 > CHF3 > CF4.
Dipole moment of disubstituted benzenes:
The dipole moment of a disubstituted banzene C6H4 AB can be given as
m AB = m 2A + m B2 + 2m A m B cos a
where mA and mB are the two group moments and a is the angle between their direction.
When both the groups are electron donating or both are electron withdrawing, the values
of a for ortho-, meta- and para-isomer are 60°, 120° and 180°, respectively. However, when
one group is electron donating and the other is electron withdrawing, the value of a for
ortho-, meta- and para-isomer are 120°, 60° and 0°, respectively. For example:
(i) Dichlorobenzenes: In each of the three isomeric dichlorobenzenes, the angle
between the vectors of the two group moment may be shown as follows:
In this case, A == B == Cl and m A = m B = mC6H5Cl = 1.55 D, The group moment of

—Cl is to be used with negative sign because it is directed away from the benzene
ring.
For o-dichlorobenzene, where a = 60°, the net dipole moment is
mO-C6H4Cl2 = (-1.55)2 + (-1.55)2 + 2(-1.55)2 cos60∞
1
= 2.40525 + 2.4025 + 2 ¥ 2.4025 ¥
2
= 7.2075 = 2.68 D
Ê 1ˆ
Similarly, for m-dichlorobenzene, where a = 120∞ Á cos120∞ = ˜ , the net dipole
Ë 2¯
moment is 1.55 D and for p-dichlorobenzene, where a = 180° (cos180° = –1), the net
dipole moment is 0 D.
(ii) Nitrotoluenes: In this case, A = CH 3 , B = NO2 , m A = mC H CH = 0.4 D and
6 5 3
mB = mC6H5NO2 = 3.95 D. The group moment of the electron–withdrawing —NO2

group is negative because it is directed away from the ring. The group moment
of the electron-releasing —CH3 group is also negative because it is assumed to
directed away from the ring.
For o-nitrotoluene, where a = 120∞, the net dipole moment is

mO-CH3 C6H4 NO2 = (-0.4)2 + (-3.95)2 + 2 ¥ (-0.4) ¥ (-3.95) ¥ cos120∞
Ê 1ˆ
= 0.16 + 15.6025 + 2 ¥ (-0.4) ¥ (-3.95) ¥ Á - ˜
Ë 2¯
= 0.16 + 15.6025 - 1.58
= 14.1825 = 3.76 D
1
Similarly, for m-nitrotoluene, where a = 60∞ ÊÁ cos60∞ = ˆ˜ the net dipole moment is 4.16 D
Ë 2¯
and for p-nitrotoluene, a = 0° (cos 0° = 1), where the net dipole moment is 4.35 D.
Formal charge: A formal charge represents the integer charges (positive or negative)
that an atom bears in a molecule or ion, assuming that electrons in a chemical bond are
equally shared between atoms irrespective of relative electronegativity. The concept of
formal charge helps us to determine which atoms bear most of the charge in a charged
molecule and also helps us to find out charged atoms in molecules that are neutral overall.
The formal charge of any atom in a molecule can be calculated as follows:
Formal Charge = number of valence electrons or periodic table group number – (number
of nonbonding electron +1/2 number of bonding electrons) = V – (N + B/2).
1. Predict the direction of dipole moment of the indicated bond in each of

the following molecules by using the symbol d+ and d– or Æ:
|
(a) H—Cl (b) CH3—Cl (c) CH3—NH2

(d) I—Cl (e) CH3—MgBr (f) CH3CH2—OH
(g) F—Br (h) HO—Cl (i) — Li
Solution
(a) (b)
(c) (d)
(e) (f)
(g) (h)
(i)
2. Which one of the following compounds has (a) the most positively charged
hydrogen, (b) the most negatively charged hydrogen and (c) a nonpolar
covalent bond
LiH, HCl, H2
Solution
(a) H—Cl has the most positively charged hydrogen because chlorine is more
electronegative than hydrogen.
(b) LiH has the most negatively charged hydrogen because lithium is less
elelctronegative than hydrogen.
(c) The H—H bond is a nonpolar covalent bond because the atoms that share the
bonding electrons are identical.
3. Give the direction of the important bond moments in each of the following
molecules (neglect C—H bonds) and also give the direction of the net
dipole moment from the molecule. If the molecule has no dipole moment,
write m = 0 D.
(a) cis-BrCH == CH Br (b) trans-BrCH == CHBr
(c) CH2 == CF2 (d) Br2C == CCl2
(e) cis-BrCl C == CClBr (f) trans-BrCl C == CClBr
Cl
Br Cl Br Br
(g) (h) (i)
Br Cl Cl Cl
Cl
Br H
Cl Cl
(j) (k) (l) Cl Cl
Br Br
Cl H
Cl F
Cl Cl
(m) (n) C
Cl F
Cl Cl
Solution Br H
Br Br C==C
net dipole
(a) C==C moment (b) H Br
H H trans-
cis- (m = 0D)
Br Cl
(c) (d) C==C
Br Cl
net dipole
moment
Cl Cl Br Cl
net dipole
(e) C==C (f) C==C
moment Cl Br
Br Br
cis- trans-
(m = 0D)
Br Cl
Cl
(g) net dipole (h) Br Cl
moment
net dipole
Cl moment
F
Br Br net dipole net dipole
(i) moment (j) moment
Cl Cl Cl Cl
Br H
Cl Cl
(k) (l) Cl Cl
Br Br
Cl H
(m = 0D) (m = 0D)
F net dipole
moment
(m) (n) C
Cl F
Cl
4. How does bond polarity differ from bond polarizability? Explain with
suitable examples.
Solution Polarization of a covalent bond (bond polarity) is a permanent phenomenon and
is caused by the inductive mechanism of electron displacement in a bond in the normal state
of a molecule. This permanent polarization arises due to difference in electronegativities of
the two dissimilar bonded atoms. Polarization influences the physical as well as chemical
properties of a molecule. The order of decreasing polarization of the C—X (X = halogen)
bond in alkyl halides is as follows:
R3C—F > R3C—Cl > R3C—Br > R3C—I
Polarizability, on the other hand, is a temporary phenomenon and is caused by the
inductomeric mechanism of electron displacement in a covalent bond during a reaction
when charged reagents are brought near the bond concerned. When the reagent is removed
from the vicinity of the bond, the electrons resume their original position. Polarizability
influences mainly the chemical properties of a molecule. Polarizability depends on the
strength of the bond. In general, the more electronegative an atom or group, the less
will be the effect of polarizability. In alkyl halides, for example, the decreasing order of
polarizability of C—X bond is as follows:
R3C—I > R3C—Br > R3C—Cl > R3C—F > (R == H or alkyl group)
Therefore, the C—I bond is more polarizable than the C—F bond and that is why in
nuclephilic displacement reactions (SN2), CH3—I is found to be more reactive than CH3—F,
even though F is more electronegative.
5. The dipole moment of a bond gives an idea about the ionic character of
that bond. Calculate the percentage of ionic character of HF. Given that
the dipole moment of HF is 1.91 D and its bond length is 0.92 Å.
Solution If HF were 100% ionic, each atom would carry a charge equal to one unit, i.e.,
4.8 × 10–10 esu. As the bond length of HF is 0.92 Å, the dipole moment of completely ionic
bond is
mionic = e × d = 4.8 × 10–10 esu × 0.92 × 10–8 cm
= 4.416 × 10–18 esu.cm = 4.416D ( 10–18esu.cm = 1 D)
Therefore the percentage of ionic character of the H–F bond
mobserved 1.91 ¥ 100
= ¥ 100 = = 43.25
m ionic 4.416
6. Explain the following observations:

(a) CO2 has no dipole moment but SO2 has.
(b) CF4 has no dipole moment but CH3F has.
(c) BF3 has no dipole moment but NF3 has.
(d) trans-BrCH == CHBr has no dipole moment but cis-BrCH == CHBr has.
Solution
(a) CO2 has no dipole moment, i.e., it is a nonpolar molecule. This triatomic molecule
with the carbon atom in the middle has linear structure (C is sp hybridized).
Since the individual C == O bond moments are equal in magnitude but opposite in
direction, therefore, they cancel each other and as a result, the molecule becomes
nonpolar, i.e., it has no net dipole moment. Viewing alternatively, the centre of
positive charge coincides with the centre of negative charge in this molecule, i.e.,
d = 0 and so m = e × d = e × 0 = 0 D.
On the other hand, the angular (bond angle 119.5°) sulphur dioxide molecule has
appreciable dipole moment. This is because there operates a resultant moment
of the two polar S == O bonds and the moment due to the unshared electron pair
acting in the direction opposite to that of the resultant moment is insufficient to
cancel the resultant moment completely.
(b) Although the C—F bond of carbon tetrafluoride (CF4) having very symmetrical
tetrahedral structure is polar, CF4 has no net dipole moment and this is due to the
fact that the net moment contributed (vectorially) by the three C—F bonds exactly
cancels the remaining C—F bond moment. In fact, the centre of positive charge
and the centre of negative charge coincide in this tetrahedral molecule, i.e., d = 0
and so m = e × d = e × 0 = 0 D.
Methyl fluoride (CH3F), on the other hand, has an appreciable dipole moment
because the C—F bond moment is not cancelled by any opposing moment. The
remaining three C—H bonds contribute a small net moment (the vectorial sum) in
the same direction and so, the overall dipole moment of the molecule is somewhat
increased.
(c) Since the boron atom in BF3 molecule is sp2 hybridized, the shape of the molecule
is planar trigonal with the boron atom at the centre of an equilateral triangle and
the three fluorine atoms at the three corners. In this symmetrical molecule, the
vectorial sum of the two B—F bond moments is cancelled by the remaining B—F
bond moment. Because of this, the molecule has no net dipole moment. In fact,
the centre of positive charge coincides with the centre of negative charge in this
triangular molecule, i.e., d = 0 and so m = e × d = e × 0 = 0 D.
Since the nitrogen atom in NF3 is sp3 hybridized, the shape of the molecule is
pyramidal with nitrogen at the apex and fluorines at the corners of a triangular
base. The net moment contributed (vectorially) by the N—F bonds opposes the
moment due to the unshared electron pair. Since these opposing moments are
approximately of the same size, therefore, the molecule has a very small dipole
moment (0.26 D) of indeterminate direction.
(d) The carbon atoms in these two isomeric dibromoethenes are trigonal
(sp2 hybridized). So, all the atoms of each molecule lie in the same plane. Since the
bromine atoms in the cis-isomer lie on the same side of the double bond, therefore,
the molecule has a net dipole moment which is the vectorial sum of two C—Br
bond moments. In the trans-isomer, on the other hand, the bromine atoms lie
on the opposite sides of the double bond and the two C—Br bond moments are
antiparallel. The individual C—Br bond moments, therefore, cancel each other and
so, the molecule has no net dipole moment.
7. The molecule of any compound composed of two dissimilar elements is

always polar–justify the statement.
Solution The statement is not always true. A diatomic molecule (A—B) consisting of two
dissimilar elements with different electronegativities is always polar because there is no
possibility of cancellation of moment of this only polar bond. HBr, HF, etc. are examples of
such molecules. A polyatomic molecule consisting atoms of two dissimilar elements may
or may not be polar. Polarity of such molecules depends on their geometrical shapes. If
the molecule is symmetrical, i.e., if the individual bond moments cancel each other, it will
be nonpolar and if not it will be polar. The linear triatomic molecule CO2, for example, is
nonpolar (m = 0 D) because the two oppositely oriented C==O bond moments cancel each
other. However, the angular H2O molecule is polar because the two O—H bond moments
and the moments caused by two unshared pairs do not cancel out each other and a resultant
moment operates in the molecule.
8. Identify the three isomeric chlorotoluenes having dipole moments:

1.35 D, 1.9 D and 1.78 D.
Solution Methyl group (—CH3) can exert a +I effect while the Cl atom can exert a —I
effect. In p-chlorotoluene, the —CH3 and —Cl group moments act linearly. So, its dipole
moment will be equal to the sum of the two group moments and hence the dipole moment
of the para-isomer will be the maximum. If the —CH3 group moment is considered to act
in a direction away from the ring, then in m-chlorotoluene, these two group moments
act at an angle of 60°. Hence, the dipole moment of m-chlorotoluene is expected to be
somewhat less than that of the p-isomer. In o-chlorotoluene, these two group moments act
an angle of 120° and hence the dipole moment of this isomer is expected to be somewhat
less than that of the m-isomer. Therefore, the dipole moment values of 1.35 D, 1.78 D and
1.90 D correspond to o-, m- and p-chlorotoluene, respectively
9. How can you account for the following fact?

(a) p-diacetylbenzene is polar but p-dinitrobenzene is nonpolar.
(b) The group moments for —NH2 and —NO2 are 1.53 D and 3.95 D, respectively,
but the measured dipole moment of p-nitroaniline is 6.20 D.
Solution
(a) In p-diacetylbenzene, the two freely rotating C==O group moments cancel each
other only in one conformation in which they are oriented anti to each other. This
particular conformation, therefore, is nonpolar. In all other conformations, the two
C==O group moments make a particular angle with each other. Thus, a resulting
moment operates in all these conformations, i.e., in all these conformations, the
compound is polar. Because of this, p-diacetylbenzene possesses an appreciable
dipole moment, i.e., it is a polar molecule.
In p-dinitrobenzene, the group moments of the two freely rotating —NO2 groups
act in the same line and thus, cancel each other. For this reason, p-dinitrobenzene
does not have net dipole moment, i.e., the molecule is a nonpolar one.
(b) The dipole moment of p-nitroaniline is expected to be the sum of the moments
caused by the —NH2 and —NO2 groups because these group moments act linearly
in the same direction. But the dipole moments of p-nitroaniline are somewhat
greater than the sum of the two group moments. In this compound, the —NO2 group
with draws electrons by its —R effect while the —NH2 group donates electrons by
its +R effect. As a consequence, there occurs a very effective resonance interaction
which results in a much higher value of e. For this reason, the dipole moment of
p-nitroaniline (6.20 D) is larger than the sum of the values of nitrobenzene (3.95
D) and aniline (1.53 D).
10. Determine the formal charge on each atom in the following species:
!
È ˘ –
Í ˙ :O:
(a) ÍH — O — H ˙ (b) O == O — O: (c)
Í | ˙ N
Î H ˚ O=
=
O:
!
H H È H ˘
| | Í | ˙
(d) H—N— B —H (e) ÍH3C — N — H ˙ (f) CH 3 - N ∫ C :
| | Í | ˙
Í ˙
H H Î H ˚
(g) :C ∫∫ O:
Solution
!
ÈH — O — H ˘
(a) Í | ˙ The formal charge on the O atom = V – (N + B/2) = 6 – (2 + 6/2)
Í ˙
ÍÎ H ˙˚
= + 1 and on each H atom = 1 – (0 +2/2) = 0
(b) =O
:O = O:
The formal charge on the left-hand O atom = 6 – (4 + 4/2) = 0, on the right-hand O atom
= 6 – (6 + 2/2) = –1 and on the middle O atom = 6 – (2 + 6/2) = + 1.
–
:O:
(c)
N
=
O= O:
The formal charge on the double bonded O atom = 6 – (4 + 4/2) = 0, on each of the
single bonded O atom = 6 – (6 + 4/2) = –1 and on the N atom = 5 – (0 + 8/2) = + 1.
H H
| |
(d) H—N— B —H
| |
H H
The formal charge on the B atom = 3 – (0 + 8/2) = –1, on the nitrogen atom
= 5 – (0 + 8/2)= +1 and on each H atom = 1 – (0 + 2/2) = 0.
H H
!
È ˘
Í | | ˙
(e) ÍH — C — N — H ˙
Í | | ˙
Í ˙
Î H H ˚
The formal charge on the C atom = 4 – (0 + 8/2) = 0, on the N atom = 5 – (0 + 8/2
= +1 and on each H atom = 1 – (0 + 2/2) = 0
H
|
(f) H — C — N ∫∫ C :
|
H
The formal charge on the left-hand C atom = 4 – (0 + 8/2) = 0, on the N atom
= 5 – (0 + 8/2) = + 1 and on the right-hand C atom = 4 – (2 + 6/2) = – 1.
(g) :C ∫∫ O: The formal charge on the C atom = 4 – (2 + 6/2) = –1 and on the O atom
= 6 – (2 + 6/2) = + 1.
1. Explain why the dipole moment p-chlorophenol is higher than that of p-fluorophenol.
2. Ethylene glycol has a higher value of dipole moment than 1,2-dimethoxy ethane
—why ?
3. Boron trifluoride (BF3) has no dipole moment (m = 0 D). Explain how this observation
confirms the geometry of boron trifluoride predicted by VSEPR theory.
4. Sulphur dioxide (SO2) has a dipole moment (m = 1.63 D), but carbon dioxide (CO2)
has no dipole moment (m = 0D). What do these facts indicate about the geometry of
sulphur dioxide?
5. Although the H—F bond length is smaller (0.92 Å) than H—Cl bond length (1.27 Å),
HF has a dipole moment larger than HCl (1.75 D compared to 1.03 D). Explain.
6. Although fluorine is more electronegative than chlorine chloromethane (CH3Cl)
has larger dipole moment (1.87 D) than fluoromethane, CH3F (1.81 D). Explain.
7. NF3 (0.26 D) has much smaller dipole moment than NH3 (1.46 D), even though
N—F bonds are more polar than N—H bonds. Explain.
8. Explain why the dipole moment of ethyl fluoride (CH3CH2F) is larger than that of
vinyl fluoride (CH2 == CH — F).
9. The dipole moment of chlorobenzene is larger than that of fluorobenze, even though
fluorine is more electronegative than chlorine. Explain.
10. Azulene is polar but its isomer naphthalene is nonpolar—why?
11. Compare the dipole moments of the following compounds with reasons:
(a)
(b)
[Hint: (a) The first compound is more polar because the electron-donating (+R)–
OMe group is attached to the positively polarized seven-membered aromatic ring
and the electron-withdrawing (–R) —CN group is attached to negatively polarized
five-memebered aromatic ring. The first ring is actually a cycloheptatrienyl cation
system and the second ring is actually a cyclopentadienyl system.]
12. 1,4-Dioxane has no dipole moment. Predict the form (boat or chair) in which it
exists.
13. Oxalic acid has dipole moment of zero in the gas phase. Explain.
14. 4-Aminopyridine has a larger dipole moment (4.4 D) than
4-cyanopyridine, (1.6 D). Explain.
15. Which one have lower dipole moment: C2H5CN or C2H5NC? Explain.
[Hint: In C2H5—C ∫∫ N, the s- and p-moments operate in the same direction, whereas
in C2H5—N==C:, the s- and p-moments operate in the opposite directions.]
16. The dipole moment of CH3Cl is greater (1.87 D) than the C–Cl bond moment (1.5
D) — why ?
17. Why 1-butyne has a larger dipole moment than 1-butene?
≈ @
18. Determine the formal charge on C in (a ) a carbocation, CH3 ; (b) a carbanion, CH 3 ;
(c) a free radical, CH 3 and a carbene, :CH2.
19. Find out the correct Lewis structure for
N2O (:N ∫∫ N— O: or :N == O == N:) and Cl2 O (:O — Cl — Cl: or :Cl — O — Cl:) by

determining formal charges of each atom.
20. Carbon dioxide molecule has no net dipole moment. How can this observation be
used to prove that the shape of the molecule is not angular?
21. The following molecule shows very high dipole moment. Explain the reason.
1.3 MOLECULAR FORMULA AS A CLUE TO STRUCTURE: DOUBLE BOND

EQUIVALENT (DBE) OR INDEX OF HYDROGEN DEFICIENCY (IHD)
The molecular formula of a compound provides more information about the structure of the
compound than might be apparent at first glance. For determining the structure of an organic
compound, it is necessary to ascertain whether the compound is an unsaturated one or not
and if it is an unsaturated compound, the degree of unsaturation is also needed to be known.
In an organic compound, the degree of unsaturation is expressed in terms of Double Bond
Equivalent (DBE). It is also called the Index Hydrogen Deficiency (IHD). If a hydrocarbon
contains two hydrogen atoms less than the alkane containing the same number of C atoms,
its double bond equivalent is 1, i.e., the compound may contain one double bond or a ring.
The compound C4H8, for example, contains two H atoms less than the alkane (butane,
C4H10) containing the same number of C atoms. So its double bond equivalent is 1. Thus,
the compound may have one double bond or it may be a cyclic one. That is, the compound
may be but-1-ene (CH3CH2CH == CH2) or but-2-ene (CH3CH == CHCH3) or cyclobutane
( ). Similarly, the double bond equivalent 2 indicates the presence of two double bond or
one triple bond or one double bond and one ring or two rings in a compound. So, the term
SODAR (Sum of Double bonds And Rings) is also frequently used. From the molecular
formula of a compound, its Double Bond Equivalent (or Index of Hydrogen Deficiency or
Sum of Double bonds And Rings) can easily be calculated.
The double bond equivalent (DBE) of a compound =

Â n(v - 2) + 1, where n is the number
2
of different types of atom present in the compound and v is the valency of each type of
atom.
It is to be noted that if the value of DBE of a compound is < 4, the compound is not a
benzenoid aromatic compound.
1. Calculate the double bond equivalent of a compound with the molecular

formula C6H8. Indicate whether the compound is an aromatic one or not.
6(4 - 2) + 8(1 - 2) 12 - 8
Solution Double bond equivalent of the compound = +1 = +1= 3
2 2
The compound is not an aromatic one because its double bond equivalent is less than 4.
2. Determine the double bond equivalent of each of the following compounds.
Each consumes 2 moles of hydrogen on catalytic hydrogenation. How
many rings are present in each of these compounds?
(a) C8H8Cl2 (b) C8H10O2 (c) C5H6Br2 (d) C8H9Cl0
Solution
(a) The DBE of the compound
8(4 - 2) + 8(1 - 2) + 2(1 - 2) 16 - 8 - 2 6
= +1= +1= +1= 4
2 2 2
The compound consumes 2 moles of hydrogen per mole on catalytic hydrogenation.
So, there are two double bonds or one triple bond and (4 – 2) or two rings in the
compound.
(b) The DBE of the compound
8(4 - 2) + 10(1 - 2) + 2(2 - 2) 16 - 10 6
= +1= +1= +1= 4
2 2 2
The compound consumes 2 moles of hydrogen per mole on catalytic hydrogenation.
So, the compound contains two double bonds or one triple bond and (4 – 2) or two rings.
(c) The DBE of the compound
5(4 - 2) + 6(1 - 2) + 2(1 - 2) 10 - 6 - 2 2
= +1= +1= +1= 2
2 2 2
Since the compound consumes 2 moles of hydrogen per mole on catalytic
hydrogenation, therefore, the compound contains two double bonds or one triple
bond and no ring.
(d) The DBE of the compound
8(4 - 2) + 9(1 - 2) + 1(1 - 2) + 1(2 - 2) 16 - 9 - 1 6
= +1= +1= +1= 4
2 2 2
Since the compound consumes 2 moles of hydrogen per mole on catalytic
hydrogenation, therefore, it contains two double bonds or one triple bond and
(4 –2) or two rings.
3. Write structures and the IUPAC name of all the isomeric compounds
(molecular formula: C4H6) by determining the double bond equivalent.
4(4 - 2) + 6(1 - 2) 8-6 2
Solution The DBE of the compound = +1= +1= +1= 2
2 2 2
The compound, therefore, contains two double bonds or one triple bond or one double bond
and one ring or two rings. The following nine isomers of the compound are possible:
1. CH2==CH—CH==CH2 (Buta-1,3-diene)
2. CH2==C==CH—CH3 (Buta-1,2-diene)
3. CH3CH2C∫∫CH (But-1-yne)
4. CH3C∫∫CCH3 (But-2-yne)
5. (cyclobutene)
6. (1-Methyleyelopropene)
7. CH3 (3-Methylcyclopropene)
8. == CH2 (Methylenecyclopropane)
9. (Bicyle[1.1.0]butane)
1. Calculate the double bond equivalent (DBE) of each of the following compounds:
(a) C13H9BrS (b) C12H16N2O4
2. Calculate the double bond equivalent (DBE) of a compound with the molecular
formula C6H10. The compound consumes 2 moles of hydrogen on catalytic
hydrogenation. Write structures and the IUPAC names of all the possible isomers
of the compound.
3. How does one distinguish between rings and double bonds?
4. Calculate the double bond equivalent (DBE) for each molecular formula:
(a) C6H11Br (b) C5H8O (c) C8H9N and propose one possible structure for each of
these compounds.
5. Which atoms can be ignored during the calculation of double bond equivalent?
1.4 ACIDS AND BASES

Several organic compounds behave as acids and bases. Two important theories have been
discussed below:
1.4.1 Brönsted-Lowry Theory of Acids and Bases

According to the Brönsted-Lowry theory, a substance that can donate a proton is called
a Brönsted acid (represented by the symbol H–A) while a substance that can accept a
proton is called a Brönsted base (represented by the symbol B:). In the following reaction,
hydrochloric acid (HCl) is a Brönsted acid because it donates a proton to water while
water is a Brönsted base because it accepts a proton from HCl by using one of its unshared
electron pairs. Similarly, in the reverse reaction H3O≈ is a Brönsted acid because it donates
proton to Cl① and Cl① is Brönsted base because it accepts a proton from H3O≈.
It is to be noted that all Brönsted acids contain a proton and the net charge may be zero,
≈ or (e.g., HCl, H3O≈, HSO4 , etc.) while all Brönsted bases contain a lone pair of
electrons or a p bond and the net charge may be zero or (e.g., NH 3 , CH2 == CH2, OH①,
etc.).
The molecule or ion that forms when an acid loses its proton is called the conjugate base
of that acid. The chloride ion (Cl①) is, therefore, the conjugate base of HCl. The molecule
or ion that forms when a base accepts a proton is called the conjugate acid of that base.
The hydronium ion (H3O≈) is, therefore, the conjugate acid of water. Therefore, in any
Brönsted acid-base reaction, there are two conjugate acid-base pairs.
Acid strength and pKa: Acidity is a measure of how easily a compound gives up a proton
while basicity is a measure of how well a compound shares its electrons with a proton.
A strong acid is one that gives up its proton easily and hence its conjugate base must be
weak because it has little affinity for the proton. A weak acid, on the other hand, is one
that gives up its proton with difficulty and so, its conjugate base is strong because it has
high affinity for the proton. Thus, the important relationship which exists between an
acid and its conjugate base is: the stronger the acid, the weaker its conjugate base.
For example, HI is a stronger acid than HCl. This means that I① is a weaker base than
Cl①. Similarly, the stronger the base, the weaker its conjugate acid. For example, OH①
is a stronger base than NH3. This means that H2O is a weaker acid than NH ! 4 . When a
strong acid, like HCl, HI, HNO3, H2SO4, etc., is dissociated in water, it dissociates almost
completely, which means that products are favoured at equilibrium. When a weaker acid,
like CH3COOH, C6H5COOH, etc., is dissociated in water, it dissociates only to a small
extent, so reactants are favoured at equilibrium.
Acidity is measured by an equilibrium constant, when a Brönsted acid H—A is dissolved
in water, an acid-base reaction occurs. The transfer of proton from the acid to water can
be represented by the following equilibrium:
!
ææ
H - A + H2O: ¨æÆ A:@ + H3 O:
æ
[A:@ ] [H3O:! ]
K eq =
[HA] [H2O]
For dilute aqueous solutions, the concentration of water remains effectively constant and
so, the expression for the equilibrium constant can be rewritten in terms of a new constant,
Ka, called the acid dissociation constant or acidity constant.
[A:@ ] [H 3O]
K a = K eq .[H 2O] =
[HA]
A large value of Ka means that the acid is a strong acid and a small value of Ka means
that the acid is a weak acid. The strong acid hydrochloric acid, for example, has an acidity
constant of 107, whereas the weak acid acetic acid has an acidity constant of only 1.74 ×
10–5. For convenience, the strength of an acid is generally indicated by its pKa value rather
than its Ka value.
pKa = – log Ka
Thus, the stronger the acid, the larger its Ka and consequently, the smaller is its pKa. Very
strong acids have pKa values < 1, moderately strong acids have pKa values 1 – 5, weak
acids have pKa values 5 – 15 and extremely weak acids have pKa values > 15.
Scale of acidi es and basici es

Acid Ka pKa Conjugate
base
CH3CH3 10–50 50 CH3CH2@
CH2 == CH2 10–44 44 CH2==CH@
Increasing Strength (basicity) of the conjugate base

H—H 10–35 35 H@
Increasing Strength (acidity) of the acid
NH3 10–33 33 NH2@

HC ∫∫ CH 10–25 25 HC ∫∫ C@
CH3CH2OH 10–16 16 CH3CH2O@
H2O 10–15.7 15.7 OH@
–4.8
CH3COOH 10 4.8 CH3COO@
CF3COOH 1 0 CF3COO@
HNO3 20 –1.3 NO3@
H3O !
50 –1.7 H2O
7
HCl 10 –7 Cl@
H2SO4 109 –9 HSO4@
HI 1010 –10 I@
HClO4 1010 –10 ClO4@
Since a strong acid readily donates a proton and a strong base readily accepts a proton,
therefore, these two species react to form a weaker conjugate acid and a weaker conjugate
base that do not donate or accept a proton readily. Thus, the equilibrium always
favours the formation of the weaker acid and weaker base.
The order of acidity of some of the important weak acids that we generally encounter is
as follow:
RH < RCH == CH2 < H2 < NH3 < RC ≡≡ CH < ROH < H2O < RCOOH
Acidity increases
Like acids, the stronger the base, the larger its Kb and consequently smaller is it pKb.
It is, however, more usual to describe the strength of base also in terms of Ka and pKa,
thereby establishing a single continuous scale for both acids and bases. To make this
possible, the following equilibrium is considered.
BH≈ ææ
+ H2O ¨æÆ B: + H3O
æ
Conjugate acid Base
of the Base
[B :][H 3O≈ ]
Therefore, Ka =
[BH≈ ]
Also, pKa = – log Ka

Thus, weaker the base B: or stronger its conjugate acid BH≈, the larger is its Ka and
consequently the smaller is its pKa.
1.4.1.1 Factors Affecting Acidity

Acid strength is determined by the stability of the conjugate base that is formed when the
acid gives up its proton. Any factor that stabilizes conjugate base A① makes the starting
acid HA more acidic.
Four factors generally affect the acidity of acid H–A and these are: (1) element effect,
(2) inductive effect, (3) resonance effect and (4) hybridization effect. In this article we will
discuss the first and last effects. We will discuss the other two in the subsequent articles.
Element effect The most important factor that determines the acidity of a Brönsted acid
is the identity of the atom (i.e., the location of the atom in the periodic table) to which the
acidic hydrogen is attached. The effect of the attached atom A on the acidity of a
Brönsted acid H—A is termed as element effect. Across a row (period) of the periodic
table, Brönsted acidity increases as the electronegativity of the atom to which the acidic
hydrogen is attached increases.
For example:
H — CH3 H — NH2 H — OH H—F
pKa 50 33 15.7 3.2
Increasing electronegativity
Increasing acidity
Down the column (group) of the periodic table, Brönsted acidity increases as the size of the
atom to which the acidic hydrogen is attached increases. For example:
H—F H — Cl H — Br H—I
pKa 3.2 –7 –9 –10
Increasing size
Increasing acidity
Explanation of these Trends: Although the ionization of H–A occurs in one step and
involves a base to accept the proton, we may think of the actual process as the sum of
three processes for understanding the observed trends in acidity. These three processes
are as follows:
Bond breaking: (1)
e – + A.Æ A:
@
Electron transfer to A.: (2)
Loss of an electron from H.: H.Æ H≈ + e – (3)

If the identical terms from either side are cancelled, the sum of these steps is the overall
dissociation reaction. Anything that makes any of these steps more fovourable tends to
increase the acidity of the Brönsted acid. Let us consider the energetics of each of these
steps. In the first step, hemolytic cleavage of the H—A bond occurs. Hence, the energy
required for this step is called the bond dissociation enthalpy. Trends in bond dissociation
enthalpy are shown by the data given below:
Within group 17 (VIII A) of the periodic table

Bond H—F H — Cl H — Br H—I
–1
Bond dissociation enthalpy (kJ mol ) 569 431 368 297
Within the second period of the periodic table

Bond H — CH3 H — NH2 H — OH H—F
–1
Bond dissociation enthalpy (kJ mol ) 439 448 498 569
Since the bond dissociation enthalpy is the amount of energy required for the dissociation
of a bond, therefore, smaller amounts represent more favorable reactions.
In the second step, an atom or group A. accepts an electron from the corresponding anion.
The energy released in this step is, in fact, the electron gain enthalpy of .A. Trends in
electron gain enthalpy are shown by the data given below:
Within group 17 (VIII A) of the periodic table
Atom :F ◊ :Cl ◊ :Br ◊ :I ◊
Electron gain enthalpy (kJ mol–1) 328 349 324 295
Within the second period of the periodic table
Group ◊CH3 ◊NH3 ◊OH ◊F:

Electron gain enthalpy (kJ mol–1) 7.7 74 177 328
Since the electron gain enthalpy is the energy released when a group combines with an
electron, therefore, larger amounts represent more favourable reactions.
In the third step, the hydrogen atom loses an electron. The energy required for this step is
the ionization enthalpy of the hydrogen atom. Since this step is the same for all the Brönsted
acids, it does not enter into comparison of different acids. From the bond dissociation
enthalpy and electron gain enthalpy data, it becomes clear that within a column (group)
of the periodic table, electron gain enthalpies do not change as much as bond dissociation
enthalpies. For example, the electron gain enthalpies of Cl and I differ by only 54 kJ mol–1
while the bond dissociation enthalpies of H — Cl and H — I differ 134 kJ mol–1. Hence,
the greater strength of the Brönsted acids H — A towards higher atomic number or larger
size within a column (group) of the periodic table is due primarily to weaker H — A bonds.
Across a row (period) of the periodic table, electron gain enthalpies change much more
than the bond dissociation enthalpies. The increase in electron affinities from •CH3 to •F is
320 kJ mol–1, whereas the increase in bond dissociation enthalpies from H — CH3 to H — F
is only 130 kJ mol–1. The direction of the change in bond dissociation enthalpies is also to
be noted. Hence, the greater strength of the Brönsted acid H — A from left to right along
a row (period) of the periodic table is due primarily to the ability of the atoms or groups A
to attract electrons. This trend, in fact is similar to the trend in electronegativities which
is actually the ability to attract electrons within a chemical bond.
Hybridization effect: The hybridization of the atom bonded to hydrogen is an important
factor affecting the acidity of H — A. Let us consider ethane (CH3CH3), ethylene
(CH2 == CH2) and acetylene (HC ≡≡ CH), the three different compounds containing C—H
bonds to illustrate the phenomenon. From the pKa values, it becomes clear that acetylene
is more acidic than ethylene which in turn is more acidic than ethane.
This order of acidities can be explained on the basis of the hybridization state of carbon
in each compound. Electrons of 2s orbital have lower energy than those of 2p orbitals
and this is because 2s electrons tend, on the average, to be much closer to the nucleus
than 2p electrons. Therefore, the electrons in an sp hybrid orbital having 50% s character
(because they arise from one s orbital and one p orbital) are closer, on the average, to the
carbon nucleus than those in an sp2 hybrid orbital having 33.3% s character. In turn, the
electrons in an sp2 hybrid orbital are closer to the nucleus than those in an sp3 hybrid
orbital having 25% s character. The closer the bonding electrons to the nucleus, the more
electronegative the atom. This means, in effect, the sp-hybridized carbon in acetylene is
more electronegative than the sp2-hybridized carbon in ethylene, which in turn is more
electronegative than the sp3-hybridized carbon in ethane. Because the electronegativity
of carbon atoms follows the order Csp > C 2 > C 3 , proton release will be progressively
sp sp
favoured in ethane, ethylene and acetylene, i.e., the order of acidity is HC ≡≡ CH > H2C == CH2
> CH3—CH3.
This order may also be explained on the basis of the stability of the carbanion, i.e.,
the conjugate base of the carbon acid. The more the carbanion is stable, the more the
hydrocarbon is acidic. As the amount of s character of the hybrid orbital occupied by the
unshared electron pair increases, the stability of the carbanion increases and this is due
to the fact that the electron pair in an orbital having good s character is held more tightly
by the nucleus and hence of lower energy.
1.4.2 Lewis Acid-Base Theory

According to G.N. Lewis, an acid is a species that can accept an electron pair and a base is
a species that can donate an electron pair. All Brönsted acids are Lewis acids because they
can lose a proton and the proton can accept an electron pair. The Lewis definition of an
acid is much broader than the Brönsted-Lowry definition and this is because according to
Lewis definition not only the compounds that can donate protons but also the compounds
like aluminium chloride (AlCl3), boron trifluoride (BF3) etc., are acids because they have
sextet of electrons (vacant orbitals in their valence shell) and thus can accept a pair of
electrons. The latter compounds are not Brönsted acids because they are not proton donors.
They react with a compound containing an unshared electron pair just like a proton reacts
with ammonia.
A curved arrow is drawn from the electron pair of the base to the electron-deficient atom
of the acid. The term ‘Lewis acid’ is generally used to refer to non-proton-donating acids
like AlCl3, BF3, FeCl3, etc. All Lewis bases are Brönsted bases and vice versa. They have
an unshared pair of electrons that they can share. They can share electrons either with an
atom like Al, Fe, B or they can share the electrons with H≈ . For example, water, diethyl
ether and ammonia are both Lewis bases and Brönsted bases.
1. Predict the relative acidity of (a) ethyl alcohol (CH3CH2OH) and ethylamine
(CH3CH2NH2); (b) ethyl alcohol (CH3CH2OH) and ethanethiol (CH3CH2SH);
≈ ≈
(c) CH 3 OH 2 and CH 3NH 3
Solution The nature of the atom holding the proton to be lost, i.e., the atom that will
hold the unshared pair of electrons in the conjugate base being formed is to be considered.
The better this atom accommodates these electrons, the greater the extent to which the
conjugate base is formed, and hence, the stronger is the acid.
H: A ææ
+ :B ¨æÆ
æ :A @ + H:B≈
Acid Conjugae base
(holds proton) (holds the unshared
pair of electrons)
Two factors that determine an ability to accommodate electrons are (i) its electronegativity,
since a more electronegative atom has a greater avidity for electrons: and (ii) its size,
since a larger atom permits greater dispersal of charge of the electrons and this tends to
stabilize a charged particle. The electronegativity of atoms increases from left to right
across a period of the Periodic Table and size of atoms increases on moving down the group
of the Periodic Table.
(a) relative acidity: CH3CH2OH > CH3CH2NH2
Nitrogen and oxygen are in the same period or row of the Periodic Table and oxygen is
more electronegative than nitrogen.
(b) relative acidity: CH3CH2SH > CH3CH2OH
Sulphur and oxygen are in the same group of the Periodic Table, and sulphur is larger in
size.
≈ ≈
(c) relative acidity: CH3 OH2 > CH3 NH2
Oxygen and nitrogen are in the same period or row of the Periodic Table and being more
electronegative oxygen is less able to accommodate a positive charge.
≈
2. Which is the stronger acid of each pair: (a) CH 3 O H 2 or CH 3 OH;
≈
(b) CH 3 N H 2 or CH 3 NH 2 ; (c) H2S or HS①; (d) OH① or H2O
Solution The accommodation of the electron pair left behind upon loss of a proton is
easiest for the neutral conjugate base formed from a positively charged acid [equation (i)],
harder for the negatively charged conjugate base formed from a neutral acid [equation (ii)]
and still harder for the doubly charged conjugate base formed from a negatively charged
acid [equation (iii)].
≈
ææ
H3O≈ + : B ¨æÆ H2O + H:B
æ … (i)
Positive Neutral
≈
ææ
H2O + :B ¨æÆ
æ OH
@
+ H:B … (ii)
Neutral Negative
≈
ææ
OH@ + :B ¨æÆ
æ O2 @ + H:B … (iii)
Negative Doubly negative
≈ ≈
Therefore, (a) CH3 OH2 is more acidic than CH3OH, (b) CH 3 NH 2 is more acidic than
CH3NH2, (c) H2S is more acidic than SH① and (d) H2O is more acidic than OH①.
3. Arrange the members of each group in order of decreasing basicity:

(a) OH①, SH①, SeH① (b) F①, Cl①, Br①, I① (c) F①, OH①, NH2①, CH3① (d) Cl①, SH①.
Solution In (a) and (b), the atoms of each set are in the same group of the Periodic Table,
and thus the negative charge will be better accommodated, i.e., the basicity of the anion
decreases with increase in atomic size down the group. Therefore, the decreasing order of
basicity is as follows: (a) OH① > SH① > SeH①; (b) F① > Cl① > Br① > I①.
In (c) and (d), the atoms of each set are in the same period of the Periodic Table, and
therefore, the accommodation of the negative charge depends upon electronegativity: the
more electronegative the element on moving from left to right in the Periodic Table, the
weaker the base. Therefore, the decreasing order of basicity is as follows: (a) CH3① > NH2①
> OH① > F① ; (b) SH① > Cl①.
4. Arrange the member of each group in order of increasing basicity:
(a) H2S, HS①, S2①; (b) H3O⊕, H2O, OH① (c) NH3, NH2①.
Solution For a given atom, the availability of electrons is greatest in an electron-
rich negatively charged species and least in an electron-poor positively charged
species. Therefore, the increasing order of basicity is as follows: (a) H2S < HS① < S2①,
(b) H3O⊕, < H2O < OH① (c) NH3 < NH2①.
5. Arrange the following ions in order of decreasing acidity and explain the
≈ ≈
order: CH 3 CH 2 N H 3 , CH 3 CH == NH 2 , CH 3 C ∫∫ N H
Solution These three ions differ in the hybridization of the nitrogen atom to which
the acidic hydrogen is attached. It is known that hybridization of an atom affects its
electronegativly (sp is more electronegative than sp2, and sp2 is more electronegative than
sp3). Again, a more electronegative atom is less willing to accommodate the positive charge
and more willing to releasing a proton. Therefore, the decreasing order of acidity of these
≈ ≈ ≈
ions is CH 3C ∫ NH > CH 3CH = NH 2 > CH 3CH 2 NH3
6. Write down the conjugate acids of (a) CH3O①, (b) CH3CH2NH2, (c) [: O — O:]2@ ,
(d) CH2 = CH2, (e) :CH2 and the conjugate bases of (f) CH3CH2NH2,
(g) CH2 = CH2, (h) HNO3.
Solution
≈
(a) CH3OH, (b) CH 3CH 2 NH 3 , (c) HO — O:@ ,
≈ ≈ @
(d) CH3 — CH2 , (e) CH3 , (f) CH3CH2 NH ,
@
(g) CH2 == CH , (h) NO3①.
7. Which of the following species are expected to be amphoteric and why?
≈
(a) HCO3①, (b) HF, (c) N H4 , (d) NH3, (e) Cl①
Solution (a) It is amphoteric because it gives H2CO3 (CO2 + H2O) and CO32①; (b) it is
amphoteric because it gives H2F⊕ and F① ions ; (c) it cannot accept an H⊕ ion because it has
no unshared pair of electrons, and hence, it is not amphoteric; (d) it is amphoteric because
≈
it gives NH4 and NH2① ions; (f) it is not amphoteric because it cannot donate H⊕.
8. Write down the net ionic reaction taking place when sodium hydride
(NaH) is dissolved in methanol.
Solution Sodium hydride contains a sodium ion (Na⊕) and a hydride ion (H①). Although
the sodium ion is a spectator ion, the hydride ion (the conjugate base of a very weak acid
H2) is a very strong base. Because of this, the hydride ion reacts readily with a molecule of
methanol, MeOH (a stronger acid), to produce methoxide ion, MeO① (a weaker base) and
H2 (a weaker acid).
CH3OH + :H@ Æ CH3O@ + H2≠
Stronger acid Stronger Weaker base Weaker acid
(pK a = 15.5) base (pK a = 35)
9. Explain why propylamine (CH3CH2CH2NH2) is more basic than pyridine

(C6H5N), which in turn more basic than ethyl cyanide (CH3CH2CN).
Solution The basicity of these nitrogenous compounds can be explained in terms of the
hybridization state of nitrogen. The nitrogen atoms in propylamine, pyridine and ethyl
cyanide are sp3, sp2 and sp hybridized respectively, and so the nonbonding pair of electrons
on nitrogen occupies sp3 (25 percent s character), sp2 (33.3 percent s character) and sp, (50
percent s character) hybrid orbitals respectively. Since s electrons are held more tightly
to the nucleus and hence of lower energy compared to p electrons, the unshared pair of
electrons in orbitals having a large amount of s character is attracted towards the nitrogen
nucleus with a greater force and is less available for forming a bond with proton. Thus,
the availability of the unshared pair of electrons on nitrogen, i.e., the basicity of these
nitrogenous compounds decreases in the order given below.
10. But-2-ynoic acid (CH3C ∫∫ CCOOH) is a stronger acid than but-2-enoic acid
(CH3CH == CHCOOH), which in turn is a stronger acid than butanoic acid
(CH3CH2CH2COOH). Explain.
Solution The electronegativity of various hybridized carbon atoms follows the order:
C sp > C sp2 > C sp3 . The a-carbon of butanoic acid, but-2-enoic acid and but-2-ynoic acid are
sp3, sp2 and sp hybridized respectively. Therefore, the conjugate base of but-2-ynoic acid in
more stable than the conjugate base of but-2-enoic acid, which in turn is more stable than
the conjugate base of butanoic acid.
CH3CH2CH2 COO①, CH3CH == CHCOO①, CH3C ∫∫ CCOO①
Stability increases
Hence, the acidity order is CH3C ∫ CCOOH > CH3CH = CH COOH > CH3CH2CH2COOH
11. How can CH3C ∫∫ CH be converted into CH3C ∫∫ CT?
Solution CH3C ∫∫ CH is first treated with sodium amide in liquid ammonia and then T2O
is added to the resulting solution containing CH3C ∫∫ C① Na⊕ to form CH3C ∫∫ CT. The
following two acid-base reaction takes place.
1. Arrange the following compounds in order increasing acidity and explain your
answer:
CH3CH2OH, CH3CH2NH2, CH3CH2CH3
≈ @
2. NaH reacts with ethanol (CH3CH2OH) not to produce Na:CH 2CH 2OH plus H2,
@ ≈
but to produce CH 3CH 2 O Na plus H2 — why?
3. Arrange the following compounds in order of decreasing basicity and explain your
answer:
CH3ONa, CH3CH2Li, NaNH2
4. Indicate which is the stronger acid in each of the following pairs and why?
≈ ≈
(a) CH 3CH 2 NH 3 or CH 3CH 2 OH 2 (b) HCl or HBr (c) CH3OH or CH3SH
5. Designate the following species as Lewis acids or bases:

(a) :H① (b) H2C == CH2 (c) CH3CH2CH2CH3 (d) BF3
!
(e) CH3OH (f) Me3 C (g) :Cl:@
6. Which of the indicated protons in each of the following molecules is more acidic and
why?
C∫∫ C—H ¨ NH—H ¨

(a) (b) (c)
S—H ¨ CH2O—H ¨
7. Predict the conjugate acid and the conjugate base of the following compound and
give your reasons:
HO NH2
8. Arrange the following species in order of increasing acidity and explain the order:
(a) H2O, H3O⊕ HO① (b) HC ≡≡ CCH2CH3, CH3CH2 CH == CH2, CH3CH2 CH2CH3
(e) HCl, H2S, NH3
9. Arrange the following ions in order of increasing basicity and explain the order:
– – –
(a) CH2CH2 , C∫∫ C, CH== CH
(b) CH3①, Cl①, HO①,
10. Which compounds are expected to be formed when OH① is allowed to react with the
≈
carbocation Me3 C as a Brönsted-Lowry base and as a Lewis base?
[Hint: 2-methylpropene (CH2 = CMe2) will be formed when OH① will react as a
Brönsted-Lowry base and tert-butyl alcohol will be formed when OH① will react as
a Lewis base.]
11. Classify each compound as a Brönsted-Lowry base, a Lewis base, both or neither:
(a) Me2C == O (b) C2H5—Br
CH3 CH3
(c) (d)
12. Classify each compound as a Brönsted-Lowry acid, a Lewis acid, both or neither:
≈
(a) BF3 (b) BF4① (c) H3O⊕ (d) (CH 3 )3 C
13. Explain why acetylene (HC ≡≡ CH) is more acidic than ethane (CH3CH3), even
though the C—H bond in acetylene has a higher bond dissociation energy than
that the C—H bond in ethane.
14. Explain why cyclopropane is more acidic than cyclohexane.
[Hint: In cyclopropane, imperfectly hybridized sp3 orbitals of carbon are involved
in bond formation. The inner orbitals forming C—C bonds have about 17 percent
s character (approximately sp5 orbitals), while the outer orbitals forming C—H
bonds have about 33% s character (approximately sp2 orbitals). The orbitals
forming C—H bonds in cyclohexane have only 25 percent s character (sp3 orbitals).
Since the outer orbitals have greater s character, therefore, the conjugate base of
cyclopropane is relatively more stable than that of cyclohexane and for this reason,
cyclopropane is more acidic than cyclohexane.
15. Glycine is an amino acid which in solution exists in equilibrium between two
forms: ≈
H2NCH2COOH H 3 N CH2COO①
Which form is expected to be favoured at equilibrium?
[Hint:
The ionic form contains the groups that are the weaker acid and weaker base. The
equilibrium, therefore, will favour this form.]
Important fact and concepts at a glance

1. Brönsted-Lowry acids: Proton donors
2. Brönsted-Lowry bases: Proton acceptors
3. Lewis acids: Electron pair acceptors
4. Lewis bases: Electron pair donors
5. The lower the pKa, the stronger the acid (pKa = –log Ka)
6. In acid–base reactions, the equilibrium favours the weaker acid and weaker
base because they are relatively more stable.
7. The conjugate base of any acid having a higher pKa can deprotonate an acid.
8. The acidity of H–A increases both left-to-right across a period and down a group
of the periodic table.
9. The acidity of H–A increases as the percent s character of the atom in A① attached
to hydrogen increases.
10. The more stable the conjugate base, the more acidic the acid.
1.5 INDUCTIVE AND ELECTROMETRIC EFFECTS

1.5.1 Inductive Effect
When a covalent bond is formed between two atoms having different electronegativities,
the bonding electron pair is not shared absolutely equally by the two atoms. The electron
pair is attracted more by the more electronegative atom and so, it becomes shifted more
towards that atom. As a consequence, the more electronegative atom acquires a partial
negative charge (i.e., d+). When an electronegative (electron-withdrawing) F atom, for
example, is attached to the end of a carbon chain (whose carbon atoms are designated as 1,
2, 3, … etc.), the s electrons of the C-1—F bond are attracted by or displaced more towards
the F atom. As a result, the F atom acquires a partial negative charge (d-) and the carbon,
C-1, acquires a partial positive charge (d+). As C-1 is now somewhat positively polarized,
it, in turn, attracts the s electrons of the C-1 — C-2 bond towards itself. This causes C-2 to
acquire a partial positive charge (dd+) smaller in magnitude than that on C-1. Similarly,
C-3 acquired a partial positive charge (ddd+) even smaller than that on C-2.
ddd + dd + d+ d-
C Æ— C Æ— C Æ— F [where, d + > dd + > ddd + ]
3 2 1
Similarly, if an element less electronegative than carbon, such as lithium (Li) (or any other
electron-releasing group or atom) is attached to the terminal carbon of a carbon chain,
then a partial positive charge (d+) is developed on the Li atom and a partial negative
charge (d-) is developed or the C-1 atom. The small negative charge developed on C-1, in
turn, repels the s electrons of the C-1—C-2 bond towards C-2. As a result, C-2 acquired
a partial negative charge (dd-) smaller in magnitude than that on C-1. Similarly, C-3
acquired a partial negative charge (ddd-) which is even smaller than that on C-2.
ddd - dd - d- d+
C Æ— C Æ— C Æ— Li [where, d - > dd - > ddd - ]
3 2 1
Definition: When some atom or group, either more or less electronegative than carbon
remains attached to a carbon chain, a permanent bond polarization caused by displacement
of electrons occurs along the carbon chain. This is what is called inductive effect.
This type of charge dispersal, which goes on diminishing rapidly as the distance from
the source increases and almost dies down after the third carbon atom. Although the
inductive effect causes some degree of polarity in the covalent bond, yet the bond is never
cleaved due to this effect. The inductive effect is represented by the symbol ( Æ— ). The
arrow always points towards the more electronegative atom or group.
Inductive effect cannot be expressed by any absolute value. Relative inductive effect of
an atom or group is measured by taking H atom of R3C—H molecule as standard. When
an atom or group Z of the C—Z bond of R3C—Z molecule attracts the bonding electrons
more strongly than hydrogen of the C–H bond in R3C – H molecule, then according to the
definition introduced by Ingold, Z is said to have negative inductive effect or electron-
withdrawing inductive effect or –I effect. On the other hand, if the atom or group Z pushes
the bonding electrons of the C—Z bond more strongly than the hydrogen atom of the C—H
bond, then it is said to have positive inductive effect or electron-releasing inductive effect
or +I effect. Some +I and –I groups are arranged below in the order of decreasing strength
of inductive effect:
@
+I groups: ① ①
- NH > –O > –COO > (CH3)3C – > (CH3)2CH – > CH3CH2 – > CH3 – > D
≈ ≈ ≈
–I group: - NR 3 > - SR 2 > - NH 3 > –NO2 > – SO2R > – CN > – COOH > –F > –Cl > –Br
> – I > – OR > – OH
The phenomenon of inductive effect may be used to explain some important properties of
organic compounds such as acidic property, basic property, bond polarity and chemical
reactivity. Stabilities of carbocation and carbanions can also be explained by inductive
effect.
1.5.1.1 Acidity of organic compounds

(1) Strength of monocarboxylic acids
Relative strengths of monocarboxylic acids can be explained in terms of inductive effect
of the substituent present in the carbon chain. If an electron-releasing group or atom
is attached directly to the —COOH group or to the carbon chain close to the –COOH
group, then the positive inductive effect (+I) of such group increases the electron density
on the oxygen atom of the O—H group and as a consequence, the shared pair of electron
of the O—H bond is less strongly attracted towards the oxygen atom. As a result, the
dissociation of the O—H bond to give H⊕ ion is less favoured. Thus, a +I group, when
present in a monocarboxylic acid molecule, decreases the strength of that acid. On the
other hand, if an electron-withdrawing group or atom is attached to the carbon chain

close to the —COOH group, the negative inductive effect (–I) of such group decreases
the electron density on the oxygen atom of the O—H group and consequently, the O—H
bonding electrons are more strongly attracted towards the oxygen atom. As a result, the
dissociation of the O—H bond to give H⊕ ion is facilitated. Thus, a –I group, when present
in a moncarboxylic acid molecule, increases the strength of the acid. Fluoroacetic acid, for
example, is stronger than acetic acid.
O O
|| ||
CH3Æ— CÆ— O Æ—H F —¨ CH2 —¨ C —¨ O —¨ H
Acetic acid (pK a = 4.76) Fluoroacetic acid (pK a = 2.66)
(Weaker acid due to + I (Stronger acid due to - I effect
effect of the - CH3 group) of the F atom)
Acid strength can also be explained in terms of relative stabilities of the acid and its
conjugate base. The more conjugate base is stable, the more the acid will be acidic.
Electron-withdrawing groups (EWGs) disperse the negative charge on the conjugate base,
thus stabilise it and hence increase acidity. Electron-donating groups (EDG’s), on the
other hand, intensity the negative charge on the conjugate base, thus destabilise it and
hence decrease acidity.
The electron-withdrawing F atom disperses the negative charge on the conjugate base
of fluoroacetic acid and thereby stabilizes it. In the conjugate base of acetic acid, on the
other hand, the negative charge on the carboxylate anion is intensified by the electron-
releasing —CH3 group and as a consequence, the anion is destabilized. It thus follows
that fluoroacetic which yields a more stable conjugate base is relatively more acidic than
acetic acid.
The strength of carboxylic acid increases as the extent of –I effect of the substituent
increases and this is because the O—H bond becomes progressively more weak or the
conjugate base becomes progressively more stable. The –I effect of halogens follow the
order: F > Cl > Br > I. So, among the halogen substituted acetic acids, fluoroacetic acid
(FCH2COOH) is the strongest while iodoacetic acid (ICH2COOH) is the weakest.
F —¨ CH2 —¨ COOH Cl —¨ CH2 —¨ COOH Br —¨ CH2 —¨ COOH I —¨ CH2 —¨ COOH
Fluoroacetic acid Chloroacetic acid Bromoacetic acid Iodoacetic acid
(p K a = 2.66) (p K a = 2.86) (p K a = 2.90) (p K a = 3.16)
With increase in the number of electron-attracting substituents, the strength of the acid
increases and this is because the O—H bond becomes progressively more weak or the
conjugate base becomes progressively more stable. For example, trichloroacetic acid is more
stronger than dichloroacetic acid which in turn is more stronger than monochloroacetic
acid.
Acid strength increases
As the distance of the electron-attracting substituent from the carboxyl group increases,
the strength of the acid decreases and this is because the O—H bond becomes progressively
more stronger or the conjugate base becomes progressively less stable. For example,
2-chlorobutanoic acid is a stronger acid than 3-chlorobutanoic acid which in turn is a
stronger acid than 4-chlorobutanoic acid.
Cl Cl Cl
|
≠ |
≠ |
≠
CH3 - CH2 – CH—¨ COOH CH3 - CH —¨CH2 —¨ COOH CH —¨ CH2 —¨ CH2 —¨ COOH
2
2-Chlorobutanoic acid 3-Chlorobutanoic acid 4-Chlorobutanoic acid
(pK a = 2.84) (pK a = 4.06) (pK a = 4.52)
Acid strength increases
(2) Acidic character of alcohols

d- d+
The acidic nature of alcohols is due to the presence of polar O — H bond. Oxygen being
more electronegative than hydrogen withdraws electrons of the O—H bond towards
itself. As a result, the O — H bond becomes weak and alcohols exhibit acidic character by
donating its proton to strong bases. In fact, alcohols are very weak acids (pKa = 16–18).
The alkyl (–R) group by its electron-donating inductive (+I) effect increases the electron
density on oxygen (R Æ— O—H) and as a result, the polarity of O–H bond decreases. As
the number of electron-releasing alkyl (–R) group increases, the polarity of the O—H bond
progressively decreases and because of this, the acidity of alcohols decreases gradually
from primary (containing on alkyl group) to tertiary (containing three alkyl groups).
(3) Gas-phase acidity of alcohols

The relative order of acidity of different types of alcohols is reversed in the gas phase
compared with the relative order of acidity in solution.
Relative gas-phase acidity: (CH3)3COH > (CH3)2CHOH > CH3CH2OH > CH3OH.
In the gas phase, the a-alkyl substituents stabilize the charged alkoxide ions by a
polarization mechanism. That is, the electron clouds of each alkyl group distort so that
the electron density moves away from the negative charge on the aloxide oxygen, leaving a
partial positive charge nearby. As a consequence, the anion is stabilized by its favouarble
electrostatic interaction (forces of attraction) with the partial positive charges. The
stabilization of the alkoxide ion by this polarization effect increases with increase in the
number of a-alkyl substituents. Therefore, a tertiary alkoxide ion having three a-alkyl
substituents is the most stable one and the methoxide ion with no a-alkyl substituent in
the least stable one, consequently, tert-butyl alcohol, (CH3)3COH, is the most acidic in the
gas phase and methyl alcohol, CH3OH, is the least acidic in the gas phase.
1.5.1.2 Basicity of organic compounds

(1) Basic strength of amines
The availability of the unshared pair of electrons on nitrogen for protonation determines
the basic strength of nitrogenous compounds. We might expect to see an increase in basic
strength in the sequence NH3 Æ CH3NH2 Æ (CH3)2NH Æ (CH3)3N. Due to the increasing
inductive effect (+I) of successive —CH3 group making the N atom progressively more
electron rich, i.e., making the unshared pair of electrons more readily available. However,
this sequence of basic strength of amines agrees with result if measurements of basicity
are made in the gas phase or in a solvent in which H-bonding does not take place.
The observed order of basic strength of methylamines in water is (CH3)2NH(2°) >

CH3NH2 (1°) > (CH3)3N (3°). In the H-bonding solvent water, the basic strength of an
amine is determined not only by the electron availability on the N atom, but also by
the extent to which the conjugate acid, i.e., the ammonium ion is stabilized by salvation
through hydrogen bonding. As the number of electron-releasing (+I) methyl group
increases, the availability of the unshared pair of electrons on nitrogen increases and so,
the amines become progressively more basic. Again, as the number of hydrogen attached
to nitrogen of the conjugate acid decreases, the extent of salvation by water molecules
through H-bonding decreases. As a result, the stability of the conjugate acid decreases,
and the amines become progressively less basic. The net effect of these two oppositely
acting factors is that the secondary amine (CH3)2NH (with two electron-releasing —CH3
groups and two hydrogen atoms in the conjugate acid for H-bonding) becomes the most
basic and the tertiary amine (CH3)3N (with three electron-donating —CH3 groups and
only one hydrogen in the conjugate acid for H-bonding) becomes the least basic.
1.5.1.3 Stabilities of carbocations

The increasing order of stability of methyl, primary (1°), secondary (2°) and tertiary (3°)
≈ ≈ ≈ ≈
carbocations is CH 3 < CH 3 CH 2 < (CH 3 )2 CH < (CH 3 )3 C. This order of stability can be
explained on the basis of inductive effect. An electron-donating (+I) methyl group tends
to stabilize an adjacent carbocation by dispersing the positive charge. Therefore, the
greater the number of methyl groups attached to the cationic carbon, the more is the
dispersal of positive charge and hence stabler will be the carbocation. Thus, tertiary butyl
cation having three methyl groups is more stable than isopropyl cation having two methyl
groups which in turn is more stable than ethyl cation having only one methyl group and
obviously, methyl cation containing only three hydrogens is the least stable one. Hence,
the increasing order of stability is
1.5.1.4 Stabilities of carbanions

An alkyl group intensifies the negative charge on the anionic carbon of a carbanion by its
electron-releasing +I effect and thereby, destabilize it. Therefore, methyl anion containing
no alkyl group is the most stable one and tert-butyl anion containing three methyl groups
is the least stable one. Hence, the order of increasing stability is
1.5.2 Field Effect

It is known that inductive effect operates through bonds. However, there is another type
of inductive effect operating not through the bond but through space or solvent molecules.
This effect is called field effect. It is not very easy to distinguish the inductive effect from
the field effect. However, an attempt has been made in this respect by taking advantage
of the fact that the inductive effect depends on the nature of the bonds, but the field effect
depends on the geometry of the molecule. For example, in the diastereoisomers A and B, the
–I effect of the Cl atoms on the position of electron in the —COOH group and therefore, on
acidity should be the same since an equal number of bond intervene. Hence, acid strength
of the two isomeric acids should be the same. However, the acid strength of the two acids
has been found to be different. The acid B is actually more acidic than the acid A. The
conjugate base of A is destabilized by repulsive interaction because the negative ends of
d+ d-
the two C — Cl dipoles are closer to the —COO① group. Hence, the isomer B is a stronger
acid than A. This is also supported by the pKa values of the two acids.
1.5.3 Electromeric Effect

Definition: The temporary effect involving complete transfer of a pair of p electrons of a
multiple bond (double bonds such as C == C and C == O and triple bonds such as C ∫∫ C and
C ∫∫ N) to one of the multiple bonded atoms (usually the more electronegative one) in the
presence of an attacking reagent is called elctromeric effect or E-effect.
As soon as the attacking reagent is removed, the transferred p electron pair again from
the bond and the molecule reverts back to its original position. Since this effect occurs by
the presence of the attacking reagent, it takes places in the direction which facilitates the
reaction. The electromeric effect may be represented as follows:
in the presence of reagent +

–
A==B A—B
in the absence of reagent
Inductive effect also plays an important role in determining the direction of the electromeric
effect. If a +I or –I group is attached to a doubly bonded carbon, the direction of the
electromeric effect is the same as that of the inductive effect. For example:
+E effect: If the electron pair of the p bond is transferred to that doubly bonded atom to
which the attacking species gets finally attached, then the effect is called +E effect. For
example:
–E effect: If the electron pair of the p bond is transferred to that doubly bonded atom to
which the attacking species does not get finally attached, the effect is called –E effect. For
example:
Distinction between inductive effect and electromeric effect

Inductive effect Electromeric effect
(1) It involves a permanent displacement of the s (1) It involves a temporary but complete transfer
electron pair towards the more electronegative of p electron pair to the more electronegative
atom, i.e., it is a permanent effect. atom, i.e., it is a temporary effect.
(2) In this effect, the displaced electron pair does (2) The transferred electron pair occupies a new
not leave its molecular orbital and only a slight orbital.
distortion in the shape of orbitals occurs.
(3) It does not need any outside reagent for its (3) It operates only in the presence of an outside
operation. reagent.
(4) There occurs a partial separation of charges. (4) There occurs a complete separation of charges.
(5) It affects both physical properties and chemical (5) It does not affect the physical properties but it
reactivity of the molecule concerned. enhances the chemical reactivity of the molecule
concerned.
(6) Electronegativity of the bonded atoms controls (6) Its direction is always such that it favours the
the direction of inductive effect. reaction. The electronegativity of the bonded
atoms or I effect of the attached group may de-
termine the direction of electomeric effect.
1. Explain why CH3CHFCH2OH is more acidic than CH3CHBrCH2OH.

Solution These two compounds differ only in the halogen atom that is attached to C-2
of these alcohols. Since fluorine is more electronegative than bromine, therefore, there
occurs greater electron withdrawal from the O—H bond in 2-fluoro-1-propanol than from
the O—H bond in 2-bromo-1-propanol and for this reason, the first alcohol is relatively
more acidic than the second one.
F Br
|
≠ |
≠
CH3 —CH—¨ CH2 —¨ O—¨ H CH3 — CH—¨ CH2 —¨ O—¨ H
2- Fluoro -1- propanol 2- Bromo -1- propanol
(more acidic) (less acidic)
2. Which one in each pair is more acidic and why?

(a) CH3CBr2CH2OH or BrCH2CHBrCH2OH
(b) CH3CH2CH2CH2OH or CH3O CH2CH2OH
(c) (CH3)3 CCH2COOH or (CH3)3SiCH2COOH
(d) CH3CHFCH2OH or CH3CHClCH2OH
Solution
(a) The two compounds differ in the location of one of the two bromine atoms. Since
the bromine atom in the first alcohol (2,2-dibromoprop-1-ol) is closer to the O—H
bond than the bromine atom in the second alcohol (2,3-dibromprop-1-ol), therefore,
it withdraws O—H bonding electrons more effectively and causes the first alcohol
to be more acidic than the second one.
Br Br
|
≠ |
≠
CH3 — C —¨ CH2 —¨ O—¨ H Br—¨ CH2 —CH—¨ CH2 —¨ O—¨ H
|
Ø
Br 2,3- Dibromoprop-1- ol
2,2- Dibromoprop-1- ol
(less acidic)
(more acidic)
(b) The electronegative ethereal oxygen present in 2-methoxy-1-ethanol
(CH3OCH2CH2OH) withdraws O–H bonding electrons and for this reason, this
compound is more acidic than 1-butanol (CH3CH2CH2CH2OH) in which less
electronegative carbon is present in the place of oxygen.
CH3CH2CH2CH2OH CH3O—¨ CH2 —¨ CH2 —¨ O—¨ H
1- Butanol 2- methoxy -1- ethanol
(less acidic) (more acidic)
(c) Silicon, being electropositive in nature, pushes electron to the O—H bond and
makes the bond more stronger. For this reason, the silicon containing carboxylic
acid is less acidic than the other containing carbon in the place of silicon.
(CH3 )3 CCH2COOH (CH3 )3 SiCH2COOH
(d) These two alcohols differ only in the halogen atom that is attached to the middle
carbon of the molecule. Because fluorine is more electronegative than chlorine, there
in greater electron withdrawal from the O—H bond in the fluorinated compound,
causing it to be the stronger acid.
CH3 — CH—¨ CH2 —¨ O—¨ H CH3 — CH—¨ CH2 —¨ O—¨ H
|
Ø |
Ø
F Cl
2- Fluoro -1- propanol 2-Chloro-1- propanol
3. Although HCl is a weaker acid than HBr, chloroacetic acid (ClCH2COOH)

is a stronger acid than bromoacetic acid (BrCH2COOH). Explain.
Solution HBr is a stronger acid than HCl because the H—Br bond is weaker than the
H—Cl bond and also Br① being larger in size than Cl① is relatively more stable conjugate
base. In both carboxylic acids, the same kind of bond is broken (an O—H bond). Therefore,
the only factor to be considered is the electronegativities (–I effect) of the halogen atoms
that are pulling the O–H bonding electron. Because Cl is more electronegative than Br,
Cl is better at pulling electrons away from the O—H bond. Thus, the conjugate base of
chloroacetic acid is relatively more stable. Hence, ClCH2COOH is a stronger acid than
BrCH2COOH.
4. How do you account for the following facts?
(a) Ethanamine is more basic than 2,2,2-trifluroro-1-ethanamine
(F3CCH NH2)
(b) The amine (C2F5)3N is virtually non-basic.
Solution
(a) It is the availability of the unshared pair of electrons on nitrogen for protonation
which determines the basic strength of nitrogenous compounds. The —C2H5 group
in ethanamine (C2H5 N H 2 ) pushes electrons towards the nitrogen atom and makes
the unshared pair of electrons on nitrogen easily available. In 2,2,2-trifluoro-1-
ethanamine (F3C CH 2 N H 2) , on the other hand, the electron-attracting fluorine
atom or trifluoromethyl group (F3C –) pulls electrons from nitrogen through
carbon and thereby decreases the availability of the unshared pair considerably.
Ethanamine is, therefore, more basic than 2,2,2-trifluoro-1-ethanamine.
This observation may also be explained in terms of the stability of the ammonium ion,
≈
the conjugate acid of the amine (R - NH 3 ) . The more the ammonium ion in stable,
the more the amine is willing to take up a proton, i.e., the more the amine is basic.
The electron-releasing —C2H5 group in ethanamine stabilizes the corresponding
≈
ammonium ion (C2 H5Æ— NH3 ) by delocalizing the positive charge, whereas the
electron-attracting –CF3 group in 2,2,2-trifluoro-1-ethanamine destabilizes the
≈
corresponding ammonium ion (CF3 —¨ CH2 —¨ NH3 ) by intensifying the positive
charge. Ethanamine is, therefore, more basic than 2,2,-trifluoro-1-ethanamine.
(b) Due to the presence of three strongly electron-attracting pentafluoroethyl (–C2F5)
groups in (C2F5 )3 N, the availability of the unshared pair of electron on nitrogen is
markedly decreased and because of this, this amine is virtually non-basic.
5. Malonic acid, HO2CCH2CO2H (a dicarboxylic acid) is stronger in its

first ionization (pKa1 = 2.83) than acetic acid, CH3COOH (analogous
monocarboxylic acid; pKa = 4.76) but weaker in its second ionization (pKa2
= 5.69) than acetic acid. Explain these observations.
Solution Malonic acid is stronger in its first ionization than acetic acid because a second
carboxyl group (—COOH) in the acid exerts its –I effect on the first and thereby facilitates
the release of the first proton by making the O—H bond weaker compared to acetic acid.
O
==
–I group HO C CH C O H
2 2
The expulsion of second proton from the conjugate base (HO2CCH2COO①) is difficult
because the electron-donating —COO① group makes the O—H bond stronger compared
to acetic acid. Furthermore, the resulting dianion is destabilized by having two negative
charges in close proximity.
Therefore, malonic acid is stronger in its first ionization but weaker in its second ionization
than acetic acid.
6. In the absence of a solvent (i.e., in the gas phase), most carboxylic acids
are far weaker than they are in solution. For example, pKa of acetic acid
in aqueous medium is 4.75 but in the gas phase, it is about 130. Explain.
Solution When an acetic acid molecule donates a proton to a water molecule in the gas
phase, the separation of the oppositely charged ions is very difficult.
In solution, solvent (water) molecules surround the ions, insulating them from one another,
stabilizing them and making it far easier to separate them than in the gas phase. It is
for this reason, in the gas phase most carboxylic acids are far weaker than they are in
solution.
7. The DG∞ values for the dissociation of acetic acid and chloroacetic acid in
water at 25°C are +27 and +16 kJ mol–1, respectively, i.e., chloroacetic acid
is a stronger acid than acetic acid. Explain these thermodynamic values
by considering the effect of solvent.
Solution Water molecule solvates both the undissociated acetic acid (CH3COOH) and its
anion (CH3COO①) by forming hydrogen bonds to them. Because the water molecules are
more attracted by the negative charge, hydrogen bonding to CH3CO2① is much stronger.
This differential solvation has important consequences for the entropy change that
accompanies the ionization. Because of solvation of any species, the entropy of the solvent
decreases and that is due to the fact that solvent molecules become much more ordered
as they surround molecules of the solute. Because solvation of acetate ion (CH3CO2①) is
stronger, the solvent molecules become more orderly around it. Therefore, the entropy
change (DS°) for the ionization of acetic acid is negative and so, the TDS∞ term in the
equation DG° = DH° –TDS° makes an acid-weakening positive contribution to DG∞. In fact,
the TDS° term contributes more to the value of DG° than does DH°, and accounts for the fact
that the free-energy change for the dissociation of acetic aid is positive, i.e., unfavourable
(DG° = + 27 kJ mol–1). The Cl atom stabilizes the chloroacetate ion by exerting its –I effect.
Thus, it makes the ion less prone to cause an ordering of the solvent because it requires less
stabilization through solvation. The entropy change (DS°) for the ionization of chloroacetic
acid is, therefore, less negative. Both DH° and TDS° are more favourable for the ionization
of chloroacetic acid. The larger contribution is clearly in the entropy term. For this reason,
the DG° value for the dissociation of chloroacetic acid is less positive than that of acetic
acid. Hence, chloroacetic acid is a stronger, acid than acetic acid.
8. Arrange 1°, 2° and 3° alcohols in order of increasing rate of reaction with
sodium and explain the order. Explain why potassium instead of sodium
is used to prepare the alkoxide salt of tert-butyl alcohol.
Solution 1°, 2° and 3° alcohols may be arranged in order of increasing rate of reaction
with sodium as follows:
R3COH R2CHOH RCH2OH
Tertiary or 3° Secondary or 2° Primary or 1°

d- d+
As the number of electron-releasing alkyl (–R) group increases, the polarity of the O — H
bond gradually decreases and O—H bond becomes progressively more stronger. Therefore,
the decreasing order of acidity of primary, secondary and tertiary alcohols containing one,
two and three alkyl groups, respectively, is
Due to this order of acidity, the rate of reaction of alcohols with sodium increases
gradually from 3° to 1° alcohol. tert-Butyl alcohol, being a very weak acid, reacts with
less electropositive sodium very slowly and therefore, relatively more electropositive and
reactive metallic potassium is normally used to prepare its alkoxide.
1. Which is the weakest base and why?

(a) FCH2COONa (b) F2CHCOONa
(c) F3CCOONa (d) CH3COONa
2. Explain why trifluoromethanol is 1000 times more acidic than ethanol.
3. Account for the difference in acidities:
(a) CH3CO CH2COOH > CH3CH2CH2COOH
(b) C6H5CH2COOH > CH3CH2COOH
4. Account for the fact that Me3CCH 2CH 2 N H 2 is a stronger base than
≈
Me3 N CH 2CH 2 N H 2 .
5. Arrange the following compounds in order of increasing acidity and explain the
order:
CH3COOH, H COOH, CH3CH2COOH, (CH3)3C COOH
6. The chloro acid I is a stronger acid than the acid without the chlorine whereas
the chloro acid II is a weaker acid than the corresponding acid with no chlorine.
Explain.
COOH Cl COOH
Cl
I II
7. Explain why 4-bromobicyclo[2.2.2]octane-1-carboxylic acid (A) is a considerably
stronger acid than 5-bromopentanoic acid (B).
Br
Br COOH COOH
(A) (B)
[Hint: Consider the possible conformation and modes of transmission of the
electrical effect of the C—Br dipole.]
8. Decide which member in each of the following pairs of compounds is the stronger
base. Give your reasoning.
(a) CF3CH2CH2NH2 or CH3CH2CH2 N H 2
≈ @
(b) (CH 3 )3 N CH 2CH 2 N H 2 or O2CCH 2 CH 2NH 2

(c) N, N-dimethylmethanamine or trifluoro-N, N-bis (trifluoromethyl) methanmine
9. Arrange the following dicarboxylic acids in order of increasing pKa1 values and
explain the order.
COOH COOH COOH
HOOC COOH
COOH
10. Arrange the following carboxylate ions in order of increasing acidity and explain
the order
HOOC CH2COO①, HOOC CH2CH2COO①, HOOC COO①
11. Arrange the following carboxylic acids in order of increasing acidity and explain
the order.
≈
O2NCH2COOH, Me3 NCH2COOH , HOCH2COOH
12. The free-energy change, DG°, for the ionization of the acid HA is 21 kJ mol–1 and
for the acid HB it is –21 kJ mol–1. Which is the stronger acid and why ?
13. In which C—C bond of CH3CH2CH2Br, the inductive effect is expected to be the
least?
[Hint: The magnitude of inductive effect decreases with distance and hence the
3 2 1
effect is least in C-2—C-3 bond (CH3Æ— CH2Æ— CH2Æ— Br) ]
14. Which one of the following two compounds is more acidic and why?
COOH COOH
C C
Cl C C
Cl
I II
[Hint: II is more acidic than I and the reason is field effect.]
1.6 RESONANCE AND RESONANCE EFFECT OR MESOMERIC EFFECT

In a number of cases, it is not possible to describe the electronic structure of a species
(molecules, ions or radicals) adequately by a single Lewis structure but by a combination
of two or more structures and this is because none of these structures conforms to all the
observed properties of these substances.
Examples
2-
(1) The Lewis structure of the carbonate ion (CO3 ), in which the position of electrons
are fixed, shows that there are two C—O single bonds and one C==O double bond.
It has been determined experimentally that all the three carbon-oxygen bond
lengths in carbonate ion are equal (130 pm). This bond length is slightly greater
than that of the double bond but less than that of the single bond. Thus, the ion is
symmetrical. In fact, three structures (I, II and III) can be written for the carbonate
ion as shown below:
These three equivalent structures have the same placement of atom but a different
arrangement of nonbonding p electrons. The electron density is shared by all the
three oxygen atoms, i.e., the electrons are delocalized over the three oxygen atoms,
the actual structure of the ions lies in between these structures. The ion is actually a
resonance hybrid (weighted average) of all these structures represented frequently
by the non-Lewis structure IV and because of this, all the carbon–oxygen bonds
are found to be equal in length. These structures are called resonance structures
or canonical structures or simply canonicals and they have no real existence. In
the hybrid structure, the negative charge is distributed equally among the oxygen
atom. The delocalization of charge leads to greater stability of the ion.
(2) Benzene molecule can be represented as a resonance hybrid (III) of two Kekule
(Lewis) structures, I and II.
Neither of the two structures can fully explain all the properties of benzene. For
example, both the structures I and II contain two types of carbon–carbon bond
such as C—C (1.54 Å) and C == C (1.34 Å). But actually, it has been found that all
the six carbon–carbon bonds in benzene are equal in length (1.34 Å). This suggests
that the actual structure of benzene can neither be represented by I nor by II but
by a resonance hybrid of these two structures in which all the six carbon–carbon
bond are equal in length (1.39 Å) and lie in between carbon–carbon double bond
length of 1.34 Å and carbon–carbon single bond length of 1.54 Å. So, benzene is
quite often represented by the non-Lewis structure III. The circle inside the ring
indicates completely delocalized 6p electrons. Since I and II are exactly equivalent,
they are of exactly the same stability and make equal contribution to the hybrid.
(1) Resonance theory

This theory states that whenever a molecule or ion can be represented by two or more Lewis
structures that differ only in the positions of electrons, two things will always be true:
(i) none of these structures, which are called resonance structures or canonical structures
or simply canonicals, is a real representation for the molecule or ion and none of them
can satisfactorily explain all its physical and chemical properties of the substance;
(ii) the actual molecule or ion is better represented by a hybrid (weighted average) of these
structures and a resonance hybrid is more stable, i.e., it has lower energy than any of its
contributing structures.
The resonance structures are represented by means of a double headed arrow (´).
Resonance structures are written by shifting the positions of electrons by using curved
arrows. In reality, no such movement of electrons takes place and the different structures
are only arbitrary.
1.6.1 Resonance Energy

The difference between the enthalpy of formation of the actual molecule (observed value)
and that of the resonance structure having the lowest internal energy (obtained by
calculation) is called the resonance energy of the molecule. Resonance energy is expressed
in kcal mol–1 or kJ mol–1. The resonance energy is greater when (i) the contributing
structures are all equivalent and (ii) the number of contributing structures of roughly
comparable energy is greater. Greater the resonance energy, more will be the stability of
the compound.
1.6.1.1 Calculation of resonance energy

Resonance energy is not a measurable quantity. It can only be estimated from
thermochemical data. Resonance energy of benzene, for example, can be calculated from
heat of hydrogenation values. Heat of hydrogenation is the quantity of heat evolved
when 1 mole of an unsaturated compound is hydrogenated. Cyclohexene containing
one double bond has a heat of hydrogenation 28.6 kcal mol–1. We might reasonably
expect 1,3,5-cyclohexatriene to have a heat of hydrogenation of about three times as
large as cyclohexene, i.e., 3 × 28.6 = 85.8 kcal mol–1 . Actually, the value for benzene is
49.8 kcal mol–1. It is 36 kcal mol–1 less than the expected value. So, benzene evolves 36
kcal less energy per mole than predicted. This can only mean that benzene is more stable
than hypothetical cyclohexatriene by 36 kcal mol–1 energy. This 36 kcal mol–1 energy is the
resonance energy of benzene.
The resonance energy may also be calculated using heat of combustion value.
1.6.2 Rules for Writing Meaningful Resonance Structures

The following rules are to be followed while writing realistic resonance structures:
1. All atoms in a resonance structure must exhibit their proper valency, i.e., each
resonance structure must be a bonafide Lewis structure. So, in any resonance
structure, a carbon atom may not have five bonds or a hydrogen atom may not
have two and so on.
2. The various resonance structures should differ only in the positions of only p and
n electrons and not in the positions of atoms, i.e., the basic structure involving
s-bonds should remain unchanged.
3. The number of paired electrons in the resonance structures must be the same.
Thus, CH2 — CH2 does not contribute towards the resonance hybrid of ethylene.
4. All the atoms involved in the process of resonance must be coplanar (or nearly
coplanar).
5. All the resonance structure should have nearly the same energy.
1.6.3 Relative Contribution of Resonance Structures towards

Resonance Hybrid
All the resonance structures are not equally significant, i.e., all of them do not contribute
equally towards the resonance hybrid. Relative contributions of resonance structures
towards the resonance hybrid depend on their relative stabilities. The more the resonance
structure is stable, the more will be its contribution towards the resonance hybrid. Various
factors which govern the stability of a resonance structure and its relative contribution
towards the hybrid are as follows:
1. Nonpolar resonance structures are more stable than the dipolar resonance
structures and contribute more towards the resonance hybrid. For example:
! @
CH3 — CH == CH — CH == CH2 ´ CH3 CH — CH == CH — CH2

(More stable: major contributor) (Less stable: minor contributor)
2. Resonance structures with a greater number of covalent bond are more stable and
contribute more towards resonance hybrid.
3. The more stable and more contributing structure of an anion is one in which the
negative charge resides on the more electronegative atom while the more stable
and more contributing structure of a cation is one in which the positive charge
resides on the least electronegative atom. For example:
4. Resonance structures in which octets of all the atoms are fulfilled are relatively
more stable and hence more contributing towards the resonance hybrid. For
example:
5. Aromatic resonance structures are more stable and hence more contributing than
the non aromatic resonance structure having the same number of covalent bonds.
For example:
6. A resonance structure having two units of charge on the same atom is less stable
and hence less contributing. Again, structures having like charges on adjacent
atoms are less stable and less contributing. However, a resonance structure
having two dissimilar charges close to each other is relatively more stable and
more contributing than the structure in which the charges are further apart. For
example:
7. Resonance energy of a system involving monopolar resonance structures is greater

than that involving bipolar resonance structures. So, the former type of systems (i.e.,
molecules or ions) are relatively more stable than the latter type. For example:
8. Equivalent resonance structures are more important and more contributing than
non-equivalent resonance structures. Fro example:
1.6.4 Resonance or Mesomeric Effect

Displacement of non bonding or p-electrons from one part of a conjugate system (having
alternate single and double bond) to the other part causing permanent polarity in the
system by creating centres of high and low electron density is called resonance effect
(R-effect) or mesomeric effect (M-effect).
There are two types of resonance or mesomeric effect:
(i) +R or +M effect: An atom or a group is said to have +R or +M effect if it causes
transfer of electrons away from the atom or group attached to a double bond or a
conjugated system.
+R or +M groups: –OH, - O R, - SH, - NH 2 - Cl:, – Br:, - I:, etc. + R effect of –Cl in
vinyl chloride (Cl – CH = CH2), for example, may be shown as follows:
(ii) –R or –M Effect: An atom or a group is said to have –R or –M effect if it causes

transfer of electrons towards the atom or group attached to a double bond or a
conjugate system.
–R or –M groups: >C = O, –CHO, –COOR, –CN, –NO2, etc.
–R effect of the –CHO group in aeraldehyde (CH2==CH — CHO), for example, may
be shown as follows:
+ –
CH2==CH—CH ==O ´ CH 2—CH ==CH—O
1.6.5 Isovalent and Heterovalent Resonance

When in a resonance hybrid, each of the resonance structures contains the same number
of covalent bonds then the resonance is known as isovalent resonance. For example:
–
O O
== ´ CH3—C ==
CH3—C (Acetate ion)
–
O O
When in a resonance hybrid, each of the resonance structures does not contain the same
number of covalent bonds it is known as heterovalent resonance. For example:
≈
CH2 == CH — CH == O: ´ CH2 — CH == CH — O:@ (Acraldehyde)
1.6.6 Effect of Resonance on the Properties of Molecules

The following properties of different molecules or ions can be explained by resonance.
(1) Bond length

(a) Because of resonance, the single bond present in a molecule or ion may acquire a
partial double bond character with consequent decrease in bond length. Similarly,
the double bond may acquire some single bond character with consequent increase
in bond length. For example, due to resonance, the C—Cl bond (1.69 Å) of vinyl
chloride (CH2 == CH — Cl) becomes shorter than the C—Cl bond (1.76 Å) of ethyl
chloride and its C == C bond (1.38 Å) becomes longer than C == C bond (1.34 Å) of
ethene.
(b) Because of resonance, some ions may have bonds of equal length. For example, the
four sulphur–oxygen bonds in sulphate ion (SO42), the three nitrogen–oxygen bonds
in nitrate ion (NO3①), the two nitrogen–oxygen bonds in nitrite ion (NO2①), the two
≈
carbon–carbon bonds in allyl cation (CH2 ==CH — CH2 ) and the two carbon–carbon
@
bonds in allyl anion (CH2 ==CH — CH2 ) have the same length. Each of these ions
can be represented as a resonance hybrid of two or more equivalent resonance
structures.
From the above resonance structures it becomes clear that the two negative charges
are distributed evenly over four oxygen atoms in SO24@ and three oxygen atoms in
NO23@ , one negative charge is distributed evenly over two oxygen atoms in NO3@
@
and two carbon atoms in CH 2 = CH - CH 2 and one positive charge is distributed
≈
evenly over two carbon atoms in allyl cation, CH2 == CH — CH2 . Therefore, the
order of bonds involved in resonance in each species taking individually is same
and hence the bonds are equal in length.
(2) Bond dissociation energy The bond dissociation energy of various compounds can be
explained in terms of resonance. The bond dissociation energy for a benzylic hydrogen is
less than that of a methane hydrogen.
From the enthalpy values it becomes clear that 19 kcal/mol less energy is needed to form
the benzyl radical from toluene than to form methyl radical from methane. This difference
in energy requirement is actually the difference in energy content between the radical and
its precursor. Both toluene (C6H5CH3) and benzyl radical (C6H5CH2) can be represented
as resonance hybrids:
Toluene Benzyl radical

(less effective resonance) (more effective resonance)
Since toluene is a resonance hybrid of only two resonance structures (I and II) and the
benzyl radical is a resonance hybrid of five resonance structures (III–IV), therefore, the
benzyl radical is much more resonance-stabilized than toluene. As a result of this, the
difference in energy content between benzyl radical and toluene becomes relatively small.
On the other hand, both methane (CH4) and methyl radical CH 3 can be represented
satisfactorily by single (localized) structures.
CH4 Æ CH3 +H
Methane Methyl radical
None of these species is resonance-stabilized and thus the difference in energy content
between the methyl radical and methane is relatively large. Because of this, it requires
higher energy to form methyl radical from methane than to form benzyl radical from
toluene, i.e., the C6H5CH2—H bond dissociation energy is less than CH3—H bond
dissociation energy.
(3) Dipole moment Because of resonance, both the magnitude of the charge separated
(e) and the distance between two charge centres (d) in any molecule may increase
and so, the value of dipole moment (m = e × d) may increase. For example, the dipole
moment of nitroethene is greater than nitroethane. Due to resonance, the amount of
charge separated and the distance between the positive and negative charge centres in
nitroethene (CH2 == CHNO2) are greater as compared to nitroethane (CH3CH2NO2) in
which the carbon chain is not involved in resonance. Consequently, the dipole moment (m)
of nitroethene is greater than that of nitroethane.
1.6.6.1 Acidity of organic compounds

(1) Acidity of carboxylic acids The acidity of carboxylic acid, for example, acetic acid,
is partly due to the significant electron-withdrawing inductive effect (–I) of the carbonyl
group which makes the hydroxyl oxygen more electron deficient and thereby makes the
proton release easier.
Resonance plays an important role in making the carboxylic acids acidic. In an aqueous
medium, acetic acid dissociates as follows:
:O : O:
|| ||
ææ
CH3 — C —O—H + H2O ¨æÆ
æ CH 3 — C — O:@ + H3 O!
Acetic acid Acetate ion
Both acetic acid and its conjugate base acetate ion can be represented as resonance
hybrids:
Acetic acid is a resonance hybrid of two non equivalent resonance structures, I and
II, of which II involves separation of charges. In structure II, the two atoms of similar
electronegativtiy carry opposite charges, therefore, II should contain more energy and
hence be less stable then I. Consequently, resonance in acetic acid is not very stabilizing.
The acetate ion, on the other hand, is a resonance hybrid of two energetically equivalent
resonance structures, III and IV. Now it is known that resonance stabilization is greatest
when the contributing structures are equivalent. Therefore, the acetate ion is much
more stabilized by resonance as compared to acetic acid. This difference in stabilization
causes the equilibrium to shift to the right. Acetic acid thus behaves as an acid. Unlike
the carboxyl group, the O—H bond in ethanol, C2H5OH (an alcohol), is not weakened
by inductive electron withdrawal. Again, both ethanol and its conjugate base, i.e., the
ethoxide ion (C2H5O①), can be represented satisfactorily by single (localized) structures, V
and VI, i.e., either of these species is not stabilized by resonance. The nonbonded electron
pair remains localized on the oxygen atom. Therefore, the driving force, i.e., the relative
stabilization of the ethoxide ion with respect to the undissociated ethanol molecule, that
promotes dissociation is absent in this case. For this reason, ethanol is much less acidic
than acetic acid.
ææ
C2 H5 —OH ¨æÆ C2 H5 — O:@ + H≈
æ
V VI
(2) Substituted benzoic acids An electron-releasing group decreases the acidity of benzoic
acid, while an electron-attracting group increases it. The acidity also depends on the position
of the substituent present. For example, p-nitrobenzoic acid (pKa = 3.43) is somewhat
stronger than the m-isomer (pKa = 3.28) but much weaker than the o-isomer (pKa = 2.17).
This can be explained as follows. In the conjugate base of a nitrobenzoic acid, the negative
charge on oxygen in the carboxylate (—COO①) group is not in proper conjugation with
the —NO2 group and so it cannot be stabilized by resonance. The –I effect of the —NO2
group is expected to fall off with distance. Therefore, p-nitrobenzoic acid is expected to be
weaker acid than the m-isomer. However, the acidity order is reverse of that predicted on
the basis of the –I effect. Although a p-NO2 group cannot enter into resonance with the
—COO① group, it exerts its –R effect to reduce the electron density of the carbon bearing
the —COO① group. A m-NO2 group also exerts its –R effect, but the carbon bearing the
—COO① group does not become positive. The conjugate base of the p-acid is, therefore,
somewhat more stabilized by resonance than the conjugate base of the m-acid and for this
reason, p-nitrobenzoic acid is somewhat stronger than its m-isomer.
Because of close proximity, the o-NO2 group exerts powerful –I effect and through –R effect
it withdraws electron from the —COO① group by making the carbon bearing this group
positive. Also, there occurs a more direct electrostatic interaction between the —COO①
and —NO2 groups. Because of all these factors, the o-nitrobenzoate ion is much more
stabilized with respect to the o-acid than the p-nitrobenzoate ion with respect to p-acid
and therefore, o-nitrobenzoic acid is a much stronger acid than its p-isomer.
Unlike halogen substituted saturated monocarboxylic acids, the effect of halogens on the
acidity of benzoic acid cannot be explained only by –I effect. Because of strong –I effect
p-fluorobenzoic acid is expected to be more acidic than p-bromobenzoic acid. However, it
has been found that p-bromobenzoic acid is actually more acidic. Halogens exert their +R
effect and tend to decrease the strength of the acid by increasing electron density of the
carboxyl group. The +R effect of the two halogens does not follow the expected order based
on electronegativity. It is known that the p-orbital overlap is much more effective when
the overlapping orbitals are of comparable size and so the +R effect of fluorine (that uses
2p orbital) is stronger than that of bromine (that uses 4p orbital). In fact, the +R and –I
effects of fluorine are nearly counter balanced. However, the +R effect of bromine is much
weaker as compared to its –I effect. Thus, although both the halogen atoms cause lowering
of electron density of the ring, the lowering is much effective in the case of bromo-acid
than in the case of fluoro-acid. For this reason, p-fluorobenzoic (pKa = 4.14) is less acid
than p-bromobenzoic acid (pKa = 4.00).
(3) Acidic Character of Phenols (Ar-OH) Greater the ease with which a compound releases
proton (H ) in its aqueous solution, the stronger it will be as an acid. That phenol is acidic
!
in nature and is a stronger acid than alcohol can be well explained in terms of resonance.
Phenol in its aqueous solution ionizes to produce phenoxide ion (the conjugate base) as
follows:
–
O—H O
+
+ H2O + H3O
Phenol Phenoxide ion

Phenol can be represented as a resonance hybrid of the following five resonance structures
(I–V):
Due to contribution of the resonance structure II, III and IV, the O atom becomes positively
polarized. Because of this, the polarity of the O–H bond increases and hence the tendency
of O—H bond fission (to release a proton) also increases. On the other hand, no such
resonance is possible in a molecule of alcohol (R—OH). Instead, a +I effect of R-group
operates which increases the electron density of oxygen. So, the alcoholic O—H bond is
relatively less prolar and the tendency of bond cleavage resulting in proton release is
actually very small. Hence, phenol is more acidic than an alcohol. This relative acidity
may also be explained by considering the phenol-phenoxide ion and alcohol-alkoxide ion
equilibria. Like phenol, phenoxide ion may also be represented as a resonance hybrid of
the following five (VI–X) resonance structures:
Three (II, III and IV) out of five resonance structures of phenol involve charge separation
and so, their contribution to the hybrid is relatively small. The resonance-structures of
phenoxide ion, on the other brand, involve no charge separation. The negative charge
is only delocalized. Because of this, phenoxide ion is more resonance-stabilized than
phenol. The effect of this difference in stabilization is that ionization energy of phenol
becomes lower than the hypothetical ionization energy when resonance is absent. As a
consequence, the equilibrium of phenol-phenoxide ion tends to shift towards the right, i.e.,
phenol exhibits acidic property by releasing proton (H!) easily.
On the other hand, both alcohol and the alkoxide ion can be satisfactorily represented by
single (localized) structures.
ææ
R — OH ¨ææÆ R — O:@ + H≈
Alcohol Alkoxide ion
None of these two species is stabilized by resonance. The driving force for ionization, i.e.,
the relative stabilization of the anion with respect to the undissociated molecule, is absent
in this case. Because of this, alcohols are much less acidic than phenols.
The acidity of phenols is partly due to the –I effect of the aryl group as compared to the +I
effect of the alkyl group in alcohols.
The energy diagram for the dissociation of an alcohol and phenol may be represented as
follows:
Ionization of both alcohol and phenol requires energy. Acidity of a compound depends on
the ionization energy, i.e., the difference in energy content of the acid and its conjugate
base. The less the energy required for ionization, the more the compound acidic. Both the
phenol and conjugate base are resonance-stabilized, but the stabilization is far greater for
the conjugate base, i.e., the phenoxide ion, than for the phenol (a > b). The effect of this
difference in stabilization is that ionization energy of the phenol becomes lower than the
hypothetical ionization energy (DE2 < DE1) and as a result, the ionization is favoured, i.e.,
a phenol exhibits acidic character.
Since either the alcohol or its conjugate base is not stabilized by resonance, therefore, the
energy required for its ionization is considerably high and in fact, it is much higher than
that required for the ionization of the phenol (DE1 > DE2). Because of this, the ionization of
a phenol is very much favoured as compared to an alcohol, i.e., phenols are much stronger
acids than alcohols.
1.6.6.2 Effect of ring substituents on the acid strength of phenols

The acidity of phenols depends on whether a ring substituent is electron-attracting or
electron-releasing and also its position with respect to the —OH group.
An electron-attracting substituents like —X, —NO2, —CHO, —CN, etc., withdraws
electrons from the ring and thereby stabilizes the phenoxide ion more than the phenol.
Thus, the energy difference between the phenol and its conjugate base decreases and
as a result, the energy required for ionization decreases. Consequently, the ionization
equilibrium becomes favoured. Therefore, an electron-attracting substituent increases
the acidity of phenol. An electron-releasing substituent, like —CH3, —OCH3, —NH2,
etc., donates electron to the ring and thereby destabilizes the phenoxide ion relative to
the phenol by intensifying the negative charge on it. For this reason, the difference in
stability between the phenol and its conjugate base, i.e., the phenoxide ion increases and
consequently, the strength of the phenol as an acid decreases. Therefore, an electron-
releasing substituent decreases the acidity of phenol.
A substituent can exert its –R or +R effect only when it is at ortho or para position. Therefore,
the acid-strengthening or acid-weakening effect of a substituent is very much pronounced
when it is present at otho or para position but not at the meta position. The acidic strength
of nitro substituted phenols, for example, increases in the following order: phenol <
m-nitrophenol < o-nitrophenol < p-nitrophenol. A nitro (—NO2) group present in ortho- or
para–position is capable of withdrawing electrons from the negatively charged oxygen
atom of the phenoxide ion by its –I and –R effects. It, therefore, stabilizes the conjugate
base by dispersing the negative charge. In fact, there is a relatively stable structure (P)
in which the negative charge is placed on the highly electronegative oxygen atom of the
—NO2 group and its contribution makes the hybrid stable. A similar resonance structure
(Q) makes the o-nitrophenoxide ion considerably stable. Hence, o- and p-nitrophenoxide
ions are relatively more stable than phenoxide ion. In fact, the relative stabilization of
the phenoxide ion with respect to the undissociated phenol is more effective with these
nitrophenols compared to phenol and for this reason, o- and p-nitrophenols are more acidic
than phenol.
The m-nitrophenoxide ion is stabilized only by the –I effect of the –NO2 group and this is
because the nitro group is unable to delocalize the negative charge on oxygen due to lack
of proper conjugation. For this reason, m-nitrophenol is a stronger acid than phenol, but
weaker acid as compared to its other isomers.
In the o-isomer, there occurs intermolecular hydrogen bonding (chelation) which, in

fact, disfavours proton release to some extent. For this reason, the o-isomer is less acidic
than the p-isomer, even though the —I effect of the relatively close —NO2 group is much
stronger than the former.
1. Acidic character of active methylene compounds An active methyl compound contains a

—CH2— or —CHR— group flanked by two strong electron-attracting (—I and —R) groups.
Acetylacetone (CH3COCH2COCH3), diethyl malonate (EtO2C – CH2—CO2Et), ethyl
acetoacetate (CH3COCH2CO2Et), ethyl cyanoacetate (EtO2C — CH2 — CN), dinitromethane
(O2N — CH2 — NO2), etc. are the most important active methylene compounds. The acidic
methylene hydrogens of these compounds can be easily abstracted by base because the
corresponding conjugate bases are well stabilized by resonance. However, their acidity
differs due to the presence of different electron-withdrawing groups. For example,
acetylacetone is a stronger acid than ethyl acetoacetate. A resonance argument can be
used to explain the difference in acidity between these two active methylene compounds.
Acetylacetone and ethyl acetoacetate ionize in the presence of a base (:B) as follows:
Charge delocalization occurs in both the conjugate bases and also the same number
of resonance structures can be written for both of them. The resonance structures of
acetylacetonate anion are all significant contributors. On the other hand, two of the three
resonance structures of acetoacetate anion (the first and second) contribute significantly
to the hybrid. The contribution of the last structure to the hybrid is, in fact, very small
because delocalization of electrons within the ester group itself lowers the ability of the
C == O group to disperse the negative charge.
Because of this, delocalization of charge is more effective in acetylacetonate anion than in

ethyl acetoacetate anion, i.e., the former conjugate base is more stabilized by resonance than
the latter conjugate base. The difference is stability between the neutral molecule and the
conjugate base is, therefore, greater in the case of ethyl acetoacetate (CH3COCH2CO2Et)
than in the case of acetylacetone (CH3COCH2COCH3) and hence the equilibrium involved
in the case of aceylacetone is relatively more favourable than that involved in the case of
ethyl acetoacetate. That is, acetylacetone is more acidic than ethyl acetoacetate.
(2) Acidic character of imides Imides are often sufficiently acidic to form alkali metal
salts. The acidity of an imide, e.g., phthalimide, is due to —I and —R effects of the two
C == O groups. The unshared pair of electrons on nitrogen is sufficiently delocalized and as
a consequence, it becomes considerably positively polarized. Because of this, the polarity
of the N—H bond increases and hence the tendency of N—H bond fission to release proton
increases. For this, imides exhibit acidic character. The acidity of phthalimide can also
be explained by considering phthalimide-phthalimide anion equilibria. Both phthalimide
and its conjugate base can be represented as a hybrid of three resonance structures such
as follows:
In phthalimde, the unshared pair of electrons on nitrogen is delocalized by the two adjacent
C == O groups. However, this resonance is less effective and less stabilizing because it
involves charge separation. On the other hand, resonance is more effective and more
stabilizing in phthalimide anion because there occurs no charge separation. Resonance
thus lowers the energy difference between the imide and its conjugate base. Because of this,
the ionization becomes much favourable, i.e., the compound exhibits considerable acidity
and dissolves readily in alkali metal hydroxide (e.g., NaOH) solution to form salts.
(3) Acidic Character of Nitroform [CH(NO2)3], Chloroform (CHCl3) and Fluoroform (CHF3)
Nitroform is more acidic than chloroform which in turn is more acidic than fluroform. These
observations can be well explained by resonance. Any thing that stabilizes a conjugate
base A①: makes the starting acid H—A more acidic. The conjugate base of nitroform, i.e.,
@
C(NO2 )3 , is stabilized by —I and —R effect (p-p p bonding) of the three —NO2 groups.
@
The conjugate base of chloroform, i.e., CCl3 , on the other hand, is stabilized by —I effect
and d-orbital resonance (p-d p bonding) involving three Cl atoms. Because of longer bond
length and the difference in size between the 2p and 3d orbitals, the p-d p bonding is far
less significant than p-p p bonding. Also, a —NO2 group exerts stronger —I effect than a Cl
@ @
atom. For these reasons, C(NO2 )3 is very much stable than CCl3 and therefore, nitroform
is a much stronger acid than chloroform.
Fluorine
@
has no d orbital. So, unlike CCl3 there is no question of resonance stabilization
@
of CF3 . It is@only stabilized by the —I effect of fluorine. Thus, CCl3 is relatively more
stable than CF3 and for this reason, chloroform (CHCl3) is more acidic than fluoroform,
even though fluorine is more electronegative than chlorine.
1.6.6.3 Basicity of Organic Compounds

(1) Basic character of arylamines (e.g., aniline, C6H5N̈H2) The basicity of nitrogenous bases
depends on the availability of the unshared pair of electrons on the N atom. Any factor that
increases the electron density on the N atom increases an amine’s basicity and any factor
that decreases the electron density on the N atom decreases an amine’s basicity. Aniline
is a weak base and its basicity is much weaker than an aliphatic amine like ethylamine
(C2H5NH2). This can be explained by resonance. Aniline can be represented as a resonance
hybrid of the following five resonance structures:
The unshared pair of electrons on the N atom of aniline is involved in resonance interaction
with the ring p electrons. As a result of this electron delocalization, the N atom acquires a
partial positive charge. Consequently, aniline exhibits little tendency to take up a proton.
So, aniline is a very weak base. The decrease in basicity is partly due to the –I effect of
the aryl group.
In the case of aliphatic amine like C2H5NH2, on the other hand, similar delocalization of
electrons by resonance is not possible. Naturally, the electron density on the N atom is not
reduced. In fact, due to +I effect of the alkyl group (–C2H5), the electron density on the N
atom is somewhat increased. As a consequence, nitrogen can easily donate its unshared
electron pair to a proton (H+) to combine with it. Thus, aniline (an arylamine) is a weaker
base than ethylamine (an alkylamine).
Apart from this, the relative basicity can also be explained by considering aniline-anilinium
ion and amine-ammonium ion equilibrium systems. In fact, the difference in stability
between a free amine and its conjugate acid is to be considered. The higher the difference,
the lesser the basicity of the amine.
Aniline is a resonance hybrid of five resonance structure (I–V) of which two are Kekule
structures (I & V). In anilinium ion, the lone pair of electrons on the N atom is localized
in the N—H bond and so only two Kekule structures (VI and VII) can be drawn for its
hybrid. Therefore, more effective electron delocalization occurs in aniline as compared to
anilinium ion and consequently, aniline is more resonance-stabilized with respect to the
anilinium ion (in the following energy diagram b > a). As a result of this, the difference in
stability between the base and its conjugate acid is increased and protonation of aniline
≈
is disfavoured. The conjugate acid of ethyl amine, i.e., CH 3CH 2 NH 3 , is stabilized only
by the weak +I effect of the —C2H5 group. None of the amine and its conjugate acid can
be stabilized by resonance. For this reason, the difference in energy content between the
conjugate acid and the base is relatively small and actually it is smaller than the difference
in energy content between aniline and anilinium ion (DE2 > DE1). The energy required
for protonation of aniline is, therefore, greater than the energy required for protonation
on for ethylamine and hence aniline (an arylamine) is less basic than ethylamine (an
alkylamine).
(2) Effect of ring substituents on the basic strength of arylamines The effect of a ring
substituent on the basic strength of arylamines depends on whether the substituent is
electron-attracting or electron-donating and also on its position with respect to the —NH2
group in the ring.
An electron-donating group, like —CH3, —OCH3, etc., increases the electron density
of the ring and thereby prevents the unshared electron pair on nitrogen of the —NH2
group, to some extent, from entering into resonance with the ring p electrons. As a result,
the unshared electron pair on nitrogen becomes more available for coordination with
a proton. Therefore, such a group increases the basic strength of aniline. On the other
hand, an electron–attracting group, like —X, —NO2, —COR, etc., favours the resonance
interaction of the NH 2 group with the ring and thereby make the lone pair on the N atom
less available for coordination with a proton. Therefore, such a group lowers the basic
strength of aniline.
The effect of a ring substituent on the basicity of anilines can also be explained in terms of
the difference in stability, i.e., the energy content, between the base and its conjugate acid,
i.e., the anilinium ion. An electron-releasing group tends to spread the positive charge
of the anilinium ion and thus stabilizes the conjugate acid but not the base. As a result,
the difference in stability between the conjugate acid and the base decreases, i.e., the
energy required for protonation decreases and consequently, the protonation equilibrium
becomes favoured. Thus, an electron-releasing group increases the base strength of the
aniline. An electron-withdrawing group causes intensification of the positive charge of the
conjugate acid and thus destabilizes the conjugate acid but not the base. As a result, the
difference in energy content between the conjugate acid and the base increases, i.e., the
energy required for protonation increases and consequently, the protonation equilibrium
becomes disfavoured. Therefore, an electron-withdrawing group lowers the basic strength
of the aniline.
A group can exert its +R and —R effects from ortho and para positions and due to that
the base-strengthening or base-weakening effect is most pronounced if the group is
present at these positions. For example, p-nitroaniline is less basic than m-nitroaniline.
This can be explained as follows. A nitro group para to the —NH2 group withdraws the
nonbonded electron pair on N of the —NH2 group by its —R and —I effects and thereby
makes the lone pair less available for protonation. An m-NO2 group, on the other hand,
cannot enter into resonance with the —NO2 group because it is not in proper conjugation
with the unshared pair of electron on nitrogen. Only a weak electron-attracting inductive
effect (—I) operates. The availability of the unshared pair is, therefore, somewhat better
in this case and so, m-nitroaniline is more basic than its para isomer.
Ê O ˆ
Á || ˜
(3) Basic character amides Ë R — C — NH2 ¯
Because of electron-attracting inductive and resonance effect (—I and —R) of the C == O
group, the lone pair of electron on amide nitrogen is very much less available for taking
up a proton. In an amine, on the other hand, the alkyl group pushes electrons towards
nitrogen by its +I effect makes nitrogen’s electron pair readily available for sharing with
a proton. Amides are, therefore, less basic than amines.
NH
||
(4) Basic character of guanidine H2N — C — NH2
That guanidine is an extremely strong base can be well explained by resonance. Both the
neutral molecule and the conjugate acid can be represented as a resonance hybrid of three
resonance structures.
The neutral molecule contains two charge-separated structures which contribute less
to the resonance hybrid. Electron delocalization is, therefore, less effective in the free
base and so, it is not much resonance-stabilized. The conjugate acid, on the other hand,
involves no charge separation; only the positive charge is delocalized. Also, the canonicals
are all exactly equivalent. So, there occurs extremely effective charge delocalization and
as a consequence, the conjugate acid is highly stabilized by resonance with respect to
the neutral molecule. The difference in stability (i.e., the energy content) between the
conjugate acid and the base is actually very small and so, protonation of the base required
very small amount of energy. For this reason, guanidine behaves as a very strong base.
1. Designate the following pairs as structural isomers or resonance structures

and give your reasoning.
(a) (b) O
==
CH3—N and CH3—O—N==O
O
(c) (d) O OH
==
CH3—C—CH3 and CH3—C==CH2
(e) CH3CH2CH==CHCH3 and CH3CH2CH2CH==CH2
Solution Two resonance structures have the same placement of atom but a different
arrangement of nonbonding and p electrons. However, two structural isomers differ in the
arrangement of both atoms and electrons. Therefore, the two structures in each of (a) and
(c) are two resonance structures, while the two structures in each of (b), (d) and (e) are
structural isomers.
2. In the following set of resonance structures, label the major and minor
contributors and give reasons for such labelling:
(a)
(b)
(c) –
O O
´
CH3CH2—C CH3CH2—C +
O—H O—H
I II
(d)
(e) + – + –
CH2—CH==CH—CH 2 ´ CH 2 ==CH—CH—CH 2
I II
(f)
(g)
(h)
(i)
(j)
Solution
(a) Resonance structures with a greater number of covalent bond are more stable and
more contributing than those with a fewer number of covalent bonds. Therefore,
the structure I with ten covalent bonds is the major contributor and the structure
II with nine covalent bonds is the minor contributor.
(b) In structure I, octets of all the atoms are filled up while in structure II, octet of
positive carbon is not filled up. Therefore, the structure I being more stable, is the
major contributor and structure II is the minor contributor.
(c) When two structures have the same number of covalent bonds, then the uncharged
structure is more stable and more contributing than the charge separated
structure. Therefore, the structure I is the major and the structure II is the minor
contributor.
(d) Resonance structures with negative charge on a more electronegative atom is more
stable and more contributing. Therefore, the structure II with a negative charge on
more electronegative oxygen atom is the major contributor and the structure I with
a negative charge on less electronegative carbon atom is the minor contributor.
(e) Resonance structures in which unlike charges are on the adjacent atom are more
stable and more contributing than those in which the charges are not adjacent.
Therefore, the structure II is the major and the structure I is the minor contributor.
(f) Since the structure II constitutes a stable aromatic system [(4n +2)p, n = 1],
therefore, it is more stable than the structure I. Therefore, the structure II is the
major contributor and the structure I is the minor contributor.
(g) The structure I (a 2° carbocation) is relatively more stable than the structure II
(a 1° carbocation) and therefore, the former structure is the major contributor and
the latter structure is the minor contributor.
(h) Resonance structures with like charges on adjacent atoms are less stable and
less contributing than those in which the charges are no adjacent. Therefore, the
structure II is the major and the structure I is the minor contributor.
(i) Less electronegative nitrogen accommodates the positive charge better than more
electronegative oxygen. Therefore, the structure II is the major contributor and the
structure I is the minor contributor.
(j) Since the structure II constitutes a stable aromatic system [(4n + 2)p, n = 1],
therefore, it is more stable than the structure I. Hence, the structure II is the
major contributor and the structure I is the minor contributor.
3. Arrange the resonance structures in each group in order of increasing
contribution to the resonance hybrid:
:O: :O:@ :O:@
|| | | ≈
(a) CH3 —C—OC2 H5 ´ CH3 — C —OC2 H5 ´ CH3 — C == OC2 H5
I ≈ III
II
@ ≈ ≈ @
(b) (C2 H5 )2 C == N — NH2 ´ (C2 H5 )2 C — N== N H2 ´ (C2 H5 )2 C — N —NH2

I II III
:O: :O:@ :O:@ :O≈

|| | ≈ | |
(c) H—C—OH ´ H —C == OH ´ H — C —OH ´ H — C — OH
I II ≈ @
III IV
@ ≈ ≈
(d) CH2 == CH— Cl: ´ CH2 — CH == C l: ´ CH2 — CH== Cl:@
I II III
≈ @ ≈ @ @ ≈ @ ≈
(e) CH2 == N == N: ´ CH2 — N== N: ´ CH2 — N ∫∫ N ´ CH2 — N== N:
I II III IV
Solution
II, III, I III, II, I
(a) æææææææææææÆ (b) æææææææææææÆ
contribution increases contribution increases
IV, III, II, I III, II, I
æææææææææææÆ ææææææææææ
(c) contribution increases (d) contribution increasesÆ
IV, II, III, I
(e) ææææææææææ
contribution increasesÆ
4. Draw the resonance structure and the resonance hybrid of phenoxide ion
(C6H5—O①) and arrange then in order of decreasing stability. Give your
reasoning. What does the resonance hybrid illustrate?
Solution The resonance structures and the resonance hybrid of the phenoxide ion my be
drawn as follows:
The resonance structures II, III and IV have the negative charge on a carbon, a less
electronegative element than oxygen. As a result, structures II–IV are less stable than
structures I and II which have the negative charge on oxygen. Moreover, resonance
structures I and V have intact aromatic rings, whereas structures II–IV do not. This,
too, make structures II–IV less stable than I and V. A resonance hybrid is more stable,
i.e., it has lower energy than any of its contributing structures. Therefore, the order of
decreasing stability is VI > I = V > II = IV ª III.
The resonance hybrid of the phenoxide ion illustrates that its negative charge is dispersed
over four atoms — three C atoms and one O atom.
5. Explain why (a) the two oxygen–oxygen bonds in ozone (O3) have the same
length and (b) the two nitrogen–nitrogen bonds in the azide ion (N3①) have
the same length.
Solution
(a) Ozone (O3) can be represented as a hybrid of the following two resonance structures:
Since the two resonance structures are equivalent, they make equal contribution
to the hybrid. Therefore, the two oxygen–oxygen bonds have the same order and
so, the bonds have the same length.
(b) The azide ion (N3①) can be represented as a hybrid of the following three resonance
structures:
+ 2– – + – 2– +
N∫∫ N—N ´ N==N==N ´ N—N∫∫ N
I II III
Each of the nitrogen–nitrogen bonds is triple in one structure, double in one
structure and single in one structure. Hence, their bond order is the same and for
this reason, they have the same length.
6. The carbon–carbon bond length in CO23 (1.31 pm) is greater than that
@
found in CH3COO (127 pm). Explain this observation.

①
Solution The CO23@ ion can be represented as a hybrid of three equivalent resonance
structures. In this ion, two negative charges are evenly distributed over three oxygen
2
atoms and so, each oxygen atom acquires 3
@ charge. On the other hand, the HCOO①
ion is resonance hybrid of two equivalent canonical and in it, one negative charge is
1
evenly distributed over two oxygen atoms, i.e., each O atom carries 2
@ charge. Since the
charge accumulation is greater in oxygen of CO23@ than in oxygen of CH3COO①, therefore

carbon–oxygen bond order in CO23@ is less than that in CH3COO① and because of this, the
carbon–oxygen bond length in CO23@ is greater than that in CH3COO①.
7. Explain why acetic acid is more acidic than phenol.

Solution Any thing that stabilizes a conjugate base A:① makes the starting acid H—A
more acidic. The acetate ion (the conjugate base of acetic acid) is a resonance hybrid of
two equivalent resonance structures, both of which places the negative charge on a more
electronegative oxygen atom. Although the phenoxide ion (the conjugate base of phenol)
is resonance hybrid of five resonance structures, the negative charge is accommodated
by three less electronegative carbon atoms and only one more electronegative oxygen
atom. Furthermore, the two resonance structures of acetate ion are equivalent. Therefore,
resonance stabilizes phenoxide ion but not as much as it stabilizes acetate ion. It is for this
reason acetic acid is more acidic than phenol.
8. Explain why picric acid is nearly as acidic as mineral acids.

Solution 2,4,6-Trinitrophenol is known as picric acid. Due to cumulative —I and —R
effects of the three electron attracting —NO2 groups placed in para and in two ortho
positions, 2,4,6-trinitrophenoxide ion, the conjugate base of picric acid, is very much
stabilized with respect to the unionized phenol. Because of this, 2,4,6-trinitrophenol or
picric acid behaves as a very strong acid and in fact, it is nearly as acidic as mineral
acids.
9. Acetic acid (CH3COOH) can act as a weak base in the presence of a strong
acid like H2SO4. Which of the two oxygen atoms is more basic and why?
Solution In the presence of H2SO4, a proton can be added either to the carbonyl oxygen
or to the hydroxyl oxygen of the carboxyl group. The two protonation equilibria are as
follows:
!
:O: :OH
|| ||
CH3 — C —OH + H2SO4 CH3 — C — OH + HSO4@
I
:O: :O:
|| || !
CH3 — C — OH + H2SO4 CH3 — C — O H2 + HSO@4
II
The conjugate acid I is a resonance hybrid of two equivalent canonicals and therefore, it is
considerably stabilized by resonance.
+
OH OH
==
CH3—C ´ CH3—C
==
+
OH OH
I
On the other hand, the conjugate acid II is not at all stabilized by resonance. The positive
charge is localized on the oxygen atom. Therefore, II is less stable than I and the first
equilibrium in more favourable than the second. That is, the carbonyl oxygen is more
basic than the hydroxyl oxygen.
(a) The central C—C bond in 1,3-butadiene (CH2 == CH—CH == CH2) is shorter in
length than the C—C bond in ethane (CH3 —CH3).
(b) The heat of hydrogenation of 1,3-cyclohexadiene ( (
in less than that of
1,4-cyclohexadiene
( ( N
(c) The doubly bonded N atom of imidazole
( N ( is relatively more basic than
the other N atom. H
(d) The a-hydrogens in propanal (CH3CH2CHO) are much more acidic than its
b-hydrogens.
(e) 1,2,3-Triphenylcyclopropane is a very much stable cyclopropane derivative.
(f) The Lewis acid character of boron trihalides decreases in the order: BI3 >
BBr3 > BCl3 > BF3.
(g) Tri-p-nitrophenylamine is not at all basic.
(h) p-Nitrotoluene is more acidic than toluene.
(i) In the following compound, Ha is more acidic than Hb which in turn is more
acidic than Hc:
(j) Sulphonic acids (RSO3H) are very strong acids (pKa ª –7).
Solution
(a) As the percent of s-character increases, the size of the hybrid orbital decreases
and consequently, the length of the bond formed by overlapping two such orbitals
decreases. Hence the C-2—C-3 bond in 1,3-butadiene, CH2 == CH—CH == CH2
(a C 2 — C 2 bond) is shorter than the C — C bond in ethane (a C 3 — C 3 bond).
sp sp sp sp
This can also be explained by resonance. 1,3-Butadiene can be represented by three
resonance structures:
The two charge separated structures II and III contain a double bond between C-2
and C-3. Therefore, the hybrid must have a partial double bond there and for this
reason, the central C—C bond is shorter than the C—C bond in ethane.
(b) 1,3-cyclohexadiene is a conjugated diene while 1, 4-cyclohexadiene is an isolated
diene. Because a conjugated diene has overlapping p orbitals on four adjacent
atoms, its p electrons are delocalized over four atoms. This electron delocalization
cannot take place in an isolated diene. Therefore, 1,3-cyclohexadiene is stabilized
by resonance.
Since on hydrogenation both the dienes produce the same cycloalkane, the more
stable diene has the smaller heat of hydrogenation. Therefore, 1,3-cyclohexadiene
has the smaller heat of hydrogenation as compared to 1,4-cyclohexadiene.
(c) The more the corresponding conjugate acid is stable, the more the N atom is basic.
No resonace structure can be written for the conjugate acid obtained on protonation
of the singly bonded N atom of imidazole, i.e., this conjugate acid is not stabilized
by resonance. Furthermore, on protonation aromaticity of the system is lost.
On the other hand, two equivalent resonance structures can be written for the
conjugate base obtained on protonation of the doubly bonded nitrogen, i.e., the
corresponding conjugate acid is well stabilized by resonance. Also, on protonation
aromaticity of the system is not lost.
It thus follows that the doubly bonded N atom which coordinates with a proton to
from a stable conjugate acid is relatively more basic than the other N atom.
(d) The anion obtained by the loss of an a-H atom of propanal (CH3CH2CHO) is
stabilized by resonance.
H :O: :O:@
a| @ || | ≈
b
CH3 — C H — CHO + B [CH3 — CH — C — H ´ CH3 — CH == C — H] + BH
Propanal Resonance-stabilized conjugate base
Owing to an intervening saturated carbon atom, the anion obtained by the loss of
a B — H atom is not stabilized by resonance.
H
| a @ !
b
CH2 —CH2 — CHO + B CH2 — CH2 — CHO + BH
Propanal Conjugate base
(not resonance-stabilized)
Therefore, the difference in energy between the anion (the conjugate base) and
the neutral molecule (the acid) is much greater in the second case than in the first
case. Hence, a-hydrogens are much more acidic than b-hydrogens in the aldehyde
molecule.
(e) 1,2,3-triphenylcyclopropane is highly stabilized by resonance because there
occurs extensive electron delocalization due to the presence of a cyclopropane ring
which behaves in some respect like a double bond and makes the system a perfect
conjugated one.
(f) This order of relative Lewis acid strength of boron trihalides, which is just the
reverse of what may be expected on the basis of the electronegativities of the
halogen atoms, can well be explained on the basis of the tendency of the halogen
atom to donate its lone pair of electrons to the boron atom through pp–pp back
bonding. Since the vacant 2p-orbital of B and the 2p-orbital of F atom containing
a lone pair of electrons are equal in size, therefore, the tendency of F atom to
donate the unshared pair by pp–pp back bonding is maximum. In fact, BF3 can
well be represented as a resonance hybrid of four resonance structures. As a result
of resonance involving pp–pp back bonding, the electron density on the boron atom
increases effectively and so its strength as a Lewis acid decreases considerably.
As the size of the halogen atom increases on going from Cl to I, the extent of
overlapping between the 2p orbital of boron and a large p orbital of halogen (3p of
Cl, 4p of Br and 5p of I) decreases. As a consequence, the electron deficiency of boron
increases and thus, the Lewis acid strength increases on going from BF3 to BI3.
(g) In tri-p-nitrophenylamine the lone pair of electrons on the
N atom is extensively delocalized over the three p-nitrophenyl groups. Because of

this, the unshared pair is not at all available for coordinating with a proton. Hence,
by ordinary standard the compound has no basic strength.
(h) Due to the presence of an electron-withdrawing —NO2 group para to the —CH3
group, the conjugate base of p-nitrotoluene is stabilized not only by the —I effect of
the —NO2 group but also by its strong —R effect. On the other hand, the conjugate
base of toluene is only less effectively stabilized by resonance involving the benzene
ring. Therefore, the difference in energy between benzyl anion (the conjugate base)
and toluene is much greater than the difference in energy between p-nitorbenzyl
anion (the conjugate base) and p-nitrotoluene. Thus, the second equilibrium is
more favoured than the first and so, p-nitrotoluene is more acidic than toluene.
@
(i) Anything that stabilizes a conjugate base A : makes the starting acid H—A
more acidic. The conjugate base obtained by the loss of Ha proton is stabilized by
resonance involving the benzene ring.
The conjugate base obtained by the loss of Hb proton is also stabilized by resonance
involving the benzene ring but it is somewhat destabilized by the +I effect of the
adjacent —CH3 group. And, the conjugate base obtained by the loss of Hc proton
is not at all stabilized by resonance due to the presence of intervening methylene

group.
Therefore, the stability of the conjugate bases decreases in the order:
Hence, Ha is more acidic than Hb which in turn is more acidic than Hc.
(j) Sulphonic acids (RSO3H) are very strong acids because their conjugate bases are
highly stabilized by resonance involving three equivalent resonance structures
containing a negative charge on the highly electronegative oxygen atom.
–
O O O O
== ==
== ==
==
–
R—S—O—H + B R—S—O ´ R—S==O ´ R—S==O
==
O O O O –
Sulphonic acid Sulphonate anion

(highly stabilized by resonance)
11. What do you mean by push-pull or captodative ethylene? Explain why

they have a very low barrier to rotation as compared to simple alkenes.
Solution The compound containing two electron-withdrawing groups on one doubly
bonded carbon and two electron-releasing groups on the other doubly bonded carbon is
called a push-pull or captodative ethylene. For example:
:N∫∫ C S Me
C==C
:N∫∫ C N Me2
The contribution of the diionic resonance structure such as shown below decrease the double
bond character of this ethylene derivative and allows easier rotation. For this reason, such
compounds have a much lower barrier to rotation compared to a simple alkene.
12. The bond ‘a’ of the following cross-conjugated compound is shorter than
the double bond of 1,3-butadiene (CH2 == CH— CH == CH2) while the bond
‘b’ of this compound is longer than the double bond of 1,3-butadiene.
Explain
a
CH2 == CH — C — CH == CH2
||b
CH2
3-methylene-1,4-pentadiene
Solution 3–Methylene–1, 4–pentadiene (a cross-conjugated alkene) can be represented

as a hybrid of the following five resonance structures:
! @
[CH2 == CH — C — CH == CH2 ´ CH2 — CH == C — CH == CH2 ´ CH2 — CH == C — CH == CH2

|| | |
CH2 ..CH2
@
≈CH2
@ ≈
CH2 == CH — C == CH — CH2 ´ CH2 == CH — C == CH — CH2 ]
| |
≈CH2 @ CH
2
Resonance in 3-methylene-1,4-pentadiene
1,3-Butadiene can be represented as a hybrid of the following three resonance structures.

≈ @ @ ≈
[CH2 == CH — CH == CH2 ´ CH2 — CH == CH — CH2 ´ CH2 — CH == CH — CH2 ]
Resonance in 1,3-butadiene
The bond ‘a’ in the cross-conjugated compound is double in three of the five resonance
structures while the double bond in 1,3-butadiene is double is one of the three resonance
structures. Therefore, the bond ‘a’ of the cross-conjugated compound possesses more double
bond character than the double bond of 1,3-butadiene and hence the bond ‘a’ is shorter
than the double bond of 1,3-butadiene. On the other hand, the bond ‘b’ is double in one of
the five resonance structures. Therefore, the bond ‘b’ of the cross-conjugated compound
possesses less double bond character as compared to the double bond of 1,3-butadiene.
Hence, the bond ‘b’ is longer than the double bond of 1,3-butadiene.
13. When 1 mole of benzene is allowed to react with 1 mole of hydrogen under
high temperature and pressure and in the presence of an active catalyst,
a mixture of 2/3 mole of benzene and 1/3 mole of cyclohexane is isolated.
The intermediated products such as 1,3-cyclohexadiene and cyclohexene
are not isolated. Account for these observations.
Solution
Because of aromatic stability, benzene is much more stable than 1,3-cyclohexadiene

and for this reason, the first reaction is endothermic and unfavourable. Cyclohexene is
thermodynamically more stable than 1,3-cyclohexadiene and for this, the second reaction
is exothermic and occurs rapidly. Cyclohexane is much more stable than cyclohexene and
the third reaction is somewhat more exothermic and occurs more readily.
From the enthalpy values it becomes clear that a hydrogen molecule reacts with a benzene
molecule under harsh conditions to yield 1,3-cyclohexadiene. The second molecule of H2
does not add to a second molecule of benzene, rather it adds to a more reactive molecule of
1,3-cyclohexadiene to yield cyclohexene and the third molecule of H2 does not add to
1,3-cyclohexadience rather it adds to cyclohexene (a thermodynamically more favourable
reaction) to yield cyclohexane. Thus, the intermediate products are not isolated.
In this hydrogenation process, 3 moles of hydrogen reacts with 1 mole of benzene or 1 mole
1
of hydrogen reacts with 1/3 mole of benzene. As a result of this, mole of cyclohexane is
3
Ê 1ˆ
obtained and Á 1 - ˜ or 2/3 mole of benzene remains unreacted. Therefore, when 1 mole of
Ë 3¯
hydrogen is allowed to react with 1 mole of benzene, a mixture of 1/3 mole of cyclohexane
and 2/3 mole of benzene is obtained.
(a)
(
Imidazole :N
(
NH is approximately 100 times more basic than pyridine
( N:(
(b) 4–Aminopyridine has a larger dipole moment (4.4 D) than 4–cyanopyridine
(1.6 D).
(c) The C—O bond length in an ester is shorter than that in an anhydride.
NH
(d) In the homologous series of compounds , those with smaller
(CH2)n
values of n are stronger bases.
(e) I is less basic than II:
(f) Benzene always reacts with O3 to give triozoride but not mono- or diozonide.
(g) A substituent such as –NO2 bonded at ortho or para position has a greater
effect on acidity of a phenol than a benzoic acid.
(h) The doubly bonded N atom in guanidine is more basic than the singly bonded
N atom.
(i) Indan-2-one is more acidic than its acyclic model C6H5CH2COCH3.
NH2
(j) Amidines (R—C==NH) are stronger bases than saturated amines.
(k) CH2(SCH3)2 is more acidic than CH2(OCH3)2
@
[Hint: The conjugate base of the thioacetal, i.e., C H(SCH 3 )2 , is stabilized by

d orbital resonance involving two S atoms (dp-pp back bonding).]
(l) The compound I will have higher electron density at the marked (*) carbon
atom than the compound II:
(m) p-Chlorophenol is more acidic than p-fluorophenol, even though fluorine is

more electronegative than chlorine.
[Hint: The corresponding p-chlorophenate ion is more stabilized by the
extended resonance involving vacant 3d orbital of chlorine atom. In the case
of p-fluorophenate, this is not possible because of the absence of the d orbital
in fluorine atom.]
(n) –COOH has –I effect but –COO@ has +I effect.
(o) p-Nitroaniline is less basic than m-nitroaniline.
(p) Benzylamine (PhCH 2NH 2 ) is more basic than aniline (Ph NH 2 ) .
(q) Benzyl alcohol is less acidic than phenol (PhOH).
(r) p-Nitrophenol is more acidic than m-nitrophenol.
2. (a) Which of the two nitrogen atoms of the following compound undergo protonation
and why?
(b) Where would protonation occur (at ‘N’ or ‘O’) in the following molecule? Give
reasons.
NH2
O==C
NH2
3. Arrange the following compounds in order of increasing basicity. Give reasons for
your answer.
4. Arrange the following acids in order of increasing acidity and explain the order:
4-Hydroxybenzoic acid, Benzoic acid, Salicylic acid
5. Which one between 1,3,5- and 1,3,6-heptatriene is more stable and why?
6. The methyl hydrogens of g-picoline CH3— ( (

N are more acidic than the methyl
( (
CH3
hydrogens of b–picoline
N
Explain.
7. Although 2-methoxyacetic acid (CH3OCH2COOH) is a stronger acid than acetic
acid (CH3COOH), p-methoxybenzoic acid is a weaker acid
than benzoic acid —COOH. Explain.
8. Explain why peroxy-acids (R COOOH) are weaker acids than carboxylic acids
(RCOOH) which are in turn weaker acids than sulphonic acids (RSO3H).
9. Arrange the anions in each group in order of increasing basicity and give your
reasoning.
– – –
O O O – –
(a) , , (b) C6H5CH2, C6H5O , C6H5NH
–
NO2
10. Arrange the compounds in each group in order of increasing acidity and explain
the order:
(a)
NO2
(b) OH OH OH
, ,
O2N Br O2N
11. When acetic acid (CH3COOH), labelled at its OH oxygen with 18O isotope, is treated
with aqueous base, and then the solution is acidified, two products having the 18O
labelled at different locations are formed. Explain this observation.
O 18
O O
|| || ||
1.NaOH
CH3 — C—18 OH ææææ 18
! Æ CH 3 — C — OH + CH 3 — C— OH
2.H3O
12. (a) Which compound has the greater electron density on its oxygen atom and why?
O O
==
==
—NH—C—CH3 or —NH—C—CH3
(b) Which compound has the greater electron density on its nitrogen atom and
why?
or
N N
H H
13. Arrange the following compounds in order of increasing acidity of the indicated
hydrogen and give your reasoning:
O O O O
|| || || ||
CH3 — C — CH2 — C — CH3 , CH3 — C — CH2 — CH2 — CH2 — C — CH3,
I II ≠
≠
O O
|| ||
CH3 — C — CH2 — CH2 — C — CH3
II ≠
14. In each of the following pairs, which species is the stronger base and why?
(a) CH 3COO@ or CH 3CH 2O@

NH2 NH
| ||
(b) CH3 — CH — CH3 or CH3 — C — NH2
@ @
(c) (CH3 )2 C — CHO or (CH3 )2 C — CH == CH2
(d)
15. Explain why N, N-di-(2,4,6-trinitrophenyl) amine is about 100 times more acidic
than acetic acid.
[Hint: The conjugate base of N, N-di-(2,4,6-trinitrophyenyl) amine is highly
stabilized by resonance.]
16. Arrange the following diazonium ions in increasing order of stability with reason:
+ +
(a) O2N— —N∫∫N (b) —N∫∫N (c)
17. Arrange the following dienes in order of increasing stability: 1,4-pentadiene,

1,3-pentadiene, 1,2-pentadiene
[Hint: The order of increasing stability: 1,2-pentadiene (CH3CH2CH==C==CH2) < 1,
4-pentadiene (CH2 == CH — CH2 — CH == CH2) < 1, 3-pentadiene (CH3CH == CH —
CH == CH2)]
18. Which compound of the following pairs will have higher electron density at the
marked (*) carbon atom and why?
Cl F
(a) and (b)
19. List the following compounds in order of increasing acidity and explain the order:
20. Which one is more acidic and why?

O S
or
O S
21. DBN and DBU are two strong sterically hindered nitrogen bases useful for E2
reactions. Explain why they are strong bases.
22. The basicity constants of N, N-dimethylaniline and pyridine are almost the same,
while 4-(N, N-dinmethylamino) pyridine is considerably more basic than either.
Which one of the two nitrogens of 4-(N, N-dimethylamino)pyridine is more basic

and why? Suggest an explanation for its enhanced basicity as compared with
pyridine and N, N-dimethylaniline.
O O
23. In R—C C == O and C — O bond distances are different but in R—C – they
OH O
are same. Give your reasoning.
24. Tricyanomethane [HC(CN)3] is as strong as mineral acids. Explain.
[Hint: The conjugate base is extensively stabilized by resonance.]
25. The dipole moment of 2-butenal (CH3CH == CHCHO) is higher than that of butanal.
Explain.
NH2
|
26. Guanidine (HN== C — NH2 ) is a much stronger base than acetamidine
CH3
|
(HN == C — NH2 ) . Explain.
27. p-Fluorobenzoic acid is less acidic than p-chlorobenzoic acid. Explain.
28. p-Nitrophenol is more acidic than m-nitrophenol which in turn is more acidic than
phenol. Explain.
29. Explain why phthalimide dissolves in alkali.
30. Explain why chloroform is more acidic than fluoroform, even though fluorine is
more electronegative than chlorine.
O
31. Ph—C –
is more stabilized by resonance compared to . Explain.
O
≈
32. Ph NH 2 is more stabilized by resonance compared to Ph N H3 . Explain.
33. Account for this increase in acidity (loss of proton):
N N O N N O N O
H H H H H
1.7 HYPERCONJUGATION
When a carbon containing at least one hydrogen atom is attached to a multiple bond
such as C == C, C == C, C == O, C == N, etc. The s-electron of the C—H bond is involved in
delocalization with the p-electrons of the unsaturated system, i.e., there occurs a s–p
conjugation. This type of conjugation may also take place when a carbon containing at
least one H-atom is attached to a carbon containing a partially filled or vacant p orbital.
This special type of resonance or conjugation giving stability to the species concerned
(molecule, free radical or carbonation) is called hyperconjugation.
For example:
(1) Hyperconjugation in propene (CH3CH = CH2) may be shown as follows:
≈
(2) Hyperconjugation in ethyl cation (CH 3 C H 2 ) may be shown as follows:
(3) Hyperconjugation in ethyl radical (CH 3 CH 2 ) may be shown as follows:
Electron displacement towards the p-MO or the vacant or partially filled p-AO caused by
hyperconjugation is called hyperconjugative effect. Since in the contributing form no bond
exists between C and H, this phenomenon is often termed as ‘no bond resonance’.
Although one C—H bond is shown to be broken in each hyperconjugation structures, H≈
never becomes free from the rest of the molecule nor does it change its position in the
molecule.
For effective hyperconjugatation to take place, the p-orbital concerned and the a C—H
bond, i.e., the sp3–s orbital must remain in the same plane. Orbital representations of
hyperconjugation in propane or in ethyl cation may be shown as follows:
The stability of a molecule, ion or free radical increases due to hyperconjugation. However,
this stability is less than that contributed, if any, by resonance. After the names of the
scientists who advanced this theory, hyperconjugation is also called Baker–Nathan effect.
The electron-releasing power of alkyl groups attached to unsaturated systems or electron-
deficient carbon atoms actually depends on the number of a–H atoms. Methyl (—CH3)
group having three a–H atoms has the highest hyperconjugative effect while this effect
is non existent with the tert-butyl (—CMe3) group having no a–H atom. So, the electron-
releasing power of various alkyl groups when attached to a double bond (or an electron
deficient-carbon) follows the order:
—CH —CH2CH3 —CH(CH3)2 —C(CH3)3
This order is exactly the reverse of the order of +I-effect of the alkyl group.
1.7.1 Sacrificial and Isovalent Hyperconjugation

Sacrificial hyperconjugation: Each of the hyperconjugative forms involved in
hyperconjugation of an alkene, for example, propene (CH3 — CH == CH2), bears isolated
charges and has one less real bond than the main uncharged form. This types of
hyperconjugation involving ‘sacrifice’ of a bond is termed as sacrificial hyperconjugation.
Isovalent hyperconjugation: Each of the hyperconjugative forms involved in
hyperconjugation of a free radical (e.g., ethyl radical, CH 3CH 2 ) and a carbocation (e.g.,
≈
ethyl cation, CH 3 C H 2 ) displays no more charge separation than the main form and has
the same number of covalent bonds as the main form.
This type of hyperconjugation involving no ‘sacrifice’ of a bond is termed as isovalent
hyperconjugation.
1.7.2 Effect of Hyperconjugation on the Physical and Chemical

Properties of Molecules and on the Stabilities of Intermediates
1.7.2.1 Heat of combustion and relative stabilities of alkenes
Stabilities of alkenes can easily be explained by hyperconjugation. Greater the number
of a-hydrogen atom (H-atom present on the carbon atom attached directly to a double
bonded carbon), i.e., greater the number of hyperconjugative structure, higher will
be the stability of the alkene due to hyperconjugation and less will be the heat of
combustion. For example, the heat of combustion of isobutene, (CH3)2C == CH2 (646.1
kcal/mol) is lower than that of 1-butene, CH3CH2CH == CH2 (649.8 kcal/mol), even though
the two isomers have the same number of C—C, C == C and C—H bonds. 1-Butene
(CH3CH2CH == CH2) contains two hyper conjugable a-hydrogen atoms while isobutene
[(CH3)2C == CH2] contains six such hydrogens. Thus, 1–butane can be represented as a
hybrid of three no bond resonance or hyperconjugation structures, whereas isobutene
can be represented as hybrid of seven no bond resonance or hyperconjugation structures.
Delocalization of s-electrons into the adjacent p-bond (s-p conjugation) is, therefore, more
effective in isobutene than in 1–butene and so isobutene is thermodynamically more
stable than 1-butene. Therefore, the heat of combustion of isobutene is less than that of
1-butene.
1.7.2.2 Directive influence of alkyl groups

The ortho- and para-directive influence of alkyl groups can be explained by hyperconjugation.
For example, the directive influence of —CH3 group in toluene can be explained on the
basis of hyperconjugation. Toluene can be represented as a hybrid of the following hyper
conjugative structures.
It is clear from these structures that hyperconjugation increases the electron density at
o- and p-positions and hence the electropohilic aromatic substitution reactions in toluene
occur mainly at these two positions. Alkyl groups are, therefore, o, p-directing.
1.7.2.3 Bond length

Shortening of carbon–carbon single bond and lengthening of carbon–carbon double bond
can be explained by hyperconjugation. For example, in propene (CH3 — CH == CH2) the
C—C bond length is 1.488 Å as compared to 1.543 Å in ethane (CH3 — CH3) and the C == C
bond length is 1.353 Å as compared to 1.334 Å in ethylene (CH2 == CH2). Propene can be
represented as a hybrid of the following hyperconjugation structures.
The C — C bond in propene is single in one structure but double in three structures while
the C == C bond in propene is double in one structure but single in three structures. Thus,
the carbon–carbon single bond possesses some double bond character and the carbon–
carbon double bond possesses some single bond character. For this reason, the C — C bond
in propene is somewhat shorter in length than the C — C bond in ethane and the C == C
bond in propene is somewhat longer in length than the C == C bond in ethylene.
1.7.2.4 Dipole moment

Hyperconjugation may influence the polarity of compounds. For example, the higher
dipole moment of acetaldehyde (2.721) as compared to formaldehyde (2.27D) is due to
hyperconjugation.
1.7.2.5 Relative stabilities of carbocations

Due to hyperconjugation, the C—H bonding electron pair is attracted towards the positively
charged C atom of the carbocation. This helps in dispersing the positive charge in different
parts of the alkyl group, i.e., charge delocalization resulting in stability of the carbocation
occurs. As the number of a–H atom increases, the number of no bond resonance structures
of cabocation increases. As a result, the extent of charge delocalization and consequent
stabilization increases. In tert-butyl cation, there are nine hyperconjugable H atoms.
Hence, nine equivalent hyperconjugation structures can be written for the cation:
È H H≈ ˘
Í | ≈ | ˙
ÍH — C — C (CH 3 )2 ´ H — C ==C(CH3 )2 ´ eight more structures ˙
Í | | ˙
Í H H ˙
Í ˙
ÍÎ Hyperconjugation in tert-butyl cation ˙˚
≈ ≈
Six and three such hyperconjugation structures can be written for (CH3 )2 CH and CH3 CH2 ,
≈
respectively, and none can be written for C H 3 . Hence, the increasing order of
hyperconjugation stability of methyl, ethyl, isopropyl and tert-butyl cation is
1.7.2.6 Relative stabilities of free radicals

Because of hyperconjugation, the odd electron of a free radical undergoes delocalization
for which it becomes stabilized. With increase in the number of a-H atom, the number
of no bond resonance structures of free radical increases. As result, delocalization of the
odd electron takes place to a greater extent and stability of free radical increases. Nine
hyperconjugation structures can be written for tert-butyl cation:
È H H ˘
Í | ˙
ÍH — C — C (CH ) ´ H — C — C (CH ) ´ eight more structures˙
3 2 3 2
Í | | ˙
Í H H ˙
Í ˙
ÍÎ Hyperconjugation in tert-butyl radical ˙˚
Six and three such hyperconjugation structures can be written for isopropyl radical
(CH3CHCH3 ) and ethyl radical (CH3CH2 ) , respectively, and none can be written for
methyl radical (CH 3 ). Therefore, the increasing order of hyperconjugation stability is
1. Which one between 2–methylbut -2-ene and 2-methylbut-1-ene has higher

heat of hydrogenation and why?
Solution 2-methylbut-2-ene contains nine hypeconjugable a-H atoms. So, this molecule
is highly stabilized by hyperconjugation and because of this, it has relatively lower heat of
hydrogenation. On the other hand, 2-methylbut-1-ene contains only five hyperconjugable
a-H atoms. So, this molecule is relatively less stabilized by hyperconjugation and because
of this, it has relatively higher heat of hydrogenation.
2. The C — C bond in acetaldehyde (CH3CHO) is shorter than that in ethane,

while the C — C bond in trifluoroacetaldehyde (CF3CHO) is essentially the
same as that in ethane. Explain.
Solution Acetaldehyde molecule contains three a-H atoms. These hydrogens is involved in
hyperconjugation with the double bond of the carbonyl group. As a result, the carbon–carbon
bond in acetaldehyde possesses some double bond character. Thus, the carbon–carbon
bond in acetaldehyde is shorter than that in ethane (CH3 — CH3). Trifluoroacetaldehyde
(CF3CHO) contains no a-H atoms. So, hyperconjugation is not possible in CF3CHO. Thus,
the C — C bond in CF3CHO is essentially the same as that in ethane.
3. Acid-catalyzed dehydration of 2-methylpent-2-ol, CH3CH2CH2C(OH)(CH3)2

gives 2-methylpent-2-ene, CH2CH2CH == C(CH3)2, as the major product
and 2-methylpent-1-ene, CH3CH2CH2(CH3)C == CH2, as the minor product.
Explain this observation.
Solution The acid-catalyzed dehydration of 2-methylpent-2-ol (a 3° alcohol) occurs
through the steps as follows:
Since the dehydration process is reversible, therefore, the more stable alkene will be formed
predominantly. The stabilization through hyperconjugation is greater for 2-methylpent-2-
ene (which offers eight hyperconjugable a–H atoms) than for 2-methylpent-1-ene (which
offers only five hyperconjugable a-H atoms) and because of this, 2-methylpent-2-ene is
obtained as the major product.
4. Arrange the following isomeric alkenes in order of increasing heat of

combustion and explain the order:
(CH 3 )2 C == C(CH 3 )2 ,CH 2 == CHCH 2CH 2CH 3 ,CH 3CH == CH CH(CH 3 )2 ,CH 3CH == C(CH 3 )CH 2CH 3
I II III IV
Solution The stability of an alkene is determined by the number of a-H atoms

present in the molecule, i.e., by the number of no bond resonance structures involved in
hyperconjugation. The greater the number of structures, the higher the stability of the
alkene. Now, the number of a-H atoms in the alkene I, II, III and IV are 12, 2, 4 and 8
respectively. Thus, the stability of the alkenes increases in the order: II < III < IV < I.
Since the heat of combustion increases with decrease in stability of the alkene, therefore,
the order of increasing heat of combustion is I < IV < III < II.
1. The carbon–carbon single bond in methylacetylene (CH3 — C ∫∫ CH) is approximately

1.46 Å in length, much less than the value (1.54 Å) found in saturated hydrocarbons.
Explain.
2. Heat of hydrogenation of ethene is greater than that of propene. Explain.
3. Alkyl groups attached to benzene ring release electrons in the order: CH3 —>
CH3CH2 — > (CH3)2CH — > (CH3)3C —. This is quite opposite to that expected on
the basis of inductive electron release (+I effect). Explain this anomaly.
4. What three alkenes yield 2-methylbutane on catalytic hydrogenation ? Match each
alkene with its correct heat of hydrogenation.
Heat of hydrogenation in kJ mol–1: 112, 118, 126
5. Which compound of the following pair will have higher electron density at the
marked (*) carbon atom ? Explain.
1.8 STERIC EFFECT

Steric effect is an effect on the relative reaction rates caused by the space-filling properties
of those parts of a molecule attached at or near the reaction site, i.e., steric effect is an
effect on the rates of chemical reactions caused by van der Waals repulsions.
The most common steric effect, however, is the classical steric hindrance. When the sheer
bulk of groups at or near a reacting site of a molecule hinders or retards a reaction, it is
called steric hindrance, i.e., a steric effect which decreases the reactivity of a compound is
called steric hindrance.
In an SN2 reaction, a backside approach by the nucleophile to a distance within the
bonding range at the carbon atom bearing the leaving group is required. For this reason,
bulky substituents on or near the carbon atom have a dramatic inhibiting effect. They
cause the energy of the transition state to be increased and consequently, they increase
the activation energy of the reaction for which the reaction becomes slow. Methyl halides
react most readily in SN2 reaction because only three small hydrogen atoms interfere with
the approaching nucleophile. Neopentyl and tert-butyl halides are least reactive because
bulky groups present a strong hindrance to the approaching nucleophile. In fact, tertiary
substrates do not react by an SN2 mechanism.
Steric hindrance increases
Steric strain is another kind of steric effect. If the constituent atoms or groups of a molecule
or ion owing to their bulky nature require more space than what is available for them, i.e.,
when they are forced too close to one another, then mechanical interference amongst the
groups or atoms takes place and the molecule or ion is then said to be under steric strain.
Steric strain makes the species unstable, i.e., its energy increases. For example, owing to
great bulk of the tert-butyl group and small available space for it, 1,2,3-tri-tert-butylbenzene
is under severe strain and so it is very difficult to prepare. Gauche conformation of butane
are higher in energy (less stable) than the anti conformation because of steric strain.
Cyclic compounds twist and bend in order to have a structure that minimizes the three
different kinds of strain that may destabilize a cyclic compound. These are
(i) Angle strain: It is the strain induced in a molecule when the bond angles are
different from the desired tetrahedral bond angle of 109.5°.
(ii) Torsional strain: It is caused by repulsion of the bonding electrons of one
substituent with the bonding electrons of a nearby susbstituent.
(iii) Steric strain.
1.8.1 Properties of Molecules Influenced by Steric Effect

1.8.1.1 Acidic character of compounds
(1) Acidic character of aromatic acids
In order to achieve maximum overlap of p-orbitals, which is necessary for delocalization,
it is essential that the skeleton should be planar in the conjugated system. In other words,
any structural feature that destroys this coplanarity of the conjugated system inhibits
resonance. This inhibition is referred to as steric inhibition of resonance (SIR). The higher
acidity of 2,4,6-trimethylbenzoic acid with respect to p-toluic acid can be explained in terms
of steric inhibition of resonance (SIR). p-Toluic acid can be represented as hybrid of the
following resonance structures:
The carboxyl group withdraws electrons from the ring by its —R effect and as a consequence,
the lone pair on hydroxyl oxygen is prevented to a greater extent from entering into
resonance with the C == O group. As a result, the OH oxygen becomes less positive by
entering into resonance with the C == O group and proton release becomes disfavoured.
For effective resonance to take place the carboxyl group must be planar or nearly planar
with the aromatic ring. In 2,4,6-trimethyl benzoic acid, the planar conformation required
for electron delocalization experiences steric strain. To avoid this strain, the carboxyl
group is forced out of the plane by the two ortho-methyl groups. As a result, the resonance
interaction of the carboxyl group with the ring decreases and consequently, the resonance
interaction of the unshared electron pair on OH oxygen with C == O increases. Because of
this, the hydroxyl oxygen becomes relatively more positive and proton release becomes
favoured.
Steric inhibition to solvation often plays an important role in determining the acidity of
aromatic acids. For example, benzoic acid (pKa = 5.05) is found to be more acidic than 2,6-
di-tert-butylbenzoic acid (pKa = 6.25) in a mixture of ethanol and water. Because of steric
inhibition of resonance, the latter acid is expected to be more acidic than the former acid.
But an opposite result is obtained when H2O-EtOH is taken as a solvent and that is due to
steric inhibition to salvation. Solvation (stabilization by solvent) in less effective for 2,6-di-
tert-butylbenzoate ion than for the benzoate ion and this is because the negative charge on
the former ion is somewhat shielded from the solvent molecules by surrounding tert-butyl
groups. Therefore, the difference in stability between the conjugate base and the acid is
less in the case of benzoic acid than in the case of 2,6-di-tert-butylbenzoic acid. Therefore,
the former equilibrium is relatively more favourable than the latter and so, benzoic acid is
found to be more acidic than 2,6-di-tert-butybenzoic acid in H2O — EtOH.
(2) Acidic character of phenols 2,6-Dimethyl-4-nitrophenol is more acidic than 3,5-

dimethyl-4-ntrophenol. This observation can be explained on the basis of steric inhibition
of resonance (SIR). In the latter compound, the oxygen atoms of the p-nitro group becomes
involved in steric interaction with the two adjacent methyl groups and as a consequence,
they are forced out of the plane of the ring. Hence, the — NO2 group cannot enter into
resonance with the — OH group. As a result, the oxygen of — OH becomes less positive
and proton release is less favoured. In 2,6-dimethyl-4-nitrophenol, on the other hand, the
—NO2 group enters into resonance with the — OH group. As a result , the — OH oxygen
becomes considerably positive and proton release in highly favoured. Hence, 2,6-dimethyl-
4-nitrophenol is more acidic than 3,5-dimethyl-4-nitrophenol.
1.8.1.2 Basic character of compounds

(1) Basic character anilines N,N,2,6-Tetramethylaniline is a stronger base than
N,N-3,5-tetramethylaniline. This can be explained by steric inhibition of resonance. In
N,N,3,5-tetramethylaniline, the unshared pair of electrons on nitrogen is well delocalized
by resonance with the benzene ring because the smaller ortho H atoms do not interfere
sterically with the methyl groups attached to nitrogen and so, the —NMe2 group is not
forced out the plane. The unshared pair of electrons on nitrogen is, therefore, not well
available for coordinating with a proton. Hence, it behaves as a weaker base.
On the other hand, in N,N-2,6-tetramethylaniline, the unshared pair of electrons on

nitrogen cannot enter into resonance with the ring because the relatively bulky ortho
methyl groups interfere sterically with the methyl group attached to nitrogen and as a
consequence, the - N Me2 group is forced out of the plane (conformation A). The unshared
pair of electrons is now localized on the N atom and is, therefore, N,N,2,6-tetramethylaniline
is stronger base than N,N,3,5-tetramethylaniline.
That 3,5-dimethyl-4-nitroaniline is stronger as a base than 2,6-dimethyl-4-nitroaniline

can also be explained on the basis of steric inhibition of resonance (SIR). In 3,5-dimethyl-
4-nitroaniline, the oxygens of the p-NO2 group is involved in steric interaction with two
adjacent methyl groups and as a consequence, they are forced out of the plane of the
ring. Hence, the —NO2 group cannot enter into resonance with the —NH2 group and the
unshared pair of electrons on nitrogen is well available for coordinating with a proton.
On the other hand, in 2,6-dimethyl-4-nitroaniline, the —NO2 group enters into resonance
with the —N H 2 group because the smaller H atoms of the —NH2 group do not interfere
sterically with the two o-methyl groups. Therefore, the unshared electron pair on nitrogen
is not well available for coordinating with a proton. This explains why 3,5-dimethyl-4-
nitroaniline is stronger as a base than 2,6-dimethyl-4-nitroaniline.
Hydrogen bonding sometime plays an important role in making electron delocalization

much more effective. 2,4,6-trinitroaniline is a very much weaker base than 2,4,6-trinitro-N,
N-dimethylaniline, even though aniline and N, N-dimethylaniline have almost same basic
strength. That 2,4,6-trinitro-N,N-dimethylaniline is a much stronger base than 2,4,6-
trinitroaniline can be well explained in terms of steric inhibition of resonance. In the
former compound, the oxygens of the two o-NO2 groups is involved in steric interaction
with the two N-methyl groups and as a consequence, the two —CH3 groups are forced out
of the plane of the ring. Hence, the —NMe2 group cannot enter into resonance with the
—NO2 groups, i.e., the —NO2 groups cannot exert their —R effect to reduce the electron
density on nitrogen. They only exert their —I effects to reduce the availability of the
unshared pair. The unshared pair of electrons is, in fact, localized on the nitrogen atom
and is, therefore, well available for coordinating with a proton. On the other hand, in the
latter compound, the —NH 2 group enters into resonance with the three —NO2 groups
because the smaller H atoms of the —NH2 group do not interfere sterically with the two
o-NO2 groups. Again, there occurs intramolecular hydrogen bonding between the oxygens
of the o-NO2 groups and the hydrogens of the —NH2 group. This help to hold the two
—NO2 groups in the planar conformation required for maximum resonance interaction.
Therefore, due to very powerful —R and —I effects of the three —NO2 groups, the
availability of the unshared pair of electrons on nitrogen is markedly reduced. Because
of this, 2,4,6-trinitroaniline behaves as a very much weaker base than 2,4,6-trinitro-N,N-
dimethylaniline. The contribution of the two electron-releasing (+I) —CH3 groups on
nitrogen in increasing the basic strength of the latter compound is actually very small.
(2) Basic character of tertiary amines (R3N) As the alkyl groups of tertiary amines
(R 3 N) become gradually more bulkier, the additional strain due to crowding is somewhat
reduced by increasing bond angles. As a consequence, the bonding orbitals acquire less p
character, while the orbital containing the lone pair acquires more p character. This type
of modification is actually possible when there is an unshared pair of electrons in one of
the hybrid orbitals. When the amine coordinates with a proton, the bonding orbitals are
forced towards tetrahedral (sp3) shape with reduction of bond angles and increased steric
strain among the alkyl groups. This type of steric strain present in ‘back’ of the amine and
away from the entering H! is called B strain (back strain). The bulkier the alkyl groups,
the more B strain there is and weaker is the base. Another factor that partially reduces
basicity of the 3° amine is the increasing steric hindrance to solvation of the conjugate
≈
acid, R 3 NH .
1.8.1.3 Bond length

The difference in length of a similar type of bonds can be explained in terms of steric
inhibition of resonance. For example, the difference in the C—N bond length o- and p-nitro
groups in picryl iodide can be explained as follows.
O O
I
+ +
– N a N –
O O
(Picryl iodide)
b
+
N –
O O
The large iodine atom in picryl iodide forces the oxygens of the o-nitro groups out of the
plane of the ring but it has no such effect on the p-nitro group. Therefore, only the p-nitro
group enters into resonance with the ring. As a result, the carbon–nitrogen bond in the
p-nitro group acquires some double bond character and hence, the C—N bond length ‘a’ is
longer (145 pm) than ‘b’ which is only 135 pm.
1.8.1.4 Dipole moment

Steric inhibition of resonance is sometime responsible for reducing the dipole moment of
compounds. For example, the dipole moment of p-nitroaniline is higher than the dipole
moment of 2,3,5,6-tetramethyl-4-nitroaniline. In the latter compound, the oxygen of the
p–NO2 group is involved in steric interaction with the two adjacent methyl groups and
as a consequence, the group is forced out of the plane of the ring. Hence, the —NO2 group
cannot withdraw electrons from the —NH2 group by exerting its —R effect. As a result,
the extent of charge separation becomes less and the magnitude of dipole moment also
becomes less. Since steric inhibition of resonance is absent in p-nitroaniline, there occurs
greater charge separation and for this, its dipole moment is much higher.
1.8.2 Proton Sponges

To become free from steric strain, some organic bases possess greater affinity for protons,
i.e., behave as much stronger bases. These compounds are known as proton sponges. For
example, 1,8-bis (diethylamino)-2,7-dimethoxynaphthalene (I) is severely strained because
the lone pairs of two N atoms are forced to be near each other. On protonation the stain
is relieved because one lone pair coordinates with proton which subsequently forms a
hydrogen bond to the other lone pair. The much higher basicities of 4,5-bis (dimethylamino)
fluorene (II) and 3,5-bis(dimethylamino)phenanthrene (III) can similarly be explained.
Compounds like I, II and III are called proton sponges.
1.8.3 Face Strain or F-Strain

When two atoms each containing three large groups approach towards each other to form
a covalent bond, a steric strain develops at the time of bond formation. This is what is
called face strain or F-strain. Because of such strain, the formation of adduct does not
always take place smoothly. For example, tripropylamine gives no detectable adduct with
trimethylborane, while quinuclidine gives a very stable adduct (both the amines have
comparable base strength). The difference of these two amines can be explained in terms
of F-strain. In tripopylamine È(CH 3CH 2CH 2 )3 N,˘ at least two of the three freely rotating
Î ˚
propyl groups remain folded towards, and partly covering the unshared pair of electrons
of nitrogen. Since this conformation causes F-strain, the approach of trimethylborane
towards nitrogen is inhibited. Tripropylamine thus gives no detectable adduct with
trimethylborane.
On the other hand, in quinuclidine, the substituents or nitrogen are held back by the ring
system so that they cannot involve in steric interaction with trimethylborane molecule.
For this reason, quinuclidine forms a very stable adduct.
1.8.4 Steric Acceleration and Steric Retardation

Steric acceleration: Substrates, which are very much compressed, always try to
avoid their steric strain. A reaction will be speeded up if the strain of the substrate is
relieved in attaining the transition state of the rate-determining step of the reaction. This
phenomenon is termed as steric acceleration. The rate of ionization and hence the rate
of solvolysis of a substrate in which there is B-strain is, therefore, expected to be much
higher. Tri-isopropylmethyl chloride, for example, undergoes solvolysis (SN1) at a much
faster rate than that expected on the basis of stabilization of the carbocation through +I
and hyperconjugative effect.
Due to sp3 angle of 109.5°, the bulky alkyl groups of this alkyl chloride are pushed together.
As a result, the compound experiences large steric strain in the ‘back’ of the molecule
(B-strain or back strain). On ionization, i.e., on going from the starting halide (with a
tetrahedral disposition of four groups about the Csp3 atom) to the carbocation (with a
planer disposition of only three group about the Csp2 atom), the bond angle increases from
109.5° to 120°. As a result, the steric strain, i.e., the B-strain resulting from the nonbonded
interactions amongst the alkyl groups, is reduced because space between two alkyl groups
increases. Consequently, the ionization (the rate-determining step) becomes favourable
and the alkyl halide undergoes solvolysis at a faster rate than expected.
Steric retardation: The sheer bulk of groups or atoms on the reacting part of the substrate
may slow down or even stop a reaction. This phenomenon is called steric retardation
or steric hindrance. Tertiary substrates do not undergo SN2 reaction because of steric
hindrance. For example, when tert-butyl bromide is treated with KI in acetone solution,
practically no tert-butyl iodide is obtained because the bulky methyl groups present a
strong hindrance to the approaching nucleophile I①.
1.8.5 Bredt’s Rule

A double bond (C== C or C== N) cannot be introduced at the bridgehead position in bridged
bicyclic compounds with small rings. This is what is called Bredt’s rule. For example,
1-chlorobicyclo [2.2.1]heptane (1) does not undergo base-promoted dehydrochlorination to
yield bicyclo[2.2.1]hept-1-ene (II)
The p-orbitals of a bridgehead double bond are not coplanar and actually these are at
right angles to each other. Because of this, significant orbital overlap is not possible
and hence the formation of a double bond is not possible. To bring the orbitals in one
plane (a condition required for maximum overlap) this regid molecule is to be distorted.
However, such distortion resulting in severe strain in this molecule is not energetically
permissible.
A number of properties of compounds can be explained in terms of Bredt’s rule. These are
as follows:
(1) Basicity of compounds That the compound I is more basic than the compound II can
be explained by Bredt’s rule.
Since the Bredt’s rule is violated in the resonance structures of I, they have practically
no contribution to the hybrid. Hence, the unshared pair of electrons on nitrogen is not
delocalized with the ring p electrons and is, therefore, more available for coordinating
with a proton.
Since II is not a bridged bicyclic compound, therefore, the resonance structures with no
bridgehead double bond are considerably stable and have significant contribution to the
hybrid. Hence, the unshared pair of electrons on nitrogen is well delocalized with the ring
and is not much available for coordinating with a proton. Thus, the compound I is more
basic than the compound II.
(2) Acidity of compounds That the compound I is more acidic and readily soluble in
alkali as compared to the compound II can be explained in terms of Bredt’s rule.
O O O O
I II
The acidity of a b-dicarbonyl compound and its solubility in aqueous alkali depend on the
stability of the enolate ion (the conjugate base). The more the anion is stable the more
the diketone is acidic and the more it dissolves in alkali. The diketone I and II react with
alkali as follows:
Three resonance structures can be written for the conjugate base of I. Two of them are
equivalent and relatively more stable because the negative charge is accommodated by
the highly electronegative oxygen atom. Therefore, these two structures have significant
contribution to the hybrid. Thus, the negative charge on carbon is highly delocalized by
resonance involving the two adjacent C== O groups and consequently, the conjugate base
is well stabilized. Because of this, the diketone I considerably acidic and dissolves readily
in aqueous alkali.
Two of the three resonance structures of the conjugate base of II are very unstable because
they have double bond at the bridgehead position, i.e., they violate Bredt’s rule and hence,
they have, in fact, no contribution to the hybrid. The conjugate base of II is, therefore, not
stabilized by resonance. Because of this, the bicyclic dicarbonyl compound II is relatively

less acidic and reacts with aqueous alkali with much reluctance, i.e., its solubility in
aqueous alkali is actually very poor.
(3) Decarboxylation of b-keto acids When b-keto acids are heated, they undergo ready
decarboxylation via a six-membered cyclic transition state to produce initially an enol
which readily tautomerizes to the more stable ketone. For example:
However, the following bridged b-keto acid does not undergo decarboxylation even when
heated to 300°C.
This observation can be explained by Bredt’s rule. This b-keto acid is expected to
decarboxylate through the formation of a transition state with the partial double bond
at the bridgehead position and thus, through an enol in which the double bond is placed
at the bridgehead position. Both the transition state and the enol are very much less
stable because they violate Bredt’s rule. Because of this, the bicyclic b-keto acid resists
decarboxylation even when heated to 300°C
(4) Formation of Grignard reagent from vicinal dibromide When a vicinal or 1,2-dibromide
is allowed to react with Mg in dry ether, it leads to the formation of an alkene along
with MgBr2. Thus, the vicinal dibromide I smoothly forms the corresponding alkene when
treated with Mg in dry ether.
However, when the bridged bicyclic vicinal dibromide II is treated with Mg in dry ether
elimination of MgBr2 to form the corresponding alkene does not take place because the
alkene and the transition state leading to it violates Bredt’s rules and as a consequence,
the corresponding Grignard reagent is obtained.
1. Explain why o-tert-butylbenzoic acid is more acidic than the para-isomer.

Solution The higher acidity of o-tert-butylbenzoic acid as compared to its para-isomer
can be explained in terms of steric inhibition of resonance (SIR). p-tert-Butylbenzoic acid
can be represented as a resonance hybrid of the following contributing forms:
The carboxyl group by its –R effect withdraws electrons from the ring. The lone pair of
electrons on hydroxyl oxygen is thus prevented to a greater extent from entering into
resonance with the C== O group. As a result, oxygen becomes less positive and proton
release becomes disfavoured. The essential requirement for effective electron delocalization
is that the carboxyl group must be planar or nearly planar with the aromatic ring. In
o-tert-butylbenzoic acid, the planar conformation needed for delocalization experiences
steric strain. To avoid this strain, the carboxyl group is forced out of the plane by the bulky
tert-butyl group. As a consequence, the resonance interaction of the carboxyl group with
the ring decreases and the resonance interaction of C== O with the unshared electron pair
on OH oxygen increases. The hydroxyl oxygen, therefore, becomes relatively more positive
and this causes a more facile proton release.
2. In the following triptycene derivative, rotation of the aryl group around

the O-aryl bond is not completely free — Why?
Me
O H Me
CMe3 H
Solution A complete 360° rotation of the aryl group around the O-aryl bond requires the
aryl group to pass over three rotational barriers; one of which is the C — CMe3 bond and
the other two are the top C — H bond of the other two rings. For this reason, the triptycene
derivative, rotation of the aryl group around the O-aryl bond is not completely free.
3. It is possible to prepare cis and trans isomers of 5-amino – 2,4,6-triiodo –
N, N, N¢, N¢-tetramethylisophthalamide. Explain.
Solution It is possible to prepare cis and trans isomers of 5-amino-2,4,6-triiodo-N, N, N¢,
N¢-tetramethylisophthalamide because each —CONMe2 group being flanked by two bulky
iodine atoms cannot rotate freely due to steric strain. This is, in fact, an example of cis-
trans isomerism resulting from restricted rotation about single bonds.
4. Explain why the compound I is more acidic than the compound II.
Solution The conjugate base of I, i.e., the carbanion derived from the compound I is
well stabilized by resonance because all the three aromatic rings are tightly held with
each other and hence coplanar. Also, the carbanion constitutes an aromatic system
[(4n + 2) p electron, where n = 1]. On the other hand, the benzene rings in the conjugate
base of II, i.e., the carbanion derived from II, prefer to exist in different planes due to
steric reason (steric interaction between ortho H atoms on adjacent rings) and so, it is
not well stabilized by resonance. Because of greater stability of the conjugate base, the
compound I is more acidic than the compound II.
5. The amide I is more basic than the isomeric amide II—Why?
Solution Nitrogen’s electron pair in amide becomes involved in resonance interaction

with the adjacent C== O group. That is, the C== O group withdraws the unshared pair
on nitrogen by is strong –R effect. In these two amides, this resonance interaction is not
equally well. Two resonance structures can be written for each of the two isomeric bicyclic
amides I and II.
The dipolar resonance structure of I with a double bond at the bridgehead position is very
unstable because it violates Bredt’s rule. So it has practically so contribution to the hybrid
of I. Hence, electron delocalization does not take place in this amide and consequently, the
unshared pair of electrons on nitrogen is well available for coordinating with a proton. On
the other hand, the dipolar resonance structure of II with a double bond not in a bridgehead
position is quite stable and so contribute significantly to the hybrid of II. Hence, effective
electron delocalization occurs in II and the lone pair on nitrogen is poorly available for
sharing with a proton. Thus, the amide I is more basic than the amide II.
6. The following b-hydroxyketone does not undergo dehydration when
heated with NaOH solution — Why?
Solution Base-catalyzed dehydration b-hydroxyketones takes place by E1CB mechanism

which involves abstraction of a proton from the a-carbon by base to form an enolate
ion in the first step followed by loss of OH① ion in the second step. When this bicyclic
b-hydroxyketone is heated in the presence of NaOH solution, an enolate is first formed
but the second step, i.e., the loss of OH① ion does not take place because this leads to
the formation of a double bond at the bridgehead position which is not feasible sterically
(violates Bredt’s rule). This explains why the given b-hydroxyketone does not undergo
dehydration when heated with NaOH solution.
1. The length of C—C bond connecting the ring in 2,2¢,6,6¢-tetramethylbiphenyl is

longer than that in biphenyl. Explain.
2. Hydroxide is a stronger nucleophile than tert-butoxide, despite its lower base
strength – Why?
3. 2,6-Di-tert-butylpyridine is a specific proton scavenger. Explain.
[Hint: Because of steric reason (F-strain), the compound does not act as a
nucleophile, but it can easily take up a small proton because the space between
the two bulky tert-butyl groups is enough for the entry of a proton. So this pyridine
derivative is a specific proton scavenger.
4. Rank the following compounds in order of decreasing basicity and give your
reasoning.
NMe N N
==
O O
O
Me
I II III
5. Solvolysis of the alkyl chloride A in aqueous ethanol proceeds about 600 times
faster than the alkyl chloride B. Account for this observation.
CH3
|
(Me3C CH 2 )2C Cl Me3C Cl
A B
[Hint: Steric acceleration]
6. It is possible to prepare cis and trans isomers of 1,8-di-o-tolylnaphthalene – Why?
[Hint: Free rotation of the o-tolyl groups about the single bond is not possible due
to steric reason.]
7. Quino[7,8–h]quinoline (I) is a much stronger base than quinoline (II). Account for
the fact.
N N
N
I II
8. Out of the following two diols (A) and (B), one forms intermolecular hydrogen bonds
while the other forms intermolecular hydrogen bonds. Identify them and give your
reasoning.
CMe3 CMe3
HO HO
OH OH
A B
[Hint:
In the stable conformations of these two diols, the bulky —CMe3 group is always
equatorial.]
9. 2,6-Di-tert-butylpyridine is a weaker base than pyridine — Why?
[Hint: This can be explained in terms of entropy effect caused by steric effect. The
conjugate acid of di-tert-butylpyridine is less stable than the conjugate acid of non-
sterically hindered pyridine because in the former conjugate acid, the bulky tert-
butyl groups restrict rotation in water molecule which forms hydrogen bond with
it and thereby lowers the entropy and decrease its stability.
10. The heat of hydrogenation of cis-2-petnene is higher (28.6 kcal/mol) than that of
trans-2-pentene (27.6 kcal/mol) — Why ?
11. The bicyclic lactam A undergoes hydrolysis 107 times faster than the monocyclic
lactam B. Explain.
12. 1,3-Cyclohexadione readily undergoes base or acid-catalyzed deuterium exchange,

whereas the following diketone does not — Why ?
[Hint: Enolate formation is not possible because electron delocalization violates

Bredt’s rule.]
13. Predict the increasing order of equilibrium constants of the reaction of the following
bases with Me3B and explain your answer.
III, II, I
[Hint: ææææææ
K eq increases
Æ (because F-strain decreases)]
14. Predict with proper reasoning which member in each of the following pairs behaves
as a stronger base towards BMe3.
N
(a) Et2NH and (b) and
15. Compare C—N bond length (a vs a¢) and (b vs b¢) in the following compounds and
justify.
16. The following bicyclo trisulfone (I) readily dissolves in aqueous sodium bicarbonate
solution, while the triketone (II) does not — Why?
[Hint: Compounds containing S atoms seem to defy Bredt’s rule. The conjugate
base of I is highly stabilized by resonance, while the conjugate base of II is not at
all stabilized by resonance.]
17. Predict whether any change will take place or not. Justify your answer.
18. Which one of the following two compounds is more acidic and why?
[Hint: The enolate of II is resonance stabilized, but the enolate of I is not. Therefore,
II is relatively more acidic.]
19. The following compound being an amide reacts more or less like a ketone — Why?
N O
==
H3 C
CH3
CH3
[Hint: It is an extreme case of a twisted amide. The overlap of the lone pair of
electrons on nitrogen with the p-system of the carbonyl group is highly prevented.
So, this compound reacts more or less like a ketone.]
20. N, N-Dimethylation triples the basicity of aniline but increases the basicity of
2,4,6-trinitroaniline by 40,000 fold. Explain.
21. Are the following two structures resonance forms? Explain why or why not ?
22. The following b-keto acid does not decarboxylate even when heated to 500°C.
Explain.
O
==
COOH
23. The following quinclidinone forms an oxime — Why?
N
O
24. The bridged bicyclic b-keto acids I, II and III undergo ready decarboxylation on
heating while the b-keto acids IV and V do not — Why?
[Hint: A bicyclic ring system [a, b, c] with bridgehead double bond is sterically
feasible when (a + b + c), i.e., the sum of the number of atoms in the bridges or the
S number is seven or more, S ≥ 7. The numerical value S is 7 for I and II, 9 for III
and 5 for IV and V.]
25. Which one of the following two carboxylic acids is more acidic and why?
COOH H
H COOH
I II
[Hint: The conjugate base (i.e., the corresponding carboxylate ion) of the cis-
isomer I is not well solvated due to steric crowding caused by C–3 and C–5 axial
hydrogens. On the other hand, the conjugate base of the trans-isomer II is well
solvated because the equatorial —COOH group is free from any steric crowding.]
1.9 INTERMOLECULAR FORCES

The forces that hold molecules together may be called intermolecular forces. There are
two kinds of intermolecular forces: (1) dipole–dipole interactions and (2) van der Waals
forces.
1.9.1 Dipole–Dipole Interactions

The attraction of the positive end of one polar molecule for the negative end of a second
polar molecule is called dipole–dipole interaction. The polar hydrogen fluoride molecules,
for example, are held together by dipole–dipole interactions that operate either head to
tail (I) or laterally (II).
This attractive force is effective over a very short distance and its strength is about
2 kcal/mol. As a result of dipole–dipole interaction, polar molecules are generally held to each
other more strongly as compared to nonpolar molecules of comparable molecular weight. The
presence of dipole–dipole interaction in a compound is reflected in its physical properties.
1.9.1.1 Hydrogen bonding

It is a specifically strong kind of dipole–dipole attraction that operates between a
hydrogen atom that is covalently bonded to an atom of high electronegativity and smaller
size such as F, O, and N and a second such atom. The energy of such a bond varies from
5 to10 kcal/mol. The covalent bond between a hydrogen atom and a highly electronegative
element is considerably polar because the electron cloud is greatly distorted towards
the electronegative atom. The partial positive charge developed on the thinly shielded
hydrogen nucleus is strongly attracted by the partial negative charge on the electron-
negative atom of a second molecule (same or different) and leads to the formation of what
is called a hydrogen bond. The dipolar HF molecules, for example, are held together by
hydrogen bonding (represented by dotted lines) as shown below:
1.9.1.2 Types of hydrogen bonding

There are tow types of hydrogen bonding:
(1) Intermolecular hydrogen bonding When H-bonding occurs between different molecules
of same or different compounds, it is called intermolecular hydrogen bonding. For
example, methanol (CH3OH) molecule contains one highly electronegative oxygen atom
bonded to H atom. Thus, the oxygen atom becomes partially negatively charged and the H
atom becomes partially positively charged. The negative end of one molecule attracts the
positive end of the other to form hydrogen bond. In this a way, a large number of methanol
molecules become associated through hydrogen bonding to form a macromolecule which
can be represented as (CH3OH)n.
(2) Intramolecular hydrogen bonding When hydrogen bonding takes place within the same
molecule, it is called intramolecular hydrogen bonding. This type of hydrogen bonding is
also known as chelation as it results in formation of a ring. It is favoured when a six or
five membered ring is formed. It is normally found in disubstituted benzene compounds in
which the substituents are attached to adjacent carbon atoms, i.e., they are ortho to each
other. For example, o-nitrophenol possesses intramolecular hydrogen bonding.
Hydrogen bonds markedly influence the physical properties like melting points, boiling
points and solubilities of compounds. Acidity of certain acids can also be influenced by
intramolecular hydrogen bonding.
1.9.2 van der Waals Forces

Since nonpolar compounds can also solidify, therefore, there must be some forces acting
between the molecules of a nonpolar compound. Such attractive forces are called van
der Waals forces. Such attractions are also referred to as instantaneous dipole-induced
dipole attractions. In nonpolar molecules (i.e., methane, CCl4, etc.), the centre of positive
charge density coincides with the centre of negative charge density and so the molecules
possess no net dipole moment. However, due to random movement of electrons around the
nucleus, a momentary distortion of their distribution may occur. This results in momentary
loss of electrical symmetry and formation of a momentary dipole in the molecule. This
instantaneous dipole induces a dipole in a neighbouring molecule. These dipoles then
attract each other and as a result, the molecules become held together. The van der Waals
forces act only between the portions of different molecules that are in close contact, that
is, between the surfaces of molecules. Therefore, the size of van der Waals forces depends
on the area of contact between the molecules. The greater the area of contact, the stronger
the van der Waals forces and the greater the amount of energy needed to overcome these
forces. Since the surface areas of heavier molecules are usually much greater, therefore,
intermolecular van der Waals attractive forces are also much larger. Another factor which
effects the strength of van der Waals forces is polarizability. Polarizability indicates the
measure of how the electron cloud around an atom can be distorted. The larger the atom,
the more loosely it holds the outermost electrons and the more they can be distorted. The
more polarizable the atom, the stronger the van der Waals attractive forces. The strength
of van der Waals forces (also called London forces) is only about 1 kcal/mol and this is
significant only at extremely short distances. These forces are, in fact, the weakest of all
kinds of intermolecular forces and these help us to understand the effect of molecular size
and shape on physical properties.
(1) Boiling Point (bp) The boiling point of a compound is the temperature at which
the liquid form of the compound becomes a gas. In other words, it is the temperature at
which the vapour pressure of the compound equals the atmospheric pressure. In order
for a compound to vaporize, the forces that hold the molecules close to each other must
be overcome. Therefore, the boiling point of a compound depends on the strength of the
attractive forces between the individual molecules in the more ordered liquid state. The
stronger the intermolecular attractive forces, the higher the boiling point and this is
because a lot of energy is required to pull the molecules away from each other.
(2) Melting point The melting point in the temperature at which a solid is converted
into its liquid phase. In melting, energy is required to overcome the attractive forces in
the more ordered crystalline state. The stronger the intermolecular forces, the higher the
melting point. Symmetry also plays an important role in determining the melting point
of compounds containing the same functional group and similar molecular masses. A
compact symmetrical molecule which packs well into a crystalline lattice has a much higher
melting point. The more the compound is symmetrical, the higher is the melting point.
(3) Solubility Solubility is the extent to which a compound (called solute) dissolves in
a liquid (called solvent). The condition for dissolution of a solute in a solvent is that the
solute–solvent interactions must be equal or larger in magnitude than the combined
solute–solute and solvent–solvent interactions. That is, the old attractive forces must be
replaced by new ones. Compounds generally dissolve in solvents possessing similar kinds
of intermolecular forces. Polar compounds dissolve in polar solvents while nonpolar or
weakly polar compounds dissolve in nonpolar or weakly polar solvents. The general rule
that explains solubility on the basis of polarity of molecules is that ‘like dissolves like’.
Table 1.1 Summary of types of intermolecular forces
Type of force Relative strength Exhibited by Example
van der Waals forces weak all molecules CH3CH2CH2CH2CH3
CH3CH2CH2CH2CHO
CH3CH2CH2CH2CH2OH
Dipole–dipole moderate molecules with a net CH3CH2CH2CH2CHO
interactions dipole CH3CH2CH2CH2CH2OH
Hydrogen bonding strong molecules with an O–H, CH3CH2CH2CH2CH2OH
N–H or H–F bond CH3CH2CH2NH2
1.9.3 Effect of Intermolecular Forces on Different Properties

of Compounds
1.9.3.1 Boiling points
(1) Boiling points of alcohols Alcohols have much higher boiling points than those
of ethers and alkanes of comparable molecular masses. For example, n-butyl alcohol,
CH3CH2CH2CH2OH (molecular mass 74) boils at 118°C, whereas the boiling points of
diethyl ether, CH3CH2OCH2CH3 (molecular mass 74) and n-pentane, CH3(CH2)3CH3
(molecular mass 72) are 35°C and 36°C respectively. Due to comparable molecular masses,
the compounds should have similar volatility and so they are expected have comparable
boiling points. But the boiling points of alcohols are much higher. Hydrogen bonding, the
strongest intermolecular force is responsible for such behaviour of alcohols. Since in alcohol
molecules a hydrogen atom remains bonded to a highly electronegative and relatively
small oxygen atom, therefore, they remain associated through intermolecular hydrogen
bonding.
Hydrogen bonding does not take place in ethers and in alkanes because hydrogen in
them is attached to less electronegative carbon atoms. In the case of ethers, relatively
weak dipole–dipole interactions operate along with very weak van der Waals forces and
in the case of alkanes only very weak van der Waals forces operate. Since at the boiling
temperature a considerable amount of thermal energy is required to separate the molecules
by breaking numerous H-bonds, therefore, alcohols boil at a much higher temperature as
compared to ethers and alkanes.
The boiling points of isomeric alcohols decrease with increase in branching of the carbon
chain. For example, the boiling points of n-butyl alcohol, CH3CH2CH2CH2OH, isobutyl
alcohol, (CH3)2CH CH2OH, sec-butyl alcohol, CH3CH2CHOHCH3 and tert-butyl alcohol,
(CH3)3COH, are 118°C, 108°C, 100°C and 83°C, respectively. As branching increases,
alcohol molecules tend to become spherical, thereby resulting in a decreased surface
area available for van der Waals interactions. Steric hindrance also affects the extent
of hydrogen bonding. As the number of alkyl groups around the carbon bearing the –OH
group increases, the extent of hydrogen bonding decreases due to steric crowding.
(2) Boiling points of carboxylic acids The boiling points of carboxylic acids are higher
than those of alcohols of comparable molecular masses. For example, the boiling point
of formic acid (100.5°C) is higher than that of ethyl alcohol (78.5°C), even though their
molecular masses are the same (46). Both formic acid and ethyl alcohol contain polar
O — H bond and remain associated through hydrogen bonding. Because of resonance, the
polarity of the O — H bond present in the —COOH group of formic acid is much higher
than that of the O — H bond in ethyl alcohol. Furthermore, the negative end of the polar
carbonyl group (—C == O) is also involved in hydrogen bonding.
Therefore, the hydrogen bond in formic acid is much stronger than that in ethyl alcohol
and for this reason, the boiling point of formic acid is higher than that of ethyl alcohol.
(3) Boiling points of nitrophenols O-Nitrophenol has a much lower boiling point than
its meta- and para-isomers. The —NO2 and —OH groups in o-nitrophenol molecule are
attached to the adjacent ring carbons. Due to close proximity of these groups, intramolecular
H-bonding giving rise to a stable six-membered ring (chelation) is possible in this isomer.
Consequently, the —OH group is no longer available to form molecular association by
intermolecular hydrogen bonding and so, the molecules exist as single unit (monomer).
However, intermolecular hydrogen bonding is not possible in the meta- and para-isomers
because in these two compounds the —OH and —NO2 groups are held farther apart. They
form molecular association by intermolecular hydrogen bonding. Therefore, it requires
a considerable amount of energy to separate the molecules by breaking hydrogen bonds.
Hence, these isomers boil at a much higher temperature than o-nitrophenol.
(4) Boiling point of nitroalkanes (R—NO2) Nitroalkane molecules are highly polar (m = 3.5
– 4.0 D) and relatively strong dipole–dipole interactions hold the molecules together. On the
other hand, weak van der Waals forces hold the nonpolar hydrocarbon molecules together.
Since the dipole–dipole interactions (dissociation energy 8.373 kJ/mol) are stronger than
the van der Waals forces (dissociation energy 4.186 kJ/mol), therefore, it requires more
thermal energy to separate the nitroalkane molecules and hence nitroalkanes possess
higher boiling points than alkanes of comparable molecular masses.
(5) Boiling point of HF That the strength of the F–H---F H-bond (41.8 kJ mol) is the
highest can be demonstrated by the following observation. Hydrogen fluoride (HF) and
ethyl fluoride (C2H5F) are equally polar and the latter is heavier than the former. Yet
the boiling point of HF (19.34°C) is much higher than that of C2H5F (–37.7°C). Since the
compounds are equally polar, the strengths of dipole–dipole interactions involved in these
two compounds are almost the same and hence the compounds are expected to boil at
nearly the same temperature. But due to large molecular mass, C2H5F is expected to boil
at higher temperature than HF. In fact, HF has a boiling point 57° higher than that of
C2H5F. This difference in boiling point can be explained in terms of hydrogen bonding.
Since in hydrogen fluoride, a hydrogen atom is bonded to the highest electronegative
fluorine atom, the molecules remain associated through strong intermolecular hydrogen
bonding.
Hydrogen bonding in C2H5F is not possible because there is no hydrogen atom bonded to
electronegative F atom. The molecules are held together by relatively weak dipole–dipole
attractions. Since it requires more energy to cleave the hydrogen bonds, therefore, HF
boils at a much higher temperature than the heavier C2H5F.
(6) Boiling points of alkanes The boiling point of isomeric alkanes decreases with increase
in branching. For example, the boiling point of n-pentane, CH3(CH2)3CH3 is 36.1°C,
whereas the boiling point of isopentane, (CH3)2CH CH2CH3 and neopentane, (CH3)4C are
27.9°C and 9.5°C, respectively. Because the strength of van der Waals forces depends
on the area of contact between molecules, branching in a compound lowers its boiling
point because it reduces the area of contact. A branched hydrocarbon molecule has a more
compact, nearly spherical shape. If we think of the unbranched alkane pentane as a cigar
and branched alkane neopentane as a tennis ball, we can see that branching decreases
the area of contact between molecules. Two cigars make contact over a greater area than
do two tennis balls. Thus, among the isomeric alkanes, the more highly branched one
will have lower boiling point and the unbranched one will have a higher boiling point. It
thus follows that n-pentane with no branching of the carbon chain boils at 36.1°C while
isopentane with one branch-chain and neopentane with two branch chain boil at 27.9 and
9.5°C, respectively.
1.9.3.2 Melting points

(1) Melting points of isomeric 1,2-dibromoethylenes The cis-isomer of 1,2-dibromoethylene
melts at a lower temperature than the trans-isomer. This can be explained in terms of
intracrystalline forces. The more symmetrical a compound, the better it fits into a crystal
lattice and hence the stronger the intracrystalline forces and the higher the melting point.
The cis-isomer of 1,2-dibromoethylene is less symmetrical than the trans-isomer. So, the
cis-isomer fits into a crystal lattice less tightly as compared to the more symmetrical trans-
isomer. Thus, the cis-isomer has lower melting point than the trans-isomer.
(2) Melting points of isomeric xylenes o-Xylene melts at a much lower temperature than
p-xylene. The highly symmetrical p-xylene fits better into a crystal lattice than the less
symmetrical o-xylene. So, the melting point of the p-isomer having stronger intracrystalline
forces is considerably higher than that of the o-isomer
(3) Melting points of carboxylic acids The melting point of a carboxylic acid containing
even number of carbon atoms is higher than the acids having odd number of carbon
atoms lying immediately below or above it in the series. This can be explained as follows.
Carboxylic acids containing even number of carbon atoms have carboxyl and terminal
methyl groups on the opposite sides of the zig-zag carbon chain and hence they fit better in
the crystal lattice thereby increasing intermolecular attractive forces resulting in higher
melting point. On the other hand, carboxylic acids containing odd number of carbon atoms
have the carboxyl and the terminal —CH3 groups on the same side of the zig-zag carbon
chain. Consequently, these molecules being less symmetrical fit less tightly in the crystal
lattice. As a result, the magnitude of intermolecular forces of attraction becomes relatively
less in this case and hence these acids have lower melting points.
1.9.3.3 Conformation Because of intramolecular hydrogen bonding some com-

pounds prefer to exist in a particular conformation. For example, 5-hydroxy-1,3-dioxane
prefers to exist in the conformation with OH axial. In this conformation, the OH hydrogen
is involved in the formation of hydrogen bond with the ring oxygen atom and so it becomes
relatively more stable.
Similarly, because of intramolecular hydrogen bonding, cis-1,4-cyclohexanediol exists

preferably in the twist-boat conformation.
1.9.3.4 Solubilities
(1) Solubility of alcohols The solubility of alcohols in water decreases as the molecular
mass increases, i.e., as the carbon chain becomes longer and among isomeric alcohols the
solubility increases with increase in branching of carbon chain. An alcohol becomes soluble
in water due to the formation of hydrogen bond with water molecules. In alcohols, the
hydrocarbon part, i.e., the alkyl group (—R) is nonpolar and hydrophobic (water avoiding)
while the hydroxyl (—OH) group is highly polar and hydrophilic (water seeking). In
lower alcohols (up to four carbons), the polar and hydrophilic —OH group constitutes
a considerably large part of the molecule and as its characteristic effect predominates
over the effect of the nonpolar and hydrophobic alkyl group, these alcohols are soluble in
water by forming H-bonds. With increase in molecular mass of alcohols, the size of the
water insoluble hydrocarbon part increases and since the hydrophilic —OH group then
contributes a very small portion of the molecule, the alcohol becomes progressively less
water-soluble. Also, for a nonpolar hydrocarbon chain to be accommodated by water, the
water molecules have to form a more ordered ‘ice-like’ structure around the chain, and for
this, the entropy change is unfavourable. Thus, solubility decreases as the hydrocarbon
chain gets larger. For example, methyl alcohol CH3OH, with a small hydrophobic part, is
highly soluble in water while n-undecyl alcohol, with a large hydrophobic part is completely
insoluble in water.
The solubility of alcohols also depends on the structure of the alkyl group. Since branching
decreases the relative volume of the hydrophobic hydrocarbon portion, therefore, the
alcohols become progressively more water soluble with increase in branching. For
example, tert-butyl alcohol (CH3)3COH is much more soluble in water than n-butyl alcohol,
CH3CH2CH2CH2OH.
(2) Solubility of amines Low-molecular-mass amines are soluble in water because amines
can form hydrogen bonds with water. Comparing amines with the same number of carbon
atoms, primary amines are more soluble than secondary amines because primary amines
have two hydrogens that can engage in hydrogen bonding. Tertiary amines, on the other
hand, have unshared pair of electrons that can accept hydrogen bonds but do not have
hydrogen to donate for hydrogen bonds. Therefore, tertiary amines are less soluble in
water than secondary amines with the same number of carbon atoms.
Amines are more water soluble than alcohols with the same number of carbons. Due to
formation of H-bonds with water molecules, both amines and alcohols are soluble in water.
However, the hydrogen bonds of the type involved in the case of amines are stronger than
those involved in the case of alcohols because amine nitrogen is a better hydrogen bond
acceptor (base) than oxygen in alcohol. It is the stronger hydrogen bonding for which
amines are more soluble in water than alcohols.
(3) Solubility of nitrophenols o-Nitrophenol has much lower solubility in water than
its meta- and para-isomers. Due to the formation of intramolecular hydrogen bond, the
molecules of o-nitrophenol cannot form hydrogen bonds with water molecules and so it
is less soluble in water. Intramolecular hydrogen bonding is not possible in the meta-
and para-isomers because in these two compounds, the —OH and —NO2 groups are held
farther apart. For these reasons, they can form hydrogen bonds with water and becomes
considerably soluble in water.
1.9.3.5 Acidic character of organic compounds

(1) Acidity of diastereoiosmeric dicarboxylic acids
Maleic acid, cis —HOOC—CH== CH—COOH (pKa1 = 1.92, pKa2 = 6.04) is stronger for the
first ionization but weaker for the second ionization than the isomeric fumaric acid, trans–
HOOC—CH== CH—COOH (pKa1= 3.03, pKa2 = 4.44). In maleic acid, the two carboxyl
(—COOH) groups are placed on the same side of the double bond and so, the maleate
monoanion is stabilized by intramolecular hydrogen bonding. In fumaric acid, on the
other hand, the two —COOH groups are placed on the different sides of the double bond
and consequently, intramolecular hydrogen bonding is not possible for stabilization of

the fumarate monoanion. Because of greater stability of the monoanion, maleic acid has
a greater tendency to ionize than fumaric acid, i.e., for the first ionization it is a stronger
acid than fumaric acid.
Due to stabilization of the maleate monoanion by intramolecular hydrogen bonding, it is
very difficult to remove a proton from it. Proton release is also disfavoured because the
field effect generated by the —COO① group on the nearby —COOH group is free and there
is no significant field effect caused by the —COO① group. On the other hand, in fumarate
monoanion, the —COOH group is free and there is no significant field effect caused by the
—COO① group. So, it easily ionizes to form the fumarate dianion, i.e., for second ionization
fumaric acid is stronger than maleic acid.
The compound I is more acidic than the compound II can also be explained in terms of
hydrogen bonding and field effect.
Each of these two diastereoisomeric acids (I and II) is locked in that particular chair
conformation in which the bulky —C(CD3)3 group is placed in an equatorial position. In
such conformation, the —COO① and —COOH groups are both axial in I and equatorial in II.
Because of close proximity of these two groups, intramolecular hydrogen bonding occurs in
II, but not in I in which they are held farther apart. Therefore, ionization of II is relatively
more disfavoured compared to I. Proton release from II is further disfavoured because of
field effect generated by the —COO① group, on the —COOH group. No significant field
effect operates in I. Hence, the compound I is more acidic than the compound II.
(2) Acidity of hydroxybenzoic acids Hydroxybenzoic acids have two acidic functional
groups; one is —COOH and the other is —OH.
The first ionization (i.e., dissociation of carboxyl proton) of o-hydroxybenzoic acid (salicylic
acid) is more favourable than that of the p-hydroxybenzoic acid, whereas the second
ionization (i.e., dissociation of hydroxyl proton) of the latter acid is more favourable than
that of the former acid. This observation can be explained by hydrogen bonding and field
effect. Since the —OH and the —COOH groups in o-hydroxybenzoic acid (salicylic acid)
are placed in adjacent ring carbons, therefore, the conjugate base, i.e., the monoanion, is
stabilized by intramolecular hydrogen bonding leading to the formation of a stable six-
membered ring (chelation). On the other hand, the two groups in the p-isomer are held
farther apart and no such intramolecular H-bonding is possible. o-Hydroxybenzoate ion
is, therefore, more stable than the p-hydroxybenzoate ion and because of this, ionization
of the —COOH group of o-hydroxbenzoic acid is relatively more favourable as compared
to that of p-hydroxybenzoic acid. Since the o-hydroxybenzoate ion is stabilized by internal
H-bonding, it is difficult to remove the —OH proton from this negatively charged cyclic
system, i.e., the second ionization does not take place easily. Also, the ionization is
disfavoured because of the field effect generated by —COO① group on the nearby —OH
group. In p-hydroxybenzoate ion, on the other hand, the —OH group is free and so the
second ionization occurs easily in this case.
[Although intramolecular H-bonding occurs in undissociated o-hydroxybenzoic acid, it

is less effective than its anion because a negative charge on oxygen leads to a stronger
H-bonding.]
1. How can o-nitrophenol be separated from its m- or p-isomer ? Give reasons.

Solution A compound can be steam-distilled if it has an appreciable vapour pressure at
the boiling point of water. Due to intramolecular H-bonding, o-nitrophenol is a low-boiling
compound and it has appreciable vapour pressure at 100°C. So it can be separated from
its m- or p-isomer by steam distillation.
2. Explain why in the vapour state acetic acid (CH3COOH) has molecular
weight of 120.
Solution In the vapour phase, acetic acid (CH3COOH) exists as dimmers in which a pair
of acid molecules are held together by intermolecular hydrogen bonds. Because of this,
acetic acid in the vapour state has a molecular mass of 120.
3. The compound II is more basic than the compound I. Explain.
Solution The conjugate acid (monoprotonated from) of the cis-diamine II is stabilized by

intramolecular hydrogen bonding while the conjugate acid of the trans-diamine I is not
stabilized by intramolecular hydrogen bonding because the groups are held further apart.
For this reason, the diamine II is relatively more basic than the diamine I.
4. Nonpolar ethane (bp – 88.2°C) boils at a temperature higher than methane

(bp – 162°C) at a pressure of 1 atm — Why?
Solution Heavier molecules require greater thermal energy in order to acquire velocities
sufficiently great to escape the surface of the liquid and since the surface areas of heavier
molecules are usually much greater, therefore, intramolecular van der Waals attractions
are also much larger. Ethane (CH3CH3) is heavier than methane (CH4) and also its surface
area is greater than methane. For these reasons, ethane boils at a temperature higher
than methane.
5. The fluorocarbon C5F12 has a slightly lower boiling point than pentane
(C5H12), even though it has a far higher molecular mass. Explain.
Solution Since the molecular mass of C5F12 is much higher than C5H12 and also its size
is somewhat larger than C5H12 (F is slightly larger than H), therefore, C5F12 is expected
to boil at a temperature higher than C5H12. This can be explained in terms of relative
polarizability of the electrons of the atoms involved. The highly electronegative fluorine
atoms show a very low polarizability because their electrons are very tightly held with the
nucleus. Because of this, C5F12 molecules are held together by very small van der Waals
forces and in fact, it is much smaller than that operates in C5H12. For this reason, C5H12
has a slightly lower boiling point than pentane (C5H12), even though it has a far higher
molecular weight.
6. Arrange the following compounds in order of increasing boiling point
and explain the order:
Solution To predict relative boiling points, we have to consider the differences in

(i) hydrogen bonding and (ii) molecular mass and surface area. 2,2-Dimethylpropane
(neopentane) is the lightest among these compounds and it has compact spherical shape for
which the van der Waals attractions are minimum. For this reason, 2,2-dimethylpropane
is the lowest-boiling compound. The other four compounds have similar molecular masses.
Neither hexane nor 2,3-dimethylbutane is hydrogen bonded, so they will be the next
higher in boiling points. Because 2-3-dimethylbutane is more highly branched and has
a smaller surface area than hexane, the former will be lower boiling than the latter. The
two remaining compounds are alcohols and they remain associated with intermolecular
hydrogen bonding. Therefore, their boiling points are higher than the three hydrocarbons.
Now, 1-pentanol has more surface area for van der Waals forces to operate and hence
its boiling point is higher than the relatively compact 2-methyl-2-butanol. Therefore, the
increasing order of boiling point is: 2,2-dimethylpropane < 2,3-dimethylbutane < hexane
>2-methyl-2-butanol < 1-pentanol.
7. The cis-isomer of 1,2-dibromethylene boils at a higher temperature than
its trans-isomer — Why?
Solution The trans-isomer of 1,2-dibromoethylene is nonpolar because the two C — Br
bond moments orienting in opposite directions cancel each other. So relatively weak van
der Waals forces hold the molecules together. The cis-isomer, on the other hand, is polar
because the two Br atoms lie on the same side of the double bond and the two C — Br
bond moments do not cancel each other. A net moment resulting from two individual
bond moments operates. As a result, there operates relatively stronger dipole–dipole
attractions amongst the molecule along with van der Waals forces. Since more energy
is needed to separate the molecules by overcoming these forces, the cis-isomer boils at a
higher temperature than the trans-isomer.
8. o-Xylene boils at a higher temperature than p-xylene — Why?

Solution o-Xylene molecules are weakly polar. So there operates weak dipole–dipole
attractions amongst the molecules in addition to the van der Waals forces. On the other
hand, p-xylene molecules are nonpolar and so only weak van der Waals forces hold the
molecules together. Because of this, o-xylene boils at a higher temperature than p-xylene,
but the difference in boiling points is actually very small.
9. The melting point of neopentane (Me4C) is much higher than that of

isopentane [CH3CH2CH(CH3)2]. Explain.
Solution The compact symmetrical molecules of neopentane (MC34C) pack well into a
crystal lattice while isopentane [CH3CH2CH(CH3)2], which has a —CH3 group dagling
from a four-carbon chain, does not. Therefore, intracrystalline forces are much stronger
in neopentane compared to isopentane. Consequently, neopentane has a much higher
melting point than isopentane.
10. Methyl fluoride (CH3F) is more soluble in water than methyl chloride
(CH3Cl) — Why?
Solution Methyl fluoride (CH3F) containing highly electronegative and small-sized F
atom can form hydrogen bond with water but methyl chloride does not. For this reason,
methyl fluoride is more soluble in water than methyl chloride.
O
CH3—F H H F—CH3
11. tert-Butyl alcohol is more soluble in water than n-butyl alcohol. Explain.
Solution Alcohols containing branched alkyl groups are more water soluble than
nonbranched alkyl groups with the same number of carbon atoms because branching
minimizes the contact surface of the nonpolar portion of the molecule and thereby decreases
the van der Waals forces of attraction. It is for this reason, the relatively more compact
tert-butyl alcohol is more soluble in water than n-butyl alcohol.
CH3
|
H3C — C — OH CH3CH2CH2CH2OH
| n-Butyl alcohol
CH3 (bp 118∞ C)
tert-Butyl alcohol
(bp 83∞ C)
12. Butane (CH3CH2CH2CH3) is soluble in carbon tetrachloride but not in

water — Why ?
Solution The molecules of both butane and carbon tetrachloride (CCl4) are nonpolar and
the molecules of each compound are held to each other by weak van der Waals interaction.
Since ‘like dissolves like’ is a rule for dissolution of a solute in a solvent and the dissolution
process involves a large increase in entropy, i.e., thermodynamically favourable, butane
dissolves in carbon tetrachloride. The highly polar water molecules are held together by
strong dipole–dipole interactions and hydrogen bonds. There could be only very weak
attractive forces (dipole-induced dipole attractions) between butane molecules on the one
hand and water molecules on the other hand. Since solute–solvent interactions cannot
outweigh or equal to the sum of solute–solute interactions (van der Waals forces) and
solvent–solvent interactions (hydrogen bonding forces), butane does not dissolve in
water.
≈ @
13. Ammonium chloride (N H 4 C l) is insoluble in the nonpolar solvent
≈ @
carbon tetrachloride while tetramethylammonium chloride (Me4 N C l) is
appreciably soluble in this solvent. Explain.
≈
Solution In nonpolar solvent CCl4, stabilization of N H 4 and Cl① ions by solvation does
not take place and because of this, ammonium chloride is insoluble in CCl4. Tetramethyl-
≈ @
ammonium chloride, Me4 NCl, on the other hand, is appreciably soluble in CCl4 because
the cation in which the positive nitrogen is surrounded by four methyl groups presents
a large nonpolar hydrocarbon surface to this solvent and becomes involved with it by
van der Waals forces of attraction.
14. Explain why glycerol (HOCH2CHOH CH2OH) is a very viscous liquid.
Solution The degree of molecular association through intermolecular hydrogen bonding
is much higher in glycerol containing three —OH groups. For this reason, this triol is a
very viscous liquid.
15. The melting point of NaCl (801°C) is very much higher than that of CCl4
(–24°C) — Why ?
Solution The crystal units in NaCl are cations and anions, which are held together by
strong electrostatic forces of attraction. Hence it requires a large amount of energy to
separate the ions. On the other hand, the crystal units in CCl4 are nonpolar molecules,
which are held together by weak van der Waals forces. Therefore, these molecules can be
separated by applying a small amount of thermal energy. This explains why the melting
point of NaCl is very much higher than that CCl4.
16. In which of the following solvents would cyclohexane have the lowest
solubility: diethyl ether, 1-pentanol, hexane and ethanol ?
Solution Cyclohexane is a nonpolar compound which is expected to be more soluble in
a nonpolar solvent like hexane or a weakly polar solvent diethyl ether (‘like dissolves
like’) than in an H-bonding and polar solvent like 1-pentanol and ethanol. Again, the
nonpolar hydrocarbon part of ethanol (C2H5OH) is smaller than that in 1-pentanol
(CH3CH2CH2CH2CH2OH). Therefore, cyclohexane is least soluble in ethanol.
17. Salicylic acid, o-HOC6H4COOH, is a stronger acid than o-CH3OC6H4COOH.
Explain why?
Solution Because of the ortho-effect of the bulkier —OCH3 group, o-CH3OC6H4COOH is
expected to be more acidic. However, the reverse is true, and this is because, the conjugate
base of salicylic acid is stabilized by internal H-bonding (a more dominating factor).
18. 2,6-dihydroxy-4-methylbenzoic acid is a much stronger acid than

2-hydroxy-4-methylbenzoic acid. Explain.
Solution The conjugate base of 2,6-dihydroxy-4-methylbenzoic acid is highly stabilized
by intramolecular H-bonding (chelation) involving both the o-hydroxyl groups and for this
reason, it is a much stronger acid than 2-hydroxy-4-methylbenzoic acid, the conjugate
base of which is stabilized by intramolecular H-bonding involving only one —OH group.
19. Arrange the following compounds in order of increasing boiling point

and explain your answer.
CH3CH2OCH2CH3 CH3CH2CHCH3 CH3CH2CH2 CH2CH3

Diethyl ether | Pentane
OH
sec-Butyl alcohol
Solution Pentane has no polar groups, so its molecules are held together only by weak
van der Walls forces. Diethyl ether has the polar ether group, so it can have dipole–dipole
interactions in addition to van der Waals forces. 2-Butanol is a polar bent molecule, so it
can have dipole–dipole interactions in addition to van der Waals forces. Because it has
an O—H bond 2-butanol molecules are held together by intermolecular hydrogen bonds.
Since hydrogen bonding is stronger than dipole–dipole attractions which in turn stronger
than van der Waals forces, therefore, the boiling point of these three compounds increases
in the order: pentane < diethyl ether < sec-butyl alcohol.
1. Which compound in each of the following pairs would have the higher boiling point?
Explain your answer.
(a) HOCH2CH2OH or CH3CH2CH2OH
(b)
(c) (d)
(e) (f)
2. Arrange the following compounds in order of increasing boiling point and explain
the order:
3. Which compound in each pair has the higher melting point ?
(a)
(b)
(c) (d)
4. Which compound in each pair is more soluble in water and why?

(a) (b)
(c) (d)
5. Symmetry affects the melting point of a compound but not the boiling point —
Why?
6. The boiling point of ethanol (78°C) is more than 100°C higher than that of dimethyl
ether (–25°C) while ethylmethylamine has a boiling point (37°C) only 34° higher
than that of triethylamine (3.5°C). Explain.
7. 1-Butanol (bp 118°C) has a much higher boiling point than its isomer ethoxyethane
(bp 35°C). However, both of them show same solubility (8 g pre 100 g) in water.
Account for these observations.
8. Comment on the solubility of the following compounds in water and in organic
solvents (such as CCl4):
NaCl, CH3CH2CH2CH3, CH3CH2CH2OH, CH3(CH2)10OH
9. What types of intermolecular forces are present in each of the following
compounds?
(a) (b) (c) (CH3)3N
(d) CH2 == CHBr (e) CH3CH2CH2COOH (f) CH3CH2C ∫∫ C CH2CH3

10. o-Hydroxybenzaldehyde has much lower boiling point and much lower water
solubility than its m- and p-isomers. Explain.
11. Which of the following compounds exhibit chelation:
(a) methyl salicylate, (b) o-iodophenol, (c) o-fluorophenol,
(d) o-cyanophenol, and (e) o-cresol?
[Hint: Chelation occurs only in (a) and (c). In (d), although the N atom of —CN is
electronegative, the linearity of the —CN group places the N atom too far away
from the —OH group to form an H-bond.]
C6H6
12. Acietic acid dimmer, (CH 3COOH)2 2CH 3COOH DS = +20 e.u.; despite
the favourable entropy of dissociation, the equilibrium lies to the left. Explain
why?
[Hint: Strength of intramolecular H-bonding in the dimer is more than enough to
comprensate for the loss of freedom in the dimer.]
13. 2-methylpyrrolidine boils at temperature higher than the boiling point of pyrrolidine
— Why.
14. Between the anti- and syn-isomers of pyridine-2-carboxaldoxime the anti-

form predominates in the isomeric distribution of the compound. Explain this
observations.
[Hint: The anti-form gets extra stability due to intramolecular hydrogen bonding
(chelation). No such chelating effect is observed in the syn-isomer. For this reason,
the anti-form predominates in the isomeric distribution of the compound.
15. Resorcinol has higher boiling point than 2-nitroresorcinol — Why?

16. 8-hydroxyquinoline can be separated from 4-hydroxyquinoline by steam distillation.
Account for this observation.
[Hint: Intramolecular H-bonding occurs in 8-hydroxyquinoline but not in
4-hydroxyquinoline.
17. Draw the H-bonding arrangements in CH3OH — H2O and CH3NH2 — HCHO systems.
[ ]
18. Why is the mp of sulphanilic acid so high?

[Hint: Sulphanilic acid exists as a salt (a dipolar ion or zwitter ion):
–
H3N— —SO3 . Because of this, sulphanilic acid has a much higher melting
point.]
19. Explain why H2O has a higher boiling point than CH3OH (65°C), NH3 (–33°C) and
HF (20°C).
20. Arrange the following compounds in order of increasing boiling point. Explain your
answer.
21. Explain why 1-pentanol has solubility of 2.7 g per 100 ml of water, where as ethanol
is completely miscible in water.
22. Why does one expect the cis-isomer of an alkene to have a higher boiling point than
the trans-isomer?
23. Alcohols with fewer than four carbons are soluble in water, but alcohols with more
than four carbons are insoluble in water — Why?
24. The boiling point of propylamine (bp 49°C) is higher than that of ethylmethylamine
(bp 37°C) which in turn is higher than that of triethylamine (bp 3.5 °C). Explain.
[Hint: Trimethylamine (MC3N) has no N — H bond and so it cannot form hydrogen
bonds with each other. Ethylmethylamine (CH3 CH2NH CH3) has one N — H bond
and as it remains associated through intermolecular hydrogen bonding. Propylamine
(CH3CH2CH2NH2) with two N — H bonds is more extensively hydrogen bonded.
This explains their boiling points]
25. There are four amides with the molecular formula C3H7 NO. Write their structures.
One of these amides has melting and boiling point that is substantially lower than
that of the other three. Identify this amide and explain your answer.
[Hint: CH3CH2CONH2, CH3CONHCH3, HCONHCH2CH3 and HCON (CH3)2. The
last one has a melting and boiling point that is substantially lower than that of
the other three because it does not have a hydrogen that is covalently bonded
to nitrogen and, therefore, its molecules cannot form hydrogen bonds to each
other. The other molecules all have a hydrogen covalently bonded to nitrogen, and
therefore, hydrogen-bond formation is possible.]
26. Which of the following compounds is expected to volatilize easily and why?
OH
NH NH
CH CH
N N
I II
[Hint: Due to intramolecular hydrogen bonding, I is expected to volatilize easily.]
1.10 REACTIVE INTERMEDIATES

A chemical reaction involves conversion of a molecule into a new molecule. The new
molecule, i.e., the product, has different arrangement of atoms compared to the starting
molecule, i.e., the reactant. A redistribution of electrons also occurs during this change. In
most of the organic reactions, conversation of reactants into products takes place through
some specific steps. The mechanism of an organic reaction is a detailed description of these
steps. It not only acquient us the number of steps involved in the reaction, but also informs
us regarding the sequence of breaking old bonds and making new bonds. Study of reaction
mechanism is of immense importance in organic chemistry as thousands of apparently
different reactions occur through a limited number of common steps.
Organic reactions usually involve fission of weaker covalent bonds and formation of
stronger ones, so that a relatively stable molecule is formed from a less stable molecule.
Breaking of bonds requires energy while formation of bonds involves release of energy.
A covalent bond is represented by a dash (—) and the transfer of electrons is shown by
using arrow signs. Curved arrow signs containing two barbs ( ) indicate the shifting
of a pair of electrons while the transfer of a single electron is indicated by curved arrow
signs containing one barb ( ) or fishhook arrow [it is to be noted that the symbol ( )
is incorrect].
Fission or cleavage of covalent bonds can take place in two ways depending on the nature
of the bond involved, the nature of the attacking agent and the conditions of the reaction.
(1) Homolytic fission or homolysis: It a covalent bond in a molecule undergoes
fission in such a way that each of the two bonded atoms gets one electron of the
shared pair, it is called homolytic fission or homolysis. This type of bond cleavage
results in formation of neutral species called free radicals, or often simply radicals,
i.e., a radical is a reactive intermediate with a single unpaired electron. Homolytic
fission is usually favoured by conditions such as nonpolar nature of the bond,
high temperature, presence of high energy (uv) radiations or presence of radical
initiators such as peroxides. The homolytic fission of a bond A—B leading to the
formation of free radicals A and B (each containing odd electrons), may be shown
as follows:
e.g.,
Homolytic bond cleavage requires less energy than heterolytic bond cleavage.
(2) Heterolytic fission or heterolysis: When a covalent bond breaks in such a way
that the pair of electrons stays with the more electronegative atom, such a fission
is called heterolytic fission or heterolysis. This type of fission results in formation
of ionic (cationic and anionic) intermediates. If in the molecule A–B, B is more
electronegative than A, the heterolytic fission of the bond leading to the formation
≈ @
of the cation A and the anion B: may be shown as follows:
e.g.,
This type of bond cleavage resulting in the formation of charged species, i.e., ions, is
favoured by conditions such as polar nature of the covalent bond and the presence of
polar solvent. If the fragments of a heterolytic fission are carbon species, then the cation
is called carbocation and the anion is called carbanion. Both of them are unstable reactive
intermediates.
Reactive intermediates: Under the influence of attacking reagent, most of the organic
compounds (substrates) undergo either hemolytic or heterolytic fission of a bond to form
certain short lived and highly reactive (hence cannot be normally isolated) chemical species
which are called reactive intermediates or reaction intermediates. Some common examples
of reactive intermediates are carbocations, carbanions, free radicals, carbenes, nitrenes,
arynes, etc. A one-step reaction is called a concerted reaction and in such a reaction, no
matter how many bonds are broken or formed, a starting material is converted directly
into a product, i.e., a concerted reaction involves no reactive intermediate. A stepwise
reaction involves more than one step. The substrate is first converted into an unstable
intermediate, which then goes on to form the product.
Concerted reaction: Substrate Æ Product
Stepwise reaction: Substrate Æ Reactive intermediate Æ Product
An understanding of the structure and properties of these intermediates (molecules,
ions or radicals) is very much important in understanding organic reaction mechanism.
Matters related to various intermediates are discussed below:
(1) Carbocations
Carbocations are a group of reactive intermediates having positively charged carbon
atom bearing only six electrons. These are represented by the symbol R!. For example,
≈ ≈ ≈
CH3 , CH3CH2 , (CH3 )2 CH, etc. Because of having a strong tendency to complete the octet
of the electron-deficient carbon, carbocations are highly reactive species.
Generation: Carbocations are generated by heterolytic fission of a bond to carbon in
which the leaving group is removed along with its shared pair of electrons.
The principal ways of generating carbocations are as follows:

(i) Heterolysis of C — L bond (L = halogen, OTs, OBs, etc.) of neutral substrates
by the influence of polar solvents like H2O, EtOH, etc. leads to the formation of
carbocations. For example:
(ii) Deamination of aliphatic amines by nitrous acid leads to the formation of

carbocation. For example:
(iii) Protonation of alcohols followed by loss of H2O leads to the formation of carbocations.
For example:
+
H + +
(CH3)3C—OH (CH3)3C—OH2 (CH3)3C + H2O
H2SO4
(iv) Protonation of alkenes leads to the formation of carbocations. For example:
(v) Carbocations can be generated by the action of Lewis acid such as AlCl3 on alkyl
halides. For example:
(vi) Carbocations can be obtained by treating alkanes with FSO3H — SbF5 called super
acid. FSO3H — SbF5 is, in fact, an extremely strong acid which donates H! even
to an alkane in order to form a carbocation by extracting a hydride ion (H①). H2 is
liberated in this reaction. For example:
(vii) A more stable carbocation can be obtained when a relatively less stable carbocation
abstracts H①. For example:
H H +
–
+ H-shift
Ph3C + Ph3CH +
Cycloheptatrienyl
cation
(aromatic)
Nomenclature: In naming a carbocation, the word ‘cation’ is added to the name of the
≈ ≈ ≈
alkyl or aralkyl group. For example, CH 3 ,(CH 3 )2 CH, and C6 H5 CH 2 are named as methyl
cation, isopropyl cation and benzyl cation, respectively.
Classification: Carbocations are classified as primary (1°), secondary (2°) and tertiary
(3°) on the basis of the number of carbon atoms (one, two or three) directly bonded to
≈
the positively charged carbon atom. Fro example, ethyl cation (CH 3 CH 2 ) is a primary,
≈
È ≈ ≈˘
isopropyl cation (CH 3CHCH 3 ) is secondary and tert-butyl cation Í(CH 3 )3 C˙ is a tertiary
≈
Î ˚
carbocation. Methyl cation (CH 3 ) with one carbon atom is a special case.
Structure: The positively charged carbon atom of a carbocation is sp2-hybridized.
Therefore, the shape of a carbocation is trigonal planar and the bond angle is 120°. The
three sp2 orbitals are utilized in making bonds to three substituents. The vacant p orbital
is perpendicular to the plane of sp2 hybridized orbitals.
Stability: Any factor which tends to delocalize the positive charge must increase the
stability of carbocations while any factor which tends to localize or intensify the positive
charge must decrease the stability of carbocation.
The stability of alkyl carbocations decreases in the following order:
Carbocations are stabilized mainly by +I, +R and hperconjugation effects. The relative
stability of carbocation can be easily assessed by determining the heterolytic R — H
Æ R! + H① dissociation energies in the gas phase. The lower the value of energy, the
greater the stability of the carbocation. The bond dissociation energies D (R+ — H–) where
R = alkyl, aralkyl or aryl are given in the following table.
Heterolytic R — H Æ R+ + H– dissociation energies in gas phase
Ion D(R+ + H–) in kcal/mol
≈
CH 3 314.6
C6 H5≈ 294
≈
CH2 == CH 287
Stability increases
C2 H5≈ 276.7
≈
CH == CH — CH2 256
≈
(CH 3 )2 CH 249.2
246
≈
C6 H5 CH 2 238
≈
(CH3 )3 C 231.9
Factors which determine the stability of carbocations:

(a) Inductive effect: An ion is stabilized by a factor that disperses its charge. The
electron-donating inductive effect (+I) exerted by an alkyl group attached to the
positive carbon of a carbocation neutralizes the charge partially. As a result, the
charge becomes dispersed over the alkyl groups and the system becomes stabilized. For
example, the methyl group in ethyl cation stabilizes the system through its +I effect.
The stabilities of carbocations increase with increasing the number of electron-

releasing alkyl groups attached to the positive carbon.
(b) Hyperconjugation effect: An alkyl group may also delocalize the positive charge
of a carbocation by the hyperconjugation effect. The charge becomes dispersed over
the a–H atoms and consequently, the system becomes stabilized. Hyperconjugation
in ethyl cation, for example, occurs as follows:
È H H≈ H H ˘
Í | ≈ | | ˙
ÍH — C — CH 2 ´ H — C == CH 2 ´ H≈ C == CH 2 ´ H — C == CH 2 ˙
Í | | | ˙
ÍÎ H H H H≈ ˙˚
Hyperconjugation in ethyl cation
As the number of a-hydrogens, i.e., the number of hyperconjugation structures,
increases, the stability of carbocations increases. Therefore, the stability of methyl
substituted carbocations increases in the following order:
≈ ≈ ≈ ≈
(least stable) C H 3 CH 3 C H 2 (CH)2 C H (CH 3 )3 C (most stable)
(3 a - H) (6 a - H) (9 a - H)
(c) Resonance effect: Resonance is the most important factor influencing the
stability of carbocations. When there is a double bond a to the positive carbon
of a carbocation, effective charge delocalization with consequent stabilization
occurs. Allyl and benzyl cations, for instance, are found to be highly stabilized by
resonance.
A nonbonding lone pair on a heteroatom which is directly attached to the electron-

deficient C stabilizes a carbocation more than any other interaction. This resonance
satisfies the octet of the cationic carbon, even though there is developed a positive
≈
charge over the heteroatom. For example, CH 3OCH 2 is a very stable carbocation
in spite of its primary nature. In fact, the methoxymethyl cation is obtained as a
≈
stable solid, CH 3OCH 2 SbF6@ .
(d) Steric effect: Stability of tertiary carbocation increases due to steric effect. For
È ≈˘
example, the alkyl groups of triisopropyl cation Í(Me2CH)3 C˙ having planar
Î ˚
arrangement with a bond angle of 120° are far apart from each other and so, there
is less steric interaction among them. However, when this carbocation undergoes
nucleophilic attack, i.e., when a change of hybridization of the central carbon
atom from sp2 (trigonal) to sp3 (tetrahedral) occurs, the bulky isopropyl groups are
pushed together. This results in a steric strain (B strain) in the product molecule.
Because of this, carbocation is not much willing to react with a nucleophile, that is,
its stability is enhanced due to steric reason.
(e) Constituting an aromatic system: The vacant p orbital of a carbocation may

be involved in constituting a cyclic and planar (4n + 2)p electron system, where
n = 0, 1, 2 …. etc., i.e., a carbocation may be stabilized by constituting an aromatic
system. In fact, it acquires maximum stability by this process. Cycloheptatrienyl
cation, for example, is unusually stable because it is a planar 6p electron system
and aromatic.
1.10.1 Nonclassical Carbocation

A carbocation in which the positive charge is delocalized by a double or triple bond that
is not in the allylic position or by a single bond is called a nonclassical carbocation.
7-Norbornenyl and norbornyl cations are two examples of nonclassical carbocation.
In a carbocation, if there is one carbon atom between the positively charged carbon atom
and the double bond, it is called a homoallylic carbocation. 7-Norbornenyl cation is also an
example of homoallylic carbocation.
1.10.2 Carbonium Ion and Carbenium Ion or Carbocation

There are intermediates in which a carbon bears a positive charge, but the formal covalency
of the carbon atom is five rather than three. The simplest example is the methanonium ion,
≈
CH5 . Such pentacoordinated positive ions are called carbonium ions. The intermediates
in which there is positive charge at a carbon atom which is trivalent are called carbenium
≈ ≈
ions or carbocations. For example, methyl cation (CH 3 ) and ethyl cation (CH 3 CH 2 ), etc.
≈
[N.B. The methanonium ion CH5 has a three-centre two-electron bond. It is not known
whether this ion is a transition state or a true intermediate, but an ion CH5≈ has been
detected by mass spectroscopy.
(2) Carbanions
Carbanions are a group of reactive intermediates carrying a negative charge on carbon
@
atom possessing eight electrons in its valence shell. For example, CH 3 (methyl carbanion)
@
and CH 3 CH 2 (ethyl carbanion), etc. They are represented by symbol R①. Their reactivity
is due to the presence of formal negative charge on the carbon atom.
Generation: Carbanions are generated by heterolytic fission of a bond to carbon in which
the bonding electron pair remains with the carbon atom.
The principal methods of generating carbanions are as follows:

(i) By an abstraction of an acidic hydrogen alpha to an electron-withdrawing group:
Compounds containing acidic hydrogen attached to carbon alpha to —NO2, —CN
or C == O groups produce carbanions when treated with a strong base. For
example:
(ii) By abstraction of hydrogen from terminal alkynes using a strong base: Terminal
alkynes being acidic produce carbanion when treated with strong bases. For
example:
–
–
H2N: + H—C∫∫ C—CH3 C∫∫ C—CH3 + NH3
(iii) By metal halogen exchange: When organic halogen compounds are treated with
strongly electropositive metals like Li, Na, etc. in an inert solvent, carbanions are
obtained in the form of organometallic compounds. For example:
@ ≈
ether
CH 3 — Br + 2Li æææ Æ CH 3 L i + LiBr
@ ≈
ether
Ph3C — Cl + 2Na æææ Æ PH 3 C Na + NaCl
@ ≈
ether
Ph — Br + 2Li æææ Æ Ph Li + LiBr
(iv) By decomposition of carboxylate ions: When metal carboxylates are heated, they
undergo decarboxylation to yield carbanions. For example:
(v) By addition of an anion to multiple bond containing electron-withdrawing groups:

Carbanions are obtained when an alkene containing electron-attracting group
undergoes nucleophilic attack by an anion. For example:
(vi) By addition of electrons to an unsaturated system: Unsaturated compounds

may accept electrons from electropositive metals to generate carbanions. For
example, when cyclooctatraene is treated with metallic sodium, it is converted to
cyclooctatetraenyl dianion which is a stable aromatic system containing (4n + 2)p
electrons, where n = 2.
Nomenclature: In naming a carbanion, the word ‘anion’ is added to the name of the
@ @ @
alkyl or aralkyl group. For example, CH 3 , (CH 3 )2 CH and C6 H5 CH 2 are named as methyl
anion, isopropyl anion and benzyl anion, respectively.
Classification: Carbanions are classified as primary (1°), secondary (2°) and tertiary
(3°) on the basis of the number of carbon atoms (one, two or three) directly bonded to
@
the negatively charged carbon atoms. For example, ethyl anion (CH 3 CH 2 ) is a primary,
@ ≈
isopropyl anion (Me2 CH) is a secondary and tert-butyl anion Me3 C is a tertiary carbanion.
@
Methyl anion (CH 3 ) with one carbon atom is a special case.

Structure: In simple carbanions, the negatively charged central carbon atom is sp3
hybridized; it is surrounded by three bonding electron pairs and one unshared pair of
electrons occupying an sp3 orbital. Therefore, a carbanion is expected to have the tetrahedral
shape. However, the shape is not exactly that of a tetrahedron and in fact, it is found to
have the pyramidal shape just like ammonia. Since the repulsion between the unshared
pair and any bonding pair is greater than the repulsion between any two bonding pairs,
therefore, the angle between two bonding pairs (i.e., between two sp3-s bonds) is slightly
less than the normal tetrahedral value of 109.5° and for this reason, a carbanion appears
to be shaped like a pyramid with the negative carbon at the apex and the three groups at
the corners of a triangular base.
The central carbon atoms in resonance-stabilized carbanions are, however, sp2 hybridized
and hence they are planar. This is due to the fact that planarity is an essential criterion
for resonance to occur. Allyl anion, for example, is a planar carbanion.
The negative carbon atom of the following conjugated anion is, however, not resonance-
stabilized because according to the Bredt’s rule a double bond at the bridgehead position
cannot be formed in bridged bicyclic compounds with small rings and for this reason, this
carbon is sp3-hybridized. Its shape is pyramidal.
Stability: Any factor that tends to delocalize the negative charge must increase the
stability of carbanions while any factor that tends to localize or intensify the negative
charge must decrease the stability of carbanions.
[W = Electron-withdrawing [D = Electron-donating
group; it disperses the group; it intensifies the
negative charge and thus, negative charge and hence
stabilizes the carbanion] destabilizes the carbanion]
@ @
The stability of carbanions follows the order: Methyl anion (CH 3 ) > R CH 2 (primary or 1°)
@ @
> R 2 CH (secondary or 2°) > R 3 C (tertiary or 3°) because an alkyl group destabilizes a
carbanion. Carbanions are stabilized mainly by –I and –R effects. Functional groups in
the a position stabilize carbanions in the following order: —NO2 > —COR > —COOR >
—CN ~ —CONH2 > —X > —H.
The structural features responsible for the increased stability of carbanions are as follows:
(a) s character of the anionic carbon atom: An s orbital being closer to the nucleus
than the p orbital in a given main quantum level possesses lower energy. An
electron pair in an orbital having large s character is, therefore, more tightly held
by the nucleus and hence of lower energy than an electron pair in an orbital having
relatively small s character. Thus, a carbanion in which the anionic carbon is sp-
hybridized (50 percent s character) is more stable than a carbanion in which the
anionic carbon is sp2 hybridized (33.33 percent s character), which in turn is more
stable than a carbanion in which the negative carbon is sp3-hybridized (25 percent
s character). Therefore, the order of decreasing stability of carbanions is
(b) Inductive electron withdrawal: Substituents possessing electron-withdrawing

inductive effects (–I) stabilize a carbanion by dispersing or delocalizing the negative
charge. Examples of some carbanions experiencing such stabilizing effect are as
follows:
(c) Resonance effect: If there is a double or triple bond conjugated with the anionic
carbon, the carbanion is stabilized by delocalization of the negative charge with the
p orbitals of the multiple bond. Thus, allylic and benzylic carbanions and carbanions
attached to the groups containing polarized multiple bond such as —NO2, —C ∫∫N,
C ∫∫ O, etc. are well stabilized by resonance. For example:
(d) Formation of an aromatic system: A carbanion becomes highly stabilized if it

is involved in constituting an aromatic system [a planar (4l + 2) p electron system,
where n = 0,1,2, …, etc.]. Cyclopentadienyl anion, for example, is unusually stable
because it is a close loop of 6p electron system and hence aromatic.
Optical inactivity of asymmetric carbanions like CH 3CH 2 CHCH 3 :

Asymmetric carbanions are found to be optically inactive and cannot be separated
into two enantiomers. This can be explained in terms of umbrella effect. There
occurs rapid interconversion between two enantiomeric pyramidal carbanions

(called umbrella effect) through a low energy planar transition state and because
of this, asymmetric carbanions exist as racemic modifications and are found to be
optically inactive. The pyramidal inversion is so rapid that such carbanions cannot
be separated into two enantiomers, i.e., cannot be resolved.
[N.B. Due to the same basic reason, a chiral tertiary amine such as CH 3CH 2 N H CH 3
is found to be optically inactive and cannot be separated into two enantiomers.]
(3) Free radicals

An atom or a group of atoms containing an odd (unpaired) electron is called a free radical
or simply a radical. Although free radicals are electrically neutral species, they are, due
to the presence of an odd electron, weakly attracted by the magnetic field and hence are
paramagnetic in nature. They are very much reactive. Some examples of free radical
are methyl radical (C H 3 ), ethyl radical (C H C H 2 ), phenyl radical (P h) , etc. They are
intermediates in a group of reactions called radical reactions.
Generation: Homolytic cleavage of a covalent bond leads to the formation of free radicals.
Homolysis of bonds to carbon leads to the formation of free radicals containing an odd
electron on carbon; these are collectively called alkyl free radicals or alkyl radicals or
simply free radicals.
Free radicals can be generated by the following three principal methods:

(a) Thermolysis: When compounds containing weak bonds are strongly heated, they
undergo hemolytic cleavage of a covalent bond to form free radicals. For example:
(i) Tetramethyl lead decomposes to form lead and methyl radicals when heated
strongly at 600°C.
600∞ C
(CH 3 )4 Pb æææÆ Pb + 4CH 3
(ii) The O–O bond of a peroxide or a peracid has a strength of only 36 kcal/mol and so
it undergoes cleavage to form radicals at room temperature. For example:
(iii) The formation of radicals is often facilitated by the simultaneous release of very
stable molecule like N2. Thermolysis of azo compounds, for example, leads to the
formation of free radicals by the expulsion of N2 molecule. For example:
(b) Photolysis: Absorption of visible or ultraviolet light provides a molecule with

sufficient energy to cleave covalent bonds and thus, photochemical dissociation
leading to the formation of radicals may take place. Acetone, for example, undergoes
photochemical decomposition in the vapour phase by light having a wavelength of
320 nm to give CO and methyl radical (CH3 ).
Azoalkanes or diacyl peroxides also undergo photolytic cleavage to form radicals.

For example:
(c) Redox reactions: One-electron transfer reactions are frequently employed to

produce radicals. For example:
(i) In Kolbe electrolysis, a reaction that has considerable synthetic importance, the
carboxylate ion gives up one electron to the anode and becomes oxidized to a radical.
The resulting radical then dissociates to give an alkyl radical which dimerizes to
yield an alkene.
(ii) Inorganic ions such as Fe2+, Cu+, etc. capable of changing their valence state by
accepting or losing a single electron are used for the generation of radicals. These,
in fact, act as a reducing agent in such reactions. For example
Fenton’s reagent behaves as an oxidant by forming another radical. For example:
A stable phenoxide radical is obtained when Fe3+ ion acts as an oxidizing agent by
accepting an electron from a phenoxide ion. For example:
Nomenclature: In naming a free radical containing an odd electron on carbon,

the word ‘radical’ is added to the name of the alkyl or aralkyl group. For example,
CH 3 , (CH 3 )2 CH and C6 H5CH 2 are named as methyl radical, isopropyl radical and benzyl
radical, respectively.
Classification: Free radicals are classified as primary (1°), secondary (2°) and tertiary
(3°) according to the number of carbon atoms (one, two or three) directly attached to the
carbon atom bearing the unpaired electron. For example, ethyl radical (CH3CH2) is a
primary, isopropyl radical (Me2CH) is a secondary and tert-butyl radical (Me3C) is a
tertiary radical.
Structure: Simple alkyl radicals may have two possible shapes. They may have a planar
triangular configuration (the carbon containing the unpaired electron is sp2-hybridized)
and the unpaired electron is accommodated in a p orbital or a rapidly inverting pyramidal
configuration (the central carbon is sp3-hybridized and the unpaired electron is in an sp3
orbital).
Available evidence indicates that simple alkyl radicals prefer a planar or near-planar
shape, even though the energy difference between a planar and a pyramidal free
radical is not large. Free radicals in which the central carbon is connected to atoms of
high electronegativity, e.g., CF3, prefer to exist in pyramidal shape. The deviation from
planarity increases with increase in electronegativity. The shape of CF3 can be well
explained by valence shell electron pair repulsion theory (VSEPR). Due to the strong –I
effect of fluorine, the electron density around the central carbon in CF3 is reduced and as
a result, the unpaired electron–bond pair repulsion becomes greater than bond pair–bond
pair repulsion. As a consequence, the angular distance between the p orbital containing
the odd electron and the C — F bonding orbital is increased to give pyramidal shape of the
radical.
When a bridgehead carbon of bicyclic system bears the unpair electron, a planar trigonal
configuration is completely prevented by the geometric requirements and a pyramidal
configuration becomes mandatory. For example, the bridgehead radicals I and II are
pyramidal. It is to be noted that bridgehead radicals are less stable than the corresponding
open-chain radicals.
Like carbanions, the central carbon atoms of resonance-stabilized free radicals are sp2
hybridized and hence they are also planar. Allyl radical, for example, is planar.
Stability: Free radicals are highly unstable and reactive species because they have a
strong tendency to gain an electron, i.e., to share with some other atom or group to have a
complete octet. The stability of alkyl radicals follows the order: R3C (tertiary or 3°) > R2CH
(secondary or 2°) > RCH2 (primary or 1°) > CH3 (methyl radical).
The factors responsible for the stability of free radicals are as follows:
(a) Hyperconjugation: Free radicals are stabilized by hyperconjugation (s–p
conjugation) involving a–H atoms.
As the number of a–H atom increases, hyperconjugation becomes progressively

more effective and consequently, the radicals become progressively more stabilized
due to more effective electron delocalization. The relative stability of simple alkyl
radicals is found to follow the order: (most stable) R3C (tertiary) > R2CH (secondary)
> RCH2 (primary) > CH3 (methyl) (least stable). For example, tert-butyl radical
(CH3)3C (with nine hyperconjugable a–H atoms) is more stable than isopropyl
radical, (CH3)2CH (with six hyperconjugable a–H atoms) which in turn is more
stable than ethyl radical, CH3CH2 (with only three hyperconjugable a–H atoms).
The methyl radical, CH3, is the least stable radical because the unpair electron is
not at all delocalized due to lack of a–H atoms.
(b) Resonance: Resonance is a major and important factor influencing the stability
of free radical. When the carbon atom bearing the odd electron is a to a double
bond or a benzene ring, effective delocalization of the unpaired electron with the p
orbital system with consequent stabilization occurs. Allyl and benzyl radicals, for
example, are found to be particularly stable because of resonance.
(c) Steric effect: Steric effect sometimes becomes very much significant in stabilizing
tertiary radicals. There occurs considerable relief of steric strain (B strain) when
an sp2-hybridized tertiary radical is formed from an sp3-hybridized substrate and
this is because the repulsion between the bulky alkyl groups is relieved to a certain
extent by an increase in bond angles from 109.5° to about 120°. As a consequence,
the radical becomes much reluctant to react further, i.e., its stability is enhanced
due to steric reason.
[Steric inhibition of resonance often decreases radical stability. For example, the
resonance stability of triphenylmethyl radical (Ph3C) is not so as expected and this
is because due to steric hindrance between ortho H-atoms of the adjacent rings,
the three benzene rings cannot be all in the same plane for effective delocalization
to take place. In fact, it exists in propeller-shaped conformation in which the three
rings being aligned at about 30° out of the common plane.
In fact, the major reason for the greater stability of Ph3C is its greater reluctance
towards dimerization because dimerization will introduce a considerable amount of
steric strain due to close proximity of the bulky phenyl groups in hexaphenylethane
(Ph3C — C Ph3), the expected dimer. Thus, it is not resonance, but steric hindrance
to dimerization is the major cause for the stability of triphenylmethane radical.
Detection: A very simple method to know whether a particular reaction proceeds through
a radical intermediate or not is to study the kinetics of the reaction (i.e., the reaction rate)
in the presence and in the absence of inhibitors such as hydroquinone. If the reaction
proceeds through radical intermediate, its rate is expected to be seriously affected by the
presence of the inhibitor.
Electron spin resonance (esr) spectroscopic studies offer the most useful method for
detecting radicals.
Bond dissociation energy (a measure of free radical stability): The bond dissociation
energy is the energy needed to cleave a covalent bond homolytically.
Since bond breaking requires energy, therefore, bond dissociation energies are always
positive numbers and homolysis is always endothermic. The shorter the bond, the higher
the bond dissociation energy. Again, shorter bonds are stronger bonds.
Bond dissociation energies (DH° values) of R—H bonds provide a measure of relative
inherent stability of free radicals R . The following table lists such values. The higher the
DH° values, the less stable the radical.
DH° values of some R—H bonds

Bond dissociation energy,
R
DH° (Kcal/mol)
111
Ph
CF3 107
CH 2 == CH 106
106
CH3 105
CH3CH2 100
Stability increases
(CH3 )3CCH2 100
CH3CH2 CH2 100
Cl3C 96
(CH3 )2CH 96
(CH3 )3C 95.8
95.5
C6 H5CH2 88
HCO 87
CH 2 == CH — CH 2 86
The bond dissociation energy for a benzylic hydrogen is less than that of a methane
hydrogen which can be explained by resonance.
From the enthalpy values, it becomes clear that 17 kcal/mol less energy is needed to form
the benzyl radical from toluene than to form methyl radical from methane. This difference
in energy requirement can be explained on the basis of the difference in energy content
between the radical and its precursor. Both toluene and benzyl radical can be represented
as resonance hybrids.
Since toluene is a resonance hybrid of only two resonance structures (I and II) and the
benzyl radical is a resonance hybrid of five resonance structures (III–VIII), therefore, the
benzyl radical is stabilized by resonance to a greater extent than toluene. As a result
of this, the difference in energy content between benzyl radical and toluene becomes
relatively small. Both methane and methyl radical, on the other hand, can be represented
satisfactorily by a single (localized) structure.
None of these species are resonance-stabilized and thus, the difference in energy content
between the methyl radical and methane is relatively large. Because of this, it requires
higher energy to form methyl radical from methane than to form benzyl radical from
toluene. This becomes clear from the following energy diagrams.
The bond dissociation energy of CH3 — H is equal to the hypothetical but greater than the
actual bond dissociation energy of PhCH2 — H (DE1 > DE2).
1.10.2.1 Radical inhibitors or ‘scavengers’

Compounds that destroy reactive radicals by creating unreactive stable radicals or neutral
molecules with only paired electrons and thereby prevent reactions that take place by
mechanisms involving radicals are called radical inhibitors. Phenol is an example of the
first type, and nitric oxide (NO) is an example of the second type.
Oxygen molecule (O2) can also function as a highly efficient radical inhibitor. This is
because it has two unpaired electrons and behaves as a diradical (O — O) . By combining
with highly reactive radical intermediates (R) , it converted into very much less reactive
peroxy radicals (R — O — O) that cannot carry out the chain reaction.
1.10.2.2 Electrophilic and nucleophilic radicals

(a) Electrophilic radicals: The radicals that have a tendency to abstract electron-rich
hydrogen atoms and which behave like electrophiles in that the rate is increased by
the presence of electron-releasing groups in the substrate are called electrophilic
radicals. For example, an electron-releasing group at the para-position of toluene
increases the rate of hydrogen abstraction by bromine while an electron-attracting
group decreases it. Therefore, B r is an example of electrophilic radical.
(b) Nucleophilic radicals: The radicals that have a tendency to abstract electron-
poor hydrogen atoms and that behave like nucleophiles that which the rate is
increased by the presence of electron-attracting groups in the substrate are called
nucleophilic radicals. For example, an electron-attracting group at the para-
position of toluene increases the rate of hydrogen abstraction by tert-butyl radical
while an electron releasing group decreases it. Therefore, (CH3)C is an example of
nucleophilic radical.
So far, we have discussed points 1, 2 and 3 in previous sections, however, the below
mentioned points 4, 5, 6 and 7 are discussed in detail in Volume-II of this book.
(4) Carbenes [see Chapter 5 of Vol.-II]
(5) Ylides [see Chapter 6 of Vol.-II]
(6) Benzyne [see Chapter 7 of Vol.-II]
(7) Nitrenes [see Chapter 8 of Vol.-II]
1.10.2.3 Electrophiles and nucleophiles

(a) Electrophiles: The reactive sites of some molecules and ions (i) are capable of
accepting more electrons or (ii) are the d+ end of polar bonds. These electron-deficient
sites are electrophilic and the species possessing such sites are ‘electron loving’ and hence
≈
called electrophiles. BF3 is an example of neutral electrophile and Me3C is an example of
ionic electrophile.
(b) Nucleophiles: The reactive sites of some molecules or ions have higher density of
electrons because the site (i) may have an unshared pair of electrons or (ii) may be the
d– end of a polar bond or (iii) may have C ∫∫ C p electrons. Such electron-rich sites are
nucleophilic and the species possessing such sites are ‘nucleus loving’ and hence called
nucleophiles. N H 3 is an example of neutral nucleophile and I@ is an example of ionic
nucleophile.
Bond formation: Like bond fission, bond formation occurs in two different ways. Two
free radicals can combine, in which each donates one electron, to form a bond. Again, two
ions with unlike charges combine to form a bond; the negatively charged ion donates both
the electrons. Bond formation is always an exothermic process.
≈
1. Triphenylmethyl fluoroborate (Ph 3C BF4@ ) can be stored indefinitely as a
stable ionic solid — Why?
≈
Solution The triphenylmethyl cation (Ph 3C) is exceptionally stabilized by resonance
involving three phenyl groups and also the BF4① ion is nonnucleophilic in nature. For
≈
these reasons, triphenylmethyl fluoroborate (Ph 3C BF4@ ) can be stored indefinitely as a
stable ionic solid.
2. Acetylene, unlike ethylene, does not dissolve in concentrate H2SO4 — Why?
Solution An unsaturated hydrocarbon is expected to dissolve in conc. H2SO4 by forming a
bisulphate salt of a carbocation. The more the s character in the positively charged carbon
atom, the less stable is the carbocation and the less likely it is to be formed, i.e., less likely
the compound is to be dissolved in conc. H2SO4. Protonation of ethylene (CH2 == CH2)
≈
produces the relatively more stable alkyl carbocation salt, CH3CH2 HSO@4 (the positive
carbon is sp2-hybridized and being relatively less electronegative it accommodates the

positive charge better). On the other hand, protonation of acetylene is expected to produce
≈
the relatively less stable vinyl carbocation salt, CH2 == CH HSO4@ (the positive carbon is
sp-hybridized and being more electronegative it is less willing to accommodate the positive
charge). For this reason, ethylene dissolves in conc. H2SO4 but acetylene does not.
3. An alkyl cation can survive long enough when it is generated from an

alkyl fluoride (RF) in the presence of Lewis super acid antimony penta-
fluoride (SbF5) dissolved in liquid SO2 — Why?
Solution It is possible for a simple alkyl cation to survive long enough if its counter ion
being very weakly nucleophilic cannot combine with it. When an alkyl fluoride is treated
with SbF5 dissolved in SO2 at low temperature, SbF6@ is obtained as the counter ion.
SbF6@ , the conjugate base of the extremely strong acid HSbF6(pKa = –25), lacks basicity as
well as nucleophilicity. So, in the presence of SbF6@ , an alkyl cation can survive long enough.
4. Explain why cyclohexyl cation is more stable than phenyl cation

, but cyclohexyl anion is less stable than phenyl anion .
Solution The stability of carbocations decreases with increase in the s character, i.e., the
electronegativity of the cationic carbon atom. The positive charge in cyclohexyl cation is
on a less electronegative sp2-hybridized (33.33 percent s character) carbon atom while the
positive charge in phenyl cation is on a more electronegative sp-hybridized (50 percent s
character) carbon atom. Therefore, the positive carbon in phenyl cation is less capable of
accommodating the positive charge compared to the positive carbon in cyclohexyl cation.
Because of this, cyclohexyl cation is relatively more stable than phenyl cation. On the
other hand, the stability of carbanions increases with increase in the s character, i.e., the
electronegativity of the anionic carbon atom. In cyclohexyl anion, the negative charge
is on a less electronegative sp3-hybridized (25 percent s character) carbon atom while
in phenyl anion, the negative charge is on a more electronegative sp2-hybridized (33.33
percent s character) carbon atom. The more the carbon is electronegative, the better it
accommodates the negative charge. Cyclohexyl anion is, therefore, less stable than phenyl
anion.
5.
Identify A, B and C and explain these observations.

Solution The strong acid HBF4 converts triphenylcarbinol, Ph3COH, to the salt of the very
≈
stable carbocation, Ph3 C (triphenylmethyl cation). Its stability is due to extensive charge
delocalization involving three benzene rings. Because of such electron delocalization, the
carbocation absorbs radiation of wavelength 480–490 nm in the visible range and as a
≈
consequence, it appears in orange colour. Therefore, A is Ph3C BF4@ .
+ – + –
Ph3C—OH + HBF4 Ph3C—OH2BF4 Ph3C BF4 + H2O
A
Triphenylmethyl
fluoroborate
(a bright orange salt)
≈
Addition of H2O or EtOH converts Ph3 C to Ph3COH or Ph3COEt and as a consequence,
the colour of the salt is discharged.
6. When a primary aliphatic amine is allowed to react with nitrous acid (i.e.,
NaNO2/HCl), it is converted into an alcohol. For example:
H NO 2
(CH 3 )2 CH NH 2 ææææææ
(NaNO 2 / HCl)
Æ (CH 3 )2 CH OH
Isopropylamine Isopropyl alcohol
(a) Give the mechanism of the reaction
≈
(b) Explain why benzenediazonium ion (Ph N 2 ) is more stable than
≈
isopropyldiazonium ion (Me 2CH N 2 ).
≈
(c) Predict the product obtained when an acidic solution of Ph N 2 Cl @ is
warmed.
Solution
(a) Isopropylamine reacts with HNO2 (i.e., NaNO2/HCl) to form isopropyl alcohol as
follows:
≈
Nucleophilic attack by isopropylamine on the nitrosonium ion (NO) obtained
from nitrous acid produces an N-nitrosoamine which, in turn, tautomerizes to a
diazohydroxide. The diazohydroxide then undergoes protonation followed by loss of
water to form a diazonium ion. The diazonium ion being very unstable dissociates
extremely readily to form isopropyl cation which reacts with water to yield isopropyl
alcohol. The instability of diazonium ion in the absence of any stabilizing structural
feature is due mainly to the extreme effectiveness of N2 as a leaving group.
(b) Benzenediazonium ion is stable if the diazotization reaction is carried out below
≈
5°C. The stability of benzenediazonium ion (as compared to Me2 CH ) is attributed
to resonance as follows:
≈
The increased stability of Ph — N2 is also due to greater difficulty of forming
≈
relatively less stable Ph! as compared to Me2 C H .
(c) Benzenediazonium ion dissociates on warming and forms N2 and phenyl cation
which subsequently undergoes nucleophilic attack by water to give phenol, after
proton loss.
7.
(a) Explain why carbocation (not all) undergoes rearrangement.
(b) What are the important nodes of rearrangement of carbocation and
why are they called 1,2-shifts?
(c) Give examples of three reactions in which different types of 1,2-shift
occur.
Solution
(a) Carbocations undergo rearrangement because they become stable as a result of
rearrangement. For example, a 1° carbocation may rearrange to a more stable 2°
or 3° carbocation and a 2° carbocation sometimes rearranges to a more stable 3°
carbocation.
(b) A less stable carbocation can rearrange to a more stable carbocation by shift of a
hydrogen atom or an alkyl group or a ring bond. These rearrangements are called
1,2-shifts because they involve migration of an alkyl group or a hydrogen atom or a
ring bond from one carbon to an adjacent carbon atom. The migrating group moves
with the two bonding electrons.
(c) (i) Electrophilic addition of HBr to 3-methyl-1-butene produces 2-bromo-2-
methylbutane as the major product and 2-bromo-3-methylbutane as the minor
product. However, according to the Markownikoff’s rule 2-bromo-3-methylbutane
is expected to be the major product.
CH3 CH3 CH3

| | |
CH3 CH CH == CH2 + HBr Æ CH3 CH CHCH3 + CH3 CCH2CH3
| |
3-Methyl-1-butene Br Br
2-Bromo-3-methylbutane 2-Bromo-2-methylbutane
(minor product) (major product)
This observation can be explained by the fact that the initially formed 2° carbocation
rearranges to a more stable 3° carbocation by a 1,2-hydride shift. As a result of this
carbocation rearrangement, two alkyl bromides are formed — one from attack of
the nucleophile (Br①) on the unrearranged carbocation and the other from attack
on the rearranged carbocation. The bromide corresponding to the more stable
carbocation is obtained as the major product. The reaction occurs as follows:
(ii) Similarly, electrophilic addition of HCl to 3,3-dimethyl-1-butene leads to the

formation of 2-chloro-2,3-dimethybutane as the major product and 3-chloro-2,2-
dimethylbutane as the minor product which, in fact, expected to be the major
product according to the Markownikoff’s rule.
(iii) Carbocation rearrangements can also take place by ring expansion involving the
shift of a ring bond. When 1-methyl-1-vinyl-cyclobutane, for example, is treated
with HBr, 1-bromo-1,2-dimethylcyclopentane is obtained. The initially formed
2° carbocation undergoes ring expansion by a 1,2- shift to form a more stable 3°
carbocation. The process is also favoured by the release of angle strain (a four-
membered ring is converted into a five-membered ring).
8. Explain why carbocations undergo rearrangement while free radicals

do not.
Solution The central carbon atoms of a carbocation and a free radical are sp2-hybridized.
The unhybridized p-orbital is vacant in carbocations but in free radicals it has one electron.
In carbocation rearrangements, the two electrons of the migrating s-bond can easily
move into the vacant p-orbital of positive carbon since in accordance with Pauli exclusion
principle, an orbital at maximum can accommodate two electrons. For this reason, less
stable carbocations rearrange to yield more stable carbocations. On the other hand, in free
radicals, the two electrons of the migrating s-bond cannot move into the p-orbital which
already has one electron because by doing so the p-orbital will have three electrons which
is not in accordance with the Pauli exclusion principle. If the rearrangement occurs at all,
the third electron has to go to the antibonding molecular orbital making the rearrangement
energetically unfavourable. Therefore, a carbocation undergoes rearrangement because it
is energetically favourable but a free radical does not undergo rearrangement because the
process is energetically unfavourable.
9. Rank the following carbocatioss in the order of increasing stability and
explain the order:
Solution The 3° carbocation I is stabilized by +I effect of the three ring bonds. However,
its stabilization by hyper conjugation is not possible because the formation of a double bond
at the bridgehead position is not possible according to Bredt’s rule. Also, this carbocation
suffers from angle strain because the angle between the bonds is somewhat less than the
sp2 bond angle of 120°. The 3° carbocation II is more stable than I because it is stabilized
by the +I effect of three methyl groups directly attached to the positive carbon and
hyperconjugation effect involving nine a-H atoms. The carbocation III is the most stable
one because it is not only stabilized by the inductive and hyperconjugation effect of two
methyl groups but also stabilized by resonance involving the adjacent double bond.
Therefore, the order of increasing stability of these carbocations is
10. Which one of the following two carbocations is more stable? Give your
reasoning.
≈ ≈
(CH 3 )3 C; (CD3 )3 C
Solution Each of these two 3° carbocations is stabilized by +I effect of the three alkyl
groups. However, the +I effect of —CH3 group is more stronger than the +I effect of —CD3
≈ ≈
group (D is more electronegative than H). Hence, (CH 3 )3 C is more stable than (CD3 )3 C
from inductive point of view. Again, both the ions are stabilized by hyperconjugation as
follows:
Since the C — D bond is stronger than the C — H bond, therefore, the hyperconjugative
≈
forms of (CD3 )3 C are relatively less stable and contribute less to the hybride. Hypercon-
≈ ≈
jugation is, therefore, less effective in (CD3 )3 C compared to (CH 3 )3 C . Hence, also from
≈ ≈
hyperconjugation point of view, (CH 3 )3 C is relatively more stable than (CD3 )3 C.
11. Which of the following carbocations would you expect to undergo
rearrangement and why?
(a) (b) C6H5CH == CHCH2CH2≈ (c)
≈
(d) (e) (f) (CH 3 )2CH CHCH 3
≈
(g) (h) (i) (CH 3 )2CHCH CHCH 3 (j)
Solution
(a) This carbocation does not undergo rearrangement because rearrangement would
yield other 2° carbocations of nearly equal stability. That is, there is no energetic
advantage in these processes.
(b) This carbocation will undergo rearrangement because a 1,2-hydride shift will
convert this primary carbocation into a more stable allylic carbocation which is
stabilized by resonance.
(c) This 1° carbocation will rearrange because a 1,2-hydride shift will convert it into a
more stable 3° carbocation.
(d) This secondary carbocation will rearrange because a 1,2-hydride shift will convert
it into a more stable tertiary carbocation.
(e) This 2° carbocation will rearrange to give another 2° carbocation because the
resulting carbocation is stabilized by resonance involving the cyclopropane ring.
(f) This 2° carbocation will undergo rearrangement because a 1,2-hydride shift will
convert it into a more stable 3° carbocation.
(g) This 2° carbocation will undergo rearrangement because a 1,2 methyl shift will
convert it into a more stable 3° carbocation.
(h) This 1° carbocation will undergo rearrangement because a 1,2-bond shift will convert
it into a more stable 2° carbocation. 1,2-Hydride shift leading to the formation of a
3° carbocation does not take place because the resulting carbocation is expected to
suffer from a good amount of angle strain (120°–90° = 30°).
(i) This 2° carbocation will not undergo rearrangement because a 1,2-hydride shift
will convert it into an equally stable 2° carbocation. Therefore, there is no energetic
advantage in the process.
(j) This 2° carbocation will rearrange because a 1,2-bond shift will convert it into a
more stable 3° carbocation.
12. Arrange the following carbocations in order of increasing stability and

explain the order:
Solution A carbocation in which the cationic carbon is directly attached to a cyclopropyl

group is especially stable because of conjugation between the bent orbitals of the cyclopropyl
ring and the vacant p orbital of the cationic carbon. Hence, III is considerably stabilized
by resonance.
The stability of the carbocation increases with each additional cyclopropyl group. Therefore,
the carbocation I bearing two such groups is more stable than III bearing only one such
group. Again, a cyclopropyl group stabilizes an adjacent positive charge even better than a
phenyl group. Hence, IV is less stable than III. II is less stable than IV because unlike IV,
this 1° carbocation is not resonance-stabilized. Therefore, the order of increasing stability
of these carbocations is
13. Rank the following carbocations in order of decreasing stability and give
your reasoning:
Solution In p-methoxybenzyl cation (III), the unshared pair of electrons on the oxygen
atom of the —OMe group is in proper conjugation with the vacant p orbital of the positive
carbon. Because of this, this carbocation is more stabilized by resonance as compared to
the benzyl cation (I).
In p-nitrobenzyl cation(II), the electron-attracting —NO2 group intensifies the positive

charge of the cationic carbon by placing a positive charge on the carbon adjacent to the
cationic carbon by its –R and –I effects and thereby makes this carbocation considerably
unstable. Hence II is less stable than I
Therefore, the order of decreasing stability of these carbocations is as follows:

+ + +
CH3O— —CH2 (III) —CH2 (I) O2N— —CH2 (II)
≈
14. Explain why the carbocation F3 C is more stable than the carbocation
≈
F3C — CH 2 .
≈
Solution In F3C — C H 2 , the electron-attracting (–I) –CF3 group withdraws electrons of
the C — C bond towards itself and thereby destabilizes the carbocations by intensifying the
≈
positive charge. On the other hand, in the carbocation F3 C , the unshared pair of electrons
on each of the three F atoms overlaps with the empty p orbital of the cationic carbon and
thereby stabilizes the carbocation by delocalizing the positive charge. Furthermore, the p
orbitals overlap is much more effective because the orbitals of fluorine and carbon are of
comparable size. In fact, the resonance stabilization outweighs the destabilization caused
≈
by the —I effect of fluorine. This explains why carbocation F3 C is relatively more stable
≈
than the carbocation F3C — C H 2 .
15. Rank the following carbocation in order of decreasing stability and

explain the order:
Solution The decreasing order of stability of these carbocations is I > II > III > IV.
In the carbocation IV, the positive charge is stabilized by the inductive (+I) and
hyperconjugation effect of one methyl group while one methyl and one ethyl group are
involved in stabilizing the charge in III. Again, an sp-hybridized doubly bonded positive
carbon is more electronegative than an sp2-hybridized singly bonded positive carbon and
hence the former carbon is less capable of accommodating a positive charge than the
latter. Because of these, the vinylic cation IV is less stable than the carbocation III. The
charge in the allylic cation II is highly stabilized by resonance and because of this, it is
more stable than the carbocation III where only weak inductive and hyperconjugation
effects are operative.
In the carbocation I (a cycloprophenyl cation derivative), the positive charge is involved in

constituting an aromating system [(4n + 2)p, where n = 0]. Hence, it is more stable than
the carbocation II. Therefore, the decreasing order of stability of these carbocations is as
given above.
16. Triptycenyl cation is less stable than trityl cation. Explain.
Solution In trityl or triphenylmethyl cation, the positive charge is well stabilized by
resonance involving three phenyl groups. However, in triptycenyl cation, the empty p
orbital is, in fact, perpendicular to the p orbitals of the three benzene rings and since bond
rotation is not possible in this rigid system; therefore, they can never become parallel
for conjugation to take place. Therefore, the positive charge in triptycenyl cation is not
at all stabilized by charge delocalization. This cation is, therefore, less stable than trityl
cation.
17.
Identity A and B and explain these observations.

Solution Sodium amalgam reduces triphenylmethyl chloride to give the salt of the very
@
stable carbanion, Ph3 C (triphenylmethyl anion). Its stability is due to extensive charge
delocalization into three benzene rings.
Because of extensive electron delocalization, the carbanion Ph3 C absorbs radiation of

wavelength 490–500 nm and as a consequence, it appears red.
The strongly basic carbanion takes up a proton from H2O or EtOH to form colourless
Ph3CH which does not absorb radiation in the visible range.
18. Rank the following carbanions in order of increasing stability and explain
the order:
Solution The order of increasing stability of these carbanions is
The external orbitals (i.e., the orbitals directed towards outside bonds) in cyclopropane (the
conjugate acid of I) have larger (33%) s character, i.e., they are approximately sp2-orbitals.
Because of this, the unshared pair of electrons on the anionic carbon in cyclopropyl anion
(I) is more tightly held with the carbon nucleus than the unshared pair in ethyl anion
(IV) that occupies an sp3 orbital (25% s character). Consequently, the former anion is more
stable than the latter. In the vinylic anion II, the unshared pair of electrons occupies an sp2-
orbital (33.33%) s character and hence this anion is somewhat more stable than cyclopropyl
anion (I). In the allylic anion III, the negative charge is delocalized by resonance with the
adjacent double bond and so it is relatively more stable than the vinylic anion II.
19. Arrange the following carbanions in order of increasing stability and give
your reasoning:
@ @ @ @
CH 3 CH NO2 CH 3 CHCOCH 2CH 3 CH 3 CH CO2Et CH 3 CH C6 H5

I II III IV
Solution The order of increasing stability of these carbanions is

@ @ @ @
C6 H5 C HCH 3 (iv) < CH3 CH CO2 Et (III) <CH 3 CH COCH 2CH 3 < CH 3 C H NO2 (I)
All these carbanions are stabilized by resonance, but the anions conjugated with a carbon–
oxygen or carbon–nitrogen double bond are relatively more stabilized than the benzylic
anion IV and this is because the more electronegative oxygen atom accommodates the
negative charge better than the less electronegative carbon atom. Therefore, the benzylic
anion IV is the least stable one.
The C == O group in CH 3 C H CO2Et (III) is less effective in stabilizing the carbanion than
@
the C == O group in CH 3 C H COC2H5 (II) and this is because unlike keto carbonyl, the
ester carbonyl is involved in resonance interaction with the unshared pair of electrons on
the oxygen atom of the –OEt group. Therefore, the anion II is more stable than III.
All the groups (except methyl) attached to the anionic carbon exert their –I effects in
addition to their resonance effects. However, the –I effect exerted by the —NO2 group is
the strongest one because the N atom in —NO2 bears a positive charge. Because of this,
the NO2 group is more effective in stabilizing the carbanion than the > C == O group.
Hence, the carbanion I is the most stable one.
20. Classify each of the following species as electrophile and nucleophile and
give your reasoning.
@ ≈
(a) C H3 (b) (CH 3 )3 C (c) :I:@ (d) :Cl≈
(e) NH 3 (f) BF3 (g) Cl 2 C: (h) : PPh 3
(i) CH3CH == CHCH3 (j) SiF4 (k) CH 3 O:@
Solution Electrophiles: (b), (d), (f), (g) and (j).
The central atoms (C, B and C) of (b), (f) and (g) have six electrons and can accept an
electron pair from a nucleophile to complete the more desirable octet. Therefore, these are
electrophiles. For the same reason, the cation (d) is an electrophile. The Si atom in (j) can
acquire more than eight electrons by utilizing its d orbitals. Hence, it is also an electrophile.
Nucleophiles: (a), (c), (e), (h), (i) and (k).
The anions (a), (c) and (k) and the neutral molecules (e) and (h) have unshared pair of
electrons. Therefore, these are nucleophiles. The alkene (i) has two available electrons in
a p bond. Hence, it is also a nucleophile.
21. Give example of a very stable salt which can be obtained by combining a
very stable carbanion with a very stable carbocation.
Solution An example of salt obtained by combining a very stable carbanion with a very
stable carbocation is as follows:
The cation (a cyclopropenyl cation) is stable because it is an aromatic system. The

anion is stabilized by resonance involving contributing structures containing aromatic
cyclopentadienyl anion systems.
22. Show how hydroquinone can act as a radical inhibitor.

Solution When hydroquinone traps a radical (R) , it forms semiquinone. Semiquinone is
stabilized by resonance and is, therefore, less reactive than other radicals. Furthermore,
semiquinone traps another radical and itself becomes converted into quinone (a compound
whose electrons are all paired). The reaction occurs as follows:
23. Give two examples of radical inhibitors that are present in biological
system.
Solution The two radical inhibitors present in the biological system are vitamin C
(ascorbic acid) and vitamin E (a-tocopherol). Like hydroquinone, they form relatively
stable radicals.
CH2OH
H OH
O
O
H
O
HO OH
Vitamin C Vitamin E
(ascorbic acid) (a-tocopherol)
24. Rank the following radicals in order of increasing stability and give your
reasoning:
i i i i
(CH3 )3 C, CH2 — CH == CH2 , CH3CH2 , (C6 H5 )2CH
I II III IV
Solution The stability of a free radical depends on the extent of delocalization of the
unpaired electron by resonance or hyperconjugation. The unpaired electron in ethyl radical
(III) is stabilized by hyperconjugation involving only three a-hydrogen atoms and so it is
the least stable among the four alkyl radicals. The unpaired electron in tert-butyl radical
(I) containing nine hyperconjugable a-H atoms is well delocalized by hyperconjugation.
Hence, its stability is higher than that of ethyl radical (III).
The unpaired electron on carbon in allyl radical (II) is delocalized by resonance interaction
with the p-electrons of the double bond and the delocalization is more effective because
the two resonance structures are equivalent. Since the stability due to hyperconjugation
is less than that caused by resonance, therefore, the allyl radical (II) is more stable than
the tert-butyl radical (I)
In diphenylmethyl radical (IV), the unpaired electron is extensively delocalized over the
p orbital system of the two benzene rings and so, delocalization is much more effective in
this case. Because of this, it is relatively more stable than the allyl radical (II).
Therefore, the order of increasing stability of these radicals is
CH 3CH 2 (III) < (CH 3 )C (I) < CH 2 — CH == CH 2 (II) < (C6 H5 )2 CH (IV)
25. Arrange the following compounds in order of increasing rate of thermal

decomposition to yield nitrogen and give your reasoning:
CH 3 — N == N — CH 3 C6 H5 CH 2 — N == N — CH 2C6 H5 CH 3CH 2 — N == N — CH 2CH 3
I II III
Solution The rate of thermal decomposition of an azo compound is much higher if it

produces radicals that have considerable stability. Dimethyldiazene (I), dibenzyldiazene (II)
and diethyldiazene (III) decompose to give methyl, benzyl and ethyl radicals, respectively.
Benzyl radical is highly stabilized by resonance and it is, therefore, more stable than
ethyl radical which is stabilized by hyperconjugation involving three a-H atoms (a less
important stabilizing effect). The methyl radical (CH 3 ) is the least stable because it is not
stabilized either by hyperconjugation or by resonance. Because of this, dibenzyldiazene
decomposes at a rate faster than diethyldiazene which in turn decomposes at a faster rate
than dimethyldiazene.
26.
Identify A and B and explain these observations.

Solution Triphenyl methyl chloride (Ph3CCl) reacts with Ag or Zn to give very stable
(stabilized by resonance) triphenyl methyl radical which gives a yellow solution in benzene.
The radical remains in equilibrium with its dimer. When the solution is evaporated, the
dimer is obtained as the residue (a white solid). Therefore, A is Ph3C and B is its dimer.
The formation of the dimer may be shown as follows:
Triphenylmethyl radical reacts with O2 to give colourless ditriphenylmethyl peroxide or

with I2 to give colourless triphenyl iodomethane. Therefore, C is Ph3COOCPh3 and D is
Ph3C — I.
27. Explain why the halogenation of alkenes in the presence of tetraethyl

lead (Et4Pb) occurs at a temperature lower than the temperature required
when halogenation is carried out in its absence.
Solution Halogenation of alkenes is normally carried out between 525 and 675K and
this is because homolysis of the X — X bond occurs to form halogen free radical (X) at this
temperature range. Tetraethyl lead decomposes at a much lower temperature of 423 K
to produce ethyl radical (CH 3CH 2 ) which subsequently reacts with halogen molecules to
form halogen free radicals.
Since in the presence of tetraethyl lead, halogen radicals are produced at a lower
temperature, therefore, halogenation also takes place at a lower temperature.
1. Arrange the following carbocations in order of decreasing stability and explain the
order:
≈ ≈ ≈ ≈ ≈ ≈
(CH 3 )3 C, (CH 3 )2 CH, C6 H5 CH 2 , CH 3 , CH 3 CH 2 , (C6 H5 )2 CH
2. Rank the following carbocations in order of increasing stability and give reasons:
≈ ≈ ≈
(a) CH 3 CH ClCH 2CH 2 C H 2 , CH 3 CH 2 CH 2 CH ClC H 2 , CH 3 CH 2 CH ClCH 2 C H 2
≈ ≈ ≈
(b) CH3 CH2CHCH2 , CH3CHCH2 CH2 , CH3 CH2 CH2 CH
| | |
OCH3 OCH3 OCH3
3. Which one is more stable and why?

≈ ≈
(a) CH 3 NH C H 2 or CH 3 OC H 2 (b)
≈ ≈
(c) CH3 OCH == CH — CH2 or CH2 == C — CH2
|
:OCH3
(d) (e)
(f)
[Hint]: (b) Consider ‘steric inhibition of resonance’ (f) Consider ‘Bredt’s rule.’]
4. Arrange the following ions in order of increasing stability. Give reasons.
(a)
(b)
≈ ≈ ≈
(c) CH3 CHCH3 , CH2 == CH — CH2 , CH3 C H2 CH2
I II III
(d)
5. Which one is more stable and why?
(a) (b)
O O
@ || @ || @ @
(c) CH3 CH — N — C — Ph or CH2 — N — CH2 — C — Ph (d) CCl3 or CF3

| |
C2H5 C2H5
(e) (f)
6. Arrange the following carbanions in order of decreasing stability and give your
reasoning:
@ @ @ @
Ph3 C , , (CH 3 )3 C, (CH 3 )2 C H, C H 3
7. Predict the state of hybridization of the anionic carbon in each of the following
carbanions:
@ @ @
(a) CH 3 , C H 2 (b) Ph C H 2 (c) CH 3 CO C H 2

@
(d) CH 3 CO C H COCH 3 (e) (f)
@ @ @
(g) O2 N C H 2 (h) Ph CH2 CH2 (i) HC ∫∫ C
@ @
(j) (k) CCl3 (l) C F3
8. Bridgehead carbocations are not much stable yet 1-tris homobarrelyl cation (I)
have been prepared in super acid solution at –78°C — Why?
9. AIBN is widely used as a free radical initiator. Write appropriate equation to show
how AIBN can produce free radicals.
CN CN
| |
[Hint: AIBN is azobisisobutyronitrile, Me2 C — N == N — C Me2 ]
10. How can chlorine atoms or radicals be generated using an azo compound?
[Hint:
11. Arrange the following free radicals in order of decreasing stability:
What is the expected common reason for the same stability order between
carbocations and free radicals?
12. Explain why the radical I undergoes ready dimerization while the radical II does not.
[Hint: The planar radical I has much less steric hindrance to dimerization than
the nonplanar (due to steric reason) radical II.]
13. Explain why acetylene (HC ∫∫ CH) is more acidic than ethane (CH3CH3), even
though the C — H bond in acetylene has a higher bond dissociation energy than
the C — H bond in ethane.
[Hint: The bond dissociation energy is a measure of how easily the bond breaks
homolytically while acidity is a measure of how
easily the bond breaks heterolytically )]
14. Rank the following free radicals in order of increasing stability and give your
reasoning:
C6 H5 — C H — CH 3 , C6 H5 — CH — C H == CH 2 , C6 H5 CH 2 CH 2 C6 H5 CH 2
I II III IV

(a) CH 3 is planar but CF3 is pyramidal.
(b) Me3C — H bond is weaker than CH3 — H bond.
(c) Toluene undergoes free radical bromination at a rate faster than ethane.
(d) Monobromination of (S)-3-methylhexane yields a racemic product.
16. Rank the following free radicals in order of increasing stability and give your
reasoning:
Benzyl, Vinyl, Methyl, Allyl
17. Mention the structure for A in the following reaction and classify the carbons in it
by type.
18. Electrolysis of an aqueous solution of CH3CH2COONa produces butane, ethane,

ethylene and ethyl propanoate suggest a suitable mechanism to account for the
formation of these compounds.
[Hint:
19. Which C—H bond in each of the following compounds is expected to be broken most
readily during radical halogenation. Give your reasoning.
20. Predict the major product obtained when 3-ethylpentane is heated with Br2.
[Hint: The more substituted the carbon atom, the weaker the C—H bond. Therefore,
the major bromination product of 3-ethylpentane will be obtained by the cleavage
of the 3° C—H bond (its weakest C—H bond), i.e., the product is (CH3CH2)3CBr
(3-bromo-3-ethylpentane).]
21. The azo compound I decomposes 20 times faster than the azo compound II. Suggest
a reason for this observation.
22. Rank the following azo compounds in order of their expected rate of thermal
decomposition to produce N2. Give your reasoning.
(a) CH3—N== N—CH3 (b)
(c) PhCh2—N == N–CH2Ph (d)
(e) (CH3)3C—N== N—C (CH3)3

23. Which of the following bicyclic alkenes would you expect to undergo allylic
bromination with N-bromosuccinimide (NBS)? Give your reasoning.
[Hint: A bridgehead radical is not much stable.]

24. What is called a ‘captodative radical’? Give an example.
[Hint: When a free radical is stabilized by the presence at the radical centre of
both, an electron-donating and an electron-withdrawing group, that radical is
called a ‘captodative radical’. For example:
25. Predict the major product of each of the following reactions and provide mechanism
that accounts for its formation:
(a) ; (b)
26. The phenoxy radical I exists not as the dimer but as such both in solution and in
the solid state. Explain this observation.
[Hint: The two bulky-CMe3 groups in the two ortho positions of one radical come
in steric interaction (F-strain) with the two-CMe3 groups of another approaching
radical and as a result, their combination to form a dimer is prevented.]
27. The radical anion of benzophenone (Ph2C== O) is more stable than that of
cyclohexanone . Explain.
[Hint: The unpaired electron in the radical anion of benzo-phenone (Ph2 C — O@ ) is

highly delocalized over the p orbital system of the two benzene rings as well as over
the C == O system. But the unpaired electron in the radical anion of cyclohexanone
is only slightly delocalized (only two resonance structures can be
written).
]
28. A benzylic hydrogen is 6.4 × 10 times more reactive in bromination than a hydrogen
atom in ethane. Explain.
29. a-Naphthoxyl radical is more stable than b-naphthoxyl radical. Explain.
30. A benzene solution of 2,4,6-tri-tert-butylphenoxy radical is decolorized in
atmospheric oxygen in 30 min, whereas 2,6-di-tert-butyl-4-phenylphenoxy radical
requires 8 hr. Explain.
31. There is no enormous energy barrier against the formation of a bridgehead radical
as there is against the formation of a carbocation. Explain.
[Hint: Unlike carbocation a radical may have pyramidal shape.]
1.11 TAUTOMERISM
Tautomerism is a phenomenon in which two structural isomers undergo rapid
interconversion and exists in dynamic equilibrium in the liquid state or in solution under
normal laboratory conditions. The two forms or isomers which exist in dynamic equilibrium
with each other are called tautomers of tautomerides. Toutomerism may be classified into
cationotropy and anionotropy depending upon whether atoms or groups of atoms shift
as cations or anions. A great majority of cases of cationotropy, however, is prototropy
in which the migrating cation is a proton. The most important and classical example of
prototropy is the keto-enol tautomerism and this can be represented as follows:
:O: OH
|| | | |
— C— C— ææ
¨æÆ
æ — C == C —
|
H
Keto form Enol form
Some other examples of prototropy are listed below:
(a) Nitro-acenitro tautomerism:
| |
(b) Amido-imidol tautomerism: —NH — C == O ææ
¨æÆ —N == C— OH
æ
Amido form Imidol form
| | | |
ææ
(c) Imine-enamine tautomerism: — C H — C == NH ¨æÆ
æ — C == C— NH2
Imine form Enamine form
| |
(d) Nitroso-oximino tautomerism: —C H — N == O ææ
¨æÆ —C == N —OH
æ
Nitroso form Oximino form
H
|
(e) Azo-hydrazono tautomerism: —N == N — C — ææ
¨æÆ —NH — N == C—
æ
| |
Azo form Hydrazono form
The essential structural need for tautomerism (prototropy) to be exhibited by compounds

like carbonyl, nitro, nitroso, etc. is that the compound must contain at least one a-hydrogen
atom. It thus follows that tautomerism is not possible in benzaldehyde (C6H5CHO),
benzophenone (C6H5COC6H5), 2,2-dimethylpropanal (Me3CCHO), 2-methyl-2-nitro
propane (Me3CNO2), etc. but possible in acetone (CH3CCH3), propanal (CH3CH2CHO),
acetophenone (C6H5COCH3), nitroethane (CH3CH2NO2), etc.
(1) Keto-enol tautomerism

Ethyl acetoacetate is the best studied classical example of keto-enol tautomerism:
O O OH O
|| || | ||
ææ
CH 3 — C — CH 2 — C — OC2H5 ¨æÆ CH 3 — C == CH — C — OC2H5
æ
Keto form of EAA Enol form of EAA
The presence of the keto form is supported by the fact that the compound forms a bisulphite
addition product with sodium bisulphite, a cyanohydrin with hydrogen cyanide, an oxime
with hydroxylamine and a phenylhydrazone with phenylhydrazine.
The presence of the enol form in the equilibrium mixture is proved by the fact that ethyl
acetoacetate gives an intense reddish violet colouration with ferric chloride solution
|
(this is characteristic of compounds containing the grouping: C == C — OH), it rapidly
decolourizes bromine and reacts with diazomethane to form an ether.
Therefore, ethyl acetoacetate is an equilibrium mixture of two distinct substances, the

keto form and the enol form. A convenient method of preparing pure samples of keto and
enol forms was devised by Meyer, who placed the equilibrium mixture in a quartz (since
glass has a catalytic effect on interconversion) distilling flask and separated the enol form,
which is more volatile than the keto form, by distillation.
The estimation of the enol content of keto-enol mixture is based on the fact that the enol
forms absorb bromine rapidly, whereas the keto form does not. Thus, if an equilibrium
keto-enol mixture is titrated rapidly with bromine, at the end point (when the colour of
Br2 is no longer discharged rapidly) all the enol form would have been converted to the
bromo compound. The a-bromo compound is then reduced by adding HI, and the liberated
iodine is titrated with thiosulphate.
R R R
| | |
Br2 HI
R — CH == C — OH æææ
fast
Æ R — CH — C == O æææ Æ R — CH 2 — C == O + I2
enol form |
Br
1.11.1 Mechanism of Keto-enol Tautomerism

Tautomerization, the process of converting one tautomeric form into another, is catalyzed
by both acids and bases.
(1) Acid-catalyzed Tautomerism
In an acidic solution, the carbonyl oxygen of the keto tautomer undergoes protonation in
the first step. In the second step (the rate-determining step), water removes a proton from
the a-carbon to form the enol.
(2) Base-catalyzed Tautomerism

In a basic solution, the hydroxide ion abstracts a proton from the a-carbon atom of the keto
tautomer to form a resonance-stabilized enolate ion in the first step (the rate-determining
step). In the second step, oxygen of the enolate ion undergoes protonation to form the enol
tautomer.
Protonation on the a-carbon reforms the keto tautomer.

In addition to its mechanistic importance, keto-enol tautomerism affects the stereochemistry
of carbonyl compounds. An a-hydrogen atom may be lost and regained through keto-enol
tautomerism. Such a hydrogen is said to be enolizable. If an asymmetric carbon atom
contains an enolizable hydrogen atom, a trace of acid or base allows that carbon to invert
its configuration through the intermediate formation of an enol or enolate ion. A racemic
mixture is the result. An optically active carbonyl compound will, therefore, be found to
have lost its optical activity if it is left standing in the presence of acid or base. For example:
1.11.2 Difference between Resonance and Tautomerism

Resonance Tautomerism
1. Resonance involves delocalization of p electrons 1. Tautomerism involves movement of both
and hence the resonance structures (canonicals electrons and atoms, and hence tautomers differ
or contributing forms) differ only in the position both in the position of electrons and atoms. For
of electrons. The position of atoms remains example:
unchanged. For example:
2. Various resonance structures are only imaginary, 2. Tautomers are real molecules, which actually
and do not exist. The real molecule is a hybrid of exist and can be isolated under suitable condi-
these structures. tions.
3. The resonance structures are not in dynamic 3. Tautomers exist in dynamic equilibrium with
equilibrium as they are not real structures. each other as they are actual compounds.
4. Resonance leads to stabilization of the molecule. 4. Tautomerism does not lead to stabilization of the
The actual molecule (i.e., the hybrid) is more molecule.
stable than any resonance structure.
5. Resonance structures are shown by double 5. Tautomeric structures are shown by using equi-
headed arrow (´). librium arrows ( or )
1.11.3 Position of the Tautomeric Equilibrium

Both the tautomers are not equally stable. The keto form is about 50 kJ/mol more stable
than the enol form largely because a C == O bond is much stronger than a C == C bond and
for this reason, under ordinary conditions mono-ketones do not exhibit much tendency to
enolize. For example, an aqueous solution of acetone exists as an equilibrium mixture of
more than 99.9% keto form and less than 0.1% enol form.
O OH
|| |
Acetone: CH3 — C — CH3 ¨æ ææ Æ CH2 == C — CH3
æ
Keto form enol fom
(> 99.9%) (<0.1%)
The fraction of the enol form is considerably greater for a b-diketone because the enol
tautomer is stabilized by intramolecular hydrogen bonding and by conjugation of the
carbon–carbon double bond with the second C == O group.
The aforesaid difference in thermodynamic stabilities of the two tautomers is not the
only factor that determines the position of equilibrium between them. The position of the
equilibrium is significantly affected by other important factors such as follows:
(1) Entropy factor
Although acyclic monoketones show negligible tendency to enolize, cyclomonoketones
(e.g., cyclohexanone) enolize appreciably and this is because introduction of a C == C bond
in enolization decreases the conformational freedom of the acyclic monoketone but no such
extra restriction is imposed in the case of cyclic monoketones.
(2) Steric factor
The enols of dicarbonyl compounds are generally stabilized by intermolecular hydrogen
bonding. Therefore, any additional substituents on the C-atoms in the ring adversely affects
ring formation by H-bonding due to steric strain. As a result, the enol content decreases.
(3) Nature of the solvent
In the enol form of dicarbonyl compounds, the intramolecular hydrogen bond is quite strong
and it is not considerably affected by solvents like water and ethanol. But the Keto form is
well stabilized by these solvents due to hydrogen bonding. Therefore, such protic solvents
tend to reduce the enol content. However, the nonpolar solvents like hexane, benzene, etc.,
which are unable to form hydrogen bond, tend to increase the enol content.
The extent of enolization in some compounds may be explained as follows:

(a) Acetone: Acetone (CH3COCH3) exists almost exclusively in the keto form.
O OH
|| |
CH3 — C — CH3 ¨æ ææ Æ
æ CH 3 — C == CH2
Keto form Enol form
(> 99%) (1.5 ¥ 10-4 %)
| |
This keto-enol tautomerization actually involves inter conversion of a — CH — C == O
| | |
group and a — C == C(OH) –group. The sum of bond energies of the — CH — C == O group
|
is 347.9 kcal/mol and that of the — C == C(OH) –group is 336.0 kcal/mol.
C — H (98.9 kcal/mol) + C — C (83.0 kcal/mol) + C == 0 (166.0 kcal/mol) = 347.9 kcal/mol
C == C (145.0 kcal/mol) + C — 0 (80.0 kcal/mol) + O — H (111.0 kcal/mol) = 336.0 kcal/mol
|
The enthalpy of the formation of the groups CH — C == O and C== C(OH)– are, therefore,
347.9 and 336.0 kcal/mol, respectively. Thus, the keto tautomer is thermodynamically
more stable than the enol tautomer by 11.9 kcal/mol and for this reason, at equilibrium
acetone exists almost exclusively in the keto form.
(b) Pentane-2,4-dione or acetylacetone: The enol content of pentane-2, 4-dione or
acetylacetone (CH3COCH2COCH3) at equilibrium is very high (76%).
Because of extended conjugation, the enol form of pentane-2,4-dione is stabilized by

resonance. Also, the enol is significantly stabilized by intramolecular hydrogen bonding.
For these reasons, the enol content of pentane-2,4-dione at equilibrium is very high.
(c) Ethyl acetoacetate: The enol content of ethyl acetoacetate (CH3COCH2CO2Et) at
equilibrium is very small (8 percent).
Since the ester carbonyl group is involved in resonance interaction with — OR oxygen
( ) it is less effective than the ketone carbonyl group in
stabilizing the enol by resonance. Also, due to lack of resonance, the O — H hydrogen is
not appreciably positive and intramolecular hydrogen bond is relatively weak. For these
reasons, the amount of enol present at equilibrium is very small in ethyl acetoacetate.
(d) 3,4-Hexanedione: 3,4-Hexanedione (CH3CH2COCOCH2CH3) exists almost exclusively

in the keto form.Conformation plays a dominant role in deciding the extent of enolization
in the a-diketone. In this compound, the two C == O groups can rotate around the single
bond. Therefore, to avoid repulsive interaction between two negatively polarized oxygen
atoms, the two C == O groups take up an anti-conformation in which the two electronegative
oxygen atoms are as far as possible from each other and in which the carbonyl dipoles are
opposed. As a consequence, the formation of intramolecularly hydrogen bonded as well
as conjugated enol through the less stable syn-conformation is not at all favoured. The
equilibrium, therefore, lies almost exclusively over in favour of the keto form.
(e) 1,2-Cyclohexanedione
1, 2-Cyclohexanedione exists almost exclusively in the enol form.
1,2-Cyclohexanedione:
In this diketone, the two carbonyl groups are syn oriented. The energy of the molecule
d+ d-
rises due to repulsion between these two polar >C == O groups. Its energy cannot be
lowered by rotating the C == O dipoles, i.e., by orienting them anti to each other because
such conformation would give an impossibly strained ring. However, the energy content of
the molecule can be lowered by the formation of an enol because enolization eliminates one
of these two polar groups. Also, the enol is stabilized by intramolecular hydrogen bonding
and conjugative effect. For these reasons, the equilibrium lies essentially completely over
in favour of the enol form.
(f) Cyclohexa-2, 4-dien-1-one: This unsaturated ketone exists totally in its enol form
which is known as phenol.
Phenol is an aromatic compound and it is, therefore, stabilized by the resonance energy
of the benzene ring. In fact, phenol is thermodynamically much more stable than its keto
form cyclohexa-2,4-dien-1-one (DH ª 20 kcal/mol) and for this reason, the equilibrium lies
exclusively over in favour of the enol form, i.e., phenol.
(g) (Me3CO)3CH: This triketone exists exclusively in the keto form. This is due to the fact
that the enol form, which suffers from severe steric strain, is energetically very much less
stable than the keto form.
(h) 2,2-Dimesitylethanal: The enol content of 2,2-dimesityl-ethanal at equilibrium is

very high.
In the enol form of this aldehyde, the two very large mesityl groups are about 120° apart,
but in the keto form they are about 109.5° apart. So, unlike in the enol form, they remain
much closer together in the keto form and become involved in severe steric interaction.
Therefore, the keto form is thermodynamically less stable than the enol form and as a
consequence, the percentage of the enol form at equilibrium is very high.
(i) 1,2-Diphenyl-1-ethanone: The enol content of this ketone at equilibrium is very

high.
Enolization, by creating a double bond, links up the two phenyl groups so as to form a
completely conjugated system which is highly stabilized by resonance. For this reason, the
enol content of PhCOCH2Ph at equilibrium is very high.
(j) 1,3,5-Trihydroxybenzene: 1,3,5-Trihydroxybezene shows more ketonic activity than

1,3-dihydroxybenzene which in turn shows more ketonic activity than phenol.
Phenol:
1,3-Dihydroxybenzene or resorcinol:
1,3,5-Trihydroxybenzene or phloroglucinol:
As the number of — OH groups in a benzene ring increases, the difference in stability

between the enol and the keto form progressively decreases because the aromatic stability
of only one ring becomes progressively less able to make the enol form very stable. As a
consequence, the keto form is progressively more favoured. And, for this reason, 1,3,5-
trihydroxybenzene shows more ketonic activity than 1,3-dehydroxybenzene which in turn
shows more ketonic activity than phenol.
1.11.4 Ring-chain Tautomerism

The type of tautomerism in which one tautomer is acyclic and the other is cyclic is denoted
as ring-chain tautomerism. These ring and chain forms of tautomers are possible where
one functional group of a bifunctional acyclic molecule reacts with other and forms a
cyclic system. Sugars like D-glucose, D-mannose, etc. exhibit ring-chain tautomerism. In
these cases, the open chain aldehydic form is converted into the cyclic hemiacetal form by
intramolecular reaction between the alcoholic–OH group and the —CHO group.
1.11.5 Valence Tautomerism

Tautomerism in which there occurs change in interatomic distances through the formation
of new bonds by redistribution of valence electrons within a molecule without the migration
of any atom from the rest in any intermediate step is termed as valence tautomerism. For
example, cyclooctatetraene and bicyclooctatriene represent a pair of valence tautomers.
1. When (R)-3-phenyl-3-butanone is dissolved in ethanol containing HCl, the

optical rotation of the solution gradually drops to zero. Explain why this
is so.
Solution The asymmetric carbon atom of (R)-3-phenyl-2-butanone contains an enolizable
hydrogen atom. When this compound is dissolved in ethanol containing hydrochloric acid,
it undergoes acid-catalyzed racemization through the intermediate formation of an achiral

enol. Because of racemization, the optical rotation of the solution gradually drops to zero.
The process may be shown as follows:
2. Explain why the enol content of 2,6-bicyclo[2.2.1]heptanedione is very

small, even though it is a b-diketone.
Solution Although the bridgehead enol is stabilized by resonance involving the
C == O group, the compound actually shows no tendency to form a bridgehead enol and
this is because it is impossible for this rigid bicyclic system to place a double bond at the
bridgehead position (Bredts’s rule). The compound, therefore, enlizes in another direction.
Since the resulting enol is not stabilized by resonance involving C == O, the enol content of
this diketone at equilibrium is very small indeed.
3. The following ketone does not undergo base-catalyzed exchange in D2O,

even though it has an a-hydrogen. Explain this observation.
Solution The ketone does not react with alkali to form the corresponding enolate ion
because the enolate ion is not stabilized by resonance which due to the presence of an
unstable structure containing a double bond at the bridgehead position is practically
insignificant. For this reason, this bicyclic ketone does not undergo base-catalyzed
exchange in D2O through intermediate formation of an enolate ion.
4. 1,3,5-Trihydroxybenzene reacts with hydroxylamine (NH2OH) to give a

trioxime. However, 1,2,3-and 1,2,4-trihydroxybenzenes do not react with
NH2OH. Account for these observations.
Solution 1,2,3- and 1,2,4-Trihydroxybenzenes do not react with NH2OH because the
triketo tautomers of these two trihydroxybenzenes are destabilized by the adjacent
C == O groups (repulsive destabilization caused by similar charges) and have an insufficient
concentration to react with NH2OH. The triketo tautomer of 1,3,5-trihydroxybenzene
suffers from no such destabilization because the C == O groups are not adjacent. For this
reason, the concentration of the triketo tautomer is sufficient to react with NH2OH to form
a trioxime.
5. 2-Ethyl-3-methylcyclobuta-2-en-1-one exists exclusively in the keto form.

Explain.
Solution The enol form of 2-ethyl-3-methylcyclobuta-2-en-1-one is very unstable because
it is an antiaromatic system (a planar closed loop of 4np electron system, where n = 1). For
this reason, the compound exists exclusively in the keto form.
2-Ethyl-3-methylcyclobuta-2en-1-one:
6. The enol form of 9-formylfluorene is favoured at equilibrium — Why?

Solution The enol form of 9-formylfluorene is favoured at equilibrium because the enol
form is considerably stabilized by resonance involving two benzene rings and one aromatic
cyclopentadienyl system [(4n + 2)p electrons, where n = 1].
7. Explain why the following compound exists exclusively in the keto form:
Solution The enol form of the this cyclic diketone is a close loop of 10 p electrons [(4n + 2)
p, where n = 2]. So, the enol is expected to be an aromatic system of much higher stability.
Although the enol has no angle strain, the hydrogens at C–5 and C–10 interfere sterically
with each other and force the molecule out of planarity. However, to achieve aromatic
stability, the system should adapt a planar shape. So, the enol does not possess aromatic
stability and, therefore, the keto form is much more stable than the enol form and the
diketone exists exclusively in the keto form.
8. Explain why cis-5-tert-butyl-2-methlcyclohexanone undergoes isomeriza-

tion to yield the trans-isomer in the presence of a base.
Solution In the cis-isomer of 5-tert-butyl-2-methyl-cyclohexanone, the bulky–CMe3
group locks the molecule in that particular conformation in which this group is placed in
an equatorial position and the methyl group is placed in an axial position. However, this
conformation with the — CH3 group axial suffers from syn-axial interactions between the
—CH3 group and H’s at C-4 and C-6. On the other hand, there is no such steric strain
in the trans-isomer because the —CH3 group is equatorial. Again, the —CH3 group is
too far to experience any appreciable steric interaction with C == O and also eclipsing of
—CH3 and C == O is electronically favourable. The trans-isomer is, therefore, relatively
more stable than the cis-isomer and because of this, when treated with a base, the cis-
isomer undergoes isomerization to yield the trans-isomer through the formation of an
intermediate enolate ion.
9. The enol content of 3,3,3-trifluoropropanal is larger than that of propanal.

Explain.
Solution The enol of 3,3,3-trfluoropropanal is stabilized by intramolecular hydrogen
bonding involving a stable six-membered ring (chelation) while the enol of propanal
(CH3CH2CHO) is not similarly stabilized. Because of this, the enol content of the former
aldehyde is larger than that of the latter.
10. At equilibrium the enol content of 3,5-heptanedione

(CH3CH2COCH2COCH2CH3) is larger than that of 4-methyl-3,5-heptanedione
[CH3CH2COCH(CH3)COCH2CH3]. Account for this observation.
Solution In the enol form of 4-methyl-3,5-heptanedione, the —CH3 group at C–4 becomes
involved in steric interaction with the adjacent —C2H5 groups. This steric strain also
makes the internal hydrogen bonding weak. No such steric strain is involved in the enol
form of 3,5-heptanedione because hydrogen is relatively smaller in size. So, the former
enol is relatively less stable than the latter enol and hence at equilibrium the percentage
of the former enol is less than the percentage of the latter enol.
11. The enol content of 2,4-pentanedione or acetylacetone at equilibrium

is only 15 percent in water, 58 percent in acetonitrile (CH3CN) and 92
percent in hexane (C6H14). Account for these observations.
Solution The keto form of a b -diketone like 2,4-pentanedione is more polar than the enol
form because intramolecular hydrogen bonding and resonance involving the other carbonyl
group results in a decrease in polar character of the enol. Therefore, the enol content is
expected to increase with decrease in solvent polarity. Hexane (CH3CH2CH2CH2CH2CH3)
with a dielectric constant 1.9 is a nonpolar solvent, acetonitrile (CH3CN) with a dielectric
constant 38 is an aprotic polar solvent and water with a dielectric constant 79 is a protic
and much polar solvent. So, the enol content in hexane is greater than in acetonitrile
and the enol content in acetonitrile is greater than in water. The percentage of enol
drops dramatically in water because water forms hydrogen bonds with C == O’s, thereby
inhibiting the intramolecular hydrogen bonding that helps stabilize the enol.
12. Account for the given trend in enol content for each of the following
reactive methylene compounds:
EtO2C — CH2 — CO2Et(<1 percent); CH3COCH2CO2Et(7.7 percent);
CH3COCH2COCH3 (76 percent); PhCOCH2COCH3 (89 percent)
Solution Cross-conjugation (delocalization of a pair of nonbonding electrons on the
oxygen of the –OR group of the ester onto the carbonyl oxygen) retards the ability of the
C = O of the ester group to help stabilize the enol. It thus follows that the enol content
will decrease with increase in the number of ester group. Therefore, diethyl malonate (I)
with two ester groups has a lower enol content than ethyl accetoacetate (II) with one ester
group which in turn has a lower enol content than acetylacetone (III) containing no ester
group. The enol content of 1-phenyl-1,3-butanedione (IV) is somewhat higher than that of
acetylacetone (III) because the –Ph group stabilizes the enol by extending the conjugation
for p-bond delocalization.
13. Which one of the following two carbonyl compounds will enolize and why?
Solution The compound I will not enolize because its enol being a flat ring containing
4n(n = 1)p electrons is an antiaromatic system and hence very unstable. On the other
hand, the compound II will enolize readily because its enol being a flat ring containing
a close loop of (4n + 2)p electrons, where n = 0, is an an aromatic system and hence very
stable.
14. When cis-1-decalone is treated with sodium ethoxide in ethanol,

it undergoes epimerization to yield trans-1-decalone. Explain this
observation.
Solution This epimerization occurs because trans-1-decalone in which no bulky groups

are axial is relatively more stable than cis-1-decalone in which a large group is axial.
1. Which compound in each of the following pairs has the greater enol content and
why?
(a) (b)
(c) CH3CH2COCH3 or CF3CH2COCH3 (d) Ph CH2 COCH3 or CH3CH2COCH3
(e) (f)
2. Compare the extents of enol content of 2,4-pentanedione (CH3COCH2COCH3) in

water and benzene.
3. Explain why the compound A exists exclusively in the enol form while the compound
B exists exclusively in the keto form
4. Label the following pairs of structures as tautomers or resonance structures:

≈ ≈
(a) CH3CH2CHOCH3 and CH3CH2CH = OCH3
(b) (c)
O OH
|| |
(d) (e) C6 H5 — C — NH2 and C6 H5 —C == NH
5. Write down the structures of all possible enol tautomers of acetylacetone
(CH3COCH2COCH3) and justify their relative stabilities. Write down the s-cis and
s-trans forms of one of the possible tautomers.
[Hint: The possible enol tautomers of acetylacetone are as follows:
The enol II is less stable than the enol I because in it the enol double bond is not
conjugated with C == O and hence it is not stabilized by resonance.
The s-cis and s-trans conformations of I are as follows:
6. State and explain which of CH3COCH2COCH3 and CH3COCHPhCOCH3 has got

higher enol content in tetrahydrofuran.
7. Which of the following has higher enol content? Give reasons for your answer.
8. Which of the following pair of compounds would predominantly remain in the enol
form? Give your reasoning.
(a) CH3COCOCH3 and CH3COCH2COCH3 (b)

9. Arrange the following compounds in order of increasing enol content. Give reasons.
CH 3COCH 2 COCH 3 ; CH 3 COCH 2CH 2COCH 3 ; CH 3COCH 2CO2Et
I II III
[Hint: II < III < I]

10. Ethyl acetoacetale exists in the enol form much more in hexane than in water — Why?
11. Explain why the following two structures are tautomers but not resonance forms:
CH 3 — C == CH 2 and CH 3 — C — CH 3
| ||
OH O
12. Draw the enol tautomers for each of the following compounds. For those compounds
that have more than one enol tautomer, indicate which one is more stable.
(a) (b) (c)
(d) —CH2COCH3 (e) CH3CH2COCH2COCH2CH3 (f) CH3COCH2CO2Et
[Hint:
(a) (b)
OH
CH3
(c) (more stable);
(d)
OH OH
| |
(e) CH 3 CH 2COCH == CCH 2CH 3 (more stable); CH 3CH == CCH 2COCH 2CH 3 (less stable)
OH OH
| |
(f) CH 3 — C == CH — CO2Et (more stable) CH 2 == C CH 2CO2Et (less stable) ]
13. Would optically active ketones like those given below undergo acid- or base-
catalyzed racemization? Explain your answer.
[Hint: In the ketone I, these is no hydrogen attached to its a-carbon atom (which
is a chirality centre) and hence enol formation involving the chirality centre is
not possible. In the ketone II, the chirality centre is a b- carbon atom and hence
it remains unaffected by enol formation. Therefore, none of these two ketones
undergo acid- or base-catalyzed racemization.]
14. Explain the following observation:
CH3 O CH3 O
| || | ||
OD /D2O
@
H5C2 — CH — C — C6 H5 æææææ ≈ Æ H 5C2 — CD — C — C6 H 5
(±) or D3O (±)
15. Write down the resonance structures and tautomers of the following compound:
O
HN NH
O N O
H
[Hint: Resonance structures:
Tautomers:
16. Explain the position of the following equilibrium:
[Hint: Ring aromaticity usually stabilizes the phenolic tautomer. However, in this
case the reverse is true. Since anthrone has two distinct intact aromatic rings in
addition to the C == O, it is relatively more stable than anthranol, which suffers a
loss of aromaticity per ring because its three rings are fused.]
17. Draw a stepwise mechanism for each of the following reactions:

–
OH
(a) (b) OH H2O
== O
18. Draw the enol intermediate and the ketone product formed in the following reaction:
19. Draw all of the enol forms of 2-butanone. Which of them is the least stable and
why?
20. Give a curved-arrow mechanism for the exchange of a-hydrogens for deuterium:
21. (a) Draw the structure of the conjugate base of each of the following compounds.
What is the relationship between the two conjugate bases?
O OH
H
H
I II
(b) Which compound is more acidic and why?
1.12 AROMATICITY
Cyclic compounds like benzene and those which resemble benzene in chemical behaviour
have unusually large resonance energies and hence possess unusual stability. These
compounds are called aromatic compounds and the quality which renders these compounds
especially stable is referred to as aromaticity.
1.12.1 Criteria for Aromaticity

For a compound to be classified as aromatic, it must fulfil the following criteria:
(1) The compound must have an uninterrupted cyclic cloud of p electron above and
below the plane of the molecule (often called a p cloud). To form a cyclic p cloud,
the structure of the molecule must be cyclic. To form an uninterrupted p cloud,
each atom in the ring must have a p orbital, i.e., the molecule must be completely
conjugated. And, to form a p cloud, each p orbital must be able to overlap with the
p orbitals on either side of it, i.e., the molecule must be planar.
(2) An aromatic compound must have an odd number of pairs of p electrons, i.e., the
uninterrupted p cloud must contain (4n + 2) delocalized p electrons, where n = 0,
1, 2, etc., (this is what is called Hückel’s rule of the 4n + 2 rule) and therefore, all
bonding molecular orbitals (and HOMOs) must be completely filled.
(3) Delocalization of the p electrons over the ring must lower the electronic energy of
the species, i.e., the species must be stabilized by delocalization.
Therefore, planar, cyclic and completely conjugated compounds containing (4n + 2)
p electrons, where n = 0, 1, 2, 3 and so on (i.e., rings containing 2, 6, 10, 14,…, etc.,
p electrons) are unusually stabilized by electron delocalization and are said to be
aromatic.
Benzene is the best known example of compounds possessing aromatic character,
since this molecule is a planar, cyclic and completely conjugated system containing
six p electrons [(4n + 2)p electrons, where n = 1]. All of its bonding molecular
orbitals (and HOMOs) are completely filled.
Some other examples of aromatic systems:
(1)
(2)
(3)
(4)
1.12.2 Antiaromatic Compounds

Planar, cyclic and completely conjugated compounds containing 4np electrons, where
n = 1, 2, 3, ... etc., are especially unstable (destabilized by delocalization). These are called
antiaromatic compounds. Cyclobutadiene or [4]annulene is an example of antiaromatic
compound, since this molecule is a planar, cyclic and completely conjugated one containing
four p electrons (i.e., 4np electrons, where n = 1). It is so highly reactive that it can only be
prepared at extremely low temperatures.
Some other examples of antiaromatic systems:

(1) Cyclopropenyl anion. A 4np electron system, where n = 1. The system is
destabilized by resonance. In fact, the unconjugated is more stable than
conjugated .
(2) Cyclopentadienyl cation. A 4np electron system, where n = 1. The system
is destabilized by electron delocalization. In fact, the unconjugated is more

stable than conjugated .
(3) Cycloheptatrienyl anion (planar). A 4np electron system where n = 2. The

system is destabilized by resonance. The unconjugated is more stable than
conjugated .
1.12.3 Nonaromatic Compounds

Compounds that lack one (or more) of the criteria fulfilled by aromatic or antiaromatic
compounds are called nonaromatic compounds or simply the compounds which are neither
aromatic or antiaromatic are called nonaromatic compounds. 1,3,5-Cycloheptatriene, for
example, is a nonaromatic compound because the ring is not completely conjugated.
A (4n + 2)p electron system that has a continuous ring of p orbitals may also be monaromatic
it overlapping of orbitals is inhibited due to nonplanarity.
1.12.4 Classification of Compounds as Aromatic, Antiaromatic and

Nonaromatic by Comparing their Stabilities with that of the
Corresponding Open Chain Compounds
If the ring is more stable (lower p-electron energy) than the open chain counterpart, the
compound is classified as being aromatic. Benzene, for example, is aromatic because it is
more stable than 1,3,5-hexatriene.
If the ring is less stable (higher p-electron energy) than the open-chain counterpart, the
compound is classified as being antiaromatic. Cyclobutadiene, for example, is antiaromatic
because it is less stable than 1,3-butadiene.
Cyclobutadiene 1,3-Butadiene
(less stable: antiaromatic) (more stable)
And, if the ring and its open-chain counterpart have the same stability (same p-electron
energy), the compound is classified as being nonaromatic. For example, 1,3-cyclohexadiene
is similar in stability to cis, cis-2, 4-hexadiene.
1.12.5 Modern Definition of Aromaticity

The modern definition of aromaticity is the ability to sustain an induced ring current
by a planar, cyclic and conjugated system with (4n + 2) delocalized p electrons, where
n = 0, 1, 2, 3, ... etc. An applied magnetic field perpendicular to a conjugated and planar
ring containing (4n + 2)p electrons will cause the p electrons to circulate around the ring
and thereby generating their own magnetic field. This circulation is called induced ring
current.
1.12.6 Molecular Orbital Energy Diagram of Some Ions and Molecules

The molecular orbital energy diagram of cyclopropenyl anion (I), cyclopropenyl cation (II),
cyclobutadiene (III), cyclopentadienyl anion (IV), cyclopentadienyl cation (V), benzene
(VI), cycloheptatrienyl cation (VII) and cyclooctatetraenyl dianion (VIII) may be given as
follows:
≠ ≠ —— — — —— —
Antibonding
—— —— —— —— ——
≠ ≠ Nonbonding
≠ ≠Ø ≠Ø ≠ ≠ ≠Ø ≠Ø ≠Ø ≠Ø ≠Ø ≠Ø
E ≠Ø ≠Ø Bonding
≠Ø ≠Ø ≠Ø ≠Ø ≠Ø ≠Ø ≠Ø ≠Ø
– –
– + +
I II III IV V VI VIII
The solution of Schrodinger wave equation for the p electron system of any cyclic, planar
and conjugated polyene reveals that energies of the molecular orbitals in such systems
have a very simple characteristic pattern. In such systems, there is always one orbital of
lowest energy which is followed by degenerate pairs of orbitals (i.e., orbitals possessing
same energy) in the order of increasing energy. The first pair of electrons is filled in the
lowest energy molecular orbital and then the rest of the electrons go to degenerate pairs
of orbitals according to Hund’s rule. In a 4np electron system, the orbital of lowest energy
will be filled by two electrons. After filling the remaining orbitals according to Hund’s rule,
there will always be two singly occupied degenerate orbitals in these systems. Because of
such diradical structure, these systems are very unstable and possess higher energy than
their open chain analogies or the hypothetical cyclic structures where delocalization does
not occur, i.e., these systems are antiaromatic. In a (4n + 2)p electron system, the lowest
orbital will be filled by a pair of electron and each of the degenerate pairs of orbitals will be
filled by four electrons. Thus, in such systems all the occupied bonding orbitals are doubly
filled. This is an energetically favourable configuration. Because of this closed shell filling
of orbitals with electrons, such systems are unusually stable, i.e., aromatic.
It thus follows that a compound is aromatic if all bonding MOs ( and HOMOs) are completely
filled and no p electrons occupy antibonding MOs. On the other hand, a compound is
antiaromatic if it has electrons in antibonding molecular orbitals or if it has half-filled
bonding or nonbonding molecular orbitals.
1.12.7 Use of Inscribed Polygon Method to Determine the Relative

Energies of p Molecular Orbitals for Cyclic Planar and
Completely Conjugated Compounds and to Classify Them as
Aromatic and Antiaromatic
In the inscribed polygon method, a regular polygon is drawn inside a circle with its
vertices touching the circle and one of the vertices pointing down. The total number of
corners is equal to the number of p molecular orbitals. The points where the corners of
the polygon touch the circle correspond to the energy levels of the p molecular orbitals
of the system. A line is then drawn horizontally through the centre of the circle and the
molecular orbitals are labelled as bonding, nonbonding and antibonding. Corners below
the horizontal diameter are bonding (p), those above are antibonding (p*) and those on the
diameter are nonbonding (pn). It is assumed that all the systems are planar. The method
may be applied to 3–, 4–, 5–, 6–, 7–, and 8-carbon systems and from the character of the p
molecular orbitals the systems can be classified as aromatic and antiaromatic as follows:
1.12.8 Classification of Some Molecules and Ions as Aromatic,

Antiaromatic and Nonaromatic
(a) The compound is aromatic because it has a continuous ring of
overlapping p orbitals (boron contains a vacant p orbital) and it has a
6p electron [(4n + 2), p, where, n = 1] electron system.
(b) The compound is nonaromatic because it does not have a continuous

ring of overlapping p orbitals (there is a saturated carbon containing no
p orbital), even though it contains 6p electrons.
(c) Since boron contains a vacant p orbital, therefore, the compound has a
continuous ring of overlapping p orbitals. However, it has
B
4p (4np, n = 1) electrons. Therefore, the compound is antiaromatic,
H
(d) Oxygen is sp2-hybridized and hence the compound has a continuous

ring of overlapping p orbitals. One unshared pair of electrons on oxygen
is used to complete the aromatic sextet, i.e., it has a 6p[(4n + 2)p, n = 1]
electron system. Therefore, the compound is aromatic.
(e) Due to the formation of dative p bond between boron and oxygen,
the molecule possesses a continuous ring of overlapping p
orbitals and the system contains 6p[4n + 2)p, where n = 1)
electrons. Therefore, this planar molecule is aromatic.
(f) The molecule contains 10p [(4n + 2)p, n = 2] electrons. However, the
molecule is nonplanar because the two hydrogens pointing to the
interior of the ring interfere sterically and for this reason effective
overlapping of p orbitals does not take place. Hence the molecule is
nonaromatic.
(g) The compound is aromatic because it is a planar close loop of 10p

[4n + 2)p, n = 2] electrons. Two interfering H’s are removed to form
a bridge.
(h) The ion is aromatic because it has a continuous ring of overlapping p

orbitals and has a 6p[(4n + 2) p, n = 1] electron system. Pyridine takes
≈
up a proton to form its conjugate acid, the pyridinium ion (C6 H5 N H) .
The additional proton has no effect on the electrons of the aromatic
sextet. It simply bonds to the nonbonding pair of electrons present in an

sp2 orbital of nitrogen.
(i) It is a planar, cyclic and completely conjugated system containing

6p [(4n + 2)p, n = 1] electrons and therefore, it is aromatic. The
uncharged nitrogen atom is sp2 hybridized, with a lone pair of
electrons in the p orbital. This p orbital overlaps with the adjacent
p orbitals to form a continuous ring.
(j) The sp3-hybridized N atom is bonded to four atoms (two carbon

+ atoms and two hydrogen atoms) and hence it contains no unhybridized
N p orbital. Therefore, the ion does not have a continuous ring of
H H overlapping p orbitals and so, it is nonaromatic.
N
(k) The compound is aromatic because it is a planar, cyclic and completely
N
conjugated 10p [(4n + 2)p, n = 2] electron system. The unshared electron pairs on
Ns do not interfere to make the system nonplanar.
Ph Ph
+
(l) The ion is aromatic because it is a planar, cyclic and completely
+
conjugated system containing 2p[4n + 2)p, n = 0] electrons.
Ph Ph
Ph Ph Ph Ph Ph Ph Ph + + Ph Ph Ph
+ +
´ ´ ´ ∫ ++
+ +
+ +
Ph Ph Ph Ph Ph Ph Ph Ph Ph Ph
Stabilized by electron delocalization Aromatic
Ph – – Ph
(m) The ion is aromatic because it is a planar, cyclic and completely
conjugated system containing 6p[(4n + 2) p, n = 1] electrons.
Ph Ph
(n) CH3 The molecule can be represented as a resonance hybrid. The

resonance structure II has a major contribution because it is a
N
== combination of two aromatic ring systems: cyclopentadienyl anion
and pyridinium ion. Therefore, the molecule is aromatic.
CH3 CH3 CH3
N + N + N
–
== ´ ´ etc. ∫ –
I II Aromatic
(o) The molecule can be represented as a resonance hybrid. The
resonance structures II and III have equal and major contribution
because in these two structures all the three rings are aromatic:
cyclopentadienyl anion, benzene and cycloheptatrienyl cation.
Therefore, the molecule is aromatic.
(p) O Since the more stable and more contributing resonance structure II of this
ion is a planar, cyclic and completely conjugated system containing 6 p [(4n
+
+2)p, n = 1] electrons, the ion is aromatic.
(q) The compound is antiaromatic because it is a planar, cyclic and completely

conjugated system containing 8p (4np, n = 2) electrons.
(r) Although the species is planar and it has a continuous ring of overlapping
p orbitals, it contains 5p electrons. Therefore, it is nonaromatic.
(s) Although the ion contains 6p [(4n + 2)p, n = 1] electrons, it does not have a
continuous ring of overlapping p orbitals (one sp3-hybridized carbon has no
p orbital). Therefore, this ion is nonaromatic.
1.12.9 Homoaromatic Compounds
A homoaromatic compound is a compound that contains one or more sp3-hybridized carbon
atoms in an otherwise conjugated cycle [a(4n + 2)p electron system].
When cyclcooctatetraene is dissolved in concentrated sulphuric acid, a proton adds to one
of the double bonds to form the homotropylium ion. In this species, an aromatic sextet
[(4n + 2)p electrons, where n = 1] is spread over seven carbons, as in tropylium ion. The
eighth carbon is an sp3-hybridized one and cannot take part in aromaticity. This ion is an
example of homoaromatic compound. In order for the p orbitals to overlap most effectively
so as to close a loop, the sp3 carbon is forced to lie almost vertically above the plane of the
aromatic atoms.
1.12.10 Some Chemical and Physical Consequences of Aromaticity

(1) Cyclopentadiene (Ka = 10–15) is much more acidic than cycloheptatriene
(Ka = 10–45).
The cyclopentadienyl anion, the conjugate base of cyclopentadiene, can be
represented as a hybrid of five equivalent resonance structures while the
cycloheptatrienyl anion, the conjugate base of cycloheptatriene, can be represented
as a hybrid of seven equivalent resonance structures.
Although a greater number of resonance structures can be drawn for the

cycloheptatrienyl anion, it is less stable than the cyclopentadienyl anion and this
is because the former anion containing 8p (4np, n = 2) electrons is antiaromatic
(destabilized by resonance) while the latter anion containing 6p [(4n + 2)p, n = 1]
electrons is aromatic (highly stabilized by resonance). Due to the formation of a
relatively stable conjugate base, cyclopentadiene is a much stronger acid than
cycloheptatriene.
(2) In hydrogen-exchange reactions, the nitrile A loses its proton at a very
much slower rate than the nitrile B.
Loss of proton from the compound A leads to the formation of the antiaromatic
(4np electron, where n = 1) and hence very unstable cyclopropenyl anion while
loss of proton from the compound B leads to the formation of a relatively stable
ordinary carbanion. For this reason, the compound A loses proton at a very slower
rate than the compound B in hydrogen exchange reactions.
(3) Croconic acid is almost strong an acid as H2SO4.

O
==
O O
==
==
HO OH
Croconic acid
The dianionic conjugate base of croconic acid is aromatic and much stable because it
is a planar, cyclic and completely conjugated system containing 2p [4n + 2)p, n = 0]
electrons and for this reason, the compound is almost as strong as acid such as H2SO4.
(4) The carbonyl oxygen of 4-pyrone is more basic than the ring oxygen and the
compound does not form an oxime or phenylhydrazone.
Protonation of the carbonyl oxygen of 4-pyrone leads to the formation of a stable
aromatic system [a(4n + 2)p electron system, where n = 1] while protonation of
the ring oxygen produces a nonaromation less stable ion with a localized positive
charge. For this reason, the carbonyl oxygen is more basic than the ring oxygen of
4-pyrone.
Because of aromatic character of 4-pyrone, its carbonyl reactivity (positive character

of the carbonyl carbon) reduces and for this reaction, the compound does not form
an oxime or phenylhydrazone.
(5) The amine B is more basic than the amine A.
The unshared pair of electrons on nitrogen in the amine A is well delocalized

because the charge-separated structures constitute a stable aromatic system
[cyclopentadienyl anion with (4n + 2)p electrons, where n = 1]. On the other hand,
the unshared electron pair on nitrogen in the amine B is not at all delocalized
by resonance because the charge separated structures constitute an unstable
antiaromatic system (cycloheptatrienyl anion containing 4np electrons, where
n = 2). Therefore, the availability of the lone pair of electrons on nitrogen in B is
greater than that in A. Hence B is more basic than A.
(6) Pyrrole is a much weaker base than pyridine.
In pyrrole, the unshared pair of electrons on nitrogen is used to complete the

aromatic sextet [a(4n + 2)p electron system, where n = 1] and so these are not
readily available for coordinating with a proton.
In pyridine, on the other hand, the nonbonding electrons on nitrogen are not used
to complete the aromatic sextet and so, these are readily available for coordinating
with a proton.
Pyrrole is, therefore, a much weaker base than pyridine.

(7) The hydrocarbon A is much more acidic than the hydrocarbon B.
==
====
A B
The conjugate base of B is resonance-stabilized because the unshared electron pair
on the negative carbon is in proper conjugation with the three double bonds. The
conjugate base of A is also resonance-stabilized. However, because of constituting
an aromatic system (a close loop of 6p electrons), it is much more stable than the
conjugate base of B. For this reason, the hydrocarbon A is much more acidic than
the hydrocarbon B.
(8) When the compound I is treated with silver perchlorate in propanoic acid,
it undergoes solvolysis while the compound II does not.
I I
I II
The reaction proceeds through the rate-determining formation of an intermediate
carbocation. The carbocation expected to be obtained from I in relatively stable 2°
carbocation while the carbocation expected to be obtained from II is a very much
unstable carbocation (cyclpentadienyl cation) which is antiaromatic [a planar
conjugated system containing 4p (4np, where n = 1) electron]. In fact, this carbocation

is not at all formed. For this reason, the compound I undergoes solvolysis when
treated with silver perchlorate in propanoic acid while the compound II does not.
(9) Cyclooctatetraene (COT) is nonplanar (actually it is tub-shaped) while its

dianion is planar.
In the planar conformation, cyclooctatetraene (COT) suffers from angle strain (a
regular octagon has a angle of 135° while sp2 angles are most stable at 120°). Also,
the molecule containing 8p (4np, n = 2) electrons is antiaromatic in planar form
and hence destabilized by delocalization. To avoid the destabilizing antiaromaticity
and strain, the molecule assumes a nonplanar shape (like a tub) which is now
nonaromatic.
Cyclooctatetraenyl dianion is a 10 p [(4n + 2)p, n = 2] electron system and is

aromatic. The delocalization energy of the system is, therefore, large enough to
overcome the angle strain in it and hence the ion exists in the stable planar state.
(10) The compound B is more stable than the compound A.
In [10] annulene (A), the C–1 and C–6 hydrogens interfere sterically with each
other and force the molecule out of planarity and for this reason, conjugation is
inhibited. So, the compound is not aromatic, even though it is a 10p [(4n +2)p,
n = 2] electron system. On the other hand, in 1,6-methano[10]annulene (B),
the internal H’s in [10]annulene are replaced by a methylene bridge above the
molecule, permitting it to be flat. So, there occurs extensive electron delocalization
in this 10p electron system and therefore, the compound is aromatic. Hence, the
compound B is more stable than the compound A.
(11) The following fulvalenes are significantly polar:
Pentafulvanlene containing 8p electrons can be represented as a hybrid of a

nonpolar resonance structure and a dipolar resonance structure which, in fact is
a hybrid of several charge-separated structures. The dipolar form which consists
of two aromatic ions (a cyclopentadienyl anion and a cyclopropenyl cation) is
considerably stable and has significant contribution to the hybrid. For this reason,
pentafulvalene is significantly polar.
The dipolar form of sesquifulvalene also containing two aromatic rings

(cyclopentadienyl anion and cycloheptatrienyl cation) is considerably stable
and has significant contribution to the hybrid. Therefore, this compound is also
significantly polar.
(12) Tropone is more polar than cycloheptanone.
==
==
O O
Tropone Cycloheptanone
In tropone, the carbonyl oxygen atom being highly electronegative takes up the
p electrons of the carbonyl group to form a dipolar structure which, in fact, is a
hybrid of several charge-separated structures that constitute an aromatic system (a
cycloheptatrienyl cation system). The dipolar structure is, therefore, considerably
stable and has large contribution to the hybrid. For this reason, tropone is
significantly polar.
Tropone:
The contribution of the charge-separated structure of cycloheptanone to its hybride

is not much significant because it is not much stable. For this reason, it is relatively
less polar than tropone.
´
Cycloheptanone:
==
+
–
O O
(less stable)
(13) Azulene is polar and it undergoes electrophilic substitution at C–1 of the
five-membered ring.
Azulene, a 10p electron system, can be represented as a resonance hybrid of two

nonpolar structures (I and II) and a dipolar structure (III) which, in fact, is a hybrid
of several charge-separated structures.
In structure III, both the rings possess a closed shell of six p electrons. The seven-
membered ring is an aromatic cycloheptatrienyl cation system while the five-
membered ring is an aromatic cyclopentadienyl anion system. For this reason, the
structure is considerably stable. Since this stable dipolar structure has significant
contribution to the hybrid, azulene possesses dipole moment, i.e., it is a polar molecule.
The five-membered ring being negatively polarized can accommodate a positive
charge while the seven-membered ring being positively polarized cannot
accommodate a positive charge. Because of this, it is the five-membered ring which
undergoes electrophilic substitution. Again, the substitution occurs at 1-position
because the 1-attack leads to a stable intermediate in which the seven-membered
ring is aromatic (a cycloheptatrienyl cation system).
(14) The rotational energy barrier around the C — N bond of the compound B is
higher than the compound A.
In the compound A, the unshared pair of electrons on nitrogen is involved in

maintaining an aromatic system of 6p electrons in the pyrrole ring and as a
consequence, its resonance interaction with the m-tolyl ring is actually very small.
Because of this, the C — N bond possesses a very small amount of double bond
character (very small contribution of the resonance structures with carbon–nitrogen
double bond).
Since the unshared pair of electrons on nitrogen in compound B is not used to

complete an aromatic sextet, it is highly delocalized with the m-tolyl ring. As a
consequence, the C — N bond possesses considerable double bond character (significant
contribution of the resonance structures with carbon–nitrogen double bond). Because
of much greater double bond character of the carbon–nitrogen bond, the rotational
energy barrier of the compound B is higher than that of the compound A.
(15) Dehydro [14]annulene is aromatic while [14]annulene is a nonaromatic

compound.
Fourteen is a Hückel number [(4n + 2)p, where n = 3]. Therefore, [14]annulene

is expected to be aromatic. However, the H’s that point to the interior of the ring
interfere sterically with each other and force the molecule out of planarity. Because
of this, it behaves like a nonaromatic compound. In dehydro[14]annulene, on the
other hand, two of the interfering Hs are removed to form a triple bond. As a result,
steric strain is released and the molecule becomes planar. One of the p bond of the
triple bond becomes involved in conjugation. The other p bond simply exists as a
localized bond because this p orbital remains at right angles to the extended p system.
Therefore, it is a 14p [(4n + 2)p, n = 3] electron system and hence it is aromatic.
(16) The following two [10]annulenes would be expected to be aromatic on the

basis of electron count. However, they are not aromatic.
The [10]annulene with all cis double bonds would, if it were planar, have
considerable angle strain because the internal bond angles would be 144° instead
of 120°. As a consequence, any stability this isomer gained by becoming planar in
order to become aromatic would be more than offset by the destabilizing effect of the
increased angle strain. Hence this isomer of [10]annulene is not aromatic. Because
of the same large angle strain, the cis-trans–cis-cis-cis-isomer of [10]annulene is
prevented from being aromatic.
1. 3-Chlorocyclopropene reacts with the Lewis acid SbCl5 to form a salt

while 3-chlorocyclopropane does not. Explain these observations.
Solution Due to the formation of aromatic cyclopropenyl cation, 3-chlorocyclopropene
reacts with SbCl5 to form the corresponding salt. Cyclopropyl cation is not an aromatic
system. Furthermore, it is much destabilized due to angle strain. For this reason, the
corresponding salt of chlorocyclopropane is not obtained.
2. Using the theory of aromaticity, explain the finding that I and II are
different compounds, but III and IV are identical.
D D D D
D D
D D
I II III IV
Solution Since cyclobutadiene is an antiaromatic compound (destabilized by resonance),
it contains localized double bonds and because of this, I and II are different compounds
(these are actually two isomers of dideuterocyclobutadiene). Since benzene is an aromatic
compound (stabilized by electron delocalization), it contains delocalized double bonds.
Therefore, III and IV are identical compounds (these are actually two resonance structures
of 1,2-dideuterobenzene).
3. The reaction (i) takes place readily to form a pale yellow precipitate of
silver bromide but the reaction (ii) does not take place at all. Explain
these observations.
AgNO3
(i) Cycloheptatrienyl bromide ææææ Æ A nitrate salt + AgBr Ø
(Pale yellow)
AgNO3
(ii) Cyclopentadienyl bromide ææææ Æ No reaction
Solution The reaction (i) takes place readily to form a pale yellow precipitate of AgBr
because the substrate cycloheptatrienyl bromide exists in the ionic form, i.e., as a salt.
This is because cycloheptatrienyl cation or tropylium ion containing 6p [(4n +2)p, where
n = 1] electrons is an aromatic species and very much stable. The reaction (ii), on the other
hand, does not take place because it leads to the formation of the antiaromatic and hence
very unstable cyclopentadienyl cation (a 4np electron system, where n = 1).
(i)
(ii)
4. The hydrocarbon A (pKa ª 14) is much more acidic than the hydrocarbon
B (pKa ª 22). Explain.
Solution In compound A, an aromatic and dipolar azulene system is, in fact, fused
with a cyclopentadiene system through the cycloheptatrienyl cation system. On the
other hand, in compound B, an azulene system is fused with a cyclopentadiene system
through the cyclopentadienyl anion system. Both of the two compounds react with a base
to form an aromatic cyclopentadienyl anion system. In the conjugate base of A, the two
cyclopentadienyl anion systems are separated by a cycloheptatrienyl cation system, while
in the conjugate base of B, the two cyclopentadienyl anion systems are adjacent. Therefore,
due to the presence of like charges on adjacent rings, the conjugate base of B is relatively
less stable than the conjugate base of A in which two negative rings are separated by a
positive ring and for this reason, A is more acidic than B.
Solution When cyclopentadiene containing radioactive carbon (14C) at C–5 is treated

with potassium hydride (KH), the potassium salt of cyclopentadienyl anion (a stable
aromatic system containing a close loop of 6p electrons) is obtained. Because of electron
delocalization, all carbons in this anion, except for the isotope, are equivalent.
In fact, the isotope makes no detectable difference in the relative importance of the
resonance structures; except for the position of isolope, all canonicals are equivalent.
Therefore, cyclopentadienyl anion undergoes protonation by water at each carbon with
equal facility (20 percent). Protonation at C–1 and C–4 (i.e., from III and IV) gives the
same labelled cyclopentadiene and protonation C–2 and C–3 (i.e., from V and II) gives the
same labelled cyclopentadiene. Hence the yield of each of these labelled cyclopentadiene
is 2 × 20 percent, i.e., 40 percent. These two remain mixed with 20 percent of the original
cyclopentadiene labelled at C–5.
6. Although there are eight free electrons in pyridine and phenyl anion
(C6H5①), they are aromatic. Explain.
Solution The unshared pairs of electrons of pyridine or phenyl anion are located in an sp2
orbital which is coplanar with the sigma skeleton of the ring but perpendicular to its p electron
cloud. Therefore, the unshared electron pair in each case cannot be delocalized through
resonance interaction with the p electron cloud of the aromatic nucleus. Therefore, each of
these species (planar) is a closed loop of six [(4n + 2)p, n = 1]p electron and hence aromatic.
7. Unlike an alkyl halide cycloheptatrienyl bromide is insoluble in nonpolar

solvents but is readily soluble in water. Explain.
Solution Cycloheptatrienyl bromide is not aromatic in the covalent form because it has
an sp3-hybridized carbon, so it does not have an uninterrupted ring of p orbital-bearing
atoms. In the ionic form, the cycloheptatrienyl cation (also called the tropylium cation) is
aromatic because it is a planar cyclic ion having an uninterrupted ring of p orbital-bearing
sp2-hybridized atoms and it has 6p [(4n + 2)p, n = 1] electrons. The stability associated
with the aromatic cation causes this alkyl halide to exist in the ionic form. Being ionic it is
insoluble in nonpolar solvents but readily soluble in highly polar solvent water (dielectric
constant = 79).
8. Predict the product of the following reaction and explain:
Solution The strong base butyllithium removes two protons from the compound and
this acid-base reaction leads to the formation of the 10p electron pentalene dianion (an
aromatic ion).
1. The following molecule shows very high dipole moment. Explain the reason.
2. [10]Annulene contains Hückel number of electrons but is not considered as an

aromatic compound. Explain.
3. Which one of the following compounds is more polar and why ?
==O ==O
I II
4. Tropolone does not form 2,4–dinitrophenylhydrazone derivative. Explain.
[Hint: Tropolone possesses aromatic character [a(4n + 2)p electron
system, where n = 1) and because of this, it shows a lack of ketonic properties like
formation of 2,4-dinitrophenylhydrazone, etc.]
5. Arrange the following carbocations in order of their increasing stability and explain
the order:
+
+ +
I II III
6. Squaric acid is almost strong as acid as H2SO4. Explain.
O OH
==
==
O OH
Squaric acid
7. Explain why it would be incorrect to write resonance structures as shown below:
8. Rank the indicated C — C bonds in order of increasing bond length, and explain
why you choose this order.
a c
b d
Ph
9. Compare the C == O bond lengths in Me2C == O and == O
Ph
10. Apply Hückel’s rule to explain why azulene is more stable than pentalene and
heptalene.
Azulene Pentalene Heptalene

11. Discuss the differences in aromatic behaviour between pyrrole and furan.
12. Coronene is an aromatic hydrocarbon, even though it contains 24p electrons. Explain.
Coronene
[Hint: It consists of two aromatic systems. An inner aromatic ring containing 6p
electrons [(4n + 2)p, n = 1] and an outer aromatic ring containing 18 p electrons
[(4n + 2)p, n = 4].]
13. Explain why borazine (also called inorganic benzene) is a very stable compound.
H
H B H
N N
B B
H N H
H
Borazine
14. The compound A is more acidic than the compound B. Explain this observation.
H H
COC2H5 COC2H5
A
15. Explain why benzene reacts with ozone to give triozonide but not mono- or diozonide.
16. Account for the following observations:
C2H5 C2H5 +
== + PhLi – C2H5 Li
C2H5 Ph
C2H5
+ PhLi No reaction
== C2H5
17. The barrier for rotation about the central double bond in the following compound
is very low (only about 14 kcal/mol). Explain.
O
==
== == CH — C
CH3
[Hint: Because of the stable (both rings are aromatic) dipolar resonance structure,
the central double bond possesses considerable single bond character.]
18. Which one of the following two molecules is more polar and why?
19. Which species is the smallest aromatic substance?

20. Compare the basicities of the following compounds:
[Hint: III > I > II]

21. Explain the following order of acidity:
22. Classify the following molecules and ions as aromatic, antiaromatic, or nonaromatic.
Give your reasoning.
(a)
– +
, ,
(b) (c) , ,
– + O O O
(d)
(e) (f)
23. Although cyclopropene is an extremely unstable compound, it is possible to prepare

its cation. Give reasons.
24. Comment on the relative stabilities of the following species:
(a) (b) +2 (c) –2
25. Compare the acidities of the compounds I and II and justify:
26. Classify each nitrogen atom as strong basic and weak basic, according to the
availability of its lone pair of electrons.
N
(a) (b)
N
H
[Hint: Azulene can be represented as a combination of two aromatic systems

such as cyclopentadienyl anion and cycloheptatrienyl cation. Therefore, the first
compound in which the electron-withdrawing —CN group is attached to the
negatively polarized five-membered ring and the electron-releasing –OMe group
is attached to the positively polarized seven-membered ring is relatively more
polar, i.e., possesses higher dipole moment than the second compound in which the
electron-attracting group is attached to the positively polarized ring and electron-
donating group is attached to the negatively polarized ring.
+ – CN + – OMe
MeO CN
More polar Less polar
28. Arrange the following compounds in order of increasing acidity and explain.
[Hint: III < II < I]

29. Predict which ones are likely to be aromatic, and explain why they are aromatic.
– +
O +
(a) (b) (c) (d) (e)
O
N
==
==
==
==
O O O H
O
H NH2
NH N N N
(f) (g) (h) (i) O
N
==
N O
==
O H
30. Which of the two ‘C == O’ bonds is shorter than the other? Give reasons.
31. Which of the following two bromo compounds would undergo silver ion-assisted
hydrolysis at a faster rate?
(a) (b)
Br
Br
32. Label the following compounds as aromatic, homoaromatic and antiaromatic:
–
+ H
(a) (If planar) (b) (c) –
H
33. Complete the following reactions:
(a) hn Br2 D
A B C
CH2N2
(b)
[Hint:
34. The compound given below is a somewhat stronger base than ammonia. Which
nitrogen do you think is protonated when it is treated with an acid and why? Write
the structure of its conjugate acid.
35. Explain why the following compounds are aromatic.
–
S
(a) (b) (c) (d)
[Hint:
(a) Sulphur uses its acant d orbital to make a cyclic conjugated system containing
(4n + 2)p electrons, where n = 2.
(b) One of the p bonds of the acetylenic function and the central p bond of the
1,2,3-triene system are perpendicularly oriented with the conjugated cyclic p
system and do not involve in electron delocalization. The conjugated system
thus contains Hückel number of electrons. Also, the system is planar because
the H’s pointing to the interior of the ring do not interfere sterically with each
other.]
36. Explain why fulvene possesses considerable dipole moment (m = 1.5 D).
[Hint: Due to greater contribution of the stable dipolar resonance
structure in which the ring is an aromatic cyclopentadienyl anion moiety
possesses considerable dipole moment.]
37. When 3-chlorocyclopropene is treated with AgBF4, AgCl precipitates. The organic
product can be obtained as a crystalline material, soluble in polar solvents such as
nitromethane but insoluble in hexane. Explain these observations.
38. Thymine is one of the heterocyclic bases found in DNA. Do you expect thymine to
be aromatic? Explain.
39. Predict the relative pKa values of cyclopentadiene and cycloheptatriene.

40. When this compound ionizes, is Cl① or Cl≈ formed? Give your reasoning.
Cl
41. Classify each species as aromatic, antiaromatic or nonaromatic.
(a) (b) (c)

N
H N
H
(d) (e) (f)

N
42. Predict the relative pKa values of cyclopropene and cyclopropane. Give your
reasoning.
43. Which is more soluble in water, 3-bromocyclopropene or bromocyclopropane?
Explain.
[Hint: 3-Bromocyclopropene.]
44. Pyrene is aromatic, even though it contains 16 p electrons. Explain.
Pyrene
45. Draw the p molecular orbitals of benzene and place them in an energy level diagram.
Explain why it is extraordinarily stable.
[Hint: The p molecular orbitals of benzene are as follows:
The energy of molecular orbitals increase with increase in the number of node. The
six electrons completely fill the three bonding MOs with two electrons in each. This
accounts for the extraordinary stability of benzene.]
46. Explain why the following compound shows a reduced reactivity towards acetylation.
O
==
N
H
47. Unlike quinoline and isoquinoline, which are of comparable stability, indole and
isoindole are quite different from each other. Which one is more stable and why?
[Hint: Indole is more stable than isoindole because the six-memebred ring in indole
corresponds to benzene while the six-membered ring in isoindole does not have the
same pattern of bonds as benzene.]
48. What does a comparison of the heats of combustion of benzene (781 kcal/mol),
cyclooctatetraene (1086 kcal/mol), [16]annulene (2182 kcal/mol) and [18]annulene
(2346 kcal/mol) reveal?
49. One of the two dipolar resonance structures I and II is expected to contribute
appreciably to the resonance hybrid of calicene. Which of these two resonance
structures is more reasonable and why?
==
+
Calicene I II
50. Which one of the following two compounds should be stabilized by resonance to a
greater extent? Give your reasoning.
51. Imidazole is a stronger base than pyrrole. Explain.

N
N N
H H
Imidazole Pyrrole
O O
==
==
OH
(a) == (b)
CH and and
I II I II
[Hint:
(b) The compound II has higher dipole moment. The ionic forms of both the
compounds have an aromatic system (cycloheptatrienyl catation). But in the
case of II, it is further stabilized by intramolecular hydrogen bonding.
53. Which of the following molecules is likely to be planar and which nonplanar?
Explain your answer.
1.13 THERMODYNAMICS, ENERGY DIAGRAMS AND KINETICS OF

ORGANIC REACTIONS
Let us consider the reaction A + B — C Æ A — B + C. Does the reaction practically occur?
It occurs, to what extent and how fast? We cannot answer these questions so easily, unless
we know properly the thermodynamics and kinetics of the reaction. The thermodynamic
factors tell us about the feasibility of the reaction, i.e., whether the reaction will take place
at all or not, while the kinetic factors tell us about the rate of the reaction, i.e., how fast
the reaction will take place.
1.13.1 Thermodynamics
Thermodynamics describes the properties of a system at equilibrium. The relative
concentrations of reactants and products at equilibrium can be expressed numerically by
the equilibrium constant, Keq, of the reaction. For example, when starting materials A and
B react to give products C and D, the equilibrium constant may be given by the following
expression.
A + B ¨æææ ÆC+D
æ
[products] [C][D]
K eq = =
[reactants] [A][B]
The value of Keq tells us about the position of equilibrium, i.e., it expresses whether
the reactants or products are more stable and therefore predominate once equilibrium
has been reached. If the products are more stable (have a lower free energy) than the
reactants, there will be higher concentration of products than reactants at equilibrium
and Keq will be greater than 1, i.e., the reaction is favoured as written from left to right. On
the other hand, if the reactants are more stable than the products, there will be a higher
concentration of reactants than products at equilibrium and Keq will be less than 1, i.e.,
the reverse reaction is favoured as written from right to left. For a reaction to be useful,
the equilibrium must favour the products and Keq > 1.
A reaction at equilibrium can be described by several thermodynamic parameters. The
difference between the free energy of the products and that of the reactants at equilibrium
under standard conditions is called the Gibbs standard free energy change (DG°). The
symbol° indicates that the reaction takes place under standard conditions, i.e., all species
at 1M, temperature at 25°C and pressure at 1 atm.
∞ ∞
DG∞ = Gproducts — Greactants
(free energy of (free energy of
the products) the reactants)
It becomes clear from this equation that DG° will be negative if the products have a lower
free energy (or more stable) than the reactants and are favoured at equilibrium. Since
the reaction will release more energy to the system than it will consume from the system,
it will be an exergonic reaction. If the products have a higher free energy (or less stable)
than the reactants, i.e., if the reactants are favoured at equilibrium, DG° will be positive.
Since the reaction will consume more energy from the system than it will release into the
system, it will be an endergonic reaction. These two terms should not be confused with
terms exothermic and endothermic (discussed later).
Whether reactants or products are favoured at equilibrium can, therefore, be indicated
either by the equilibrium constant (Keq) or by the change in free energy (DG°). DG° is
related to Keq by the following equation:
DG∞ = – RT ln K eq = – 2.303 RT log K eq
where R is the gas constant (1.986 kcal/mol/K or 8.314 × 10–3 kJ/mol/K) and T is the
absolute temperature in Kelvin (K).
The equation can be used to determine the relationship between the equilibrium constant
and the free energy change between reactants and products. When Keq > 1, i.e., when log
Keq is positive, DG° is negative. Thus, for reactions with a negative DG°, the formation of
products at equilibrium is favourable. When Keq < 1, i.e., when log Keq is negative, DG° is
positive. Thus, for a reaction with a positive DG°, the formation of products at equilibrium
is unfavourable.
Since DG° depends on the logarithm of Keq, a small difference in DG° gives rise to a
large difference in the relative concentrations of products and reactants. For example, a
difference in energy of only ~ 4.184 kJmol–1 or ~1 kcal mole–1 means that there is 10 times
as much of the more stable species at equilibrium. A difference in energy of ~16.592 kJ
mol–1 or ~ 4.2 kcal mol–1 means that there is essentially only one species (99.9%), either
reactant or product, at equilibrium.
1.13.1.1 Enthalpy and entropy

The free energy change (DG°) has two components. These are enthalpy change (DH°) and
entropy change (DS°). These three thermodynamic quantities are related as follows:
DG∞ = DH ∞ - T DS∞ (at a fixed temperature T)
The enthalpy change (DH°) is the heat liberated or the heat consumed during the course
of a reaction. Heat is liberated when bonds are formed and heat is consumed when bonds
are broken. If the bonds that are formed in a reaction are stronger than the bonds that are
broken, energy will be released and DH° will be negative. A reaction with a negative DH°
is called an exothermic reaction. If the bonds that are formed are weaker than those that
are broken, energy will be consumed and DH° will be positive. A reaction with a positive
DH° is called an endothermic reaction.
DH° = (Energy of the bonds broken during the course of reaction) – (Energy of the bonds
formed during the course of the reaction)
For an exothermic reaction (DH° is negative) DG° is always negative at all temperatures if
DS° is positive, i.e., if the number of species increases on going from reactants to products.
For an exothermic reaction (DH° is negative) if the number of species decreases on going
from reactants to products, i.e., if DS° is negative, DG° is negative only when the term TDS°
is numerically less than DH°, i.e., at low temperature. If DS° is positive for an endthermic
reaction (DH° is positive), then the reaction becomes feasible (DG° is negative) only at
high temperature. On the other hand, if DS° is negative for an endothermic reaction (DH°
is positive), it is not feasible at all. Therefore, the feasibility of a reaction is based on the
thermodynamic parameter, DG° and is temperature dependent. It is relatively easy to
calculate the values of DH° which is the largest factor in the driving force in most organic
reactions. If the reaction involves only a small change in entropy, the TDS term will be
small and the value of DH° will be very close to the value of DG°. However, we must be
cautious in making this approximation because many organic reactions have relatively
small changes in enthalpy, larger changes in entropy or occur at high temperatures and
so have significant TDS° terms.
1.13.2 Energy Diagram

A schematic representation of the energy changes that take place as reactants are converted
into products is called an energy diagram or a reaction coordinate diagram. From an
energy diagram we can know how readily a reaction proceeds, how many step are involved
in a reaction and how the energies of the reactants, products and intermediates compare.
In an energy diagram, energy is plotted on the y-axis and the progress of the reaction is
plotted on the x-axis. As the reactants are converted into products, the reaction passes
through a maximum energy state which is known as transition state. The reactants attain
the transition state by acquiring the energy supplied by collisions. The transition state is
a hypothetical arrangement of atoms possessing a definite shape and charge distribution.
Its structure is somewhere in between the structures of the reactant and product. Any
bond that is partially cleaved or formed is drawn with a dashed line. Any atom that loses
or gains a charge contains a partial charge in the transition state. Transition states are
often drawn in brackets, with a superscript double dagger (=| ). Because of high energy and
very short life time, it can never be isolated or trapped.
@
Let us consider a concerted
@
reaction between the molecule B—C and the anion A : to form
products A — B and C: . Since the reaction occurs in a single step, the bond between A and
B is formed as the bond between B and C is broken. Let us also assume that the reaction
is exergonic, i.e., the products are lower in free energy than the reactants.
The anion A①: attacks the molecule B — C from the side remote from C and along the
bonding line of B — C. This mode of attack requires minimum energy because the
electrostatic repulsion between the entering and leaving groups is minimum. As the bond
A — B begins to form, the bond B — C begins to break. Since the energy released by bond
formation (which lags behind) is less than that required for bond cleavage, the energy of
the system increases on going from reactants to the transition state which is represented
d- d-
by A B C (B is partially bonded to A and C). The energy of the system then begins to
decrease on the going from transition state to products and reaches a minimum when C is
completely displaced by A. The energy diagram of this reaction is given below. Since DG°
values are used, y-axis is free energy.
In this exergonic reaction, the energy difference between the reactants and products is
DG°. The energy difference between the transition state and the reactants is called the
free energy of activation, symbolized by DG=. DG° relates to the equilibrium constant of the
|
=
|
reaction, while DG relates to the rate of the reaction.
If a reaction involves only a small change in entropy the TDS° term will be small and the
value of DH° will be close to the value of DG°. In that sense the above energy diagram may
also be represented as follows. Since DH° values are used, the y-axis is here potential energy.
The energy difference between the transition state and the reactants is called energy of
activation, symbolized by Ea. The reaction is now called an exothermic one (DH° = –ve).
The two variables, DG=| and DG° (or Ea and DH∞), are independent of each other. Two
reactions can have identical values of DG° (or DH°) but very different DG= (or Ea) values.
|
A multistep reaction proceeds through transient chemical species which unlike transition
state possesses some degree of stability and can be isolated and detected in many cases. These
are known as intermediates. Examples are carbocations, carbanions, free radicals, etc.
If the same exergonic (or exothermic) reaction occurs by a two-step pathway and bond
breaking occurs before bond making, it can be represented as follows:
Step 1: Cleavage of the B — C bond
Step 2: Formation of the A — B bond
Each step has its own energy barrier, with a transition state at the energy maximum.
Step 1 is endergonic (or endothermic) because energy is required to break the B — C bond,
making DG° (or DH∞) a positive value and placing the products at higher energy than
reactants. In the transition state, the B — C bond is partially broken. Step 2, on the other
hand, is exergonic (or exothermic) because energy is released due to formation of the
A — B bond, making DG° (or DH°) a negative value and placing the products at lower
energy than the reactants. In the transition state the A — B bond is partially formed. The
species B! is a reactive intermediate which is formed in step 1 and consumed in step 2.
The two transition states are separated by an energy minimum, at which the reaction
intermediate B! is located. Since the step 1 has the higher energy transition state
(DG1|= > DG2|= or Ea1 > Ea2 ), it is the slower, that is, the rate-determining step of the
reaction. The energy diagram of the reaction can be represented as follows:
1.13.3 Kinetics
From a knowledge of DG° (or DH°) we cannot predict how fast a reaction occurs, i.e., it does
not tell us any thing about the energy barrier or the energy hill that must be climbed for
the reactants to be converted into products. The higher the energy barrier, the slower the
reaction. Kinetics is the field of chemistry that describes the reaction rate and the factors
that affect them. The energy barrier of a reaction, indicated by DGG=, is called the free
|
energy of activation. It is the difference between the free energy of the transition state and
the free energy of the reactants.
DG=| = (free energy of the transition state) – (free energy of the reactants)
The rate of a chemical reaction is determined by measuring the decrease in the concentrations
of the reactants or the increase in the concentrations of the products over time. The rate
of a reaction depends on the following three factors: (i) the number of collisions that take
place between the reacting molecules in a given period of time, (ii) the fraction of the
collisions that occur with sufficient energy to get the reacting molecules over the energy
hill and (iii) the fraction of the collisions that occur with proper orientation. A rate law or
rate equation is an equation that shows the relationship between the rate of a reaction
and the concentration of the reactants. A rate law can be determined experimentally, and
it depends on the mechanism of the reaction. The two important terms of a rate law are
the rate constant k and the concentration of the reactants. All reactant concentrations
may not appear in the rate equation. The rate law or rate equation of a reaction may be
represented as
rate = k [reactants]
The rate constant k (a fundamental characteristic of a reaction) is a complex mathematical

term that takes into account the dependence of the rate of a reaction on temperature and
the experimental energy of activation, Ea, and this is because the Arrhenius equation
relates the rate constant of a reaction to Ea and the temperature at which the reaction is
carried out.
The Arrhenius equation: k = Ae– Ea / RT
The rate constant k and DG=| (or Ea) are inversely related. A high DG=| (or Ea) corresponds
to a small k. Fast reactions have large rate constants, while slow reactions have small
rate constants. A rate equation contains concentration terms for all reactants in a one-
step (concerted) reaction, while a rate equation contains concentration terms for only the
reactants involved in the rate-determining step in a multistep reaction.
For the following one-step reaction, both reactants appear in the transition state of the
rate-determining step (the only step here) of the reaction. This type of reaction involving
two reactants in the transition state is said to be bimolecular.
A: + B—C Æ A—B + C:
The order of a reaction is the summation of the exponents of the concentration terms
appearing in the rate equation. The rate equation of the above reaction is
rate = k [A①:] [B — C]
In this rate equation, there are two concentration terms, each with an exponent of one.
Therefore, the sum of the exponents is two and it is a second-order reaction. Since the rate
of the reaction depends on the concentration of both reactants, doubling the concentration
of either A①: or B —C doubles the rate of the reaction. If the concentrations of both A①:
and B — C are doubled, the rate of the reaction will be quadruple. However, the situation
is different if the same reaction occurs in two steps such as follows:
In multistep reactions, only the concentrations of the reactants involved in the rate-
determining step appear in the rate equation. The rate of this reaction, therefore, depends
only on the concentration of B — C only, because only B — C is involved in the rate-
determining step.
rate = k [B — C]
Since there is only one concentration term (raised to the first power), it is a first-order
reaction. Also, it is an unimolecular reaction because only one reactant is involved in the
transition state of the rate-determining step. Since the rate of the reaction depends on
the concentration of only B — C, doubling its concentration doubles the reaction rate, but
doubling the concentration of A①: has no effect on the reaction rate. In fact, neither its
concentration nor its identity affects the reaction rate.
The rate constant of a reaction should not be confused with the rate of a reaction. The rate
constant informs us how easy it is to reach the transition state, i.e., to cross the energy
barrier. Low energy barriers (low DG=| ) are associated with large rate constant while high
energy barriers (high DG=| ) have small rate constants. The rate of a reaction is a measure of
the amount of reactant disappeared or the amount of product formed per unit of time. The
reaction rate is the product of the rate constant and the concentration of the reactant(s).
Thus reaction rates depend on concentration, while rate constants are independent of
concentration. Therefore, when two reactions are compared to see which one occurs more
easily and at a faster rate, their rate constants and not their concentration-dependent
rates of reaction are to be compared.
Again, the free energy of activation, DG=| , should not be confused with the experimental
energy of activation, Ea. The free energy of activation has both enthalpy and entropy
components (DG=| = DH=| – TDS=| ), whereas the experimental energy of activation has
only an enthalpy component (Ea = DH=| + RT). Ea is an approximate energy barrier to
a reaction, while DG=| is a true energy barrier to a reaction and this is because some
reactions proceed by a change in enthalpy, some by a change in entropy, but most by a
change in both entropy and enthalpy. Without going to detail suffice it to say that the two
quantities DG=| and Ea are closely related and that both measure the difference in energy
between the reactants and the transition state.
1.13.4 Catalysis
Some reactions proceed faster when certain substances are added in small amounts to the
reaction. These substances are called catalysts. A catalyst may be described as a substance
that increases the rate of a reaction, i.e., speeds up a reaction without appearing in the
stoichiometric equation of the reaction (i.e., by remaining unaltered in mass and chemical
composition). A catalyst operates in such a way that it combines with one of the reagents
so as a provide an alternative pathway for the reaction and is regenerated in a later stage
of the reaction. The position of equilibrium of the reaction remains unaffected because
the catalyst increases the rate of the forward as well as the reverse reaction to the same
extent, i.e., a catalyst does not change the amount of reactant and product at equilibrium
(DH° of the reaction remains unchanged). In fact, a catalyst helps to achieve equilibrium
quickly.
The energy of activation (Ea) is related to the rate constant (k) as follows:
k = A. e– Ea / RT
The rate of the reaction can thus be accelerated either by increasing the value of A or
by decreasing the values of Ea. Since the value of A is usually constant, therefore, in the
presence of a catalyst a reaction follows a course of lower Ea. This is shown in the following
energy diagram. The uncatalyzed reaction (solid line) has the transition state denoted by
TS1, and the catalyzed reaction (dotted line) by TS2. A catalyst thus speeds up the reaction
by lowering the activation energy of both the forward reaction as well as the backward
reaction.
1.13.5 Hammond Postulate

According to the Hammond postulate, the transition state of a reaction resembles the
structure of the species (reactant or product) to which it is closer in energy. In exothermic
reactions, the transition state is closer in energy to the reactants. Therefore, the structure
of the transition state will more closely resemble the structure of the reactant than that
of the product. In endothermic reactions, the transition state is closer in energy to the
products. Therefore, the structure of the transition state will more closely resemble the
structure of the product.
That chlorination is less selective than bromination can be well understood by the help of
Hammond postulate. The first propagation step, i.e., abstraction of a hydrogen atom by a
bromine or chlorine radical, is the rate-determining step, in bromination or chlorination.
This step is endothermic for bromination but exothermic for chlorination. According to
the Hammond postulate, the product-like transition state for endothermic bromination
has a great deal of radical character on the carbon atom. Therefore, the energy difference
of the two transition states forming the 1° and 2° radicals reflects most of the energy
difference of the radical products. On the other hand, the reactant-like transition state for
exothermic chlorination has a little radical character on the carbon atom. So, the energy
difference of the two transition states forming 1° and 2° radicals reflects only a small
part (about one-third) of the energy difference of the radical products. Consequently, the
transition states leading to the formation of 1° or 2° radical for bromination have a larger
energy difference than those for chlorination, even though the difference in energy of the
products is the same in both reactions. Chlorination of propane is, therefore, less selective
(product ratio 60:40) than bromination of propane (product ratio 97:3).
hn
CH3CH2CH3 + Br2 ææææ
125° C
Æ CH3CH2CH2 Br + CH3CHBrCH3 + HBr
Propane 1-Bromopropane (3%) 2-Bromopropane (97%)
hn
CH3CH2CH3 + Cl2 ææææ
125° C
Æ CH3CH2CH2Cl + CH3CHClCH3 + HCl
Propane 1-cholorpropane (40%) 2-chloropropane (60%)
1.13.6 Kinetic Control versus Thermodynamic Control of a

Chemical Reaction
When a reaction produces more than one product, the product that is formed most readily
is called the kinetically controlled product or simply kinetic product. Reactions that produce
the kinetic product as the major product are said to be kinetically controlled reactions.
The most stable product is called the thermodynamically controlled product or simply
thermodynamic product. Reactions that lead to the formation of the thermodynamic
product as the major product are said to be thermodynamically controlled reactions.
Example: Electrophilic addition of HBr to 1,3-butadiene is an example of a reaction
in which the kinetic product and the thermodynamic product are different. When the
reaction is carried out at low temperature (–80°C), the major product (80%) is formed by
1,2-addition and the minor product (20%) is formed by 1,4-addition.
-80∞ C
CH2 == CH — CH == CH2 + H Br ææææ Æ CH3CH BrCH == CH2 + CH3CH == CH CH2 Br
1,3-Butadiene 1,2 addition product 1,4-addition product
(80%) (20%)
At lower temperature, the relative amounts of the products are determined by the relative
rates at which the two additions occur. The 1,2-addition occurs at a faster rate (TS is stable
and Ea is lower because the +ve charge is on a secondary allylic carbocation) and so, the
1,2-addition product is the major product. In fact, at lower temperature, there is enough
energy for the reactants to overcome the energy barrier of the first step of the reaction
to form the carbocation, and enough energy for the carbocation to form the two addition
products. However, there is not enough energy for the reverse reaction to take place.
When the reaction is carried out at higher temperature (40°C), the major product (80%) is
formed by 1,4-addtion and the minor product (20%) is formed by 1,2-addition.
40∞ C
CH2 == CH — CH == CH2 + H Br æææ Æ CH3CH BrCH == CH2 + CH3CH == CH CH2 Br
1,3-Butadiene 1,2-addition product 1,4-addition product
(20%) (80%)
At higher temperature, the relative amounts of the products are determined by the position
of an equilibrium. The 1,4–addition product is thermodynamically more stable (stabilized
more by hyperconjugation), so it is the major product. In fact, at higher temperature
(40°C), there is enough energy for the reactants to get over the energy barrier to form the
products and enough energy for one or more of the products to go back to the intermediate
carbocation. Since the product can interconvert, the relative amounts of the two products at
equilibrium will depend on their relative stabilities. The thermodynamic product reverses
less readily because it has higher energy barrier (larger Ea) to the common intermediate.
As a result, it gradually becomes the predominant product in the product mixture. The
reaction will be more clearly understood if we examine the energy diagram given below.
1. How can the thermodynamic stability of the product compared with the
reactants be predicted from the DG° value of the reaction?
Solution If DG° of the reaction is negative, the product is thermodynamically more stable
compared with the reactant and if DG° is positive, the product is thermodynamically less
stable compared with the reactant.
2. Explain why the standard free energy change (DG°) of a reaction can
often be approximated by the change in bonding energy, i.e., the enthalpy
change (DH°) of the reaction.
Solution If a reaction involves only a small change in entropy and it is not carried out
at high temperature, the TDS° term is small compared to the enthalpy term (DH°) and it
can be neglected. For this reason, the standard free energy change (DG°) of a reaction can
often be approximated by the change in bonding energy, i.e., the enthalpy change (DH°) of
the reaction.
3. Draw the energy diagram of (i) an exothermic reaction that takes place
at room temperature; (ii) an exothermic reaction that cannot take place
without adding energy above that provided by the external thermal
conditions; (iii) an endothermic reaction having small experimental
energy of activation and (iv) an endothermic reaction having large
experimental energy of activation.
Solution
(a)
(b)
(c)
(d)
4. Mention the factors on which the rate of reaction depends.

Solution The rate of a reaction depends on the following three factors:
(i) The number of collisions taking place between the reacting molecules in a given
period of time. The greater the number collisions, the faster the reaction.
(ii) The fraction of the collisions that occurs with sufficient energy to get the reactant
molecules over the energy ‘hill’ or energy barrier. If Ea is small, more collisions will
lead to reaction than if Ea is large.
(iii) The fraction of collisions that occur with proper orientation. For example, in the
reaction of propene with HBr, the reaction will take place only if the molecules
collide with hydrogen of HBr approaching the p bond of propene. If collision occurs
with hydrogen approaching the methyl group of propene (CH3CH== CH2), no
reaction will take place regardless of the energy of the collision.
5. Explain why the rate of a reaction increases on increasing the concen-
tration of the reactants and on increasing the temperature at which the
reaction is carried out.
Solution Increasing the concentration of the reactants increases the reaction rate
because it increases the number of collisions that occur in a given period of time. The rate
of reaction also increases on increasing the reaction temperature because an increased
temperature increases the frequency of collisions and the number of collisions that have
sufficient energy to get the reactant molecules over the energy barrier.
6. What do you mean by the kinetic stability of a compound ?
Solution A compound is said to be kinetically stable if it is not very reactive, i.e., if DG=|
is large.
7. Write down the Arrhenius equation and from this predict how the
experimental activation energy (Ea) and temperature affect the rate
constant of a reaction.
Solution The Arrhenius equation: k = A. e– Ea /RT where k is the rate constant, Ea is the
experimental energy of activation, R is the gas constant (1.986 × 10–3 kcal mol–1 K–1 or
8.314 × 10–3 kJ mol–1 K–1), T is the absolute temperature (K) and A is the frequency factor.
The frequency factor accounts for the fraction of collisions that occur with the appropriate
orientation for the reaction to take place. The factor, e– Ea / RT , accounts for the fraction of
collisions having minimum energy (Ea) needed to react. Taking logarithms on both sides,
the Arrhenius equation becomes
ln k = ln A - Ea /RT
From this equation, it becomes clear that the rate constant decreases on increasing the
experimental activation energy (Ea) and increases on increasing the temperature of the
reaction.
8. What is called free energy of activation (DG=| )? How does it govern the rate
of a reaction?
Solution The free energy of activation (DG=| ) is the difference between the free energy of
the transition state and the free energy of the reactants.
According to the transition state theory, the reactants are in equilibrium with transition
state and the equilibrium constant K=| is related to DG=| by the equation:
DG=| = –2.303 RT log K=|
This theory further tells that all the transition states collapse into the products at the
same rate. Therefore, the rate constant of the reaction depends only on the position of
equilibrium between the reactants and the transition state, that is, the rate constant
depends on the value of the equilibrium constant, K=|. A lower value of DG=| indicates a
higher rate constant, i.e., a faster reaction. Conversely, a higher value of DG=| means a
slower reaction.
9. What are intermediates? How does transition states differ from interme-
diates?
Solution Multistep reactions proceed through transient chemical species which possess
some degree of stability and can be isolated and detected in many cases. These are known
as intermediates. Carbocations, carbanions, free radicals, arynes, carbenes, etc., are
examples of intermediates.
The transition state, on the other hand, are hypothetical arrangement of atoms possessing
a definite shape and charge distribution. Because of high energy and very short lifetime,
they can never be isolated or trapped.
10. The energy diagram of a reaction is as follows:
Answer the following questions by considering the energy diagram:

(a) Mention the number of steps involved in this reaction.
(b) Label DH°, Ea and transition states for each step and labels the DH∞ of
the reaction.
(c) At which point on the graph the reactive intermediate is located.
(d) Which is the rate-determining step of the reaction ?
(e) Predict whether the overall reaction is exothermic or endothermic.
Solution
(a) The reaction involves two steps.
(b)
(c) The two transition states are separated by an energy minimum, at which the
intermediate is located.
(d) Since Ea1 > Ea2 , the first step is slower than the second step and hence, it is
the rate-determining step of the reaction.
(e) Since DH° of the overall reaction is +ve, the overall reaction is endothermic.
11. Draw an energy diagram for the following reaction in which C is the most
stable and B is the least stable of the three species and the transition state
for the conversion of A Æ B is more stable than the transition state for the
conversion of B Æ C.
k1 k2
A B C
k -1 k -2
(a) How many intermediates are involved in this reaction ?

(b) How many transition states are there in the overall process?
(c) Identify the faster step in the forward and reverse directions.
(d) Which out of the four steps is the fastest one?
(e) Mention the rate-determining steps in the forward and reverse
processes.
Solution The energy diagram of the reaction is as follows:
TS2
TS1
Potential energy
Progress of the reaction

(a) The reaction involves only one intermediate (B).
(b) There are two transition states in the overall process.
(c) A Æ B is the faster step in the forward direction and B Æ A is the faster step in the
reverse direction.
(d) Out of the four steps B Æ A is the fastest one.
(e) B Æ C is the slower step, i.e., the rate-determining step in the forward process and
C Æ B is the rate-determining step in the reverse process.
12. Write a rate equation for each of the following reactions:
ii ii
(a) (CH 3 )2 CHBr + Et O : @ Æ CH 3 CH = CH 2 + ii Br:@ + EtOH
ii ii
≈ @
slow OH
(b) (CH 3 )3 CBr æææ Æ Br @ + (CH 3 )3 C æææÆ
fast
(CH 3 )3 COH
Solution
(a) rate = k [(CH3 )2 CHBr][EtO@ ] (b) rate = k [(CH3)3CBr]
13. In which two cases the entropy changes of reactions are very important?
Solution Entropy changes are important in the following two cases:
(i) When the number of molecules of reactant differs from the number of molecules
of product in the balanced chemical equation.
(ii) When an acyclic molecule undergoes cyclisation, or cyclic molecule is converted
to an acyclic one.
14. For which reaction DS∞ is expected to be more significant and why?
(a) A B + C or A + B C+D (b) A B or A + B C
Solution
(a) DS∞ is more significant for the first reaction because the number of particles
increases, i.e., disorder increases on going from starting material to products.
(b) DS∞ is more significant for the second reaction because the number of
particles decreases, i.e., disorder decreases, on going from starting material
to product.
15. What do you mean by each of the following values of Keq, DG∞, DH∞ and DS∞?
(a) Keq > 1 (b) DG∞ < 0 (c) DH∞ < 0 (d) DS∞ > 0
Solution
(a) Keq > 1 means that more products than reactants are present at equilibrium.
(b) DG∞ < 0 means that the free energy of the product is lower than the free
energy of the reactants.
(c) DH∞ < 0 means that bonds in the products are stronger than the bonds in the
reactants.
(d) DS∞ > 0 means that the products are more disordered than the reactants.
16. Predict whether the value of DS∞ for the dissociation of Cl2 is positive, i.e.,
favourable or negative, i.e., unfavourable. Give an idea about the sign of
the value DG° of this reaction.
hn
Cl 2 ææ Æ 2Cl.
æ
Solution Two isolated chlorine atoms have much more freedom of motion than a single
Cl2 molecule. Thus, the disorder of the system increases, and therefore, the change in
entropy is positive. However, the positive (unfavourable) value of DH° of this reaction is
much larger than the positive (favourable) value of TDS°. Since DG° = DH° – TDS°, the
value of DG° of this reaction is positive, i.e., the reaction is unfavourable.
17. Using bond dissociation energies predict whether the following reaction
is endothermic or exothermic.
[Bond dissociation energies: Cl — Cl: 58 kcal/mol; CH3 — H: + 104 kcal/mol;
CH3 — Cl: –84 kcal/mol and H — Cl: –103 kcal/mol]
Solution Bonds broken DH ∞ (per mole)
Cl — Cl + 58 kcal
CH 3 — H + 104 kcal
Total : + 162 kcal
Bonds formed DH ∞ (per mole)

H — Cl - 103 kcal
CH3 — Cl - 84 kcal
Total : - 187 kcal
DH∞ (The overall enthalpy change) = Sum of DH∞ of bonds broken + (–) sum of DH° of bonds formed
= + 162 + (–187) kcal/mol
= –25 kcal/mol
Since the value of DH∞ is negative, the reaction is an exothermic one.
18. The experimental energy of activation (Ea) of the following reaction is
+4 kcal/mol and the enthalpy change (DH∞) is +1 kcal/mol. Draw an energy
diagram for this reaction. What are the activation energy and heat of
reaction (DH∞) for the reverse reaction.
CH + Cl. Æ .CH + HCl
4 3
Solution The energy diagram of this reaction is as follows:
The activation energy for the reverse reaction is +3 kcal/mol and DH° for the reverse
reaction is –1 kcal/mole.
19. Explain why the tert-butyl cation is formed at a rate faster than isobutyl
cation when 2-methylpropene is subjected to react with HBr. Give an
energy diagram.
Solution
In between the two resulting carbocation tert-butyl cation (a tertiary or 3° carbocation) is

more stable than the isobutyl cation (a primary of 1° carbocation). Since the formation of a
carbocation is an endothermic reaction, the transition state resembles the carbocation and
it possesses a significant amount of positive charge on a carbon atom. The same factors
that stabilize the positively charged product, i.e., carbocation, will stabilize the partially
positive charged transition state. Therefore, the transition state leading to the formation
of the tert-butyl cation is more stable than the transition state leading to the formation
of isobutyl cation and so, the tert-butyl cation (lower Ea) is formed at a faster rate than
isobutyl cation (higher Ea).
1. Using the bond dissociation energies, calculate the DH° for the addition of HBr to
ethene.
[DH°: p-bond of ethane = 61 kcal/mol; H — Br = 87 kcal/mol, C — H = 101 kcal/mol
and C — Br = 69 kcal/mol]
Do you expect the reaction to exergonic or endergonic?
2. Draw an energy diagram for a reaction in which the product is both

thermodynamically and kinetically unstable and for a reaction in which the product
is thermodynamically unstable and kinetically stable.
3. The DG° for the conversion of ‘axial’ phenylcyclohexane to ‘equatorial’ phenylcy-
clohexane at 25°C is –2.9 kcal/mol. Calculate the percentage of phenylcyclohexane
molecules that have the phenyl substituent in the equatorial position.
4. State whether you expect the entropy change (DS°) to be negative, positive or
approximately zero for each of the following reactions:
(a) A Æ B + C (b) A + B Æ C (c) A + B Æ C + D
5. What is the value of DG° for a reaction where Keq = 1 and where Keq = 10? Assuming
DS° ª 0, what change in DH° is required to produce 100-fold increase in the Keq?
6. (a) Which Keq corresponds to a negative value of DG°, Keq = 100 or Keq = 0.01?
(b) Which Keq corresponds to a lower value of DG°, Keq = 10–5 or Keq = 10–3?
7. Considering the following values predict whether the reactant or product is lower
in energy?
(a) Keq = 102 (b) DG° = 3.0 kcal (c) Keq = 10–3 (d) DG° = –3.5 kcal
8. Neglecting entropy and considering each of the following values predict whether
the reactant or product is favoured at equilibrium:
(a) DH° = 30 kcal/mol, (b) DH° = –20 kcal/mol,
(c) DS° = 2 k cal/mol, (d) DS° = –2 kcal/mol.
9. Which value indicates a faster reaction:
(a) Ea = 15 kcal or Ea = 3 kcal;
(b) a reaction temperature of 5°C or a reaction temperature of 30°C;
(c) Keq = 10 or Keq = 150;
(d) DH° = –4 kcal/mol or DH° = 4 kcal/mol?
10. Consider the following energy diagram for the conversion of M Æ S and answer the
following questions:
P
N
Potential energy
O R
M Q

(a) Mention the number of steps present in the reaction?

(b) Which point on the graph corresponds to reactive intermediates?
(c) Which point on the graph corresponds to transition states?
(d) Label each step of the reaction and the overall reaction as endothermic or
exothermic.
11. A reaction will not take place at all if DH° is +ve and DS° is –ve — Why?
12. For a reaction DH° = – 14 kcal/mol and DS° = 0.01 kcal/K/mol, calculate DG° and
Keq at 20°C and at 100°C. How does DG° and Keq change as T increases?
13. Which of the following factors affect the rate of a reaction?
(a) DH°, (b) DG° (c) temperature,
(d) DS°, (e) Ea (f) concentration,
(g) Keq, (h) catalysts (i) k
14. In esterification reaction, a carboxylic acid is subjected to react with an alcohol
in the presence of sulphuric acid to form an ester with less of water. An example
of intermolecular esterification reaction and an example of intramolecular
esterification reaction are given below. Explain why Keq is different for these two
apparently similar reactions.
15. At 298K the enthalpy change, DH° for the ionization of trichloroacetic acid
(Cl3C COOH) is +6.3 kJ.mol–1 and the entropy change, DS°, is +0.0084 kJ mol–1K–1.
Calculate the pKa value of trichloroacetic acid.
[Hint:
DG∞ = DH ∞ - T DS°= 6.3 kJ mol -1 - (298 K)(0.0084 kJ mol -1K -1 ) = 3.8 kJ mol -1
– DG∞
– = log K eq = log K a = -pK a
2.303 RT
DG∞ 3.8 kJ mol -1
Therefore pK a = = = 0.66]
2.303 RT (2.303) (0.008314 kJ mol-1 K -1 ) (298K)
16. Label each product in the following reaction as a 1,2- or 1,4-product, and decide
which is the kinetically controlled product and which is the thermodynamically
controlled product.
H3C H3C H3C
+ HBr +
Br
Br
1.14 METHODS OF DETERMINING MECHANISMS OF REACTIONS

A mechanism is the actual process by which a reaction takes place. There is no specific way
to know how exactly a reaction proceeds. There are a number of commonly used methods for
determining mechanism. In most cases, one method is insufficient, and in most cases, the
problem is approached from several directions. The techniques for determining mechanisms
of reactions include: (i) identification of products (ii) stereochemical evidence, (iii) isotope
labelling, (iv) determination of the presence of intermediates along with spectroscopic, and
crystallographic analysis, (v) crossover experiments and (vi) kinetic evidence.
(1) Identification of Products
Any mechanism proposed for a reaction must account for all the products obtained and for
their relative proportions, including the by-products (if any). A proposed mechanism cannot
be correct if it fails to predict the products in approximately the observed proportions. For
example:
hn
(1) Any mechanism for chlorination of methane (CH4 + Cl2 ææ Æ CH3Cl + HCl),
æ
that fails to account for the formation of a small amount of ethane (CH3CH3) cannot
be correct. Since the following mechanism shows the formation of ethane, it is the
actual mechanism of chlorination.
(2) One of the oximes of 2-bromo-5-nitro acetophenone readily undergoes ring closure
to benzisoxazole, thus establishing its structure as I. The other isomer II does
not undergo ring closure because —OH and —Br are not close to each other.
However, it readily gives the Beckmann rearrangement product, the amide III. The
identification of III thus leads to the conclusion that the rearrangement involves a
trans migration of the substituted phenyl group.
CH3
O2N CH3
O2N C==
N OH
–
N
O
Br OH
I A benzisoxazole
Therefore, the mechanism of Beckmann rearrangement is as follows:
(2) Stereochemical Evidence

The stereochemical course of a reaction, which can be obtained from a knowledge of
the stereochemistry of the products, is often helpful in determining the mechanism of a
reaction. For example, bromination of trans-2-butene yields meso-2,3-dibrombutane and
not (±)-2, 3-dibromobutane.
The formation of optically inactive meso-product indicates that the two bromine atoms
add to the double bond from the opposite sides, i.e., it is a case of trans-addition. This
trans-mode of addition rules out the possibility of one-step (concerted) mechanism because
the atoms in a Br2 molecule are very close and unable to add simultaneously in an anti
fashion. Hence, the addition is a two-step process such as follows:
(3) Isotopic Labelling

The mechanism of a reaction can be ascertained by carrying out the reaction with
isotopically labelled compound and then identifying the position of the isotope in the
products. The mechanism of the following reaction, for example, can be ascertained by
isotopic labelling experiment.
Direct substitution of chlorine with an amide ion (NH 2@ ) appears to be an attractive

possibility, but experiments performed with 1-14C-chlorobenzene provided a sample of
aniline, about half of which was labelled in the 1-position, and remaining half in the
2-position. A two-step mechanism involving the formation of an intermediate known as
benzyne is consistent with the experimental observations.
The mechanism of Cannizzaro reaction, which involves transfer of hydrogen from one
molecule to another, has been investigated by the use of deuterated benzaldehyde
(PhCDO), instead of benzaldehyde.
O
|| @ +
50% NaOH
2PhCDO æææææÆ Ph — C — O Na + PhCD2OH
Since the deuterium is exclusively located on the carbon atom of benzyl alcohol bearing the
hydroxyl group, the possibility of exchange of hydrogen with the hydrogen atoms in the
solvent (H2O) is ruled out. Thus, there occurs a direct transfer of hydrogen (or deuterium)
between the carbonyl carbons of two aldehyde molecules. A plausible mechanism for
Cannizzaro reaction can thus be written as follows:
(4) Determination of the Presence of Intermediates

A lot of organic reactions proceed through the formation of one or more intermediates.
A knowledge of the structures of that intermediates (isolated, detected by spectroscopic
methods or trapped) gives useful information about the mechanistic course of a reaction.
(a) Isolation of intermediates: Sometimes an intermediate can be isolated from
a reaction mixture by stopping the reaction after a short time or by use of very
mild conditions. For example: (1) the formation of cyclobutanone from a ketene
and diazomethane (CH2N2) at low temperature has been known for a long time.
It has been demonstrated by 14C labelled diazomethane that cyclopropanone is
the most likely intermediate. Recently, direct evidence for its formation has been
provided by preparing cyclpropanone and then allowing it to react with CH2N2 to
yield cyclobutanone.
(2) In Hofmann rearrangement in which amides are converted into amines by Br2
NaOH
and alkali (RCO NH 2 + Br2 ææææ Æ RNH 2 ), three intermediates, e.g., an N-bro-
@
moamide (RCONHBr), its anion (RCO N Br) and an isocyanate (R — N == C == O),

have been isolated. When these intermediates are subjected to normal reaction
conditions, they lead to the formation of usual reaction products at a rate not slow-
er than the rate of the overall reaction. Therefore, the mechanistic pathway to be
sujected for this reaction is that which accounts for the formation of these interme-
diates. The mechanism of the reaction is as follows:
(b) Detection of intermediates: An intermediate, which cannot be isolated, can often

be detected by spectroscopic studies, such as IR, NMR, ESR or other spectra. The
detection of nitronium ion (NO≈ 2 ) by Raman spectra, for example, was regarded as
a strong evidence that this is an intermediate in the nitration of benzene.
(c) Trapping of intermediates: An intermediate can often be trapped by running
the reaction in the presence of a compound with which the intermediate reacts.
The compound obtained as a result of this is isolated and from a knowledge of its
structure the structure of the intermediate is predicted. For example, arynes react
with dienes to give adducts (a Diels-Alder reaction). In any reaction where benzyne
is a suspected intermediate, the addition of anthracene (which acts as a diene)
to the reaction mixture and the detection of triptycene (the Diels-Alder adduct)
indicate that the reaction involves benzyne as an intermediate.
(5) Crossover Experiments

Intramolecular rearrangements are those in which a group migrates from one position
to another without detaching itself completely from the molecule. On the other hand, the
rearrangements in which the group becomes completely free during its migration are
known as intermolecular rearrangements.
Whether a particular rearrangement is intra- or intermolecular can be known by crossover
experiments. In such experiments, the rearrangement is carried out with a mixture of
two structurally related reactants and the products are then identified. If the reactant
X—Y—Z is allowed to rearrange in the presence of a structurally related reactant A—Y—B
(where X is closely related to A and Z is related to B), a mixture of four or a mixture of two
rearrangement products is expected to be obtained.
Y—Z—X + Y—B—A
Normal products
[X—Y—Z + A—Y—B] + Y—Z—A + Y—B—X

Crossover products
Y—Z—X + Y—B—A
Normal products
It only Y—Z—X and Y—B—A (the normal intramolecular products) are obtained, the
process is intramolecular, but if Y—Z—A and Y—B—X (the crossover products) are
obtained along with the normal products, the process is intermolecular.
The crossover experiment can be used, for example, to establish that the pinacol–pinacolone
rearrangement is intramolecular and the Fries rearrangement is intermolecular. The
pinacol–pinacolone rearrangement involves conversion of a pinacol (a 1, 2-diol) to a
pinacolone in the presence of moderately concentrated H2SO4. For example:
≈
H
Me2 C —CMe2 ææææ H2SO4
Æ Me3C — C — Me + H2O
| | ||
OH OH O
When two structurally related pinacols are allowed to rearrange in the same solution no
crossover products are obtained. For example
This observation suggests that this rearrangement is an intramolecular process.

The Fries rearrangement involves conversion of a phenyl ester into a hydroxyl-ketone in
the presence of the Lewis acid AlCl3
COCH3 OH
COCH3
1. only AlCl3
2. H3O +
C2H5 C2H5
When this rearrangement is carried out with a mixture of two structurally related
phenyl esters, two crossover products along with two normal products are obtained. For
example:
Formation of crossover products indicates that this rearrangement is an intermolecular

one.
(6) Kinetic evidence

The order of a reaction (determined by kinetic methods) gives us an important information
about which reactants and how many of them take part in the rate-determining step of
the reaction. The benzidine rearrangement, for example, has been found to follow a third-
order kinetics (second-order in [H+] and first-order in [PhNHNHPh]).
From the rate law it can be suggested that the reaction involves rate-determining
protonation of hydrazobenzene followed by rapid rearrangement of the hydrazonium
cation to benzidine.
+ +
+ 2H + slow fast
NH—NH NH2—NH2 N2 N NH2
An indirect method is quite helpful in formulating the most plausible mechanism
of a reaction. In this process, the rate law of a reaction is determined experimentally.
Subsequently, various mechanisms are postulated intuitively for the reaction. The rate
equations are then derived for each of these individually postulated mechanisms. The
calculations, which may sometimes be quite complex, can be simplified by the steady-state
assumption. Any derived rate law that corresponds to the experimentally observed rate
law is then selected. The mechanism corresponding to the law is then suggested to be
the most plausible mechanism for the reactions. For example, the mechanism for the
hydrolysis of tert-butyl chloride in aqueous acetone can be established by this procedure.
H2O
(CH 3 )3CCl ææææ
acetone
Æ (CH 3 )3 COH + HCl
The reaction is found to be first-order and obeys the following rate law:
d[(CH 3CCl]
- = k[(CH 3 )3CCl]
dt
The rate law suggests that the rate-determining step of the reaction involves only the
concentration of tert-butyl chloride. The reaction sequence can then be written as
Step 1:
Step 2:
If we apply the steady-state hypothesis assuming the tert-butyl carbocation in a steady-
state concentration, we can arrive at a rate law similar to the experimental rate law
written above.
Hence, the proposed mechanism involving rate-determining cleavage of the C — Cl bond
is the most plausible one.
1.14.1 Kinetic Isotope Effects

Changes in reaction rates brought about by isotopic substitution are known as kinetic
isotope effects. They provide important information about the mechanism of the reaction.
(1) Primary kinetic isotope effect: A large difference in the rate of a reaction is observed
when one of the atoms of a bond that breaks in the rate-determining step is replaced by
its isotope (the rate of a reaction for a compound containing a heavy isotope is slower than
that of the compound with a lighter isotope). This is what is called primary kinetic isotope
effect. If a C — H bond cleaves in the rate-determining step of a reaction, substitution of
D for H results in decrease in the reaction rate and that is due to the fact that a C — D
bond is stronger than a C — H bond. For example, the rate of dehydrobromination of

2-bromopropane (CH3CHBrCH3) on treatment with NaOEt in EtOH has been found to
become slow when the hydrogens in the methyl group of the alkyl halide are replaced
by deuteriums. A large kinetic isotope effect (kH/kD = 6.7) is observed. This observation,
therefore, suggests that the rate-determining step of the reaction involves cleavage of the
C — H (or C — D) bond. The reaction thus proceeds through the E2 mechanistic pathway
and not through the E1 mechanistic pathway.
E2 pathway:
E1 pathway:
(2) Secondary kinetic isotope effect: There are many reactions in which substitution
of hydrogen by deuterium in a bond which is not even broken in the rate-determining step
may still bring about a measurable change in the reaction rate by isotopic substitution.
These are known as secondary kinetic isotope effects. The rate of solvolysis of tert-butyl
chloride in 60% aqueous ethanol is found to be faster than the rate of solvolysis of its
nonadeuterio analog, i.e., (CD3)3CCl. The value of kH/kD was found to be 2.32.
C H OH/ H O
(CH3 )3Cl ææææææ
2 5
kH
2
Æ (CH3 )3COC2 H5 + (CH3 )3 COH
tert-Butyl chloride
C H OH/ H O
(CD3 )3CCl ææææææ
2 5
kD
2
Æ (CD3 )3 COC2H5 + (CD3 )3 COH
tert-Butyl chloride-D9
This secondary kinetic isotope effect indicates that the reaction proceeds through the rate-
determining formation of a 3° carbocation which is differently stabilized by hyperconjugation
involving — CH3 and — CD3 groups (because of greater strength, the C— D bond is not
a good hyperconjugative participant when compared to C—H). The reaction, therefore,
proceeds through SN1 pathway.
SN1 pathway:
1. a-Methylallyl chloride and crotyl chloride yield a mixture of ethyl

a-methylallyl ether and ethyl crotyl ether when treated with ethanolic
sodium ethoxide under SN1 conditions. Predict the mechanism of these
two reactions from the knowledge of the products obtained.
Solution Since each of the isomeric allylic chlorides gives the same mixture of isomeric
ethers, the two substitution reactions proceed through a common intermediate carbocation
which is a hybrid of two resonance structures.
2. How would you establish by stereochemical studies that the acid-catalyzed

hydrolysis of the following ester involves alkyl-oxygen bond cleavage?
Solution This dextrorotatory ester of acetic acid undergoes hydrolysis to give racemic
alcohol.
Since racemization occurs, therefore, an intermediate 3° carbocation must have been

formed by alkyl-oxygen bond cleavage. This may be shown as follows:
3. The base-promoted hydrolysis of an ester involves acyl-oxygen bond

cleavage:
Establish this mechanistic course by a labelling experiment.

Solution When ethyl propanoate labelled with 18O in the ethyl-type oxygen of the ester
is subjected to hydrolysis with aqueous NaOH, all the 18O shows up in the ethanol that is
produced. None of the 18O appears in the propanoate ion (CH 3CH 2COO@ )
This labelling result is completely consistent with the mechanism given above, i.e., the
hydrolysis involves acyl-oxygen bond cleavage.
4. How can you establish that the alkaline hydrolysis of chloroform involves
dichlorocarbene (:CCl2) as the reactive intermediate?
Solution That the alkaline hydrolysis of choloroform involves dichlorocarbene (:CCl2) as

the reactive intermediate can be proved by trapping this intermediate.
When cis-2-butene is added to the reaction mixture, a cyclopropane derivative is
obtained. This observation suggests that the reaction proceeds through the formation of
dichlorocarbene (:CCl2).
5. 3-Pentanol undergoes chromic acid oxidation to yield 3-pentanone. Give

the mechanism of the reaction.
OH O
| ||
Na2Cr2O7 /H2SO4
CH 3CH 2 CH CH 2 CH 3 æææææææ Æ CH 3CH 2 — C — CH 2CH 3
3- Pentanol 3- Pentanone
Which step of this reaction is rate-determining? Give evidence in favour

of your answer.
Solution When sodium dichromate (Na2Cr2O7) is dissolved in a mixture of sulphuric acid
and water, chromic acid (H2CrO4) is obtained.
O
|| ≈
Na 2Cr2O7 + 2H 2SO4 + H 2O Æ 2HO — Cr — OH + 2Na + 2HSO@4
||
O
The oxidation of 3-pentanol by chromic acid proceeds through the steps as follows:
Step 1:
Step 2:
The chromate ester undergoes E2 elimination to give 3-pentanone. Water acts as a base
and chromium readily accepts an electron pair and is thus reduced from Cr (VI) to Cr (IV).
The decomposition of the chromate ester (step 2) is the rate-determining step of the reaction.
When the rate of oxidation of 3-pentanol to 3-pentanone is compared with the rate
of oxidation of its deuterated derivative (Et2CDOH), an isotope effect is observed
(kH/kD = 6.6).
This observation suggests that step 2, i.e., the decomposition of the chromate ester (an
E2 elimination reaction) involving C—H bond cleavage is the rate-determining or rate-
limiting step of the reaction.
6. What is solvent isotope effect? Mention the factors responsible for such
effect.
Solution Reaction rates often change when the solvent used is isotopically labelled, i.e.,
when the solvent is changed from H2O to D2O or from ROH to ROD. This kind of isotope
effect is called solvent isotope effect.
This isotope effect may be due to any of the following three factors or a combination of all
of them:
(1) If the solvent is a reactant and an O — H bond of the solvent is cleaved in the
rate-determining step, there will be a primary isotope effect. There may also be a
secondary effect (caused by the O — D bonds that are not breaking) if the molecules
involved are D2O or D3O≈ .
(2) Labelling of substrate molecules may take place by rapid hydrogen exchange, and
then the resulting labelled molecule may become cleaved in the rate-determining
step.
(3) The nature and extent of solvent–solute interactions may be different in the
deuterated and nondeuterated solvents. This may change the energy of the
transition state, and hence the activation energy (Ea). These are secondary isotope
effects.
7. Base-promoted hydrolysis of methyl mesitoate occurs through attack on
the alcohol carbon instead of the acyl carbon:
O O
Me Me
C C
O—CH3 :O + CH3OH
–
Me Me + :OH Me Me
Suggest an experiment with labelled compounds that would confirm this
mode of attack.
Solution When the above reaction is carried out in weakly alkaline solution in water
enriched with 18O, it resulted in the formation of methanol containing 18O label.
This observation, therefore, confirms the given mode of attack.

Solution The mechanism of the reaction in which phenoxide group acts as a neighbouring
group explains the scrambling of 14C label in the products.
9. Account for the following differences in the primary kinetic isotope effect:
kH/kD ª 7; kH/kT ª 10 – 20, where T is tritium.
Solution Isotope effects are due solely to the mass differences, which are manifested in
the vibrational frequencies of the C — H, C—D, and C — T bonds. The greater the mass of
the isotope whose bond is broken in the rate-determining step, the greater the activation
energy of the bond breaking and so, the greater is the primary kinetic isotope effect. Since
the mass of T is greater than the mass of D, therefore, kH/kT > kH/kD.
1. Salts of aliphatic or aromatic carboxylic acids can be converted to the corresponding

nitriles by heating with BrCN.
250 – 300∞C
RCOO@ + Br CN æææææ Æ RCN + CO2 + Br @
By isotopic labelling prove that the CN group in the product does not come from
the CN in BrCN.
* * 14
[Hint: R COO@ + Br CN ææÆ R CN + CO2 + Br @ (* = C)
2. Using water enriched in 18O how will you prove that the hydration equilibrium
3.
( C == O + H2O C
OH
OH ( is operative in the case of acetone?
In the oxidation of Ph2CHOH and Ph2CDOH by alkaline permanganate, the

kH/kD ratio was found to be 6.6. From this kinetic isotope effect, suggest an acceptable
mechanism for this reaction.
4. How the general mechanism of electrophilic aromatic substitution can be estab-
lished by the kinetic isotope effect?
5. The relative rates of chromic acid oxidation of isotopically labelled isopropyl alcohol
are as follows:
relative rates: CH3CHOHCH3 CH3CHODCH3 CH3CDOHCH3 D3CCHOHCD3
1.0 1.0 0.16 1.0
Comment on how these observed relative rates relate to the mechanism of oxidation.
18
6. Oxidation of Ph2C = 18O by m-chloroperbenzoic acid yields only Ph COO Ph , i.e.,
there is no scrambling of the 18O label in the product ester.
18 18
m-ClC H COOH
Ph2C == O æææææææ
6 4
Æ Ph COOPh
Suggest a mechanism for this reaction.
7. Where would you expect to find the labelled oxygen if you carry out an acid-
catalyzed hydrolysis of methyl benzoate in 18O-labelled water?
[Hint: The labelled oxygen atom should appear in the carboxyl group of the acid.]
8. Predict the product and give the mechanism of the following reaction:
[Hint: The oxirane ring undergoes SN2 attack at the less crowded labelled carbon
to give an alkoxide. The alkoxide then undergoes intramolecular SN2 displacement
to give the product.
9. When chlorobenzene is treated with sodamide in liquid ammonia in the presence

of furan, a 1,4-addtion product (I) is obtained. Explain this observation
10. Explain the results of the following crossover experiment:

2
CHAPTER
PRINCIPLES OF
STEREOCHEMISTRY
Chapter Outline
Introduction 2.5 Configuration and Configurational
2.1 Projection Formulas of Stereoisomers Nomenclature
2.1.1 Flying-Wedge Projection Formula 2.5.1 D, L-System of Configurational
Designation
2.1.2 Fischer Projection Formula
2.5.2 Specification of Configuration: The
2.1.3 Sawhorse Projection Formula
R, S-System
2.1.4 Newman Projection Formula
2.5.3 Erythro and Threo Nomenclature
2.1.5 Interconversion of Projection of Compounds with Two Adjacent
Formulas Chiral Centres
2.2 Symmetry Elements 2.5.4 The E-Z System of Designating
2.2.1 Simple Axis of Symmetry or Alkene Diastereoisomers
Rotational Axis of Symmetry (Cn) 2.5.5 R, S and E, Z Assignment in
2.2.2 Plane of Symmetry (s) the Same Molecule (Geometric
2.2.3 Centre of Symmetry (i) Enantiomerism)
2.2.4 Alternating Axis of Symmetry (Sn) 2.5.6 Syn-anti Nomenclature for Aldols
2.2.5 Symmetric, Asymmetric and 2.5.7 Number of Stereoisomers for
Dissymmetric Molecules Compounds with Chiral Centres
2.5.8 Chirotopic and Achirotopic Atom in
2.3 Isomerism: Constitutional Isomers
a Molecule
and Stereoisomers
2.5.9 Prostereoisomerism and Topicity
2.3.1 Constitutional Isomers
2.3.2 Stereoisomers 2.6 Optical Activity of Chiral Compounds
2.3.3 Enantiomers and Diastereoisomers 2.6.1 Plane-polarized Light
2.6.2 Optical activity
2.4 Molecular Chirality
2.6.3 Polarimeter
2.4.1 Chiral and Achiral Molecules
2.6.4 Specific Rotation
2.4.2 Source of Chirality
2.6.5 The Necessary and Sufficient
2.4.3 Stereocentre or Stereogenic Centre
Condition (the ultimate criterion)
2.4.4 Meso Compounds for Optical Activity
2.6.6 Racemic Modification 2.7.3 Conformations of Butane

2.6.7 Enantiomeric Excess(ee) or Optical (CH3CH2CH2CH3)
Purity (op) 2.7.4 Conformations of Chloroethane
2.6.8 Racemization (CH3CH2Cl)
2.6.9 Resolution of Racemic Modification 2.7.5 Conformations of
1,2-dichloroethane (ClCH2CH2Cl)
2.7 Conformation of Acyclic Organic
Molecules 2.7.6 Conformations of Some Typical
Acyclic Molecules
2.7.1 Conformations of Ethane
(CH3—CH3) 2.7.7 Invertomerism
2.7.2 Conformations of Propane
(CH3CH2CH3)
INTRODUCTION
The nature of the functional groups does not solely determine the properties of organic
compounds. The difference in orientation of the atoms or groups in space also responsible
in determining molecular properties. If molecules having the same molecular formula are
superimposable, then these are called homomers, and those that are not superimposable
are called isomers. Isomers having different atomic connectivities are called constitutional
isomers, and isomers having same atomic connectivities, but they differ in orientation of their
atoms in space are called stereoisomers. This phenomenon is known as stereoisomerism.
The part of chemistry that deals with the properties of stereoisomers and the chemical
effects of stereoisomerism is called stereochemistry. In fact, stereochemistry is the study
of the three-dimensional structure of molecules. It is not possible to understand organic
chemistry, biochemistry or biology without the knowledge of stereochemistry. Biological
systems are very much selective, and they often discriminate between molecules with very
small stereochemical differences.
Difference in spatial orientation seems apparently to be unimportant, but stereoisomers
often process remarkably different physical, chemical and biological properties. For
example, the cis and trans-isomers of butenedioic acid are a special type of stereoisomer
called cis- trans or geometric isomers. Both compounds have the same structure (HOOC—
CH==CH—COOH), but they differ in how these atoms and groups are arranged in space.
The cis-isomer in which the two –COOH groups are on the same side of the double bond is
called maleic acid, and the trans-isomer in which the two –COOH groups are in opposite
sides of the double bond is called fumaric acid. Fumaric acid is an essential metabolic
intermediate in both animals and plants, whereas maleic acid is toxic and irritating to
tissues.
Principles of Stereochemistry 2.3
O O
H C cis-isomer H C
C OH C OH trans-isomer
C OH HO C
H C stereoisomer C H
O O
Maleic acid Fumaric acid
(mp 138°C, a toxic and (mp 287°C, an essential
irritating substance) metabolite)
The discovery of stereochemistry is, in fact, one of the most important breakthroughs in
the structural theory or organic chemistry. In this chapter, we will study the principles
of stereochemistry and in the following chapters, we will see how stereochemistry plays a
major role in the properties and reactions of organic compounds. Stereochemistry is now
an overgrowing subject that helps one to understand and explain the chemical, physical
and biological properties of molecules. However, the subject is so vast that it is not possible
to cover it in one or two chapters. An attempt has been made in this chapter to discuss the
basic principles of organic stereochemistry.
2.1 PROJECTION FORMULAS OF STEREOISOMERS

It is inconvenient to represent three-dimensional (3D) molecules in two-dimensional (2D)
surface of paper or blackboard. So several 2D-formulas have been developed to represent
the molecules having 3D-structures and these are known as projection formulas. The
important projection formulas that are used to represent organic molecules are (i) Flying-
wedge projection formula, (ii) Fischer projection formula, (iii) Sawhorse projection formula,
and (iv) Newman projection formula.
2.1.1 Flying-Wedge Projection Formula

This is a three-dimensional representation of a three-dimensional molecule on a two-
dimensional surface and is generally used for compounds containing chiral carbon (often
marked * ), i.e. a carbon which is attached to four different substituents, such as in alanine,
*
CH3 CH(NH2)CO2H. In this representation, the ordinary lines represent bonds in the
plane of the paper, a solid wedge ( ) represents a bond above the plane of the paper and
a hashed wedge ( ) represents a bond below the plane of the paper. The Flying-wedge
projection formula of (R)-alanine, for example, can be shown as follows:
bonds in the CO2H bond below the
plane of the paper plane of the paper
C
H3C NH2
H bond above the
plane of the paper
(R)-Alanine
Compounds containing two or more chiral carbons may be expressed similarly. The
following figure represents the Flying-wedge formula (2S, 3S)-tartaric acid.
2.1.2 Fischer Projection Formula

Although the Flying-wedge formula represents the actual three-dimensional structure of
a molecule, it is not convenient to draw this formula especially for a molecule containing
a number of chirality centres. In that case, three-dimensional chiral molecules are
represented with two-dimensional formulas which are called Fischer projection formulas.
In this representation, the chiral carbon lies in the plane of the paper and the four bonds
are shown by two vertical lines representing bonds projecting behind the plane of the
paper (i.e., pointing away from the viewer) and two horizontal lines representing bonds
projecting above the plane of the paper (i.e. pointing towards the viewer). The point of
intersection of these lines represents the chiral carbon. The Fischer projection, therefore,
looks like a cross. To represent the 3D-structure of a molecule, for example, (S)-alanine,
in a Fischer projection, the asymmetric or chiral carbon is viewed in such a way that the
two of the bonds to the carbon are vertical and pointing away from the viewer, and two
are horizontal and pointing towards the viewer. When this view is projected on the plane
of the paper, the Fischer projection of the molecule is obtained.
S S
Fischer projection is very much useful for representing molecules containing two or more
chiral carbons that are the part of a carbon chain. In such cases, formula for Fischer
projection of molecules is written with the main carbon chain extending from top to bottom
(usually according to IUPAC numbering), i.e., along the vertical line with all groups
eclipsed. For example, the Fischer projection formula for (2R, 3R)-3,4-dichlorohexane may
be written as follows:
CH2CH3 CH2CH3
H
CH3CH2 Cl Cl H
C Cl C H
90° rotation
H C Cl H Cl
C Cl
CH3CH2
H CH2CH3 CH2CH3
3D-representation of Fischer
(2R,3R)-3,4-dichlorohexane projection of
(2R, 3R)-3,4-
dichlorohexane
A 3D-structure can be written in Fischer projection directly as follows:

hashed wedge to right
horizontal line
COOH upper full line to upper

COOH
vertical line
C H
3D to Fischer H2 N H
H3C NH2
CH3 Fischer
3D- projection
representation of (S)-alanine
of (S)-alanine solid wedge to left
horizontal line
lower full line to lower

vertical line
The 3D-molecule is to be viewed between the solid and the hashed wedges. Then the solid
wedge bond (on the left hand side of the viewer) is written as the left horizontal line in
the Fischer projection, the hashed wedge bond (on the right hand side of the viewer) is
written as the right horizontal line, the upper full line bond as the upper vertical line and
the lower full line bond as the lower vertical line.
Similarly, if the solid and hashed wedges are on the left side of the 3D-figure, then the
corresponding Fischer projection can be written as follows:
CO2H CO2H
C 3D to Fischer
H2N H 2N H
CH3
H
CH3
Fischer projection
3D-representation of (S)-alanine Fischer projection
of (S)-alanine of (S)-alanine
General rules for exchanging ligands or rotation of Fischer structures

(a) An even number of interchanges result in no change in the configuration of a
molecule, i.e., an identical structure is obtained.
2 3
Allowed
(i) 2 (exchanging ligands
across the horizontal bonds
3 and the vertical bonds) 2
R R
(ii)
(b) An odd number of interchanges convert the molecule into its enantiomers (non
superimposable mirror image). Therefore, this operation is not allowed. For
example, (R)-lactic acid becomes its enantiomer (S)-lactic acid when only one
exchange is made.
2 2
Forbidden
(i) (interchanging ligands
across the horizontal bonds)
3 3
(ii)
(iii)
(c) 180° horizontal or vertical rotation of a Fischer formula outside of the plane of the
paper changes the configuration of the molecule, i.e. converts it into its enantiomer.
Therefore, this operation is not allowed. For example:
(i)
(ii)
R S
(d) 90° or 270° rotation of a Fischer formula in the plane of the paper is not allowed
because this operation does not obey the convension ‘horizontal lines forward,
vertical lines back’ and hence, the configuration of the molecule is changed, i.e.,
the molecule is converted into its enantiomer. For example:
not identical
but enantiomers
CO2H OH
Forbidden
H OH (90° rotation in the
HO2C CH3
plane of the paper)
CH3 H
( )-Lactic acid ( )-Lactic acid
(e) 180° rotation of a Fischer projection formula in the plane of the paper is permissible
because this operation follows the convension ‘horizontal lines forward, vertical
lines back’ and hence, there occurs no change in configuration, i.e., an identical
molecule is obtained.
identical
CO2H CH3
Allowed
(i) H Br (180° rotation in the
Br H
plane of the paper)
CH3 CO2H
(R)-2-Bromopropanoic (R)-2-Bromopropanoic
acid acid
Cl C2H5
Allowed
H C2H5 (180° rotation in
Cl H
(ii) the plane of the paper)
H Cl H5C2 H
C2H5 Cl
(3R, 4R)-3, 4-Dichlorohexane (3R, 4R)-3, 4-Dichlorohexane
(f) Clockwise or counterclockwise rotation of any three groups is allowed because it

leads to the same molecule.
3 3
Allowed Allowed
(clockwise rotation (clockwise rotation 3
of F, Cl and Br) of CH3, F, Cl)
(i)
A A A
OH C2H5 C2H5
Allowed Allowed
(ii) C2H5 H H OH H OH
(clockwise rotation (anticlockwise rotation
of C2H5, OH and H) of Br, H and C2H5)
H Br H Br Br C2H5
C2H5 C2H5 H
(3S, 4R)-4-Bromo-
3-hexanol
2.1.3 Sawhorse Projection Formula

When it is required to know the spatial relation between two adjacent C-atoms, both
chiral or achiral, Sawhorse representation is quite convenient. In this projection, the
bond between the two key C-atoms is joined by a diagonal line (slightly elongated) in
the plane of the paper. The remaining bonds are shown by smaller lines projected above
and below the plane of the paper, and is a free rotation around the diagonal line either
1 2 3 4
clockwise or anticlockwise. For example, 2,3-dibromobutane (CH3 CHBr CHBr CH3 ) can
be represented by the following Sawhorse projections when C-2 and C-3 carbon atoms are
taken as the two key carbon atoms.
180° rotation of the front
carbon around the diagonal
line
Br
H
back carbon CH Br H
Br CH
H
CH
front carbon CH H Br
Configuration
I remains II
(2R, 3S)-2, 3-Dibromo- unchanged (2R, 3S)-2, 3-Dibromo-
butane butane
(eclipsed conformation) (staggered conformation)
In conformation I, two Br and

two H atoms are above the
plane of the paper and two
–CH groups are below the
plane of the paper. In II, one
of each groups above the plane
and one of each group below
the plane.
Both I and II represent two conformations of (2R, 3S)-2,3-dibromo-butane. The conformation

II in which the groups attached to C-2 and C-3 are oriented as far away from each other as
possible is known as the staggered conformation and the conformation I in which the groups
attached to C-2 and C-3 are nearest is known as the eclipsed conformation. Transformation
of one Sawhorse projection into another can be effected simply by rotating one of the two
key carbon atoms around the diagonal line either clockwise or counterclockwise. However,
the configuration of that carbon (if chiral) remains unchanged. It is to be noted that the
conformation with the dihedral angle (q) = 0° is called the eclipsed conformation and the
conformation with the dihedral angle (q) = 60° is called the staggered conformation.
2.1.4 Newman Projection Formula

This is also a perspective formula which is drawn from Sawhorse projection formula by
viewing the molecule along the bond joining the two key C-atoms of the molecule, both
chiral or achiral. The key C-atoms are represented as superimposed circles. Only one circle
is drawn the centre of which represents the front C-atom, and the circumference of the
circle represents the rear C-atom. The three s-bonds of the front carbon are represented
by three small lines drawn from the centre of the circle. The angle between any two bonds
is 120°. The three s-bonds of the back carbon are drawn from the circumference of the
circle. The Newman projection of n-butane, for example, can be shown as follows:
n n
n n
2.1.5 Interconversion of Projection Formulas

(a) (i) Fischer projection to Flying wedge projection This transformation may be carried
out as follows:
The above figure demonstrates the conversion of the Fischer projection of (R)-mandelic
acid to its Flying-wedge projections (A and B). This conversion involves placing the vertical
bonds in the Fischer projection in the plane of the paper and the horizontal bonds above
and below the plane of the paper. If the lower vertical bond in Fischer projection is bent on
the right side with respect to the vertical line (as in projection A), the group on the right
side horizontal bond in Fischer projection is to be placed above the plane of the paper
(to be represented by solid wedge), and the group on the left side is to be placed below
the plane of the paper (to be represented by hashed wedge). If the lower vertical bond in
Fischer projection is bent on the left side with respect to the vertical line (as in projection
B), the group on the left horizontal bond in the Fischer projection is to be placed above the
plane of the paper and the group on the right horizontal bond is to be placed below the
plane of the paper.
(ii) Flying-wedge projection to Fischer projection The reverse method is to be followed to

convert a Flying-wedge representation to Fischer projection. For example:
bent on the left
with respect to COOH
the vertical line COOH
H N H
C H
H C NH CH
vertical line
Fischer projection
Flying-wedge projection of ( )-alanine
of ( )-alanine
In (S)-alanine molecules, the —COOH and —CH3 groups are in the plane of the paper.
Since the —CH3 group is bent on the left side with respect to the vertical line, the
—NH2 group, which is above the plane in the Flying-wedge structure, is placed on the left
horizontal bond in the Fischer projection and the H atom, which is below the plane, is put
on the right horizontal bond.
(b) (i) Fischer projection to Sawhorse projection This conversion in a molecule, for
example in (2R, 3R)-2,3-dibromobutanoic acid, may be shown as follows:
R R
The Fischer projection is to be transformed into that eclipsed form of Sawhorse projection
in which the horizontal groups of the back and front carbon remain above the plane of
the paper. In this conversion the last chirality centre (C-3 in this example) in the Fischer
projection (counting from the top) is considered as the front carbon in sawhorse projection
(ii) Sawhorse projection to Fischer projection For this conversion, the Sawhorse projection
of the molecule is to be first transformed into that eclipsed conformation in which two
groups of the front carbon and two groups of the back carbon remain above the plane of the
paper. The groups above the plane occupy the horizontal position (front carbon becomes
the last chiral centre) and the groups below the plane become the two vertical groups. The
Sawhorse projection of meso-2,3-dichlorobutane, for example, can be converted into the
Fischer projection as follows:
clockwise
60° rotation of 3
3 front carbon 2 2
3
3 3 3 3
3
(c) (i) Sawhorse projection to Newman projection The Sawhorse projection (staggered
and eclipsed form) of (2S, 3R)-3-bromobutan-2-ol, for example, may be converted into the
Newman projection (staggered and eclipsed form) as follows. The front and back carbon
in Sawhorse projection becomes the front and back carbon, respectively, in the Newman
projection. Groups attached to each key carbon are placed accordingly.
S R
(ii) Newmann projection to Sawhorse projection The reverse method is to be followed to

convert a Newman projection to a Sawhorse projection. The bond between front and back
carbons is drawn diagonally and the groups attached to them are placed as follows:
Br H
Br H
Cl H
CH3
Cl H
CH3 CH3
CH3
Newman projection Sawhorse projection
of (2 , 3 )-2-bromo-3- (eclipsed)
chlorobutane
(eclipsed)
CH3 Br H
Br H CH3
Cl
H Cl
H Cl
Sawhorse projection
CH3 (staggered)
(Staggered)
(d) Conversion of a Flying-wedge projection to Newman projection through Fischer and

Sawhorse projection This process can be illustrated by taking active tartaric acid as an
example:
H OH CO2H
H OH
HO H HO H OH
H CO2H
HO H
CO2H CO2H CO2H

COOH Fischer
Newman projection Sawhorse projection projection
(eclipsed) (eclipsed)
COOH HOOC H OH
H OH
OH CO2H
H
H OH
COOH Sawhorse projection
Newman projection (staggered)
(staggered)
1. Convert the following Flying-wedge projections to Fischer projections:

3
3
(a) (b)
2 2
2
Cl
(c) (d) F C Br
CH3
Br H CH2 CH2
H
Cl
(e) C C (f) HO C C
H3C H
H 3C C CH3 CO2H
H Br
HO H
Solution
H
H CH3
(a) C HO2C CH3
HO CO2H
OH
3
3
(b) 2
2
2 2
CHO CHO
(c) H C Br H Br
CH3
CH3
Cl F F
90° rotation H3 C Cl
(d) F C Br H3C C Cl
(in-plane)
Br
CH3 Br
Viewing
direction
H OH H CH3
H H
H Br Br H
Cl Br C Cl
C C C C HO H
(e)
CH3 Cl H
H 3C C CH3 H3C
CH3
HO H
OH
HO2C CO2H
C H
H OH
CH2 CH2
CH2 CH2
(f) C C H
HO CH2 CH2
CO2H Br
H3C Br H
H C H
H 3C CH3
Br
2. Convert the following Flying-wedge projections to Fischer projections:

H
O Cl
O
(a) CH3 (b) F (c) H
H Br Br
CH2 6
(d) H (e) H C C
NH2
OH
CH3 H
Solution
O
1 2 1
O C
3
(a) CH3 H CH3 (b) 5
H 4
H 3
Cl 2 3
4 H Cl 4
(c) 2
5
H 1 6 H 1 Br 5
Br 6
4
5 4 3
6 3 5
(d) HO 2 H
7
8 1 2 H 1
OH 6
8 7
NH2
1 1
H H NH2
(e) CH2 6 C
(CH2)6 (CH2)6
C2 C
1 C H3C 2 H
H NH2 2 H
CH3 H CH3
3. Convert the following Fischer projections to Flying-wedge projections:
CO2H CH3 OH
(a) H OH (b) Cl Br (c) CH3
CH3 F H
CO2H CH3
H 2N H
(d) H Br (e) (f)
H OH
H OH Cl
CH3 CHO
C2H5 H OH
H Cl
(g) H OH (h) HO H (i) H Br
(CH2)6 H OH
H OH
H Cl
CH3 CH2OH
Solution
CO2H CO2H CO2H
(a) H OH C or H C
H3C OH CH3
CH3 H HO
CH3 CH3
(b) Cl Br C
F Br
F Cl
OH
OH
(c) CH3 C
H CH3
H
Br
CO2H
HO2C H
H Br C
(d)
H OH C OH
H3C
CH3 H
CH3 CH3
(e) C
Cl Cl
H
NH2
H 2N H
(f)
H OH OH
H
2 5 5 2
(g)
2 2 2 6
3 3
H OH
CHO HO
(h) H OH H C
C CHO
HO H
H OH
H OH H C CH OH
C
CH OH HO
H OH
Cl
H Cl H
(i) H Br
Br
H
4. Draw a Fischer projection for each of the following compounds:
* *
(a) CH3 C HOHCOOH (b) CH3 C H(NH2)CO2H
* * * *
(c) CH3 C HBr C HClC2H5 (d) CH3 C H(OH) C H(OH)C2H5
* *
(e) CH3CH2 C HFCH2 C HOHCH3
Solution
CO2H CO2H CH3
(a) H OH (b) H NH2 (c) H Br
CH3 CH3 H Cl
CH3 CH2CH3 C2H5

(d) H OH (e) H F
H OH CH2
H OH
C2H5 CH3
[Other Fischer projections, which may be the enantiomer or the diastereoisomer,

may also be written.]
5. Convert the following Fischer projections to Sawhorse projections:

CO2H CH3 CO2H
(a) H OH (b) H Cl (c) H OH

H Br H NH2 H OH
CH3 CH3 CO2H

CHO CHO
H Br
(d) H Br (e) H (f)
Cl H
H Cl H NO2
CH2OH
Solution
CO H H OH H
H OH
(a) H OH H
Br CO H CO H
H Br Br
H C
CH
CH
(eclipsed) (staggered)
1 3 2
2
2
(b)
3
3 3 3
2 3
2 2
3
4 3
3
1 2 2
2
2
(c)
3
3 2 3
2
2
4 2
2
1 CHO H Br H
2
2 Br
2 H
(d) H Br H
3
3 Cl CHO 3
CHO
H Cl Cl
HOH2C
4 CH2OH
CH2OH
CHO H H
(e) H H H
NO2 CHO CHO
H NO2 NO2
H
H Br
(f) H Br Cl Br
H
Cl H Cl
(eclipsed) H
(staggered)
6. Convert the following Sawhorse projections to Newman projections:
H H
OH CH2Br Br
(a) HO (b)
H CO2H CHO
H
Cl
CH2OH
NH2
H Br
H
CH3
(c) H (d) H Cl
OH
CO2H
H2N Cl
Ph H Br
(e) CH3 (f) H
Br CH3
OH Br
CH3
Solution
H
OH
H OH
(a) HO HO H
H CO2H
CH2OH CO2H
CH2OH
CH2Br
H
(b) BrH2C Br H Br
CHO H Cl
H
Cl CHO
NH
NH
H H OH
(c) CH
H
OH H CH
CO H CO H
H Br
(d) H Cl H Br
H Cl
Ph
H2N Cl H2N Cl
Ph
(e) CH3
Br Br OH
OH
CH3
H Br
H Br H Br
(f) H CH3
Br CH3
CH3 CH3
7. Draw meso-2,3-dibromobutane as indicated below: Flying-wedge (stag-

gered form) Æ Fischer Æ Sawhorse Æ Newman
Solution
1
1 3
3 2
3 2
2 3
4 3 3
1 4
3 3 3
3
4
3
3 3
3
3
3
3
3
3
8. Convert the following Fischer projections into Newman projections:

CO2H CHO
(a) H D (b) F CH3
Cl Br H NO2
CH2OH CH2Cl
Ph CH2Br
(c) H Cl (d) H Br
H Br H Cl
CO2H CO2Cl
Solution
CO2H H D
(a) H D Cl Br
Cl Br
CO2H
CH2OH
CH2OH
CHO F CH3
(b) F CH3 H NO2
H NO2
CHO
CH2Cl
CH2Cl
Ph H Cl
(c) H Cl H Br
H Br
Ph
CO2H
CO2H
(d)
1. Draw Fischer projection of any stereoisomer of each of the following compounds:

(a) Lactic acid (b) Alanine
(c) Atrolactic acid (d) Mandelic acid
(e) Glyceraldehyde (f) Tartaric acid
(g) Amphetamine (h) Malic acid
(i) 2-Bromo-3-chlorobutane (j) 2-Fluoro-3-iodobutanoic acid
(k) 2,3,4-Trichloropentane (l) 2-Bromo-4-chloropent-3-one
2. Convert the following Flying-wedge projections into Fischer projections:
NH2 CN
(a) H C (b) H C OH
CO2H
H3C
Ph
C2H5
H3C CO2H
(c) C (d) H C OH
Ph OH
Ph
Cl 3
(e) Ph C Br (f)
2
H
NH2
HO2C CO2H H3C H
C
(g) C C (h)
H H
H C
OH OH HO CH3
3. Convert the following Fischer projections into Sawhorse projections (eclipsed and
staggered):
CH3 CO2H Br
(a) F Br (b) H D (c) H3C F
Cl H F H H3 C Br
C2H5 CH3 F
Ph
Br H CH3
(d) (e) Et CO2H (f) H3C H
Cl CO2H Ph
2 CH3
CHO
(g) H Cl (h) H NH2
H Br H OH
CH2Cl
4. Convert the following Sawhorse projections into Newman projections:

CHO
H D
H Br
(a) (b) H Cl (c) H Cl
CH3
Ph
Br
Ph Br H
CH3
CH3 H2 N
HO Br H
(d) Br H (e) Br (f) Br
Cl CH2F Ph
5. Convert the following Fischer projections into Newman projections:
CH3 CO2H
(a) H Cl (b) H OH
H Br HO H
CH3 CO2H
CHO NH2
(c) H OH (d) Ph Cl
HO H Ph Br
CH2OH CH3
6. Convert the following Flying-wedge projections into Newman projections through
Fischer and Sawhorse projections:
H
H3C CH3 OH
H3C
(a) C C (b) C C
H D H
HO C2H5
Br H
H H
Br Cl H Ph
(c) C C (d) H C C C
3 H
Ph Ph HO CHO
H3C
F H H
(e) C (f) HO2C Cl
C
C CH3
H C H
Cl
HO2C
D C2H5
(g) H CO2H
C
H C
HO Ph
7. Identify the equivalent Fischer projections in each case:

CO H CH CO H CH OH
(a) H OH ; H OH ; HO CH ; HO CO H ; H CH
CH CO H H H CO H
I II III IV V
CO2H Cl CO2H CO2H Cl
(b) H Cl H CO2H H Cl H Cl HO2C H
; ; ; ;
H Br H Br H3C H Br H H3C H
CH3 Cl Br CH3 Br
I II III IV V
3 3
(c)
3 3
8. Identify the equivalent Sawhorse projections:

H Cl H CO H
H Cl HO C H
Br Br
H ; Br Cl ; CO H ; Cl
Br CO H CH
H CH
CH CH H C H
I II III IV
CO H
Cl
H
Br
CH
H
V
9. Identify the equivalent projections in each case:
F CO2H
CO2H CH3
H Br Br H Br
(a) H Br F ; H H ; Br H
; ; H CH3 ;
H F
H CH3 H
CH3 F F
CO2H CO2H
CH3 CO2H
I II III IV V
(b)
2.2 SYMMETRY ELEMENTS

Organic molecules, because of directional property of covalent bonds, assume definite
geometric shape. On the basis of the geometric distribution of atoms and groups they
can be classified as symmetric molecules and nonsymmetric molecules. Four fundamental
elements of symmetry are encountered in organic molecules and these are: (a) simple axis
of symmetry or rotational axis of symmetry (Cn), (b) plane of symmetry (s), (c) centre of
symmetry (i), and (d) alternating axis of symmetry (Sn).
2.2.1 Simple Axis of Symmetry or Rotational Axis of Symmetry (Cn)

When rotation of a molecule by an angle of 360°/n around an imaginary axis passing through
the molecule results in an equivalent structure or orientation that is superimposable with
the original one, that axis, denoted by the symbol Cn (Latin word Circulate), is called the
n-fold simple axis of symmetry. Iodoform, for example, has a three-fold (n = 3) simple axis
of symmetry. When the molecule is rotated by an angle of 120° (360°/3) about the axis
passing through the C—H bond, a structure indistinguishable from the original is obtained.
H indistinguishable H
C 120° rotation C
I I
I about the C3 axis I
I I
120°
C3 axis in iodoform molecule
C3
Some other examples:

1. Deuterium oxide (D2O): The molecule has a two-fold simple or rotational axis of
symmetry (C2) which bisects the D—O—D bond angle.
indistinguishable
O 180° rotation O
Da Db about the C2 axis Db Da
180°
C2
2. cis-1,2-Dideuterioethene: The molecule has a C2 axis that passes through the

molecular plane and the mid-point of the C==C bond.
indistinguishable
H H H H
C C C C
D D D D
180°
C2
3. trans-1,2-dideuterioethene: The molecule has a C2 axis that is perpendicular to

the molecular plane and passes through the mid-point of the C==C bond.
indistinguishable
H H 180° rotation H H
C C C C
D D about the C2 axis D D
180°
C2
4. Difluoromethane (CH2F2): The molecule has a C2 axis that passes through the
two H and two F atoms bisecting H—C—H and F—C—F bond angles.
indistinguishable
F
F
C
H 180° rotation C
F H
H about the C2 axis F
H
C2
5. Carbontetrachloride (CCl4): The molecule has four C3 axes each of which is

perpendicular to the plane of three Cl atoms and passes through the fourth Cl
atom and the carbon atom. Also, the molecule has three C2 axes each of which
bisects two pairs of Cl—C—Cl bond angles.
Cl Clc Cld Clc

C 120° C C C
Cl Cla Clb Cld
C3 180° 180° 180°
Cl Cl Clb Cld Clc Cla Cla Clb
120°
C2 C2 C2
120°
C3 120° C3
C3
6. Copper acetylide (Cu – C ∫ C – Cu): The molecule is linear. It has a C axis

coincident with the internuclear axis since rotation around it by any angle gives an
indistinguishable structure. The molecule also has an infinite number of C2 axes
perpendicular to the centre of the C axis.
Cu C C Cu C
180°
180°
C2
C2
7. cis-1,3-Dibromocyclobutane: The molecule has a C2 axis that passes vertically

through the centre of the molecule.
indistinguishable
Br H Br H
3 4 1 2
H Br H Br
about the C2 axis
2 1 4 3
H H H H
180°
C2
8. trans-1,3-Dibromocyclobutane: The molecule has a C2 axis that passes through

C–2 and C–4.
indistinguishable
Br H Br H
3 4 1 4
H H H H
about the C2 axis
2 1 2 3
C2 180° H Br H Br
9. Boron trifluoride (BF3): Trigonal planar BF3 molecule has one C3 axis that
passes through the B atom and perpendicular to the molecular plane. Also, it has
three C2 axes each of which passes through the B–F bond and bisects the angle
between two other B–F bonds.
F 180° F
F B C2 F B
F F
180° 180°
120° C2
C3 C2
10. Hexadeuteriobenzene (C6D6): The molecule has a C6 axis passing through the
centre of the regular hexagon. It has six C2 axes. Three of them pass through the
mid-point of two opposite bonds and the other three pass through the two opposite
corners of the molecule. The molecule has also a C2 and a C3 axis collinear with C6
axis, i.e., C6, C2 and C3 are on one and the same axis. C6 is the principal axis of the
molecule.
D D D D D D
180°
C2 D D D D D D
180° 180°
D D 180° C2 D D D D
C2 60°
180° 180°
C2 C2 C2 C6(C3, C2)
Three relevalent topics

(i) Principal axis: If a molecule is found to possess Cn axis with different values
of n, then the Cn axis having maximum value of n (fold or order) is called the
principal axis. For example, C6 in the principal axis of C6D6 or C6H6 molecule.
When a molecule has several Cn axes with the same value of n, then the principal
axis is the one that passes through maximum number of atoms of the molecule.
(ii) Trivial axis: Rotation through an angle of 360° about any axis through any
molecule gives back the identical structure. C1 axis is, therefore, known as the
trivial axis. It is not included in the category of Cn axis.
(iii) Indistinguishable and identical structures: The terms indistinguishable and
identical have different meaning in the present context. The former refers to any
equivalent structure arrived at by exchanging similar atoms or groups while the
latter refers strictly to the original.
2.2.2 Plane of Symmetry (s)

An imaginary plane that bisects a molecule into two equal halves that are mirror image of
each other is called the plane of symmetry. This is indicated by s (sigma, originated from
the German word Spiegel which means mirror). The plane of symmetry is also called a
mirror plane. For example, the eclipsed conformation of meso-2, 3-dibromobutane has a
plane of symmetry. This may be shown as follows:
3 3
meso
It is to be noted that every planar molecule has a plane of symmetry coinciding with the
molecular plane. The molecular plane of H – I, for example, is the s-plane.
H — I Æ Molecular plane (s)
1. 2-Bromo-2-chloropropane: The molecule has a plane of symmetry passing
through Br, C and Cl atoms.
plane of symmetry
Br Br
C H3C CH3
Cl CH3
Cl
CH3
Flying-wedge
Fischer
projection
projection
2. cis-1,3-Dichlorocyclobutane: The molecule has two s-planes passing through

C-1 and C-3 and through C-2 and C-4.
Cl H s-plane passing
3
4 through C-2 and C-4
H H
H Cl
2 1
H H s-plane passing
through C-1 and C-3
3. trans-1,3-Dichlorocyclobutane: The molecule has a plane of symmetry which

passes through C-1 and C-3.
H H
3 4
Cl H
H Cl
2 1
s-plane passing
H H through C-1 and C-3
4. 1-chloro-3-methyl-1,2-butadiene: The molecule has a plane of symmetry passing

through Cl, H and the three double bonded carbons.
Cl CH3
C C C s-plane
H CH3
5. Water (H2O): The molecule has two s-planes. One is the molecular plane and the
other is perpendicular plane bisecting the O atom.
perpendicular
plane (s)
O molecular
H H
plane (s)
6. (2R,4S)-2,4-Dibromopentane-1,5-diol: The molecule has a plane of symmetry

which passes through C-3, H and the –OH group (the designations R and S have
been discussed later in this chapter).
HOH C CH OH
CH OH
C Br H
H s C
C R H Br H
Br
Br Br H
H OH
s-plane
CH OH
3D-representation Fischer
or Flying-wedge projection projection
sh, sv and sd The symbol s is usually found to carry three subscripts indicating the
position of the symmetry plane relative to the principal axis (Cn).
(i) sh: It represents the s-plane perpendicular to the principal axis (h stands for
horizontal).
(ii) sv: It represents the s-plane containing the principal axis (v stands for vertical).
(iii) sd: It represents the s-plane which contains the principal axis and bisects the
angle between the two C2 axes (d stands for diagonal).
Tetrachloroethene, for example, has both sh and sv and tetrabromoallene, for example,
has two sd.
sh sd
sd
Cl Cl Br 2 Br principal
sv C C C axis
2 C C 2
Cl Cl Br
Br 2
Tetrachloroethylene Tetrabromoallene
The presence of sh, sv and sd in a single molecule, for example, in square plane XeF4
molecule may be shown diagrammatically as follows:
C
sd sv
sh
F F
Xe C
F F
C
principal axis
Square palanar XeF (Xenon tetrafluoride) molecule
2.2.3 Centre of Symmetry (i)

The centre of symmetry or a point of symmetry or an inversion centre (i) is a point within a
molecule such that if an atom or a group is joined to it by a straight line and then extended
an equal distance towards the opposite direction, it meets an identical atom or group. It
is symbolized as i because inversion of all the atoms or groups in the molecule through
this point provides a structure which is indistinguishable from the original. For example,
trans-1,3-dimethylcyclobutane has a centre of symmetry.
Some other examples.

1. trans-1,2-Dichloroethene:
i
2. meso-Tartaric acid:
3. trans-(1R,3S)-Di-sec-butylyclobutane:
R
2.2.4 Alternating Axis of Symmetry (Sn)

If a molecule is rotated by an angle 360°/n around an axis and then reflected on a plane
perpendicular to that axis, a structure indistinguishable from the original is obtained,
the molecule is said to have an ‘n’-fold alternating axis of symmetry (Sn). According to
the mode of operation it may also be called rotation–reflection symmetry. For example,
a-truxillic acid has a two-fold alternating axis of symmetry.
S2 S2
2
180° 2 1
4 3
2
180° rotation about S2
3 4 2
1 2
ind
2 isti
ngu
ish
a abl
e
2
Some other examples

1. 1,3-Dichloro-2,4-difluorocyclobutane (I): The molecule has a two-fold
alternating axis of symmetry.
S2 H H F
Cl
180° 1 2
3 4 H
F Cl
H 180° rotation about S2 Cl
H 4 H
2 1 F 3
Cl ind F H mirror
H isti
ngu
I ish
abl
e Cl H
F
H
H
F
H Cl
I
2. 5l5-Azaspiro[4.4]-2,3,7,8-tetramethylnona-5-ylium ion: The molecule has a
four-fold alternating axis of symmetry.
S4-axis
90° Me H
H H Me
Me 2 3
Me
H 4 90° rotation about S4 N
1 N
9 5 6 H
Me Me
Me H H
8 7
ind mirror
H Me
ist
ing
uis H
ha Me Me
ble H
N
Me H
H Me
It is to be noted that a plane of symmetry (s is equivalent to one-fold alternating axis of

symmetry (S1) and a centre of symmetry (i) is equivalent to two-fold alternating axis of
symmetry (S2).
2.2.5 Symmetric, Asymmetric and Dissymmetric Molecules

A molecule is said to be symmetric if at least one of the symmetry elements s, i or Sn
is present in it. Molecules in which there is no element of symmetry, i.e., Cn (except C1
which is present in all objects) and Sn axes (i.e. s, i and other Sn axes) are absent are
called asymmetric molecules. On the other hand, molecules, in which Sn axis is absent,
but Cn axis may be present or absent are known as dissymmetric molecules. Therefore,
all asymmetric molecules are dissymmetric, but not all dissymmetric molecules are
asymmetric. For example, trans-1,2-dibromocyclopropane is a dissymmetric molecule and
not an asymmetric molecule because it has a C2 axis. However, the alanine molecule is
both an asymmetric and dissymmetric one because both Sn and Cn are absent in it.
H
CO2H
Br Br C
180° H NH2
C2
H CH3
Alanine
trans-1, 2-Dibromocyclopropane (an asymmetric or a
(a dissymmetric molecule) dissymmetric molecule)
1. Indicate the simple axis of symmetry present in the following molecules

or ions:
(a) Ammonia (NH3) (b) trans-1,2-Dimethylcyclobutane
(c) CO32@ (d) 1,3-Dibromoallene
(e) cis-(1R, 3R)-Di-sec-butylcyclobutane
(f) trans-(1R, 3R)-Di-sec-butylcyclobutane
(g) Chair form of dodecadeuterocyclohexane
(h) Boat form of dodecadeuterocyclohexane
(i) Anthracene
Me
(j) (k) (l) N
Me
Me
Br Cl Cl
(m) (n) (o)

Br Br Cl
Cl Cl
(p) (q) Tetramethylallene
Cl
(r) trans-1,2-Dichlorocyclopropane
H
HOOC OH
(s) NH 2 (t) COOH
H OH
Solution
(a) Ammonia ( NH 3 ): The shape of the ammonia molecule is pyramidal. It has a C3
axis that passes through N atom and the perpendicular to the plane of three H
atoms.
N H
H
H
120°
C3
(b) trans-1,2-Dimethylcyclobutane: This molecule has a C2 axis that passes

through the midway of C-1—C-2 and C-3—C-4 bonds.
H3C H
H CH3
C2
(c) CO32 (carbonate ion): The planar carbonate ion has one C3 axis that passes
through the C atom and is perpendicular to the plane of the species. It has three
C2 axes, each of which passes through the C atom and one O atom and bisects the
angle between two other C O bonds.
2– 2–
O3 O3 180°
2– 2–
C O3 C O3 C2
2– 2–
3 O 3 O
120°
C2
C2
C3
(d) 1,3-Dibromoallene (BrCH=C=CHBr): This molecule has only one C2 axis that
passes through the central sp-hybrizided carbon. It can be clearly demonstrated by
its Newman projection.
Br
H
Br
C C C H Br
H 180°
180°
Br H
2
2
(e) cis-(1R, 3R)-Di-1-methylpropylcyclobutane: This molecule has a C2 axis that

passes vertically through the centre of the molecule.
Me
H
Et H C Et
H 4 3
Me
C H H H
1 2
H H
180°
C2
(f) trans-(1R, 3R)-Di-1-methylpropylcyclobutane: This molecule has a C2 axis

which passes through C-2 and C-4.
Me
H
H C Et
4 3
H H H H
1 2
C H
Me H
Et C2
(g)Dodecadeuterocyclohexane (chair form): The chair conformation of cyclo-
hexane has a C3 axis passing vertically through the centre and three C2 axes
passing through the centre and the mid-points of two opposite bonds.
D
D
D D = Carbon atoms below
D D the plane of the paper
D D 180° 180°
D = Carbon atoms above
D C
D C2 2 the plane of the paper
180°
D
120°
C2
C3
(h) Dodecadeuterocyclohexane (boat form): The boat conformation of cyclohexane
has a C2 axis passing vertically through the centre of the boat.
D D
D D D D
D D D D
D D
180°
C2
(i) Anthracene: Anthracene molecule has two C2 axes one of which passes through
C-9 and C-10 and the other passes through the mid-points of C-2—C-3 and C-6—C-7
bonds.
8 9 1
7 2 180°
3
C2
6
5 10 4
180°
C2
(j) The molecule has three C2 axes passing through the mid-point of C-2—C-3, C-5—C-6
and C-7—C-8 bond and one C3 axis passing through C-2 and C-4 carbon atoms.
8
7
4
5 3
180° 180°
1
C2 6
120° 2 C2
C3 180°
C2
(k) The triphenylene molecule has three C2 axes passing through the mid-point of
C-2—C-3, C-6—C-7 and C-10—C-11 bonds. It has also a C3 axis passing vertically
through the centre of the middle ring.
10
9 11
180° 8
7 12
C2
6 1
5
4 2 120°
180° 3 180°
C2 C3
C2
(l) The molecule has only one C3 axis passing through nitrogen and the carbon opposite
to it (C-4).
Me 8
7
4
5 3
6
N
Me
120° 2
Me
C3
(m) The molecule has three C2 axes each of which pass through Br—C and the opposite
C—H bonds. It has also a C3 axes passing vertically through the centre of the ring.
Br Br
Br
Br Br 180°
Br
C2 180° 180° C2 120°
C2
C3
(n) The molecule has two C2 axes passing through two Cl atoms and the mid-points of
C-2—C-3 and C-5—C-6 bonds. It has also a third C2 axis passing vertically through
the centre of the ring.
Cl
1 2 180°
6
C2 Cl Cl
5 3
4
Cl 180°
180°
C2 C2
(o) The molecule has only one C2 axis passing through C-2 and C-5.
Cl
1 2
6
180°
5 3 Cl
4
C2
(p) The molecule has only one C2 axis passing through the mid-points of C-1–C-2 and
C-4—C-5 bonds.
6 1 Cl 180°
5
4
C2
2
3 Cl
(q) Tetramethylallene (Me2C=C=CMe2): The molecule has three C2 axes. One of

them passes through the C atoms and each of the rest two axes passes diagonally
through the central sp-carbon atom.
CH3
H H3C
CH3 CH3
C2 C C C C C C H3C CH3
180° CH3 CH3
180° 180°
18°
H 3C H3C 18° CH3 C2
C2
C2 C2
(r) trans-1,2-Dichlorocyclopropane: The molecule has only one C2 axis passing
through C-3 and the mid-point of C-1—C-2.
3 H
1
Cl Cl
180°
2
C2
H
(s) Amide ion ( NH 2 ): The ion has a C2 axis that bisects the H—N—H bond.
N
–
H H
180°
C2
(t) This anti-staggered conformation of the active tartaric acid molecule has a C2 axis
that passes through the mid-point of C-2–C-3 bond.
indistinguishable
H OH H 3 OH
HO2C 2 CO2H
CO2H 180° rotation CO2H
3 about the C2 axis 2
H OH 180° H OH
C2
2. Indicate the element(s) of symmetry (other than Cn) present in each of the
following molecules:
(a) cis-1,2-Dichlorocyclopropane (b) a-Truxillic acid
CH3
Et H H Et
(c) C C C C (d) CH3 H Br
H C H C
Br
Br
H H
H O
H N C CH3 D
H
(e) H3C C N H (f) CH2OH
O H
H D
CH2OH
O
H H
NH C
C C
(g) H3C CH3 (h) D D
C
H 3C CH3
C NH Cl
O H
H
H
Cl
(i) Cl (j) CH3 CO2H
H H
CH3 CH3
H3C
O O O
(k) (l) (m)
O H3 C
CH3 H3C O CH3
H H
H H3 C H
H
C
(n) C O (o) (p)
Me C Et
Me Et
O
H H
Cl H
O
(q) Br H H Br (r) Me Me
H Cl
Me H +
(s) H (t) H3C CH3

Me
H C C H
N
CH3 CH3
Me H
H Me
Cl
(u) Br
H
O
Cl
H
H
Solution
(a) cis-1,2-Dichlorocyclopropane: The molecule has a plane of symmetry passing
through C-3 and the mid-point of C-1—C-2 bond.
Cl
1
H
Cl
H plane of symmetry (s)
2
3
H
H
(b) a-Truxillic acid: The molecule has a centre of symmetry.

HO2C H Centre of symmetry (i)
Ph H Ph
H
H CO2H
(c) The molecule has a plane of symmetry passing through Br, C and H atoms.
s-plane
Et H H
C C Et
H C C
C H
Br H
(d) The molecule has a plane of symmetry passing through the two Br atoms, two
doubly bonded carbons and the mid-point of a ring bond.
CH3
H
H3C Br plane of symmetry (s)
C
Br
H
(e) The molecule has a centre of symmetry. It has also an S2 axis which is equivalent
to (i).
Centre of symmetry (i)
H O
H N C CH3
C N H
H3C
O H
(f) The molecule has a plane of symmetry passing through the mid-points of C-2—C-3,
C-5—C-6 and C-7—C-8 bonds
8
7
plane of symmetry (s)
D
5 3 H
H CH2OH
6 D 2
CH2OH
(g) The molecule has a centre of symmetry.

H O
s-plane N C
H 3C CH3
C N
centre of symmetry
H
O
It has also a plane of symmetry which is the molecular plane and an S2 axis which
is equivalent to (i).
(h) The molecule has a plane of symmetry passing through Cl, C and H atoms.
H H
C C D
D C
CH3
H3C
H Cl plane of symmetry (s)
(i) The molecule has a centre of symmetry and a plane of symmetry passing through
C-1, C-4, two H and two Cl atoms.
H
H
H H H
Cl
Cl Cl
H
H H
Cl H
H
H centre of
H
symmetry (i)
(j) The molecule has a plane of symmetry passing through C-1 and C-4.
H
4
H3C
CO2H
1
H
(k) The molecule has a centre of symmetry and it also has an S2 axis.
H 3C O H O
H3C
O CH3
CH3 O H
centre of symmetry (i)
(l) The molecule has a plane of symmetry passing through C-2 and C-5.
H
CH3 1
H3C 6 5 plane of symmetry (s)
H
3
H3C
2 CH3 4
(m) The molecule has a plane of symmetry passing through C-1 and C-4.
O
CH3 CH3 1 O plane of
O symmetry (s)
O O H
H3C CH3
4
H
H3 C O CH3
H
(n) The molecule has a plane of symmetry passing through H—C— C —O—C—H
system. ||
O
s
(o) The molecule has a plane of symmetry passing through the mid-points of C-2–C-3
and C-5–C-6 bonds and C-7 including the —CH3 group and H atom.
(p) The molecule has a centre of symmetry and an S2 axis. There is also a plane of
symmetry passing through the two methylene carbons and two opposite ring bonds.
H H
H H
(q) The molecule has a centre of symmetry and so also it has an S2 axis.
Cl H
Br H Br
H
H Cl
(r) The molecule has a plane of symmetry passing through C=O and the mid-point of
C-3–C-4 bond.
Me
O
Me Me O H 3
Me
4 3 4
H
(s) The molecule has a four-fold alternating axis of symmetry.
Me H
H Me Me H
H Me
N
90° rotation about
N
Me H the S4 axis
H Me Me H
90° Me
H mirror
S4 ind
ist
ing
uis H
ha Me
ble
H Me
N
Me H
H Me
(t) The molecule has a plane of symmetry passing through two phenonium ion ring
carbons and two cyclopropane ring carbons.
+ plane of symmetry (s)

H3 C CH3
C C H
H CH3
H3C
(u) The molecule has a plane of symmetry passing through C==O and C–3 carbon
including Br and H.
Cl
Br H plane of symmetry (s)

O
H Cl
H
3. Which of the following molecules have plane of symmetry?
H H H
(a) HO CO2H (b) Br CH3 (c) OHC OH
H OH H Br HO H
CO2H CH3 HO CHO
H
CHO CH3
CH3
(d) HO H (e) Br H (f) H Br
H OH Cl Cl
OHC OH H3C Br C2H5
H H
CH3
H C
C H CO2H
H CH3 CH3
(g) C CH3 (h) C C
H3C C H H
Br D CH3 D
H
CO2H
HO H
(i) H OH
H OH
HO H
CO2H
Solution
H CO2H
(a) HO CO2H H OH
H OH H OH
CO2H CO2H
H H
(b) Br CH3 Br CH3
H Br Br CH3
CH3 H
(c) The molecule has no plane of symmetry (s).
(d)
s
CH3 CH3
(e) Br H Br H
Cl Cl Cl Cl plane of symmetry (s)
H3 C Br Br H
H CH3
(f) The molecule has no plane of symmetry.

CH3
H C
H C H H Br H
(g)
C CH3 C C
H3C C H3C C C CH3
Br CH3
H H H
CO2H
CH3 CH3
(h) C C
H H
H CH3 D

CO2H
(i) HO H
H OH
H OH
HO H
CO2H
4. Draw various conformations of n-butane in Newman projection considering
rotation about the C-2—C-3 bond and indicate the conformation that has
a plane or centre of symmetry.
Solution The various conformations of n-butane obtained by rotating the front carbon
about the C-2—C-3 bond may be shown as follows:
The conformation I has a plane of symmetry. The conformations II, III, V and VI have
neither a plane nor a centre of symmetry. The conformation IV has a centre of symmetry
which exists in the mid-point of C-2—C-3 bond.
5. Identify the elements of symmetry (if any) in the following molecules as
shown below.
F
Cl H
(a) (b) (c) C C C C
Cl H Cl
H Cl
Me Br H
(d) Me (e) CHCl3
H
Br
Cl H
H H
(f) C C C (g) CH4
H
H
Solution
(a) It has no element of symmetry.
(b) It has only one plane of symmetry, i.e., the molecular plane.
(c) It is a cumulene with odd number of double bonds and so, the molecule is planar.
It has one C2-axis perpendicular to the molecular plane and passing through the
centre of the molecule, a sh bisecting the C2 and a centre of symmetry (i).
180°
Cl H Cl H
C C C C sh C C C C
H Cl H Cl
(d) It has a centre of symmetry.

(e) It has one C3 axis and three sv planes. The C3 axis and one sb plane is shown in the
following figure.
H sv
C
Cl
Cl Cl
120°
C3
(f) It has three C2 axes and two sd planes (i.e., planes that contain the principal axis
and bisect the angle between the two C2 axes.
C2
H H sd
C
C
C2 C2
C
H H sd
(g) It has three C2 axes, four C3 axes, three S4 axes, and six sd planes.
6. Identify the elements of symmetry (Cn and s) of o-, m- and p-dichloro-
benzenes.
Solution o-Dichlorobenzene: It has a C2 axis passing through C-1—C-2 and C-4—C-5
bonds. Also, it has two sv planes.
180°
Cl sv (molecular plane)
Cl
sv
C2
m-Dichlorobenzene: It has a C2 axis passing through C-2 and C-5. Also, it has two sv
planes.
s
Cl Cl
s (molecular plane)
180°
C2
p-Dichlorobenzene: It has three C2 axes. Also, it has two sv and one sh planes.
sv
sv (molecular plane)
Cl sh
C2 Cl Cl
180°
180°
Cl
C2
180°
C2
7. Identify which of the following objects possess at least one plane of

symmetry: (a) an empty spool, (b) a hand, (c) a cup with DAD written
opposite the handle, (d) a cup with MOM written opposite the handle, (e)
a spoon, (f) a six-pointed star-like cake, (g) a bicycle, (h) a spectacle, (i) a
sock.
Solution The objects with at least one plane of symmetry are (a), (d), (e), (f), (h) and (i).
8. Draw a square planar molecule with the formula X2Y2A (A is the central
atom) which has a centre of symmetry and at least two planes of symmetry.
How many planes of symmetry are actually present in the structural
formula you have drawn?
Solution The structural formula of the molecule (X2Y2A) having a centre of symmetry
and at least two planes of symmetry is as follows:
X Y s (molecular plane)
A
s-plane passing
Y X through Y-A-Y
s-plane passing
through X-A-X
There are three planes of symmetry in the molecule. They are: (i) the plane through X–A–X
and perpendicular to the molecular plane, (ii) the plane through Y–A–Y and perpendicular
to the molecular plane and (iii) the plane that cuts through all five atoms of the molecule,
i.e., the molecular plane.
1. Indicate the simple axis of symmetry present in each of the following molecules/
ions:
≈
(a) Cyclohexane (b) CMe3 (c) CH3COO@
O O
N
(d) BBr3 (e) (f)

N
N
O O
≈
Et F
(g) H3O (h)
Et F
H
H
Br
(i) Br (j) (k)
H Br
Br H
≈ H F
(l) NH4 (m) (n) C C
F H
O
OCH3 O CH3
(o) (p) C C (q) C
D D
H3C O
H
(r) (s) (t)
O
Cl Cl
(u) (v) (Basketane)
O
2. Indicate the element(s) of symmetry (other than Cn present in each of the following
molecules:
(a) 3-bromopentane (b) 1-chloro-3-methyl-1,2-butadiene
(c) quinuclidine (d) 4-bromopiperidine
(e) trans-2,5-dimethyl-1,4-dioxan (f) meso-3,4-dibromohexane
(g) cis-1,3-dibromocyclohexane (h) 8-iodospiro [4,5] decane
(i) trans-1, 4-dimethylcyclohexane (j) trans-diethyldiketopiperazine

(k) 1,1-dichlorocyclopropane
(l) cis-cyclobutane-1, 2-dicarboxylic acid (m) cis-1,3-dibromo-cyclopentane
(n) cis-2,3-dimethylcyclohexane-r-1-ol
(o) cis-1 4-ditrichloromethyl-cyclohexane
(p) cis-3, cis-5-dimethylcyclohexane-r-1-ol
(q) trans-3, trans-5-dimethylcyclohexane-r-1-ol (r) 1,3-dioxan
Cl Cl
(s) cubane (t) C C C C
F F
CH3
Br
Br
C H
CH3 H
(u) (v) H ,
Br Br
H Br
H
Cl Br
Cl CH3
(w) (x) C C C
H CH3
Br Cl
Br Cl
(y) N C Br
C (z)
H3 C H
H Br CH3
Cl Br
3. Which of the following molecules may have a plane of symmetry?
H H
(a) HO CH3 (b) Cl CH2CH3
H OH H Cl
CH3 CH2CH3
H CCl3
(c) HO2C OH (d) Cl H

HO H H Cl
HO CO2H Cl3C Cl
H H
Et CH3
(e) I H (f) H Cl
C O CH(CH3)2
Et I
H
Et
CHO
H C Et Et
C H
(g) H (h) C C
H H
Et C CH3 CH3 CH2OH CH3
C
H Cl
H
2
(i) (j) H B H
N N
B B
H N H
2 H
4. Identify which of the following objects possess at least one plane of symmetry: (a)
a chair, (b) a glove, (c) a nail, (d) a cup, (e) a nose, (f) a shoe, (g) a tree, (h) a seven-
pointed star, (i) a plate, (j) a book, (k) a gas cylinder.
5. Identify which of the following objects possess a centre of symmetry: (a) five-pointed
star, (b) a dumb-bell, (c) a carom board (d) an eight-point star, (e) a tennis ball.
6. Out of twenty six letters in English language how many of them are symmetric and
how many of them are non-symmetric (considering them two-dimensional)?
[Hint: F, G, J, L, P, Q and R are non-symmetric and remaining nineteen letters are
symmetric.]
2.3 ISOMERISM: CONSTITUTIONAL ISOMERS AND STEREOISOMERS

2.3.1 Constitutional Isomers
Isomers are different compounds that have the same molecular formula. Molecules
having the same molecular formula but different molecular constitution, i.e., different
atom-to-atom bonding sequence or atomic connectivity are called constitutional isomers or
structural isomers and the phenomenon is called constitutional or structural isomerism.
Examples of constitutional isomers are as follows:
Molecular Constitutional isomers

formula
CH3
C4H10 |
CH3 —CH 2 —CH 2 —CH3 and CH3 —CH — CH3 (chain isomers)
Butane 2-Methylpropane
OH
|
C3H8O CH 3 —CH 2 —CH 2 —OH and CH3 —CH — CH3 (position isomers)
1-Propanol 2-propanol
CH3CH 2OH and CH 3 —O—CH3 (functional group isomers)

C2H6O 1-Propanol Dimethylether
2.3.2 Stereoisomers
Isomers having similar molecular constitution but differing in the spatial arrangement of
the various groups or atoms about a rigid part of the molecule (e.g., an asymmetric carbon
atom, a double bond, etc.), i.e., having different configurations are called stereoisomers and
the phenomenon is called stereoisomerism. Some examples of stereoisomers are given as
follows.
CO2H CO2H
C and C (two optical isomers)

(i) HO OH
H H
CH3 CH3
(R)-Lactic acid (S)-Lactic acid

(ii) H3C H H3C H
(two cis-trans isomers)
H3C H H CH3
cis-2-Butene trans-2-Butene
(iii) HO2C CO2H HO2C H
(two cis-trans isomers)

H H H CO2H
cis-Cyclopropane-1, trans-Cyclopropane-1,
2-dicarboxylic acid 2-dicarboxylic acid
2.3.3 Enantiomers and Diastereoisomers

Stereoisomers can be classified into two categories: (i) those that are enantiomers of each
other, and (ii) those that are diastereoisomers of each other.
2.3.3.1 Enantiomers
Stereoisomers whose molecules are related to each other as non-superimposable mirror
images are called enantiomers. For example, alanine, CH3CH(NH2)CO2H, has two
stereoisomers having mirror image relationship, but one is not superimposable on the
other. To superimpose an object on its mirror image means to align all parts of the object
with its mirror image. With molecules, this means aligning all atoms and bonds. Therefore,
they represent a pair of enantiomers.
Since enantiomers are usually optically active, i.e., they turn the plane of polarized light
to an equal degree but in opposite directions, they are also called optical isomers or optical
antipodes.
2.3.3.2 Diastereoisomers
Stereoisomers whose molecules are not mirror images of each other are called
diastereoisomers or diastereomers. The simplest example of diastereoisomers may be cited
in the isomers of 2-butene.
By examining the structural formulas for cis- and trans-3-butene, we see that they have
the same molecular formula (C4H8) and the same connectivity (both compounds have two
central carbon atoms joined by a double bond and both compounds have one methyl group
and one H atom attached to each carbon atom). But their groups/atoms have different
arrangement in space (around the regid C==C bond) that is not interconvertible from
one to another (because of large barrier to rotation of the C==C bond), making them
stereoisomers. They are different compounds, as revealed from their physical and chemical
properties. Furthermore, they are stereoisomers that are not mirror images of each other.
Therefore, they are diastereoisomers and not enantiomers. In fact, they represent a pair
of cis-trans isomers and since they are not optically active, they are not included in the
class of optical isomers.
Cis- and trans-isomers of cycloalkanes provide another example of stereoisomers that are
diastereoisomers. For example:
diastereoisomers
Et Et Et H
H H H Et
-1,2-Diethylcyclopentane -1,2-Diethylcyclopentane
(C8H16) (C8H16)
These two compounds have the same molecular formula (C8H16), the same atom-to-atom
bonding sequence or atomic connectivity, but different arrangement of their atoms/groups
in space, i.e., different configurations. In one compound, both ethyl groups are bonded to
the same face of the ring, while in the other compound, the two ethyl groups are bonded
to opposite faces of the ring. Furthermore, interconversion of the positions of the ethyl
groups cannot be possible by conformational change. Therefore, these two compounds are
stereoisomers and since they are not mirror images of each other, they can be further
classified as diastereoisomers.
[Geometric or cis-trans isomerism Isomers having same molecular constitution, i.e., the
same atom-to-atom bonding sequences or atomic connectivity, but differing in spatial
arrangement of the groups or atoms owing to the restricted rotation about a double
bond (C==C, C==N, etc.) or about a ring are called geometric or cis-trans isomers and
phenomenon of existence of such isomers is called geometric or cis-trans isomerism. For
example, 1,2-dibromoethene exists in two stereoisomeric forms (I and II) as shown below,
one being different from the other in the spatial arrangement of the two bromine atoms
and the two H atoms. The isomer I with the similar groups on the same side of the double
bond is called the cis-isomer and the isomer II with similar groups on opposite sides of the
double bond is called the trans-isomer. Similarly, III is the cis-1,2-dimethylcyclopropane
and IV is the trans-1,2-dimethylcyclopropane.
The condition for an alkene to exhibit geometric or cis-trans isomerism, i.e., to exist as
cis- and trans- stereoisomers, is that each of the doubly bonded carbon atoms must be
attached to two different atoms or groups. Therefore, an alkene Cab=Ccd will exist in two
stereoisomeric forms if a π b and c π d. It does not matter whether a and/or b are the same
as c and/or d. Similar condition is applicable for cyclic systems for showing geometric or
cis-trans isomerism.
In case of polyenes, the number of cis-trans isomers depends on the number of double bonds
in the molecule as well as the terminal substituents. If the general formula of a polyene is
R—(CH==CH)n—R¢ (with different terminal substituents), the number of stereoisomers is
2n and if the general formula is R—(CH==CH)n—R (with identical terminal substituents),
the number of stereoisomers is (2n–1 + 2p–1) where p = n/2, when n is even, and p = (n + 1)/2
when n is odd. For example, four cis-trans isomers are possible for octa-2,4-diene (with
different terminal groups) and three cis-trans isomers are possible for octa-3,5-diene (with
identical terminal groups).
1
CH3 2 3 H CH3 H
C C 4 5 H C C CH2CH2CH3
H C C 6 7 8 H C C
H CH2CH2CH3 H H
(2E, 4E)-Oct-2, 4-diene (2E, 4Z)-Octa-2,4-diene
H H CH3 H
C C H C C CH2CH2CH3
H 3C C C H C C
H CH2CH2CH3 H H
(2Z, 4E)-Octa-2, 4-diene (2Z, 4Z)-Octa-2, 4-diene
1 2
CH3CH2 3 4 H H H
C C 5 6 H C C CH2CH3
H C C 7 8 CH3CH2 C C
H CH2CH3 H H
(3E, 5E)-Octa-3,5-diene (3Z, 5Z)-Octa-3,5-diene
CH3CH2 H H H
C C CH2CH3 C C H
H C C CH3CH2 C C
H H H CH2CH3
(3E, 5Z)-Octa-3,5-diene (3Z, 5E)-Octa-3,5-diene
Identical
There are some other molecules (acyclic systems containing no C==C bond) that can exist
as diastereoisomers. For example, active tartaric acid and meso-tartaric acid represent a
pair of diastereoisomers.
diastereoisomers
CO2H CO2H
H OH H OH
HO H H OH
CO2H CO2H
Active tartaric acid meso-Tartaric acid
Acylic diastereoisomers are possible if a compound contains more than one asymmetric or
chiral carbon.
2.3.3.3 Determination of the stereochemical relationships between molecules

A series of questions may be framed with yes or no answers in such a way that the
relationship between two compounds may be traced through no stereochemical relationship
to a diastereoisomeric relationship.
Do the molecules
have the same No They are not isomers
molecular formula?
Yes
They are homomers, i.e.,

Are they superimposable? No identical molecules which
are superimposable
No
They are isomers
Do they have the same

molecular constitution, They are constitutional
No
i.e., atom-to-atom bonding or structural isomers
sequence or connectivity?
Yes
They are stereoisomers
Are they non superimposable

mirror-images?
Yes No
They are enantiomers They are diastereoisomers

2.3.3.4 Chemical and physical properties of enantiomers and diastereoisomers

The atoms in two enantiomers have exactly the same spatial arrangements with respect
to distance and dihedral angles (interactions). Two enantiomers are, therefore, isomeric
to each other. Because of this, enantiomers exhibit the same reactivity (Eact values
are the same) towards achiral reagents, solvents, catalysts and conditions. However,
two enantiomers react at different rates, i.e., show different reactivity towards chiral
reagents, solvents, catalysts and conditions. The transition states developed from the
individual enantiomer and the chiral reactant are not enantiomeric. They are, in fact,
diastereoisomeric and hence possess different energy of activation (DG=|) due to different
enthalpies of activation (DH =|). Because of this, the two enantiomers react with a chiral
reagent at different rates.
The two enantiomers have identical physical properties like melting points, boiling
points, solubilities, densities, refractive indices, dipole moments, etc. because they are
isometric to each other. However, they differ in their behaviour towards plane-polarized
light (rotation in two opposite directions but in the same magnitude) and this is because
a plane-polarized light is composed of two dissymmetric components, a right-handed and
a left-handed circularly polarized light beam and thus constitutes a chiral environment.
The atoms in two diastereoisomers have different spatial arrangements of atoms with
respect to distance and dihedral angles (interaction) and are, therefore, anisometric
relative to one another. For this reason, neither any two diastereoisomers nor their
transition states have the same energies (Eact values are different) and as a consequence,
they react at different rates with both chiral and achiral reagents.
Because any two diastereoisomers are anisometric to each other, they also differ in
physical properties like melting points, boiling points, refractive indices, dipole moments,
specific rotations, etc. They may have the same or opposite sign of specific rotation or some
may be optically inactive (do not rotate in plane-polarized light). They can be conveniently
separated from each other by fractional distillation or fractional crystallization because
their boiling points and solubilities are different. Again due to differences in molecular
shape and polarity, they undergo differential adsorption on an adsorbent and can be
separated easily by chromatography.
1. How are the compounds in each pair related to of each other? Are they
homomers (i.e., identical), enantiomers, diastereoisomers constitutional
isomers or not isomers of each other?
H
H3C CH3 H3C Br
(a) C C and C C
H Br H CH3
Br H Br
Cl H Cl Cl
(b) C C and C C
Cl H H H
H CD3 H H
(c) C C and C C
D3 C H D3C CD3
Br
Cl
(d) and
Br
Cl
3
(e)
3
(f) and
HO
CH3 H3 C CH2OH
(g) C and C
H CH2Br
Br H
Cl Cl
(h) C and C
Br H Br
I H I
C2H5 C2H5
(i) N and N
H Ph (CH3)2CH Ph
CH(CH3)2 H
(j) H H and H Br
Br Br Br H
(k) H3C OCH3 CH3O CH3
and
CH3O CH3 H3 C OCH3
(l) S and S
H Me
O O
Me H
O O
(m) P and P
EtO D EtO OMe
OMe D
O O
(n) Cl Cl
and
Solution
(a) Homomers:
H H
H 3C 2 3 CH3 H 3C Br H3C Br
C C 180° rotation C C C C
H Br of C-3 H CH3 H CH3
Br H Br Br
1st structure 1st structure 2nd structure
Superimposable
and therefore, identical
(b) Constitutional isomers: (1,1-dichloroethene and 1,2-dichloroethene)

(c) Diastereoisomers: (they are geometric or cis-trans isomers)
(d) Constitutional isomers: (1-bromo-2-chloronaphthalene and 2-bromo-1-
chloronaphthalene)
(e) Constitutional isomers: (1,1-Dimethylcyclopentane and cycloheptane)
(f) Not isomers of each other: (naphthalene and decaline)
(g) Constitutional isomers: (1-bromo-2-propanol and 2-bromo-1-propanol)
(h) Enantiomers:
Mirror plane
Cl Cl Cl
C Br Br C C Br
I I H H I
H
1st isomer 2nd isomer
1st isomer
nonsuperimposable
mirror images
(i) Enantiomers:
Mirror plane
C H C H C H
N N (CH ) CH N
H Ph Ph CH(CH ) Ph
CH(CH ) H H
1st isomer
nonsuperimposable
mirror images
(j) Diastereoisomers: (they are geometric or cis-trans isomers)

(k) Homomers:
H C OCH CH O CH CH O CH
CH O CH H C OCH H C OCH
superimposable and
therefore, identical
(l) Enantiomers:
Mirror
plane
180° rotation
about the axis
S S S
H Me Me
O O O
Me H H
1st isomer 1st isomer 2nd isomer
nonsuperimposable
mirror images
(m) Enantiomers:
Mirror
plane
O O O
180° rotation
P D MeO P P OMe
about the axis
EtO OEt EtO
OMe D D
nonsuperimposable
mirror images
(n) Enantiomers: (nonsuperimposable mirror images)

2. Label the following pairs of structures as homomers, constitutional
isomers, diastereoisomers or enantiomers:
H F
F H
(a) and
F H
H F
Cl Cl
(b) and
Cl
Cl
Cl
Cl
(c) and
Cl
Cl
O I O I
(d) and
I
I
CH3 H H H
(e) C C C and C C C
H CH3 H3C CH3
CH3 H Br H
(f) Br H H Br and H3 C H H CH3
H CH3 H Br
5 2 2 5
(g)
and CH3
(h)
CH3
Br Br
F F
(i) and
F F
Br Br
H Et H H
Et and H
(j) H Et
H H Et H
H
OH
H3C H H C
(k) and CH3
H C CH3 H 3C H
HO
H
Br Br Br H
(l) C C C C and C C C C
H H H Br
Solution
(a) Enantiomers:
Mirror
plane
H F F
F 180° rotation H H
F about the axis H H
H F F
nonsuperimposable
mirror images
(b) Diastereoisomers: (they are geometric or cis-trans isomers: cis-1,2-

dichlorocyclobutane and trans-1,2-dichlorocyclobutane)
(c) Homomers:
Cl
180° rotation Cl Cl
Cl about the axis
Cl Cl
1st structure
1st structure 2nd structure
superimposable and
(d) Diastereoisomers: (they are geometric or cis-trans isomers)

(e) Enantiomers:
Mirror
plane
CH3 H H H H H
90° rotation
C C C about the axis C C C C C C
H CH3 CH3 CH3 CH3 CH3
non-superimposable
mirror images
(f) Homomers:
CH3 H Br H Br H
Br H H Br 180° rotation H3C H H CH3 CH3 H H CH3

about the axis
H CH3 H Br H Br

1st structure
superimposable and
(g) Homomers:
H 5 C2 H H C2H5 H C2H5
180° rotation
about the axis

1st structure
superimposable and
(h) Enantiomers:
nonsuperimposable
mirror images
(i) Homomers:
Br Br Br
F F
180° rotation F
F
about the axis F F
Br Br Br
superimposable and
(j) Homomers: [1st structure on 180° rotation in the plane of the paper gives the
second structure.]
(k) Homomers:
HO H H
OH
H3C H 180° rotation H C H C
about the axis CH3 CH3
H C CH3 CH3 H CH3 H
HO
H 1st structure 2nd structure
1st structure
superimposable and
(l) Diastereoisomers: (cumulated trienes with two different terminal groups exist
as cis-trans or geometric isomers)
3. Tell whether the compounds of each pair are enantiomers, diastereo-
isomers, constitutional isomers or homomers.
CO2H OH
(a) H OH and HO2C CH3
CH3 H
CH2OH CH2OH
(b) H OH HO H
and
H OH HO H
CH2Br CH2Br
CO2H CH2OH
(c) H NH2 H OH
and
HO H H2 N H
CH2OH CO2H
CHO CHO
(d) HO H H OH
and
H OH H OH
CH2CH3 CH2CH3
CO2H CO2H
(e) H Br Br H
and
H Br Br H
Br H H Br
Br H H Br
CO2H CO2H
H CHO
(f) HO CHO H OH
and
HO H H OH
HO H HO H
H OH H OH
CH2OH CH2OH
Solution
(a) Enantiomers:
Mirror
plane
CO2H OH OH
H OH H3C CO2H HO2C CH3
CH3 H H
nonsuperimposable
mirror images
(b) Enantiomers: (nonsuperimposable mirror images)

(c) Enantiomers:
Mirror plane
COOH CH2OH CH2OH

180° rotation
H NH2 in the plane H OH HO H
of the paper
HO H H2 N H H NH2
CH2OH COOH COOH

nonsuperimposable
mirror images
(d) Diastereoisomers:
CHO CHO
HO H H OH
H OH H OH
CH2CH3 CH2CH3
stereoisomers
which are not mirror
images of each other
(e) Enantiomers: (nonsuperimposable mirror images)

(f) Diastereoisomers:
H CHO CHO
HO CHO H OH H OH
HO H HO H H OH
HO H HO H HO H
H OH H OH H OH
CH2OH CH2OH CH2OH
stereoisomers which
are not mirror images of
each other
4. Label the following pairs of structures as homomers, constitutional

isomers, enantiomers or diastereoisomers:
C2H5 C2H5
Cl H
(a) H Cl
and
H Cl Cl H
C2H5 C2H5
(b) H3C H H3 C Br
H3C Br H3C H
and
Br H
H Br
I H
(c) F C2H5 F I
and
F CH3 F CH3
H C2H5
CHO CH2OH
(d) H OH H OH
and
H OH HO H
CH2OH CHO
H H H H
(e) H H H H
and
Cl H H Cl
H H H H
CH3 Cl
H H3C
F H
(f) Cl and F
H H
C2H5 C2H5
CH3
CH3 H OH
(g) HO H and
H OH HO H
CH3
CH3
CH3 Cl H
(h) H Cl and
H Br H CH3
CH3
CH3
Br
Solution
(a) Homomers:
C H C H C H
H Cl Cl H Cl H
H Cl Cl H Cl H
C H C H C H
1st structure
superimposable and
(b) Enantiomers:
Mirror
plane
H3C H Br CH3 H3 C Br
H3C Br 180° rotation H CH3 H
about the axis
H3C
Br H H
H Br Br
1st isomer
nonsuperimposable
mirror images
(c) Constitutional isomers: [2,3-Difluoro-2-iodopentane and 2,3-difluoro-3-iodo-pentane]

(d) Diastereoisomers:
3 2 2
180° rotation
about the axis
2
(e) Homomers:
H H H H H H
H H H H H H
180° rotation
about the axis
Cl H passing through the H Cl H Cl
H H centre of the molecule H H H H
1st structure
superimposable
and therefore, identical
(f) Constitutional isomers: [3-chloro-2-fluoropentane and 2-chloro-3-fluoropentane]

(g) Diastereoisomers:
3 3 3
180° rotation
of back carbon
3 3
3 3
3
(h) Enantiomers:
Mirror
plane
CH H Cl H Cl
Cl H
H Cl
H Br H CH H C CH H CH
Br H CH
CH Br Br
CH
1st isomer 1st isomer 1st isomer 2nd isomer
non-superimposable
mirror images
5. How is each compound related to the sugar D-erythrose? Is it an

enantiomer, diastereoisomer, or homomer?
H
HOCH2 CHO OHC OH
(a) C C (b) C C
HO OH HO CH2OH
H H H
H
2 OHC OH
(c) (d) C C
H CH OH
HO
Solution
OH
(a) HOH C CHO OHC CH OH OHC H
C C C C C C
HO OH H H H CH OH
H H HO OH OH
identical with
D-erythrose, i.e.,
homomer
H HO
OHC OH H CHO
(b) C C C C
HO H
CH OH HOH C
OH
H
nonsuperimposable
mirror-image of D-erythrose,
i.e., enantiomer
(c)
(d) This stereoisomer is not the mirror image of D-erythrose, i.e. it is a diastereoisomer
of D-erythrose.
1. Label the following pairs of structures as enantiomers, diastereoisomers,

constitutional isomers or homomers.
H Br H H
(a) and
Br H Br Br
(b)
2 2 3
Br
(c) and
Br
CH3 CH3
(d) N and N
Ph D Ph CMe3
CMe3 D
(e) and
H H
(f) C CH3 and H3C C
Br Br
(g) and
CH3 CH3
H Br Br H
(h) and
H3C H F CH3
F H
CH3 CH3
O
(i) CH3 and CH3
O
Br Br
(j) Br and
Br
Cl Cl
(k) and
Cl Cl
Br Br Br Br
(l) and
Br Br Br Br
2. How is compound I related to compounds II to IV? Choose from diastereoisomers,
enantiomers, constitutional isomers or homomers (i.e., identical molecules).
OH OH
OH OH
OH
I II II III IV
3. How each pair of the following compounds related: (a) A and B; A and C; (c) A and
D; (d) C and D? Choose from homomers, enantiomers or diastereoisomers.
CHO CHO CHO CHO
H CH2OH Br H H Br H CH2OH
H Br H Br Br H Br H
Br CH2OH CH2OH Br
A B C D
4. How are the compounds in each pair related to each other? Are they homomers,
enantiomers, diastereoisomers, constitutional isomers or not at all isomers of each
other?
Me Br Br
(a) and
Me
Me
(b) and H Me
Me H
Me
Cl Cl
CH2CH3
(c) and
CH3
H H
(d) Cl and Cl
H H
Cl Cl
Br H
H Br
(e) C C C and C C C
Br H
H Br
H H Et H
(f) C and C
Et Et
Et H
3
3
(g)
3 3
Br
(h) and
Br
H H H H H H
C C C C C C D
(i) H H and H
C C C C H
D D D C C
H H
O
(j) C and C
O O O
H OH H
(k) N and N
H3C CH3 OH
F F F H
(l) C C C C and C C C C
H H H F
Me H H Me
D
O
(m) and
D
O
5. Label the following pairs of structures as homomers, constitutional isomers,
enantiomers or diastereoisomers:
CHO CH2OH
(a) H CH3 H NH2
and
H2N H H CH3
CH2OH CHO
COOH COOH
(b) H OH HO H
and
H OH HO H
COOH COOH
Br Cl
(c)
F Cl and F Br
I I
CO2H CO2H
(d) H OH H OH
H Cl and Cl H
H OH H OH
CO2H CO2H
CH3 Br
(e) H3C H HO H
and
H OH H CH3
Br CH3
(f)
F
H Me
(g) Me F
H and
H Cl
Me
Me H
Cl
Ph
H CH3
(h) Ph Br
Cl and H3 C H
H
H Br
Cl
Br H
H Br
(i) and
Br Br
H H
O O
(j) H CH3 H CH3
and
CH3
H
H3 C
H
H3C H H H
(k) C C C and C C C
H3C CH3
H CH3
Br Br
H Br
(l) and
H Br H H
6. Predict the product (A and B) of the following reaction and mention how they are
related to each other.
3 3
3
2
2.4 MOLECULAR CHIRALITY

2.4.1 Chiral and Achiral Molecules
A molecule is called a chiral molecule if its mirror image is not superimposable on it. On
the other hand, if the mirror image of a molecule is superimposable on the original, the
molecule is called an achiral molecule. Enantiomers only occur with compounds whose
molecules are chiral. A chiral molecule and its mirror image are called pair of enantiomers
giving rise to a type of isomerism known as enantiomerism. The word chiral comes from
the Greek word cheir, meaning ‘hand’. Chiral objects are said to possess ‘handedness’. The
term chiral is used to describe molecules of enantiomers because they are related to each
other in the same way that a left hand is related to a right hand. Alanine, for example, is
chiral, whereas glycine is achiral.
enantiomers identical
Mirror Mirror
plane plane
CO2H CO2H CO2H CO2H
C C C C
H2 N H H NH2 H 2N H H NH2
CH3 CH3 H H
Mirror images of alanine. Mirror images of glycine.

They are not superimposable They are superimposable and
and therefore are chiral and both of these two structures
enantiomeric. represent the same, achiral
molecule.
Similarly, the cis- and trans-isomers of 1,2-dibromoethene are both achiral because each
isomer is superimposable on its mirror image, as illustrated by the following formulas.
Mirror Mirror
plane plane
Br H H Br Br H H Br
C C C C
C C C C
Br H H Br H Br Br H
Mirror images of cis- Mirror images of trans-

1,2-dibromoethene 1,2-dibromoethene
The mirror images of the cis-isomer are superimposable on each other (one structure on
180° rotation gives the other structure) and therefore, the cis formulas both represent the
same, achiral molecule. This analysis is true also for the trans-isomer.
Chirality The property of any molecule of being nonsuperimposable on its mirror image
is called chirality.
Chiral centre or chirality centre or asymmetric centre A chiral centre or chirality centre or
asymmetric centre is an atom bonded tetrahedrally to four different atoms or groups (also
called ligands). A chiral centre is usually a C atom, but may also be N, P, S, Si, etc. A chiral
centre is usually indicated with an asterisk (*).
2.4.2 Source of Chirality

The source of chirality may be as follows:
1. The presence of chiral centre A molecule must be chiral if it contains a chiral centre
*
or chirality centre or asymmetric centre. Thus, the chirality of 2-butanol, CH3CH(OH)
CH2CH3, for example, is due to the presence of a chiral centre that bears four nonidentical
ligands: CH3, H, OH and CH2CH3. 2-Butanol, thus exists in two enantiomeric forms.
It is to be noted that if a molecule contains more than one chirality centre, it may or may
not be chiral. For example, D-erythrose containing two chiral centres is chiral, whereas
meso-tartaric acid containing two chiral centres is achiral.
It thus follows that the presence of more than one chiral centres is not always sufficient to
induce chirality in a molecule. A molecule may even be chiral without having any chiral
centres and this may happen when a molecule contains a chiral axis or a chiral plane.
2. The presence of a chiral axis This is an important source of chirality which may be
observed in some allenes, alkylidenecycloalkanes, spiranes, adamantanes, biphenyls,
etc. if they are properly substituted. These constitutionally nonplanar compounds show
enantiomerism if they are devoid of any Sn axis. Elongated tetrahedron approach can be
applied to explain the principle of axial chirality. When a chiral centre is replaced by a
linear grouping, e.g., C—C or C==C==C, the tetrahedron becomes elongated, i.e., extended
along the axis of the grouping. Such an elongated tetrahedron having lesser symmetry
than a regular tetrahedron will be chiral if only the two lingands at each end of the axis are
different, i.e., the minimum condition for chirality of an allene, abC==C==Cab, for example,
is that ligand a π b. Thus, the structure IA, which represents an elongated tetrahedron
(a desymmetrised tetrahedron of the type Caabb), becomes chiral and have enantiometric
relationship with its mirror image (IB). The axis along which the tetrahedron is elongated
(shown by dotted line) is called the chiral axis or the stereoaxis and the molecular chirality
of this type is called axial chirality.
3 3
3 3
Orbital structure of an allene The central carbon atom in an allene is sp-hybridized (linear)
and the two-terminal carbon atoms are sp2-hybridized (trigonal). The sp-hybridized
central carbon atom uses its two mutually perpendicular p orbitals to form two p bonds
with two outer carbon atoms. Therefore, the two p bonds must also be perpendicular. As a
consequence, the terminal ligands lie in mutually perpendicular planes. This arrangement
places the four ligands in the positions of an elongated tetrahedron.
p-bond
a a
C C C sp2 orbital picture
sp2 of an allene
b
sp b
The two mutually perpendicular planes of the terminal lingand can be shown , for example,
by a molecule of 1,3-dibromoprop-1,2-diene.
orthogonal planes
H H
C C C
Br Br
1,3-Dibromoprop-1,2-diene
Atropisomerim If the two ortho positions of a biphenyl are occupied by sufficiently large
groups, free rotation about the single bond joining the two benzene rings (the pivotal
bond) is no longer possible because the planar conformations are destabilized due to
steric repulsion. As a result, the two ring planes remain approximately perpendicular to
each other. Provided each ring has no vertical plane of symmetry (i.e., dissymmetrically
substituted), a nonplanar combination of two such phenyl groups would give rise to axial
chirality. For example, 6, 6¢-dinitrodiphenyl-2, 2¢-dicarboxylic acid having a chiral axis
exists as two enantiomers.
Mirror
plane
chiral axis NO2 HO2C CO2H NO2 chiral axis
CO2H NO2 NO2 CO2H
Enantiomers of 6,6¢-dinitrodiphenyl-2,2¢-dicarboxylic acid

(atropisomers)
This type of stereoisomerism due to restricted rotation around a single bond is known as
atropisomerism and the isomers are called atropisomers.
3. The presence of chiral planes: Various cyclic molecules like paracycloplanes, ansa
compounds and trans-cycloalkenes are chiral due to presence of chiral planes. For example:
Mirror
plane
(CH ) (CH )
(i) H C CH H C CH
O O O O
Cl Cl
Enantiomers of an compound ( = 8)
Mirror
plane
H2 C CH2 H2 C CH2
(ii) H2 C CH2 H2 C CH2

HO2C CO2H
R S
enantiomers of paracyclophane containing

two aromatic rings
Mirror
plane
(iii) H H
C C C C
H H
R S
enantiomers of trans-cyclooctene
2.4.3 Stereocentre or Stereogenic Centre

If two ligands attached to an atom of a stereioisomer are interchanged spatially to produce
a new stereoisomer, the atom is called a stereocentre. For example, C-2 of 2-butanol
*
(CH3CHOHCH2CH3) is a stereocentre as well as a chiral centre. But the C-2 and the
C-3 carbons of cis- and trans-1, 2-dichloroethene are stereocentres but not chiral centre
because interchange of H and Cl atoms on any of these C-atoms gives a new stereoisomer
(a diastereoisomer) which is not an enantiomer. Therefore, all stereocentres are not
chiral centres but all chiral centres are stereocentres. The stereocentre of 2-butanol is a
tetrahedral stereocentre, whereas that of cis- and trans- dichloroethene is a trigonal planar
stereocentre.
chiral centre and chiral centre and

stereogenic centre stereogenic centre
OH OH
2C C
3 4 –CH3 and –CH2CH3 groups
H 1CH2CH3 H CH3
are interchanged spatially
CH3 CH2CH3
enantiomers of 2-butanol
stereogenic centre
but not chiral centre
1 2 1 2
H H H Cl
C C H and Cl atoms of C2 C C
Cl Cl are interchanged spatially Cl H
-isomer -isomer
Diastereoisomers
of 1,2-dichloroethene
2.4.4 Meso Compounds

A meso-compound is an achiral compound containing chirality centres. Because they are
achiral, they are optically inactive (does not rotate the plane of polarized light). Meso-
compounds may be acylic as well as cyclic stereoisomers. A meso-compound is achiral
because at least one of its conformer has either a plain or a centre of symmetry. For
example, the following stereoisomer of 2,3-dibromobutane (CH3CHBrCHBrCH3) is a
meso-compound because it is achiral (a plain of symmetry is present in the fully eclipsed
form and a centre of symmetry is present in the anti-staggered form) even though it
contains two asymmetric centres or chiral centres.
Chiral Identical
Plane of symmetry centre
Centre of symmetry Mirror
H H H CH3 plane CH3
H3C Br H Br Br H
Br Br
C C C C
Br H Br Br H
CH3 H3C CH3
H CH3 CH3
Chiral
2,3-Dibromobutane (fully-eclipsed centre
(anti-staggered conformation) superimposable on each
conformation) other after 180° rotation
in the plane of paper
cis-1,3-Dichlorocyclohexane is an example of a cyclic meso compound. It contains two

chiral centres (C-1 and C-3) and it has a plane of symmetry passing through C-2 and C-5
for which it is achiral.
H plane of symmetry (s)

1
Cl 5
H 6
2 Cl 3
4
In fact, a meso compound is possible if a molecule of a compound with two or more

asymmetric atoms or chiral centres can be divided into structurally identical halves. For
example:
HO2CCH(OH) CH(OH)CO2H CH3CH2CHCl CHBr CHClCH2CH3
structurally structurally
identical halves identical halves
1. Identify the chiral centre(s), if any in each of the following compounds

and indicate them with an asterisk (*):
(a) CH3CHClCH2CH2CH3
(b) CH3CH2CH=CH—CHBrCH(CH3)2
≈
(c) CH3 CH2 N H(CH3) CH2 CH(CH3)2
O
≠
(d) CH3CH2CH2 N (CH3)CH2CH(CH3)CH2CH3
(e) CD3 SiCl (CH3) CH(CH3)2
≈
(f) CH3CH2CH2CH2 P H (CH3) CH(CH3)2
D D D
(g) O (h) (i)
D
O CH3
H3C Br H3C Br
16
Br
(j) (k) (l) 2
CH3 O 18
D18O
CH3
D D
(m) (n) (o)
H
O D
H H H CH3
(p) C C C C
H3C CHCl H
CH3 OH
O O CH3
(q)
Solution
* *
(a) CH3 C HClCH2CH2CH3 (b) CH3CH2CH=CH— C HBrCH(CH3)2
*≈
(c) CH3 CH2NH (CH3) CH2 CH(CH3)2
* *
(d) CH3CH2CH2 N (CH3)CH2 C H(CH3)CH2CH3
Ø
O
* *≈
(e) CD3 S iCl (CH3) CH(CH3)2 (f) CH3CH2CH2CH2P (CH3) CH(CH3)2
O
CH
(g) There is no chiral centre in the molecule (two of the substituents at
H
O
C-2 are the same).
D D
2
1
(h) There is no chiral centre in the molecule (two of the substituents at C-4
5 3
are the same).
H3C 4 Br
D Br
*
(i) (j)
* D * CH3
H 3C Br
H H 16
O
* *
CH2 *S
(k) (l)
* * O 18 O
H H
(m) There is no chiral centre in the molecule (cubane). Each of the eight carbons is
bonded to three identical substituents.
CH3
(n) * (o) There is no chiral centre in the molecule.
*
H
O
CH3 OH
3
(p) (q) O *
O CH3
*
3
2. Classify the following objects as to whether they are chiral or achiral.

(a) scissors, (b) knife, (c) shoe, (d) child’s block, (e) cricket bat, (f) a six-
pointed star, (g) hand (h) foot (i) a p orbital (j) a spiral staircase (k) helix,
(l) spool of thread, (m) spoon, (n) a golf club, (o) a tennis racket.
Solution (a) chiral, (b) achiral, (c) chiral, (d) achiral, (e) achiral, (f) achiral, (g) chiral, (h)
chiral, (i) achiral, (j) chiral, (k) chiral, (l) chiral, (m) achiral, (n) chiral, (o) achiral.
3. Write names and structures of the two lowest molecular-weight alkanes
(with no isotopic atom) that are chiral.
Solution The two lowest molecular-weight chiral alkanes (with no isotopic atom) are:
*
(i) 3-methylhexane [CH3CH2 C H(CH3)CH2CH2CH3] and 2,3-dimethylpentane [CH3CH2
*
C H(CH3)CH(CH3)2]. Each of them contains one chiral centre.
4. In each of the following pairs of compounds one is chiral and the other is
achiral. Identity them and give your reasoning.
CH3 CH3
H Br H Br
(a) and
Br H H Br
CH3 CH3
(b) Br CH2 CHBr CH2CH3 and Br2 CH CH2CH2 CH3
(c) and
Cl
Cl
H
F Br F Br
(d) H H H H and
H H H H
Br F F Br
H H
Cl
Cl Cl
(e) and
Cl
H
H
Br Br
(f)
H and H
O O
CH3CH2 O H H O H
(g) and
H O CH3CH2 CH3CH2 O CH2CH3
O O
(h) H2C C CHCl and ClCH C CHCl

O
Cl and Cl OH
(i)
OH OH Cl
CH3 CH3
HO H H OH
(j) H OH and H OH
H OH H OH
CH3 CH3
Solution
CH3
H Br
(a) Br H : Chiral (It has no plane, centre and other
Sn axis of symmetry.)
CH3
s plane CH3
H Br
: Achiral (It has a plane of symmetry.)
H Br
CH3
*
(b) BrCH2 C HBrCH2CH3 : Chiral (It contains one chiral C-atom.)
Br2CHCH2CH2CH3 : Achiral (It has a plane of symmetry.)
(c) : Achiral (It has a plane of symmetry

passing through C-1, C-4, C-7 and C-8.)
Cl
: Chiral (It has no plane, centre and other

Cl Sn axis of symmetry.)
H
F Br
(d) H H : Achiral (It has a centre of symmetry.)

H H
Br F
F Br
H H : Chiral (It has no plane, centre and other

H H
F Br
H
4
Cl
(e) : Achiral (It has a plane of symmetry
Cl
1 passing through C-1 and C-4 including
H two Cl and two H atoms.)
H
Cl
Cl : Chiral (It has no plane, centre and other
H
Br
(f) * : Chiral (It contains one chiral C-atom.)
H
Br
H
O
HO CH2CH3
(g) : Achiral (It has a centre of symmetry.)
CH3CH2 OH
O
O
CH3CH2 O CH3CH2
H OH Sn axis of symmetry.)
O
H H
(h) H2C C CHCl C C C : Achiral (It has a plane of symmetry.)
H Cl
H H
C C C : Chiral (It has no plane, centre and other
Cl Cl Sn axis of symmetry.)
1
(i) 4 : Achiral (It has a plane of symmetry
passing through C-1 and C-4.)
Cl OH
H Cl Sn axis of symmetry).
(j) : Chiral (It has no plane, centre and other

3
3
s-plane
*
5. Compare the two enantiomers of 2-butanol (CH3 CHOHCH2CH3) with
respect to:
(a) melting point, (b) boiling point, (c) relative density, (d) refractive index,
(e) specific rotation, (f) solubility in 200 g of water, (g) rate of reaction
with HBr, (h) adsorption on alumina (Al2O3), (i) rate of reaction with an
enantiomer of CH3CH2CH(CH3) COOH to form an ester, (j) specification as
R or S.
Solution They have equal but opposite specific rotation, opposite R/S specification or
designation and different rate of esterification with an enantiomer of 2-methylbutanoic
acid. The different rate of esterification is due to formation of two diastereoisomeric
transition states of different energies, i.e., the Eact values are different. All other properties
are the same.
6. What is the necessary and sufficient condition for enantiomerism?
Solution Chirality (i.e., nonsuperimposability on mirror image) is the necessary and
sufficient condition for enantiomerism.
7. What do you mean by epimers? Give examples.
Solution Two diastereoisomers which are different in the configuration of a single chiral
centre are called epimers. For example, D-ribose and D-xylose (two carbohydrates) are
two epimers because they differ only in the configuration about C-3.
CHO CHO
H OH H OH
H OH HO H
H OH H OH
CH OH CH OH
D-Ribose D-Xylose
8. Draw the eight constitutional isomers having molecular formula C5H11Br.

Label the isomers as chiral and achiral.
Solution The isomeric alkyl bromides having molecular formula C5H11Br are as follows:
*
CH3CH2CH2CH2CH2Br, CH3CH2CH2 C HBrCH3, CH3CH2CHBrCH2CH3,
I II III
*
CH3CH2 C H(CH3)CH2Br, (CH3)2CHCH2CH2OH, CH3CH2CBr(CH3)2,
IV V VI
*
(CH3)2CH C HBrCH3 and (CH3)3CCH2Br
VII VIII
Among these eight alkyl bromides, only three II, IV and VII are chiral because each of
them contains a chiral carbon atom. Other five alcohols (I, III, V, VI, and VIII) are achiral
because each of them has a plane of symmetry (s).
9. Which pairs of the salts shown below are expected to have identical
solubilities in methanol? Explain.
6 5 6 5 3
– –
6 5
3 6 5
6 5 6 5 3 6 5
3 6 5
– –
Solution
Mirror
plane
C6H5 C6H5 NH3 C6H5 C6H5 NH3
–
H Et Et COO H C6H5 H Et Et H C6H5 H
– –
NH3 H Et COO COO Et
I I IV
enantiomers
Mirror
plane
C6H5 C6H5 NH3 H H NH3
– –
H3N Et H Et C6H5 H OOC Et Et COO H C6H5
H COO
–
Et C6H5 C6H5 Et
III III II
enantiomers
Since the salts I and IV are enantiomeric, therefore, they have identical solubilities in
methanol. Similarly, III and II have identical solubilities in methanol.
10. Identify the stereocentres in each of the following compounds and predict
whether each compound is chiral or not.
Et Et
(a) Et H (b) H
Et Et
H H H
Et Et
(c) H (d) Et H
Et Et
Et H Et
Solution
(a) The compound is chiral and it contains four stereocentres.
stereocentres
Et
Et H
stereocentre Et
H H
(b) The compound is chiral and it contains three stereocentres.
stereocentres
Et
H
Et
H
(c) The compound is chiral and it contains one stereocentre.
Et stereocentre
H
Et
Et
(d) The compound is achiral and it contains two stereocentres.
Et
stereocentres
Et H
Et
H Et
11. Give an example of compound that contains a tetrahedral stereogenic

centre and a trigonal planar stereogenic centre. How can you verify that
each of these carbons is a stereogenic centre?
Solution The following alkylidenecycloalkane contains one tetrahedral stereogenic
centre and two trigonal planar stereogenic centre.
H Et
stereocentre C
H
Et
stereocentres
Whether any one of these carbons is a stereogenic centre or not can be verified by
interchanging any two groups bound to it. This interchange may generate the other
enantiomer. For illustration, the two ring bonds at the middle stereogenic centre are
interchanged. This results in the formation of an enantiomer.
bond cleavage
H Et H 180° Et Et Et
C C
H
Et Et H H H
(positions of the groups
are interchanged ring is reformed
enantiomers
by rotating 180°)
Et H 180° rotation Et Et
about the axis
H Et H H
A group that interchanges at each of the other two stereogenic centres also gives
enantiomers.
12. Which of the following compounds has a stereoisomer that is a meso
compound?
(a) Cyclohexane-1, 4-dicarboxylic acid (b) 4,5-Dipropyloctane
(c) 1,2,5-Tribromopentane (d) 2,3,4-Trihydroxybutanal
(e) 1-Bromo-2-fluorocyclopentane (f) 1,3-Dichlorocyclopentane
(g) 1,2-Dimethylcylobutane (h) 2,4-Dibromo-3-pentanol
(i) 3,5-Dichloroheptane (j) 3-Heptene
(k) 4-Methyl-2-pentyne
Solution A compound must have a stereoisomer that is a meso compound if it contains at
least two chiral atoms which are bonded to the same four substituents.
Compounds (a), (b), (j) and (k) do not have a stereoisomer that is a meso compound because
they do not have any chiral centre.
PRINCIPLES OF STEREOCHEMISTRY
CH2CH2CH3
CO2H
CH3CH2CH2CH CHCH2CH2CH3
HO2C
CH2CH2CH3
Cyclohexane-1,4- 4,5-Dipropyloctane
dicarboxylic acid
(b)
(a)
CH3CH2CH CHCH2CH2CH3 CH3 C C CH(CH3)2
3-Heptene 4-Methyl-2-pentyne
(j) (k)
The compound (c) does not have a stereoisomer that is a meso compound because it has
only one chiral centre.
Br
|
*
BrCH2 —CHCH2CH2CH2Br
1,2,5-Tribromopentane
Each of the compounds (d) and (e) contains two chiral atoms. But each of the chiral atoms
is not bonded to the same four substituents. Therefore, they do not have a stereoisomer
that is a meso compound.
Compounds (f), (h) and (i) have a stereoisomer that is a meso compound because each of
them has two chiral atoms which are bonded to the same four substituents.
Br Br
* *
* * CH3 CH CH(OH) CH CH3
Cl Cl
1,3-Dichlorocyclopentane 2,4-Dibromo-3-pentanol
(f) (h)
13. Identify the meso compounds and give your reasoning:
H H Br
CH3 C C CH3
(a) C C (b) H H
H H
Cl Cl Br
H3C CH3
H H
D3 C CD3
C C
(c) (d) CH3
H5C2 C C2H5
H F CH3
N O
Et H H
(e) Cl (f)
H N Et
O H
CH3
H OH H COOH
(g) H OH (h)
H OH HOOC H
CH3
H
H3 C Br
C
HO2C CO2H
(i) (j)
H H
C H
H3C Br
Solution
(a) It is a meso compound because it is achiral due to the presence of a plane of
symmetry (s) and contains two chiral centres.
H H
CH3 C C CH3
C C
H H
Cl s-plane Cl
(b) It is not a meso compound because it contains no chiral centre. However, it is
achiral due to the presence of a s-plane and a point of symmetry.
Br
H H
s-plane Br point of symmetry
(c) It is a meso compound because it contains two chiral centres and is achiral due to
the presence of a plane of symmetry.
H H
D3 C CD3
C C
* *
H5C2 C C2H5
H F
(d) The compound has a plane of symmetry and so, it is achiral. However, it contains
two chiral centres. Hence, it is a meso compound.
3 3
*
*
3
Me Me s
*
*
3
(e) It is not a meso compound because the molecule is chiral.
(f) The compound is achiral because it has a centre of symmetry (i). However, it
contains two chirality centres. Therefore, it is a meso compound.
N O
Et H H
H N Et
O H
(g) It is a meso compound because it is achiral due to the presence of a plane of
symmetry (s) and it contains two chiral centres.
CH3
H * OH
H OH s-plane
H * OH
CH3
(h) The molecule is chiral and so, it is not a meso compound.
2
Sn
2
(i) The molecule is achiral due to the presence of a plane of symmetry (s) and also, the
molecule contains two chirality centres. Therefore, it is a meso compound.
HO C CO H
* *
H H
s-plane
(j) It is a meso compound because it contains two chiral centres and is achiral due to
the presence of a plain and a centre of symmetry.
1. Identify the chirality centre(s), if any, in each of the following compounds and
indicate them with an asterisk (*):
CH3
(a) CH3CH2CH(CH3) (b)
CH3 O
OH
(c) (d) CH3CH2O P(OCH3)NMe2
O CH3
Me
(e) Me OMe (f) Ph N CH2 CH2 CH CH2
Me CH2Ph
O OH
(g) HO OH (h) 3 2 2 3
HO OH
CH3
(i) CH3CH2CH2Si CH2CH(CH3)2 (j)
Cl
O
(k) (l) CH3CH2N(CH3)CH(CH3)2
Cl
(m) C6D6 CH C6H6 (n) 3
(o) H3C OH (p) 3
O
Cl
(q) H Cl (r)
Cl
H H
2. Classify the following objects as to whether they are chiral or achiral and give your
reasoning:
(a) a bucket, (b) a brick, (c) an s orbital, (d) a double helix, (e) a tennis racket,
(f) gloves, (g) a scarf tied around your neck, (h) a filled spool of thread, (i) an empty
spool, (j) a five-pointed star, (k) a balance, (l) a boot, (m) a hammer, (m) a mug with
HAT written opposite the handle, (n) a mug with TAT written opposite the handle.
3. The positive carbon of a carbocation cannot be a chiral centre – Why?
4. Write the structural formulas and names for the lowest molecular-height chiral:
(a) alkene, (b) alkyne, (c) alcohol, (d) aldehyde, (e) ketone, (f) carboxylic acid,
(g) amine and (h) ester.
* *
[Hint: (a) CH3CH2 C H(CH3)CH==CH2 (3-methylpent-1-ene), (b) CH3CH2 C H
*
(CH3)CH∫∫CH (3-methylpent-1-yne), (c) CH3CH2 C HOHCH3 (2-butanol),
* *
(d) CH3CH2 C H(CH3)CHO (2-methylbutanal), (e) CH3CO C H(CH3)CH2CH3
*
(3-methyl-2-pentanone), (f) CH3CH2 C H(CH3)CO2H (2-methylbutanoic acid), (g)
* *
CH3CH2 C H(NH2)CH3 (2-aminobutane), (h) CH3CH2 C H(CH3)COOCH3 (methyl-
2-methylbutanoate)].
5. Locate the stereogenic centre(s), if any, in each of the following compounds:
H
HO
CH3 OH Me H
(a) (b) HO (c)
C
H CH2Br Me Cl
H
H Cl H CH3
CH H Cl C C CH3
(d) CH (e) C C (f) H C C
Br H Br H H
H
CH3
H CH3 H OH
(g) CH3 (h) (i) H Br
Br
H3C CH3
C2H5
Cl H H Br
(j) C C C (k)
H H
Cl Br
6. Indicate whether each compound is chiral. Identify the chiral carbons and
stereocentres (if any) in each:
Br H
Et CH Et C N
(a) (b) Et
C C
Et Et Br
CH3CH2
CH2CH3
(c) (d)
CH2CH3 H
H
H H
Cl C C CH3
(e) (f) CH3 (g) C
CH3 H CH3
H
CH3
Br Br
(h) O (i) H (j) H
H
Cl O
7. Two stereoisomers of the compound (H2N)2 PtCl2 have different physical properties.
From this observation predict whether the shape of the molecule is tetrahedral or
square planar.
8. Indicate whether each of the following statements is true or false. If false, explain
why?
(a) Constitutional or structural isomers may be chiral.
(b) If a compound has an enantiomer it must be chiral.
(c) Every chiral compound has a diastereoisomer.
(d) Each molecule containing one or more chiral carbons is chiral.
(e) Any molecule containing a stereocentre must be chiral.
(f) Any molecule with stereogenic centre must have a stereoisomer.
(g) Mirror-image molecules are in all cases enantiomers.
(h) All chiral molecules have no plane of symmetry.
(i) If a structure has no plane of symmetry it is chiral.
(j) Some chiral compounds are optically inactive.
9. Draw the structure of the chiral cyclic alkane (with no isotopic atom) of lowest
molecular mass.
10. Draw structures of all compounds having molecular formula C6H12Br2 that exist
as meso compounds. Indicate how many meso compounds are possible for each
structure.
11. Neither cis-2-hexene nor trans-2-hexene (both achiral) a meso compound, why?
12. Identify the meso compound and give your reasoning.
H3C Br
CH3 Et
C C C C H
(a) (b) C C C C
Cl H
Et H
H Cl Br
Cl3C CCl3
(c) C C (d)
H H
D D O O
PRINCIPLES OF STEREOCHEMISTRY
CH3
Br F3C CF3
(e) Br (f)
H H
CH3
Cl
F3C H
(g) (h) Cl H
H CH3
H
Ph Et
Cl H
(i) Ph H (j) (CH2)3
H Et
H Cl
CH3 CH3
CH3 H C
D H Br
(k) C H Br (l) H Br
H
CH3
Cl
2.5 CONFIGURATION AND CONFIGURATIONAL NOMENCLATURE

The arrangement of atoms that characterizes a particular stereoisomer is called its
configuration. Using the test of superimposability, we may conclude, for example, that
there are two stereoisomeric 2-methyl-1-butanols. Their configurations are I and II.
CH3 enantiomers CH3

C C
H CH2OH HOH2C H
C2H5 C2H5
I II
2-Methyl-1-butanol
In the laboratory samples of two compounds of formula CH3CH2CH(CH3)CH2OH, we find

that one rotates the plane of polarized light to the right and the other to the left. These are
taken in two bottles and labelled as (+) -2-methyl-2-butanol and (–) -2-methyl-2-butanol.
Now, an important question arises, which configuration does each isomer have? Does the
(+) isomer, say, have the configuration I or the configuration II ? How can it be known?
Which structural formula, I or II, is to be drawn on the label of each bottle? That is to say,
how can the configuration of each isomer be assigned. For any optically active compound,
the question of configuration could not be answered in an absolute sense until 1951. In
that year, J.M. Bijvoet, the director of the Van’t Hoff Laboratory of the University of
Utrecht in the Netherlands, reported by using a special technique of X-ray diffraction that
he had determined the actual arrangement, i.e., absolute disposition in space of the atoms
of the optically active sodium rubidium (+) -tartaric acid. Over the years, prior to 1951,
the relationships between the configuration of (+) -tartaric acid and the configurations of
hundreds of optically active compounds have been determined (through reaction of known
stereochemistry, e.g. SN2, or reactions in which no bond to the chirality centre is broken).
The configurations obtained by this way is called relative configurations. When the actual
or absolute configuration of (+) -tartaric acid became known, the absolute configurations
of other compounds also became known. (In the case of 2-methyl-1-butanol, for example,
the (+) -isomer is known to have the absolute configuration I and the (–) -isomer is
known to have the absolute configuration II.) Therefore, the absolute configuration of a
chirality centre is the actual arrangement of groups in space at the chirality centre that
distinguishes it from its mirror image and can be designed by R or S.
2.5.1 D, L-System of Configurational Designation

The relationship between the configurations of two chiral molecules is the basis of D,
L-system of designation. When two chiral molecules can be interconverted chemically
without breaking any bond to the chirality centre, they are said to have the same relative
configuration independent of the direction of rotation of the plane polarized light. The
oldest system of nomenclature of optical isomers containing chiral centres is the D,
L-system, introduced by as early as 1890s, by German chemist Emil Fischer, who worked
extensively with carbohydrates and established the relative configuration of (+) -glucose.
Glyceraldehyde (HOCH2CH(OH)CHO), the simplest carbohydrate molecule, was chosen
by Fischer as the standard for defining configurations of molecules with chirality centres.
The (+) -enantiomer (that rotates the plane polarized light clockwise) was arbitrarily
labelled as D-glyceraldehyde and the other enantiomer was L-glyceraldehyde. According
to Fischer, the projection formulas of D- and L-glyceraldehydes are as follows:
CHO CHO
H OH HO H
CH2OH CH2OH
D-(+)-Glyceraldehyde L-(–)-Glyceraldehyde
Any molecule containing a single asymmetric centre was assigned as D or L on the basis
of a resemblance between the group on its chirality centre and those in glyceraldehyde,
i.e., the enantiomer containing ‘similar’ groups in the same place, as D-glyceraldehyde,
was designated as ‘D’ and its mirror image as ‘L’. Thus, the naturally occurring amino acid
alanine, for example, was labelled as L-enantiomer.
CO2H
H2N H
CH3
L-Alanine
In fact, L-alanine can be chemically correlated to L-glyceraldehyde. This generic

nomenclature, however, did not work since in many cases, both of the enantiomers of a
compound may be chemically correlated to the same glyceraldehyde. For example, both
(+)- and (–)-lactic acid can be correlated to D-glyceraldehyde because the former can be
obtained by reduction of —CHO to —CH3 and oxidation of —CH2OH to —COOH and
the latter by oxidation of —CHO to —COOH and reduction of —CH2OH to —CH3. To
circumvent this drawback, Rosanoff (1906) modified the system and suggested a projection
nomenclature according to the following conventions.
In this modified system of D, L-nomenclature, the molecule of the type RCHXR¢ is written
in Fischer projection such that the carbon atoms of the longest chain R—C—R¢ are placed
on the vertical line with C-1 (according to the IUPAC system of nomenclature) at the top.
Then, if X (which is taken to be a negative group) is on the right horizontal bond, the
molecule is given D designation and if it is on the left, the molecule is given L designation.
R R
H X X H
R¢ R¢
D-enantiomer L-enantiomer
Examples:
[It is to be noted that the D, L-symbols have no relation with the sign of rotation of the
optically active compound.]
When a chirality centre remains attached with a ring system, then the total number of
carbon atoms including the ring carbons are to be considered for writing the main chain
in Fischer projection and for this; a C6H5-group attached to a chirality centre is generally
placed at the lower vertical bond in Fischer projection. For example:
CH2CH2CH3 CO2H CH2CH2CH3

H CH3 CH3 OH H3C H
C6H5 Ph C6H5
D-2-Phenylpentane D-Atrolactic acid L-2-Phenylpentane
When the chiral centre contains two different substituents of comparable electronegativity,
then D- and L-symbols are separately used for each chiral centre to specify the configuration.
This is also true when two different alkyl groups are present in horizontal bonds. For
example:
CH3 CO2H
Br F H3C C2H5
C2H5 Ph
2L-Bromo-2D- 2D-Ethyl-2L-methyl-2-phenyl
fluorobutane -butanoic acid
D, L-designation to compounds containing more than one chirality centre involves

specifying each of the chirality centres by D, L-notations. For example:
1
CH3 CO2H H 1
OH
2 2
Et CO2H
HO H H Br Cl H
5
H 3 OH HO CH3 H 3 CH3 Me HH OH
4 5 6
CH2CH2CH3 C2H5 4
Hexane-2L, 2D-Bromo-3D-methyl 2L-Chloro-3D- 1L-Ethyl-3D-

3D-diol -3L-hydroxypentanoic methyl-1D- hydroxy-5L-methyl
acid cyclopentanol 1D-carboxylic acid
2.5.2 Specification of Configuration: The R, S-System

The two enantiomers of bromochloroiodomethane are the following:
I I
C C
H Br Br H
Cl Cl
I II
If we name these two enantiomers using only the IUPAC system of nomenclature,
both enantiomers will have the same name: bromochloroiodomethane. This is not
desirable because each compound must have its own distinct name. Moreover, we must
be able to write the structure of the compound from its name alone. Given the name
bromochloroiodomethane, we could write either structure I or structure II.
Three chemists, R.S. Chan (England), Sir Christopher Ingold (England) and V. Prelog
(Switzerland), devised a system of nomenclature that, when added to the IUPAC system,
solves both the problems. This system of configurational nomenclature is called the R,
S-system or the Cahn-Ingold-Prelog (CIP) system and since this system of specifying
configurations is independent of any reference compound, it is often termed Absolute
Configuration assignment. In this system, configurations of stereoisomers are designated
on the basis of individual chiral centres.
According to this system, one enantiomer of bromochloroiodomethane should be
designated (R)-bromochloroiodomethane and the other enantiomer should be designated
(S)-bromochloroiodomethane. [(R) and (S) are from the Latin words rectus and sinister,
meaning right and left, respectively.] These molecules are said to have opposite
configurations.
2.5.2.1 Assigning (R) and (S) configurations

For assigning R or S designation to any chiral centre of a molecule, certain conventions
are to be followed. These are given below:
1. The four different groups or atoms attached to each of the chiral centres are
ranked in order of priority and a priority symbol 1, 2, 3, 4 or a, b, c, d is assigned
to them based on some sequence rules such that the decreasing order of priority is
1 > 2 > 3 > 4 or a > b > c > d (1 or ‘a’ having the highest priority and 4 or ‘d’ the
lowest).
2. The molecule is then oriented with the lowest priority group (4 or ‘d’) back (bonded
by a hatched wedge), and the relative positions of the remaining three groups are
visualized when a hypothetical path is traced from the first-priority group, through
the second, to the third, i.e., 1 Æ 2 Æ 3 or a Æ b Æ c. If the path describes a
clockwise motion, then the chiral centre is said to have R configuration and if the
path describes a counterclockwise or anticlockwise motion, then the chiral centre
is said to have S configuration. The designation (R) or (S) is written in italics
within parentheses followed by a hyphen before the name of the compound (when
necessary with appropriate locants).
Æ Æ
Æ Æ
If in the figure, the group or atom of lowest priority is shown above the plane of the
paper or in the plane of the paper, the molecule is to be rotated by 120° about the axis
passing through the chiral carbon and a group in the plane of the paper, so as to place
the lowest priority group back (bonded by hatched wedge). Then, the configuration of the
chiral centre is determined by usual process. For example:
120° rotation
Æ Æ
120° rotation
Æ Æ
120°
120° rotation
Æ Æ
R
The lowest priority group (4) may also be bonded by a hatched wedge by simply
interchanging group 4 with the group that is bonded by a hatched wedge. However, in
that case the actual configuration will be opposite to that determined.
Æ Æ
Example: The chiral centres of the two enantiomers of bromochloroiodomethane are

designated as R or S as follows. The atoms bonded to the chiral centre are ranked in
order of priority. The atomic number of atoms attached to the chiral centre determines
the relative priorities. The higher the atomic number of the atom, the higher the priority
(given in rule 1 for assigning priority). Therefore the priority order is:
–I –Br –Cl –H
1 2 3 4
highest lowest
priority decreases
Then, the chiral centre of each enantiomer is viewed from the opposite side of the lowest
priority group (4 or ‘d’). In the enantiomer A, 1 Æ 2 Æ 3 traces a clockwise path. Hence,
its configuration is R, i.e., it is (R)-bromochloroiodomethane. On the other hand, in the
enantiomer B, 1 Æ 2 Æ 3 traces a counterclockwise path. Hence, its configuration is S, i.e.
it is (S)-bromochloroiodomethane.
1
I
viewer 1 1 Æ 2 Æ 3 traces a clockwise
C path. Hence the configuration is R, i.e.
4 H Br
3 2 the IUPAC name of this enantiomer
Cl 2 is (R)-bromochloroiodomethane.
3
Enantiomer A of
bromochloroiodomethane
1
1 1 Æ 2 Æ 3 traces a counterclockwise
I
path. Hence the configuration is S, i.e.
viewer C the IUPAC name of this enantiomer
H 4 2 3
Br is (S)-bromochloroiodomethane.
2 Cl
3
Enantiomer B of
bromochloroiodomethane
Sequence rules or CIP chirality rules for determining priority of ligands attached to a chiral
centre.
Rule 1. If the four atoms bonded to the chiral centre and are all different, priority depends
on atomic number, with the atom of higher atomic number getting higher priority. An
example has already been discussed.
Rule 2. In cases where two of the attached atoms are isotopes of each other, higher atomic
mass has higher priority over lower. For example, the priority order of the isotopes of
hydrogen is T > D > H. However, —12CH2NH2 will get preference over —14CH2CH3 because
the atomic number of the second atom (N) in the former group is higher than that of the
second atom (C) in the latter group, even though the mass number of methylene carbon
in the latter group is higher than that in the former. Again, —CH218OH has the higher
priority than —CD216OH because 18O has highest atomic number as well as higher mass
number, even though D is present in the latter group. Similarly, —CH2OH has priority
higher than —CD2NH2.
Rule 3. When the chirality centre is attached to two or more identical atoms, i.e.,
if the priority remains undecided, then the atomic numbers of next sets of atoms are
to be considered and exploration is continued until a point of difference is reached. The
priorities are then assigned at the first point of difference. For example, in 2-butanol
1 2 3 4
(CH3 CHOHCH2 CH3 ) , the oxygen atom gets highest priority (1) and H gets lowest
priority (4) using Rule 1. The two remaining groups are —CH3 and —CH2CH3. The first
point of difference is at the two carbons attached to the chiral centre. The carbon of the
methyl group, i.e., C-1 is bonded to three H atoms, i.e., to (H, H, H), while the first carbon
of the ethyl group, i.e., C-3 is bonded to one carbon and two H atoms, i.e., to (C, H, H).
Since carbon has higher atomic number than hydrogen, ethyl (—C2H5) gets higher priority
over methyl (—CH3).
Rule 4. When multiple bonded groups like H C=O, —CH=CH2, —C∫∫N, etc. are found to
H
be attached with the chiral carbon, both atoms attached to the multiple bond are treated
as if they are duplicated or triplicated single bonds. This pretreatment for assigning
priority of groups containing multiple bonds is termed as ‘replication’. The replicated
atoms are enclosed in parentheses in the expanded form of the group. Each replicated
atom, except for H atom, is converted to single bond tetracovalency by adding the so-called
‘phantom atoms’. Phantom atoms are denoted by the cipher zero which are often shown as
subscript. These are used to bring the valency up to four. Phantom atoms are imaginary
atoms having an atomic number zero. A few examples of replications of multiple bonds
are shown below:
2 00
000 000 000 000
000 000 000 000
000 000 000 000
000 000
000
000
000 000
The priority sequence of the four groups —CHO, —CH==CH2, —C∫∫CH and —C6H5 may
be determined as follows. The first atoms are connected, respectively, to (H, O, O), (H, C,
C), (C, C, C) and (C, C, C). Therefore, —CHO gets highest priority and —CH==CH2 gets
lowest priority because only one oxygen outranks three carbons and three carbons outrank
two carbons and a hydrogen. To rank the remaining two groups, we have to proceed further
along the chains. The phenyl group (—C6H5) has two of its (C, C, C) carbons attached to
(C, C, H), while the third carbon is attached to (000) and is, therefore, gets priority over
the —C∫∫CH group, which has one of its (C, C, C) carbons is connected to (C, C, H), while
the other two are connected to (000)s.
the carbons bonded
to the chiral centre
C CH
(C, C, C) (C, C, C)
(C, C, H) (C, C, H) (000) (C, C, H) (000) (000)

two one
for this –Ph group has higher

priority than –C ∫ CH group
Rule 5. When two enantiomeric groups are found to be attached with the chiral centre,
the group with R configuration gets priority over the group with S configuration. Again,
(R, R) or (S, S) gets priority over (R, S) or (S, R) configuration.
Rule 6. When the chiral centre is attached to two groups having cis and trans configuration,
then cis gets priority over trans. In case of E and Z groups, Z gets priority over E.
A list of common groups and atoms with increasing priority
1. H 10. CH=CHR 19. CHO 27. NHR 36. F

2. D 11. CR3 20. COR 28. NR2 37. SH
3. CH3 12. C(R) = CR2 21. CONH2 29. NHCOR 38. SR
4. CH2CH3 13. C∫CH 22. COOH 30. NO 39. SOR
5. CH2CH=CH2 14. C6H5 23. CO2R 31. NO2 40. SO2R
6. CH2C∫CH 15. C∫CR 24. COX 32. OH 41. SO3H
7. CH2C6H5 16. CH2OH 25. CX3 33. OR 42. Cl
8. CHR2 17. CH(OH)R (X = halogen) 34. OPh 43. Br
9. CH=CH2 18. CR2OH 26. NH2 35. OCOR 44. I
H gets priority over (..) lone pair of electrons
R and S assignments in compounds with two or more stereogenic centres When a compound
has more than one stereogenic centre, the R or S configuration must be assigned to each
of them. For example, in the following stereoisomer of 2,3-dichloropentane, C-2 has the S
configuration and C-3 has the R. So, the complete IUPAC name of the compound is (2S,
3R)-2, 3-dichloropentane.
According to the sequence rules, the priority order of groups around C-2 is —Cl —
CHClCH2CH3 —CH3 H and around C-3 in —Cl —CHClCH3 —CH2CH3 H.
Configuration of C-2:
3 2 3 Æ Æ
2 3
S.
H3C CH2CH3 3 3 1 Æ 2 Æ 3 traces a clockwise path.
2 3
C C Hence, the configuration of C-3 is .
H H4
2
2 1
Cl Cl 1
Assigning R, S-designation to chiral centres when a compound is represented by Fischer

projection The procedure to determine the configuration of a compound drawn as a
Fischer projection involves two steps: (i) the groups or atoms bonded to the chiral centre
are to be ranked in order of priority and (ii) a semicircle is to be traced joining 1 Æ 2 Æ
3 (or a Æ b Æ c), ignoring 4 (or d), the lowest priority group. When the group with lowest
priority (4 or d) is on a vertical bond in Fischer projection, the sequence gives the correct
configurational descriptor. That is, if 1 Æ 2 Æ 3 (or a Æ b Æ c) traces a clockwise path, the
configuration is R, but if the sequence traces counterclockwise path, the configuration is S.
For example:
(i) According to the sequence rules, the priority order of groups in 2-chloropropanoic
*
acid (CH3 C HCl COOH) is —Cl —COOH —CH3 H.
2
COOH 2
3 H3C Cl 1 3 1 (the lowest priority group
is on a vertical bond)
H 4
4
Since 1 Æ 2 Æ 3 traces a counterclockwise path, the configuration of this enantiomer
of 2-chloropropanoic acid is S, i.e., the compound is (S)-2-chloropropanoic acid.
(ii) According to the sequence rules, the priority order of groups in 2-butanol
(CH3CH2CHOHCH3) is —OH —CH2CH3 —CH3 —H.
4
H 4
1 HO CH2CH3 2 1 2 (the lowest priority group is
on a vertical bond)
CH3 3
3
Since 1 Æ 2 Æ 3 traces a clockwise path, the configuration of this enantiomer of
2-butanol is R, i.e., the compound is (R)-2-butanol.
If the group or atom with the lowest priority is on a horizontal bond, the answer we get
from the direction of the arrow will be opposite of the correct answer. For example, if
the arrow points clockwise, suggesting the R configuration, then the compound actually
has the S configuration and if the arrow traces a counterclockwise path, suggesting the
S configuration, then it actually has the R configuration. For example, in the following
enantiomer of lactic acid, the group with the lowest priority is on a horizontal bond, so a
clockwise arrow signifies the S configuration.
The configuration of a molecule drawn in Fischer projection with the lowest priority group is
on a horizontal bond and may also be determined by bringing the lowest priority group into a
vertical bond, either above or below, by changing the positions among three groups sequentially
on the chiral centre without disturbing the position of the fourth or by exchanging two pairs
of groups simultaneously. Now, if 1 Æ 2 Æ 3 traces a clockwise path, the configuration is
R and if the arrow traces a counterclockwise path, the configuration is S. For example:
2
COOH 3 4
interchange of
4H NH2 1 4 1 positions of 1, 4 1 3 (lowest priority
and 3 sequentially group is on a
CH3 2 2 vertical bond)
3
Alanine
Since 1 Æ 2 Æ 3 traces a clockwise path, the configuration of this enantiomer of alanine is
R, i.e., the compound is (R)-alanine.
This procedure of assigning R, S-designation can also be applied to Fischer projections
with more than one chiral centre. For example:
CHO
2
H Br
3
H Cl
CH3
2-Bromo-3-Chlorobutanal
In 2-bromo-3-chlorobutanal, the priority order of groups attached to C-2 chiral centre is:
—Br CHClCH3 —CHO —H and the priority order of groups linked to C-3 chiral
centre is: —Cl —CHBrCHO —CH3 —H.
Since 1 Æ 2 Æ 3 traces a counterclockwise path, the configuration of C-2 is S.

2
CHBrCHO 2 2
interchange of
4H Cl 1 4 1 positions of 1, 4 1 3 (the lowest
and 3 sequentially priority group
CH3 3 4 is on a
3 vertical bond)
Since the arrow from 1 Æ 2 Æ 3 points clockwise, the configuration of C-3 is R. Therefore,
the complete stereochemical name of the compound is (2S, 3R)-2-bromo-3-chlorobutanal.
A ‘very good’ (mnemonic) procedure for assigning absolute configuration to chiral centres
is due to Epling (1982). The procedure, applicable to Fischer projections only involves
two operations: (a) assigning priority symbols 1, 2, 3, 4 or a, b, c, d to each of the groups
attached to the chiral centre and (b) tracing a semicircle joining 1 Æ 2 Æ 3 or a Æ b Æ
c ignoring 4 or d, the lowest priority group. If 4 or d is found on either of the vertical
bonds, the path traced gives the correct description, i.e., R for clockwise motion and S for
counterclockwise motion. But if 4 or d remains on a horizontal bond, the counterclockwise
path indicates R configuration and the clockwise path indicates S configuration, i.e., the
descriptors arrived at from the sequence 1 Æ 2 Æ 3 or a Æ b Æ c have to be reversed.
This procedure is very helpful for assigning absolute configuration to chiral centres in a
molecule containing any number of chiral centres. For example:
2 3
Priority of groups at different chiral centres is as follows:

C-2 : —OH —CClFCHOHCH2CH3 —CHO H
C-3 : —Cl —F —CHOHCHO —CHOHCH2CH3
C-4 : —OH —CClFCHOHCHO —CH2CH3 —H

At C-2 and C-4, the lowest priority ligand (H) is on the horizontal bond and therefore,
the absolute configuration of C-2 is S, because 1 Æ 2 Æ 3 or a Æ b Æ c traces a clockwise
path and the absolute configuration of C-4 is R, because 1 Æ 2 Æ 3 or a Æ b Æ c traces
a counterclockwise path. Since the lowest priority group, i.e., —CHOHCH2CH3 on C-3 is
on a vertical bond and the sequence 1 Æ 2 Æ or a Æ b Æ c traces a clockwise path, the
absolute configuration of C-3 is R. Therefore, the complete IUPAC name of the compound
is (2S, 3R, 4R)-3-chloro-3-fluoro-2,4-dihydroxyhexanal.
Ra, Sa-nomenclature of allenes

For assigning Ra, Sa designations to allenes having chiral axis, the structure of the allene
is to be regarded as an elongated tetrahedron and the molecule is to be viewed along the
axis. It is immaterial from which end the allene is viewed. The four groups (two on each
terminal sp2 carbon) are projected onto a plane right angles to the chiral axis. In the
resulting Newman projection formula, the near groups are to be given priority over the
far. For better visualization, the near groups are to be placed on a thick line (horizontal or
vertical). In the chiral allene xyC = C = Cxy, if the group x has priority higher than y, then
the horizontally placed (nearest to the viewer) x and y are to be numbered 1 and 2 and the
vertically placed (rear) x and y are to be numbered 3 and 4, respectively. Then, if 1 Æ 2 Æ
3 traces a clockwise path, the configuration is Ra (as shown in the figure below) and if 1
Æ 2 Æ 3 arrow points counterclockwise, the configuration is Sa (‘a’ stands for ‘axial’). The
same procedure is followed to assign R, S-designations to cumulenes having even number
of double bonds (4, 6, 8, etc.).
Æ Æ
It is to be noted that if in the projection formula the lowest priority group (4) remains in
a horizontal bond, it is not necessary to transfer it to any vertical bond before tracing the
path in the sequence 1 Æ 2 Æ 3.
Example:
3
CO2H
viewed from
the side A CH2CH3 1
2H
H
Viewer H 4
CO2H
A B
C C C
H
CH3CH2 Viewer 1
I CO2H
Hexane-2,3-dienoic acid
(CH3CH2– H– ; –COOH –H) viewed from H 4
the side B 3 CH3CH2
H
2
1 Æ 2 Æ 3 traces a clockwise
path. Hence, the configuration
of the molecule is Ra.
viewed from 2 3
the side A
2
A B
2 3 2
viewed from
the side B 3 2
Æ Æ
Sa
Ra, Sa-nomenclature of biphenyls

Ra, Sa-descriptors to the atropisomers of biphenyl derivatives can also be assigned by
the same method as in the cases of substituted allenes with chiral axis. In this method,
the four carbon atoms C-2, C-6, C-2¢ and C-6¢, which correspond to the four vertices of
an elongated tetrahedron are to be taken into considerations for the application of the
sequence rules according to CIP conventions. In the following biphenyl derivative, for
example, the priority sequence in the left ring is C-2 C-6 (–NO2 –COOH) and in the
right ring is C-2¢ C-6¢ (–NO2 –COOH). Therefore, the configuration is Sa when the
molecule is viewed from either direction (A or B).
1
NO2
viewed from
the side A NO2 3
4 HO2C
along 1-1¢
bond
CO2H
Viewer NO2 CO2H 2
5 6 2¢ 3¢
A B
4¢
4 1 1¢
3 2 6¢ 5¢ Viewer 3
HO2C NO2
NO2
viewed from
( )-6,6¢-Dinitrobiphenyl-2,2¢- the side B
dicarboxylic acid 1 O2 N CO2H 2
along
1¢-1 bond
CO2H
4
The traced path from 1 Æ 2 Æ 3
is counterclockwise. Hence,
the configuration is .
When C-2 and C-6 in a ring are attached to identical groups and C-3 and/or C-5 are also
substituted, the priority order of the ortho carbons is then determined through exploration
around the ring or side chain. Thus, in the following example, the priority order of the
ortho carbons in the right ring is C-2¢ C-6¢ (–Me –H).
2
viewed from
the side A
2
2
6¢
A 5¢ B
4¢
1 1¢
2¢ 3¢
2
2
viewed from
the side B
Æ Æ
a
2.5.3 Erythro and Threo Nomenclature of Compounds with Two

Adjacent Chiral Centres
The terms erythro and threo are commonly used to describe pairs of enantiomeric molecules
of the type abxC—Caby. That enantiomeric pair is designated by the prefix erythro in
which at least two pairs of similar groups eclipse each other and that enantiomeric pair is
designated by the prefix threo in which only one pair of similar or like groups eclipse each
other. Therefore, A and B are the erythro-forms because in these structures a eclipses a
and b eclipses b and C and D are the threo-forms because in these structures a eclipses a
and b eclipses b but not both a, a and b, b at the same time.
Mirror Mirror
plane plane
a a a a b a a b
b b b b a b b a
C C C C C C C C
y x x y y x x y
a b b a a b b a
a b b a
b x a x a x b x
y y y y
a b b a a b b a
a b b a b a a b
x x x x
y y y y
x x x x
a b b a a b b a
a b b a b a a b
y y y y
erythro enantiomers threo enantiomers
4-Bromo-3-hexanol containing two chiral centres, for example, can exist in two pairs of
enantiomeric forms. One pair is erythro-4-bromo-3-hexanol and the other pair is threo-4-
bromo-3-hexanol. This may be shown as follows:
Mirror Mirror
plane plane
CH CH CH CH CH CH CH CH
H Br Br H Br H H Br
H OH HO H H OH HO H
CH CH CH CH CH CH CH CH
(±)-erythro-4-Bromo-3-hexanol (±)-threo-4-Bromo-3-hexanol
An erythro-isomer is diastereoisomeric with a threo-isomer.
2.5.4 The E-Z System of Designating Alkene Diastereoisomers

In article 2.3.3, we learned the terms cis and trans to designate the stereochemistry
of alkene diastereoisomers (cis-trans-isomers). The terms cis and trans are unambiguous
only when used to designate the stereochemistry of alkenes of the type Cab==Cab or
Cab==Cac. When, however, all the four groups attached to the two doubly bonded carbons
are different, cis and trans prefixes cannot be applied at all. For example, it is not possible
to designate two isomeric 1-chloro-1-fluorobut-1-enes (I and II) as cis and trans since no
pairs of substituents are the same.
Cl C2H5 Cl H
C C C C
F H F C2H5
I II
To avoid this difficulty, E-Z system of nomenclature has been introduced.

The E-Z system is used as follows:
1. The two groups attached to one doubly-bonded carbon are examined and relative
priorities of the groups are determined according to the CIP sequence rules.
2. The operation is repeated at the other carbon atom.
3. The group of higher priority on one carbon atom is compared with the group of
higher priority of the other carbon atom. If the two groups of higher priority are
on the same side of double bond, the alkene is said to have the Z-configuration
(Z from the German word Zusammen, meaning ‘together’). If the two groups of
higher priority are on opposite sides of the double bond, the alkene is said to have
E configuration (E from the German word entgegen, meaning ‘opposite’). These
configurational designations are added to the name of the alkene as an italics
prefix in parentheses.
higher priority higher priority higher priority lower priority

C C C C
lower priority lower priority lower priority higher priority
Z-configuration E-configuration
(groups of higher priority are (groups of higher priority are
on the same side of the double bond) on the opposite sides of the double bond)
Example:
higher priority higher priority higher priority
F I F H F CH3
C C F CH3 C C
H CH3 H I
higher priority
(Z)-1-Fluoro-2-iodopropane (E)-1-Fluoro-2-iodopropane
Although in most cases, Z corresponds to conventional cis and E corresponds to trans, this
may not always be the case. For example:
F F F>H F Cl
C C Cl > F C C
H Cl H F
(E)-1-chloro-1, 2-difluoroethene (Z)-1-chloro-1, 2-difluoroethene
[cis-1-chloro-1,2-difluoroethene] [trans-1-chloro-1,2-difluoroethene]
When cis-trans- isomers contain two or more double bonds, the configuration of each double
bond is to be specified with appropriate locants before the designations E and Z. For example:
H 3 4 H
1 2 C C 5 6 H
CH3CH2 C C 7 8 9
H CH2CH2CH3
(3Z, 5E)-3,5-Nonadiene
When there is a choice in counting the carbon atoms in a molecule having both (Z) and (E)
descriptors, Z gets preference over E. For example:
1 2
CH3CH2 3 4 H 7 8
C C 5 6 CH2CH3
H C C
H H
(3Z, 5E)-3,5-Octadiene
[Lower numbers are assigned
to the (Z) double bond]
It is to be noted that E-Z descriptors are not used for disubstituted cycloalkanes.
2.5.5 R, S and E, Z Assignment in the Same Molecule

(Geometric Enantiomerism)
Geometric isomers are optically inactive. However, if they develop molecular chirality
(with or without chiral centre), then they also exhibit optical isomerism (rotates plane
polarized light). In such cases, the isomerism is known as geometric enantiomerism. If
a molecule contains both a chiral centre or chiral axis and an appropriately substituted
double bond (C = C, C = N) to give E, Z-isomers, then four stereoisomers are possible.
According to the CIP system, they may be designated as (R, E), (S, E), (R, S) and (S, Z).
For example:
(i) 1-Phenylpent-2-en-1-ol (PhCHOHCH=CH CH2CH3), which contains a chiral centre
and a suitably substituted double bond capable of giving cis-trans isomers, exists
as four stereoisomeric forms as given below:
2 5 5 2
2 5 5 2
(A, B) and (C, D) are two pairs of enantiomers. But each of (A, C), (A, D), (B, C) and
(B, D) represents a pair of diastereoisomers.
(ii) 1,5-Dichloropent-1,2,4-triene (ClCH=C=CH–CH=CHCl) shows geometric enantio-

merism because the allene moiety is chiral and the double bond is appropriately
substituted to give cis-trans isomers. Its four possible stereoisomers are as follows:
Mirror
plane
Cl H Cl
H
C C C H H C C C
R S
C C C C H
H E Cl Z H
H Cl
A B
(1 , 4 )-1,5-Dichloropent- (1 , 4 )-1,5-Dichloropent-
1, 2, 4-triene 1, 2, 4-triene
A pair of enantiomers
Mirror
plane
Cl H Cl
H
C C C Cl Cl C C C
R S
C C C C H
H Z H Z H
H H
C D
(1 , 4 )-1,5-Dichloropent- (1 , 4 )-1,5-Dichloropent-
1, 2, 4-triene 1, 2, 4-triene
A pair of enantiomers
(A, B) and (C, D) are two pairs of enantiomers. But each of (A, C), (A, D), (B, C) and
(B, D) represents a pair of diastereoisomers.
2.5.6 Syn-anti Nomenclature for Aldols

Aldol condensation can create two asymmetric and also stereogenic centres in the products
(i.e., aldols) depending on the nature of the reactants. Many aldol condensations are found
to be diastereoselective.
OH OH
O O
R¢ R≤ + R¢ R≤
O O R R
Base
R¢ + R≤ I I¢
H
R OH
OH O
O
+ R¢ R≤
R¢ R≤
R
R
II II¢
Diastereoisomers of the type (I, I¢) are commonly referred to as syn or erythro and those of
the type (II, II¢) as anti or threo.
2.5.7 Number of Stereoisomers for Compounds with

Chiral Centres
The number of stereoisomers (both chiral and achiral, i.e., meso compound) for a compound
having chiral centres depends upon the number as well as the nature of chiral centres
present in the compound.
Case 1: If a molecule contains even or odd number of ‘n’ different chiral centres and
the molecule cannot be divided into two mirror-image halves, the number of possible
stereoisomer = 2n, all of which are chiral. For example:
*
(a) Lactic acid (CH3CHOHCOOH) has n = 1 and therefore, the number of stereoisomer
is 21 = 2, which are enantiomers.
Mirror
plane
COOH COOH
HO CH3 H3C OH
H H
(R)-Lactic acid (S)-Lactic acid
Enantiomers
* *
(b) 2-Bromo-3-chloropentane (CH3 C HBr C HClCH2CH3) has two different chiral
centres and hence, the number of stereoisomers is 22 = 4. I and II and III and IV
are two pairs of enantiomers, while each of I and III, I and IV, II and III and II and
IV represents a pair of diastereoisomer.
Chiral Chiral
CH3 Mirror CH3 CH3 Mirror CH3
plane plane
H Br Br H H Br Br H
H Cl Cl H Cl H H Cl
C2H5 C2H5 C2H5 C2H5
(2S, 3R) (2R, 3S) (2S, 3S) (2R, 3R)
I II III IV
Enantiomers Enantiomers
Case 2: If the molecule contains an even number (n) of chiral centres, and the molecule
can be divided into two mirror-image halves in one of the possible conformations, it can
have 2(n–1) chiral stereoisomers and 2(n–2) achiral stereoisomers, i.e., meso compounds.
* *
For example, tartaric acid (HOOC C HOH C HOHCOOH) has two similar chiral centres
and can be divided into two mirror-image halves. Therefore, it has 2(2–1) = 21 = 2 chiral
stereoisomers (V and VI) and 2(2–2)/2 = 2° = 1 achiral stereoisomer (VII). In fact, VII and
its mirror image are identical structures (superimposable on each other by 180° rotation
of the Fischer projection in the plane of the paper).
identical
Chiral and achiral
COOH Mirror COOH COOH Mirror COOH
plane plane
H OH HO H H OH HO H
s-plane s-plane
HO H H OH H OH HO H
COOH COOH COOH COOH

(2R, 3S) (2S, 3R)
(2S, 3S) (2R, 3R)
V VI VII
Enantiomers meso compound

(optically active)
V and VI represent a pair of enantiomers (optically active) and VII represents an achiral
(optically inactive) meso compound having a plane of symmetry (s).
Case 3: If a molecule contains an odd number (n) of chiral centres and the molecule can be
divided into two mirror-image halves by a plane passing through the central chiral centre in
one of the possible conformations, it can have 2(n–1)–2(n–1)/2 chiral (optically active) isomers
and 2(n–1)/2 achiral (optically inactive) meso compounds. For example, 4-bromoheptane-3,
* * *
5-diol (CH3CH2CCHOHCHBr C HOHCH2CH3) has three chiral centres and can be divided
into two mirror-image halves by passing a plane through C-4 chiral centre. Therefore, it
has 2(3–1)–2(3–1)/2 = 4 – 2 = 2 chiral isomers (VIII and IX) and 2(3–1)/2 = 21 = 2 achiral isomers
(X and XI).
chiral achiral
C2H5 Mirror C2H5 C2H5 Mirror C2H5

plane plane
H OH HO H H OH H OH
H Br Br H s-plane H Br Br H s-plane
HO H H OH H OH H OH
C2H5 C2H5 C2H5 C2H5
VIII IX X XI
Enantiomers meso compounds
VIII and IX represent a pair of enantiomers (optically active) and X and XI represent a
pair of achiral (optically inactive) meso compounds.
2.5.8 Chirotopic and Achirotopic Atom in a Molecule

The site symmetry of atoms in molecules may be classified as chiral and achiral.
A chirotopic atom is one whose site symmetry is chiral, i.e., which resides in a chiral
environment. Chirality is an all-inclusive property (as it affects all parts of a chiral
molecule) and so, all the atoms in a chiral molecule are chirotopic. For example, all the five
atoms in bromochlorofluoroiodomethane (CFClBrI) are chirotopic. [The molecules bearing
chirotopic atom or centre need not be as a whole chiral.]
An achirotopic atom is one whose site symmetry is achiral, i.e., the atom lies on a plane
of symmetry (s) or on a centre of symmetry or an alternating axis of symmetry (Sn)
passes through it that intersects its reflection plane. For example, in chlorofluoromethane
(CH2ClF), the carbon atom is achirotopic since a s-plane passes through it.
These terms including pseudoasymmetric atom or centre can be well illustrated by 2,
* * *
3, 4-trihydroxyglutaric acid (HOOC C HOH C HOH C HOHCOOH). This dicarboxylic acid
containing three chiral centres exists as two chiral and two achiral isomers.
enantiomers
1 Mirror
COOH plane COOH COOH COOH s-plane
2
s-plane
H OH HO H H OH H OH
3
HO H H OH H OH OH H
4
HO H H OH H OH H OH
5
COOH COOH COOH COOH
I II III IV
(chiral) (chiral) (achiral) (achiral)
chirotopic achirotopic
and and
(R) C-2 C-2 (S) (R) C-2 C-2 (R)
nonstereogenic stereogenic
HO H H OH H OH HO H
(R) C-4 C-4 (S) (S) C-4 C-4 (S)
I II III IV
[C-3 is achiral [C-3 is achiral [C-3 is chiral and [C-3 is chiral and
because C-2 and because C-2 and the configuration the configuration
C-4 are both R] C-4 are both S] is (r)] is (s)]
[priority order of the groups at C-3 is:

–OH R S –H]
Though the isomers I and II are chiral, C-3 in both of them is achiral. However, C-3 in them
is chirotopic, as it resides in a chiral environment. Again, C-3 in them is nonstereogenic
since interchange of ligands (say –H and –OH) bonded to C-3 does not produce a new
stereoisomer (in fact, interchange of ligands followed by 180° rotation in the plane of the
paper produces the original).
interchange 180° rotation

of H and –OH in the plane
at C-3 of the paper
I I
Identical
C-3 in III and IV is chiral. However, it is achirotopic because a s-plane passes through
it. Again, C-3 in them is stereogenic because interchange of –H and –OH at this centre
generates new stereoisomer (III generates IV and II generates III).
A pseudoasymmetric carbon atom or centre is one which is attached to two enantiomorphic
ligands having opposite chirality sense and two other ligands different from the previous
ones. A pseudoasymmetric carbon or centre is stereogenic but chiral even though the
molecule is achiral. In spite of the chiral centre as C-3, the molecules III and IV are
achiral and their C-3 is designated as a pseudoasymmetric centre. The configuration of
a pseudoasymmetric centre is designated by ‘r’ or ‘s’ (instead of R or S). Therefore, the
complete IUPAC name of III is (2R, 3s, 4S)-2, 3, 4-trihydroxyglutaric acid and of IV is (2R,
3r, 4S)-2, 3, 4-trihydroxyglutaric acid.
1 1
COOH COOH
2 OH 2
H H OH
3 3
H OH HO H
4 4
H OH H OH
5COOH 5COOH
R s S R r S
2.5.9 Prostereoisomerism and Topicity

Ligands (atoms or groups in a molecule) are called homomorphic (Greek homos means same
and morphe means form) if they are indistinguishable when considered in isolation. In the
case of atoms, they must be the atoms of the same element, e.g., two Cl atoms, two H atoms,
etc. and in the case of groups they must have the same constitution and configuration, e.g.,
two Et or two Ph groups or two secondary butyl groups of same configuration (R or S). In
an intact molecule, if the homomorphic ligands are bonded to constitutionally different
ligating centres, the ligands are called constitutionally heterotopic. For example, H’s at
C-2 in heptane are constitutionally heterotopic compared to H’s at C-4.
constitutionally
heterotopic H’s
1CH 2 3 4 5 6 7
3— CH2— CH2— CH2— CH2— CH2— CH3
Heptane
When the spatial relation of two ligands with the rest of the molecule is different, the
ligands are called stereoheterotopic. For example, the geminal H’s at C-2 (designated as
HA and HB) of heptane are stereochemically not equivalent, i.e., stereoheterotopic (actually
enantiotopic).
Topicity describes the relationships of two or more homomorphic ligands (or faces) which
together constitute a set. Homomorphic atoms or groups can be classified as homotopic,
enantiotopic or diastereotopic. Ligands, which are not homomorphic, cannot be related
topically.
2.5.9.1 Homotopic ligands

A pair of homomorphic ligands (atoms or groups) in a molecule are called homotopic
if substitution (replacement) of each of them in turn by a ligand, which is not already
attached to the ligating centre, gives identical product. For example, the three hydrogens
of acetic acid (CH3COOH) are homotopic because replacement of any of these H’s by atom,
one at a time, produces the same molecule of bromoacetic acid (BrCH2COOH).
H.V.Z.
reaction
2.5.9.2 Enantiotopic ligands

A pair of ligands (atoms or group) in a molecule is called enantiotopic if replacement of
one or the other of them by a different achiral ligand, which is not already attached to the
ligating centre, gives rise to one or the other of a pair of two enantiomers. For example,
the two methylene hydrogens (HA and HB) or propanal (CH3CH2CHO) are enantiotopic
because replacements of these hydrogens in turn, say by F, gives rise to two enantiomers
of 2-Fluoropropanal (CH3CHFCHO).
enantiotopic H’s
HA HB
C
OHC CH3
Propanal
CHO
replacement replacement
of HA by F
HA HB of HB by F
CH3
Mirror
plane
CHO CHO
F HB HA F
CH3 CH3
Enantiomers
2.5.9.3 Diastereotopic ligands

Two ligands (atoms or groups) in a molecule are called diastereotopic if replacment of
one or the other of them by a different ligand produces one or the other isomer of a set
of diastereoisomers. For example, the two H atoms on the same carbon (C-1) of propene
are diastereotopic because sequential replacements of these H’s, say by Br, result in the
formation of a pair of diastereoisomers (two cis-trans or geometric isomers).
disastereotopic
hydrogens
Br H HA HB H Br
C replacement of C replacement of C
HA by Br HB by Br
C C C
CH3 H CH3 H H3C H
cis-1-Bromopropene Propene trans-1-Bromopropene
diastereoisomers
The two H atoms on C-2 in 3-bromobutanoic acid are also diastereotopic because sequential
replacements of these hydrogens, say by Cl, give rise to a pair of diastereoisomers.
replacement replacement
A B
of HA by Cl of HB by Cl
3 3 3
R R S R
R
2.5.9.4 Prochiral and prostereogenic centres

A chiral centre is created when one of the pair of enatiotopic or diastereotopic ligands
is replaced by a ligand different from the ligands already present on the chiral centre.
Therefore, an achiral centre bearing enantiotopic or diastereotopic ligands is called
a prochiral centre. For example, C-2 of acetic acid, C-3 of 2-bromobutane, etc. are
prochiral centres. Again, if the two faces of an sp2-hybridized carbon are enantiotopic or
diastereotopic, that carbon is also called a prochiral centre. For example, the carbonyl
carbon of chloral and the carbonyl carbon of 3-hydroxy-2-butanone are prochiral centres.
chiral
prochiral centre
H centre CH3 prochiral
S centre
C CH3 H C
H CO2H CH2 CH3
Br
(S)-2-Bromobutane
chiral
centre
prochiral
centre CH3 prochiral
H centre
C O H C CH3
Cl3C HO C
O
Chloral (S)-3-Hydroxy-2-butanone
A prostereogenic centre is a centre which can be converted into a stereogenic centre by
replacing one of the homomorphic ligands (atoms or groups) by a different ligand. A
prostereogenic centre may or may not be prochiral. For example, the C-2 of bromoethene
is prostereogenic because on replacement of HA or HB by Br it becomes converted into a
stereogenic centre. However, C-2 is not prochiral because it cannot be converted into a
chiral centre by replacing HA or HB.
A replacement
of HA by Br
B
4 3 2 1
However, when one of the homomorphic H atoms on C-3 of 2-butanone (CH3 CH2 COCH3 )
is replaced by F atom, the centre C-3 becomes stereogenic as well as chiral. Therefore, in
this case, C-3 is both prostereogenic and prochiral.
prochiral and
prostereogenic centre
chiral and
CH3 CH3 stereogenic centre
H (or H ) is
HA HB F H
replaced by F
COCH3 COCH3
2-Butanone (R)-3-Fluorobutan-2-one
2.5.9.5 Pro-R and pro-S designation of enantiotopic and

diastereotopic ligands (groups or atoms)
Enantiotopic and diastereotopic pair of atoms or groups on a prochiral centre in a molecule
is designated as pro-R and pro-S. If replacement of one of them by a nonequivalent ligand
with higher priority than the other (without disturbing the priority order with respect to
the remaining ligands) gives (R)-configuration of that chiral centre, that ligand is specified
as pro-R and if (S)-configuration, then as pro-S. For example, replacement of HA in
chloroiodomethane (CH2ClI) by D gives (R)-chlorodeutereoiodomethane and replacement
of HB by D gives (S)-chlorodeutereoiodomethane. Therefore, HA is pro-R and HB is pro-S.
pro-R
prochiral centre
D H H
A B
C I C I C I
H H D
Cl Cl Cl
pro-S (S)-Chlorodeutereoiodo-
(R)-Chlorodeutereo- Chloroiodomethane methane
iodomethane
2.5.9.6 Homotopic faces

The two faces of a planar group like C=O, C=C, etc., are said to be homotopic if addition
of the same reagent to either face generates the same product. Thus, the two faces of
formaldehyde (HCHO) are homotopic because addition of MeMgI to either face of it leads
to the formation of ethanol.
d- d+ top face H3O

addition
bottom face
d- d+ addition
H3O
2.5.9.7 Enantiotopic faces

The two faces of a planer n-orbital system like C = O, C = C, etc. are said to be enantiotopic,
if attack of a reagent on one side of the p-bond produces one enantiomer, whereas attack on
the opposite side leads to the other enantiomer. For example, the two faces of acetophenone
(PhCOCH3) are enantiotopic because reaction with the achiral reagent LiAlH4 at one face
gives a molecule of phenylmethylcarbinol, whereas reaction at the other face produces the
enantiomeric phenylmethylcarbinol.
Acetophenone top face H
attack
Me C
H
–
S OH
Ph
Me 1. LiAlH4 Mirror
C O 2. H O≈ plane enantiomers
3
Ph bottom face Me
attack OH
Ph
H
– C
R
(hydride ion) H
Phenylmethylcarbinol
2.5.9.8 Diastereotopic faces

The two faces of a double bond, i.e., a p-system like C=O, C=C, etc. are said to be
diastereotopic, if attack of a reagent on one side of the p-bond leads to one diastereoisomer,
whereas attack on the opposite side produces the other diastereoisomer. For example,
the two faces of the carbonyl group of chiral 2-phenylpropanal are diastereotopic because
reaction with MeMgI at one face affords a molecule which is diastereoisomeric with the
molecule formed by reaction at the other face.
top face H Me
d– d+ addition
CH3 MgI Ph C C
R S OH
H H3C H
O 1. ether
C C 2. H3O≈
diastereoisomers
H H OH
Ph bottom face
H
H3C addition Ph C C
d– d+ R R
H3 C MgI H 3C Me
(R)-3-Phenylpropanal 3-Phenyl-2-butanol
2.5.9.9 Re and Si designation of enantiotopic and diastereotopic faces

The enantiotopic and diastereotopic faces can be specified by an extended use of CIP
convention of assigning configurational designation based on priority sequence rules in
two dimensions. If the three groups of the trigonal centre as arranged by the sequence
rules have the order a b c, that face (when viewed from the top of the molecule) in
which the groups in this sequence are clockwise (as in X) is Re face (from Latin rectus) and
that face in which the groups in this sequence are counterclockwise (as in Y) is the Si face
(from Latin Sinister).
c b [The structure Y is actually the

C a C a bottom face of structure X. This
b c can be obtained by 180° out-of-plane
X Y rotation of structure X.]
Re face Si face
For example, the face of the carbonyl group of benzaldehyde as shown in structure I is
the Re face because the ligands (=O Ph H) trace a clockwise path, whereas the face
as shown in structure II is the Si face because the ligands trace a counterclockwise path.
3 H 2 Ph
C O 1 C O 1
2 Ph 3 H
I II
1Æ 2 Æ 3 clockwise 1Æ 2 Æ 3 counterclockwise
Hence, it is Re face Hence, it is Si face
Two opposite faces of

benzaldehyde
The two carbonyl faces of (S)-3-chloropentanal are diastereotopic (due to presence of a

chiral centre). The face shown in structure III is Re (1 Æ 2 Æ 3 clockwise) and the other face
shown in structure IV is Si. A further subscript S (the chirality description of C-3) may be
added so that the double-lettered subscripts ReS and SiS are in true sense diastereotopic.
3
C2H5 H C2H5
C 2 C O 1 C 2
H S H S
CH2 CH2
Cl Cl
III C O 1
ReS face 3 H
IV
SiS face
(S)-3-Chloropentanal
2.5.9.10 Retention of configuration

If the spatial arrangement of bonds with respect to a chiral or asymmetric carbon in a
chiral molecule remains the same before and after the reaction, retention of configuration
occurs.
In the conversion of the substrate XCabc into the product YCabc, if the hypothetical path
traced by the ligands a, b and c (a Æ b Æ c) remains the same (clockwise as shown in
the following figures) in both the substrate and the product, the reaction is said to have
taken place with retention of configuration. XCabc and YCabc have the same relative
configuration.
c c
+Y –
C –X –
C
b X b Y
a a
(a Æ b Æ c traces (a Æ b Æ c traces
a clockwise path) a clockwise path)
configuration
retained
In general, during a reaction, if none of the bonds to the chiral centre is broken, the product
has the same relative configuration of the ligands around to the chiral centre as that of the
substrate. Such a reaction is said to proceed with retention of configuration. For example,
when 2-methyl-butanol is heated with concentrated HCl, 1-chloro-2-methylbutane is
obtained. In this reaction, no bond to the asymmetric carbon is broken. Therefore, the
reaction must take place with retention of configuration.
C2H5 C2H5
HCl
H C D H C + H 2O
CH2 OH CH2Cl
CH2 CH3
(–)-2-Methyl-1-butanol (+)-1-Chloro-2-methylbutane
Configuration
retained
Mechanism:
C2H5 C2H5 C2H5
H Cl SN2
H C H C + Cl – H C
CH2 OH CH2 OH2 CH2Cl + H2O
CH3 CH3 CH3
2.5.9.11 Inversion of configuration

If the spatial arrangement of bonds with respect to an asymmetric or chiral carbon in
a chiral compound suffers a change after the reaction, the compound is said to have
undergone inversion of configuration.
In conversion of the substrate XCabc, into the product YCabc, if the hypothetical path traced
by the ligands a, b and c (a Æ b Æ c) changes (clockwise in XCabc but counterclockwise in
YCabc), the substrate is said to have undergone the reaction with inversion of configuration.
c c
+ Y
–
Y C X Y C + X
– –
– X
–
b b
a a
(a Æ b Æ c traces (a Æ b Æ c traces
a clockwise path) a counterclockwise path)
Inversion of configuration
When 2-chlorobutane, for example, is subjected to alkaline hydrolysis under SN2 conditions,
the product 2-butanol is obtained with inversion of configuration.
H H
– NaOH/H2O
HO C Cl HO C + Cl
–
SN2
Et Et
Me Me
2-Chlorobutane 2-Butanol
Inversion of configuration
1. Write down the Fischer projections of the following compounds:

(a) L-Glyceraldehyde
(b) D-2-Hydroxybutanedioic acid
(c) D-2-Phenylbutan-2-ol
(d) L-2-Fluoro-2-phenylpropanoic acid
(e) 2D-Ethyl-2L-methyl-2-phenylbutanoic acid
(f) D-2-Butylmethylether
(g) 2L-Chloro-2D-fluoropentane
(h) 5D-Ethyl-3D-hydroxy-1L-methylcyclohexane-1D-carboxylic acid
Solution
CHO COOH
(a) HO H (b) H OH
CH2OH CH2COOH
L-Glyceraldehyde D-2-Hydroxybutanedioic
acid
C2H5 COOH
(c) H 3C OH (d) F CH3
C6H5 Ph
D-2-Phenylbutan-2-ol L-2-Fluoro-2-phenyl-
propanoic acid
COOH CH3
(e) CH3 C2H5 (f) H OCH3
Ph CH2CH3
2D-Ethyl-2L-methyl-2- D-2-Butylmethylether
phenylbutanoic acid
1
CH3 Me COOH
(g) Cl F (h)
H Et H OH
CH2CH2CH3 5 3
2L-Chloro-2D- 5D-Ethyl-3D-hydroxy-
fluoropentane 1L-methylcyclohexane-
1D-carboxylic acid
2. Assign R or S designations for each of the following compounds:
CH2OH
(a) C* (b) * H
H O
Br
CH3
COOH Br
(c) *C (d) C*
CH CH2 HC C
(CH3)2CH CN–
OH O2 N
CH(CH3)2
(e) N* (f) Me S*
H
Ph CH3 CH2COOH
CH3
*
Me
(g) (h) *
Me
O
(i) Br (j) C C
H * * Cl
Br
* *
Cl H
H H
O C CH
(k) (l) *C
CH2Cl Ph
* H3 C C CCH3
H Cl
NH2
Ph OH
* *
(m) *C
CD2OH (n) C C
HSH2C CH2OH H CH3
D
CH(OCH3)CH(CH3)2 H H R OH
S
(o) C* (p) C C
H HO *
H 3C CH(OCH3)CH2CH3 C CH3
H3C
H
CH3
CH3
S R
(q) S C
R
OH H HO
H H
Solution
(a) Priority order of ligands: CH2OH CH3 > H
1 CH2OH 1
The arrow (1 Æ 2 Æ 3)
C traces a clockwise
4 H 2 3 2 path, suggesting
3 CH3 the R configuration.
1
or,
CH2OH 1
Allowed
4 H CH3 3 interchange 3 2
2
(in Fischer)
O
||
(b) Priority order of ligands: Br CH2— C — CH2CH2— H
3
The arrow from 1 Æ 2 Æ 3 traces
H 4 a clockwise path, suggesting the
O
2 Br 1 R configuration
or,
3 3
1 4 2 1
2 4
(c) Priority order of ligands: OH COOH CH=CH2 CH(CH3)2
*
120° rotation
2
3 2 3 2
2
Æ Æ
(d) Priority order of ligands: Br NO2 CN C∫CH
Æ Æ
S
(e) Priority order of ligands: Ph CH(CH3)2 CH3 H

2
2
CH(CH3)2
The arrow from 1 Æ 2 Æ 3
N H 4 traces a clockwise path,
1 3
1 Ph suggesting the R configuration.
CH3 3
or,
2 2
Allowed
3 4 interchange 1 3
1 4
(f) Priority order of ligands: CH3 CH2COOH Me (◊ ◊) lone pair of
electrons
4
2 2
120° rotation
3 Me S S
1 4
1 3 1
CH2COOH CH3 3
2
or,
4 The arrow from 1 Æ 2 Æ 3
3 2 points counterclockwise,
suggesting the S configuration
1
Et H
(g) Priority order of ligands: Cl Cl C Cl C H
H Z H H E Et
H
H C
C 2 Et 2 2
120° rotation
C H C
1 Cl 4 1
C C 3 1
4 H H 3 Et 3
or,
2 2 The arrow from 1 Æ 2 Æ 3
Allowed points counterclockwise,
1 4 interchange 3 1
suggesting the S configuration.
3 4
C
(h) Priority order of ligands at C-1 and C-4: C(Me) = C(Me) — CH2CH
CH2CH2— 4 C
7 C-4:
*
5 4
3
*
6 1
2
Æ Æ
180° rotation
Allowed
interchange
7 C-1:
1
2
6
3 4 Æ Æ
5
Allowed
interchange
C
(i) Priority order of ligands at C-2 and C-6: Br CH CH2CH2— H
C
8
7
5 4
3
Br 1
*6
Br
2 *
H H
3 3
3 3 C-2:
4H 6 2 C The arrow from
H 4 4
1 Br 2
2 2 1 1 Æ 2 Æ 3 points
Br 1
1 clockwise, suggesting
the R configuration
or, of C-2.
3 4
Allowed
1 4 interchange 3 1
2 2
3 3 The arrow from

C-6:
1 Æ 2 Æ 3 points
4 C counterclockwise,
2 1 2
1 suggesting the S
configuration
or, of C-6.
3 3
Allowed
4 1 interchange 1 2
2 4
(j) Priority order of ligands at C-2 and C-3: Cl O— CHCl— H
Æ Æ
∫∫
R
Æ Æ
R
O
||
(k) Priority order of ligands: CH2Cl C —} CH2CH2—} H
O 2 O
2
∫ ∫
CH Cl H 4
* 3 CH Cl 1 3 1
H
2
2 The arrow from
1 Æ 2 Æ 3 traces
or, 1 4 3 1 a clockwise path,
suggesting the
4
3 configuration.
(l) Priority order of ligands: C∫CCH3 Ph C∫CH CH3
3 3
C∫CH 3
120° C ∫
*C Ph 2 rotation 4 1 2
4 H3C 1
C∫CCH3 1 2
3 The arrow from

1 Æ 2 Æ 3 traces
or, 1 2 a clockwise path,
suggesting the S
configuration.
3
(m) Priority order of ligands: NH2 CH2SH CD2OH CH2OH

Æ Æ
(n) Ph Cl OH Priority order of ligands at C-2: Cl > OH > CHDPh > CH3
* *
H C—C
3 2 Priority order of ligands at C-3: C(OH)ClCH3 > Ph > D > H
D CH3
Configuration of C-2:
120°
rotation
∫
2
3
∫
Æ Æ
R
Configuration of C-3:
2 2
2 Ph 1
4 H C—C(OH)ClCH3 ∫ C ∫
4 1 3 1
3 D 3
2 2
The arrow from
or, 1 Æ 2 Æ 3 traces
Allowed
4 3 interchange 3 1 a clockwise path,
suggesting the
1 4 S configuration
of C-3
(o) Priority order of ligands: CH(OCH3) CH(CH3)2 CH(OCH3)CH2CHCH3 CH3 H
1
CH(OCH3)CH(CH3)2 1 1
4 H C* ∫ C ∫
CH(OCH3)CH2CH3 4 2 3 2
3 H3C 2 3
1 4
The arrow from
or, 1 Æ 2 Æ 3 traces
Allowed
4 3 interchange 3 1 a clockwise path,
suggesting the
2 2 R configuration.
H H
S 2 1 R
HO C 3 C OH
4 5
2 CH
(p)
1CH *C 3 Priority of ligands at C-3:
3 H 4
(R) CH(OH)CH3 (S) CH(OH) CH3 CH3 H
CH3
3
∫∫
The arrow from
∫
1 1 Æ 2 Æ 3 traces
1 an anticlockwise
C 4 ∫
path, suggesting the
2 2 3 S configuration.
3
1 1
or, Allowed 2 3
3 4
interchange
2 4
(q) Priority order of ligands at the central chiral carbon: (R, R) (S, S) CH3 H
3
Æ Æ
S * R
* *
* * R
S R
∫∫
3. Assign priorities to the following pairs of groups according to the sequence

rules:
(a) —CH2CH2CH2CH3 and —CH2CH(CH3)2
(b) —CH2CH2Br and —C(CH3)3
C(CH ) CH(CH )
(c) —CH and —CH
CH CH F CH CHF
12 14
(d) — CH2NH2 and — CH2CH3
(e) —Ph and —C∫∫C—CH2CH3
CH2CH3
(f) C==CH2 and —C∫∫CH
(g) —CD216OH and —CH218OH
(h) and (CH3)2CH—
Solution
(a) The priority sequence is: —CH2CH(CH3)2 —CH2CH2CH2CH3.
The first point of difference is at the carbons b to the chiral centre. The b carbon
a b a b
of — CH2 CH(CH3 )2 and — CH2 CH2 CH2CH3 are attached to (C, C, H) and (C,
H, H), respectively. Since, C H, therefore, —CH2CH(CH3)2 gets priority over
—CH2CH2CH2CH3.
(b) The priority sequence is: —C(CH3)3 —CH2CH2Br.
The first point of difference is at the two carbons attached to the chiral centre.
The first carbon (Ca) in —C(CH3)3 is attached to (C, C, C) whereas the first carbon
—CH2CH2Br is attached to (C, H, H). Since C H, (CH3)3C— gets priority over
a b
— CH2 CH2 Br , even though Cb bears an atom (Br) having higher atomic number.
C(CH3)3 CH(CH3)2
(c) The priority sequence is: —CH (I) —CH (II)
CH2CH2F CH2CHF2
The first point of difference is found at the branch carbon at Ca which is attached to
(C, C, C) in the case of I, but (C, C, H) in the case of II. Since C H, I gets priority
over II.
(d) The priority sequence is: —12CH2NH2 —14CH2CH3.
Although the mass number of carbon in —14CH2CH3 is greater than that in
—12CH2NH2, the latter group is preferred because the atomic number of the second
atom (N, AN = 7) is higher than that of the former group (C, AN = 6).
(e) The priority sequence is: —C∫∫C—CH2CH3 —C6H5
(C)000 (C)000
replication
—C ∫∫ C—CH2CH3 C C CH2CH3
(C)000 (C)000
(C)000 H (C)000
C C H H
replication
C C
(C)000 C C (C)000
H H
(C)000 (C)000
In each of these two groups, the carbon bonded to the chirality centre is attached to
(C, C, C). 1-Butenyl (—C ∫ C—CH2CH3) has two of its (C, C, C) carbons connected
to (000), while the third is (C, C, C) and is thus preferred to phenyl (—C6H5), which
has only one (0, 0, 0) and two (C, C, H)s.
CH CH
(f) The priority sequence is : —C ∫ CH C==CH .
(C)000 (C)000 CH2CH3 C2H5 H
replication replication
—C ∫ CH C C H ; —C CH2 C C H
(C)000 (C)000 (C)000 (C)000
In each of these ligands, the first carbon is connected to (C, C, C). In ethynyl
(—C∫∫CH), two of these carbons are connected to (0, 0, 0) and one is connected to
CH2CH3
(C, C, H), whereas in 1-ethylethenyl —C==CH2 , one of these carbons is connected to

(0, 0, 0) and two are connected to (C, H, H). Since C > H, ethynyl gets priority over
1-ethylethenyl.
(g) The priority sequence is: —CH218OH —CD216OH
This is due to the fact that in —CH218OH, 18O has higher atomic number as well as
higher mass number.
(h) The priority sequence is: Cyclopentyl isopropyl (Me2CH–).
In each of these two groups, the first carbon is attached to (C, C, H). Both of these
carbons in cyclopentyl are connected to (C, H, H), whereas in isopropyl, they are
connected to (H, H, H). Since C > H, cyclopentyl gets priority over isopropyl.
4. Give flying wedge or 3D-representations of the following compounds:
(a) (S)-Alanine (b) (R)-2-Pentanol
(c) (R)-3-Bromocyclopentanone (d) (2R, 3S)-3-Chloro-2-hexanol
(e) (1S, 2R)-1-Ethyl-2-fluorocyclohexane
Solution
*
(a) In alanine [CH3 CH(NH2)COOH], the priority order of ligands is: NH2(1) > COOH(2)
> CH3(3) > H(4).
The flying wedge structure (perspective formula) of (S)-alanine can be drawn by
putting the lowest priority ligand (4) on the hatched wedge and other ligands on
the remaining bonds such that the arrow from 1 Æ 2 Æ 3 traces a counterclockwise
path.
3
2
S S
*
(b) In 2-pentanol [CH3 C H(OH)CH2CH2CH3], the priority order of ligands is: OH (1) >
CH2CH2CH3(2) CH3 (3) H (4).
The flying wedge structure (perspective formula) of (R)-2-pentanol can be drawn
by putting the lowest priority ligand (4) on the hatched wedge and other ligands on
the remaining three bonds such that the arrow from 1 Æ 2 Æ 3 traces a clockwise
path.
3
3 2 2
O
(c) In 3-bromocyclopentanone , the priority order of ligands is: Br (1)
O Br
||
CH2— C — (2) CH2CH2— (3) H(4).
Therefore, the flying wedge or 3D-representation of (R)-3-bromocyclopentanone is:
* *
(d) In 3-chloro-2-hexanol (CH3 C HOH C HClCH2CH2CH3), the priority order of ligands
at C-2 is: OH(1) CHClCH2CH2CH3 (2) CH3 (3) H(4) and at C-3 is: Cl (1)
CH(OH)CH3 (2) CH2CH2CH3 (3) H(4).
The flying wedge formula of 3-chloro-2-hexanol can be drawn by putting the
lowest priority ligand (4) on the hatched wedges and other ligands (1 and 3) on
the remaining bonds (2 becomes automatically fixed after putting 1) such that the
arrow from 1 Æ 2 Æ 3 traces a clockwise path around C-2 and a counterclockwise
path around C-3.
R S R S
3 2 2 3
R S
1 Et
H
(e) H
2 F The priority order of ligands at
1-ethyl-2-fluoro- C-1: CHFCH2— (1) > CH2CH2CH2— (2) > Et(3) > H(4) and
cyclohexane at C-2: F(1)> CH(Et) CH2— (2) > CH2CH2CH2— (3) > H(4).
The flying wedge formula of (1S, 2R)-1-ethyl-2-fluorocyclohexane is as follows:

H
2 3 Et
1 S
4H C C
S 2 R H 4 H R
3 Et F 1 F
(1S, 2R)-1-Ethyl-3-fluorocyclohexane
5. Draw Fischer projections for the following molecules:

(a) (S)-Lactic acid (b) (R)-2-Butanol
(c) (2R, 3S)-3-Iodo-2-hexanol.
Solution
*
(a) In lactic acid (CH3CHOHCOOH), the priority order of ligands is: OH (1) COOH
(2) CH3 (3) H (4).
Now, the Fischer projection of (S)-lactic acid can be drawn by putting the lowest
priority ligand (4) on any of the bonds on the vertical line and other ligands on the
remaining bonds such that the arrow from 1 Æ 2 Æ 3 traces a counterclockwise
path.
2 COOH 4 H
3 1 H C OH or, 2 1 HOOC OH
4 H 3 CH
(S)-Configuration (S)-Lactic acid (S)-Configuration (S)-Lactic acid
*
(b) In 2-butanol (CH3CHOHCH2CH3), the priority order of ligands is: OH (1)
CH2CH3 (2) CH3 (3) H (4).
Now, the Fischer projection of (R)-2-butanol can be drawn by putting the lowest
priority ligand (4) on any of the vertical bonds and other ligands on the remaining
bonds such that the arrow from 1 Æ 2 Æ 3 traces a clockwise path.
2 3
3 2 3
3
R R R R
* *
(c) In 3-iodo-2-hexanol (CH3 C HOH C HI CH2CH2CH3), the priority order of ligands
at C-2 is: OH (1) CHICH2CH2CH3 (2) CH3 (3) H (4) and at C-3 is: I (1)
CH(OH)CH3 (2) CH2CH2CH3 (3) H (4).
For each chirality centre, the chiral ligand (2) is automatically placed on the
vertical bond (lower for C-2 and upper for C-3). The Fischer projection of (2S, 3R)-
3-iodo-2-hexanol can be drawn by putting the lowest priority ligands (4) on the
vertical bonds and two other ligands (1 and 3) on the horizontal bonds such that
the arrow from 1 Æ 2 Æ 3 traces a counterclockwise path around C-2 (i.e., 1 is on
the left bond and 3 is on the right bond) and a clockwise path around C-3 (i.e., 1 is
on the left bond and 3 is on the right bond).
4
H
1 HO CH3 3
2
1 I CH2CH2CH3 3
H
4
(2 , 3 )-3-Iodo-2-hexanol
6. Designate R or S description for each chiral centre of the following

compounds using the so-called ‘very good’ procedure:
Cl
CH3 I Br CH3
(a) Cl CO2H (b) Br F (c) H OH
EtO H Cl I H OH
H C2H5 Br CH H OH
OH F CH3
Solution
(a) 4 C-2: Since the lowest priority group (4) is on
CH3 a vertical bond, clockwise motion of the arrow
from 1 Æ 2 Æ 3 indicates R configuration.
1 Cl CO2H 2
2 C-3: Since the lowest priority group (4)
3 is on a horizontal bond, clockwise motion
2 of the arrow from 1 Æ 2 Æ 3 indicates S
1 EtO 3
H 4 configuration.
3 C-4: Since the lowest priority group (4)
2 is on a horizontal bond, clockwise motion
4H 4
C2H5 3 of the arrow from 1 Æ 2 Æ 3 indicates S
configuration. Therefore, it is the (2R, 3S,
OH
4S)-enantiomer.
1
(2R, 3S, 4S)
(b) 3 C-1: The lowest priority group (4) is on a

Cl vertical bond. Therefore, counterclockwise
motion of the arrow from 1 Æ 2 Æ 3 indicates
1
1 I Br 2 S configuration.
4 C-2: The lowest priority group (4) is on a
3 vertical bond. Therefore, counterclockwise
2
1 Br F 2 motion of the arrow from 1 Æ 2 Æ 3 indicates
4 S configuration.
3 C-3: The lowest priority group (4) is on a
2 Cl 3 I 1 vertical bond. Therefore, clockwise motion
4 of the arrow from 1 Æ 2 Æ 3 indicates R
3 configuration.
4
1 Br CH3 4 C-4: The lowest priority group (4) is on a
horizontal bond. Therefore, counterclockwise
F
2
R configuration. Therefore, it is the (1S, 2S,
(1S, 2S, 3R, 4R)
3R, 4R)-enantiomer.
(c) 3 C-2: The lowest priority ligand (4) is on

CH3 a horizontal bond. Therefore, a clockwise
4 H * 2 OH 1 S configuration.
2 C-4: The lowest priority ligand (4) is on
3
H OH a horizontal bond. Therefore, a clockwise
2 motion of the arrow from 1 Æ 2 Æ 3 indicates
1 HO * H 4
4 S configuration.
CH3 Since both C-2 and C-4 have S configuration,
3 C-3 is achiral. Hence, it is the (2S, 4S)
(2 , 4 ) enantiomer of chiral pentane-2,3,4-triol.
7. Specify the configuration (E or Z) of each of the following compounds

containing various types of double bonds:
H3C H BrCH2 CH2I
(a) C C (b) C C
DH2C CH3 ClCH2 CH2F
H CH CH2 Ph H
(c) C C (d) C C
H 3C CH(CH3)2 H COOH
N N CH3 Cl H H
(e) (f) C C C C
CH3 CH3 Ph C C COOH
H H
H5C2 H C5H5 OH
(g) C C CH2COOH (h) C N
H C C C2H5
H H
18
CH3 2
(i) C (j) 16
CHO 2
O
D3 C COOH
(k) C C C C C C
H CH3
Solution
3
(a) CH2D CH3 ; CH3 H
2 3
Higher priority ligands on the same side: Z configuration

2 2
(b) CH2Br CH2Cl; CH2I CH2F
2 2
H CH CH higher
(c) C C priority CH3 H; CH = CH2 CH(CH3)2
H C CH(CH )
higher
priority
Higher priority ligands on different sides: E configuration.

higher
priority Ph H
(d) C C higher Ph H; COOH H
H COOH priority
Higher priority ligands on different sides: E configuration
(e) N N 3 (..) unshared pair of electrons
CH3 CH3
higher higher
priority priority

higher C-2==C-3 : COOH H; CH==CH — > H
(f)
priority C-4==C-5 : CH==CHCOOH H;
CCl==C(CH3)Ph H
3 7 6 3 2 C-6==C-7 : Cl CH==CH — ; Ph CH3
5 4
higher higher
priority priority
Higher priority ligands on the same side of C-2==C-3 bond: Z configuration
Higher priority ligands on the same side of C-4==C-5 bond: Z configuration
Higher priority ligands on different sides of C-6==C-7 bond: E configuration
(g) higher
priority C-3==C-4 : CH2COOH H;
CH = CHC2H5 H
5 2 6 5
C-5==C-6 : CH = CHCH2COOH H;
4 3 2
higher C2H5 H
priority
higher
priority
Higher priority ligands on the same side of C-3 == C-4 bond: Z configuration
Higher priority ligands on different sides of C-5 == C-6 bond: E configuration
C6H5 OH higher
(h) higher C N priority C6H5 C2H5; OH (..) unshared
H5C2 pair of electrons
priority
Higher priority ligands on the same side : Z configuration.
higher
(i)
priority O
O
CH3 CHO CH3 ; CH2C— CH2CH2—
C
CHO
higher
priority
Higher priority ligands on different sides : E configuration.
(j) 18
2
16
∫ 2
18
2
16
2
(k) higher higher COOH > CH3 ; CD3 > H

priority priority
D3 C COOH
C C C C C C
H CH3
Higher priority ligands on the same side : Z configuration
(This compound containing five cumulated double bond is planar and gives rise to
cis-trans isomerism.)
8. Write down the structural formula for each of the following compounds:
(a) (Z)-1-Bromo-2-methylbut-l-ene (b) (E)-2-Chloro-1-nitropent-1-ene
(c) (Z)-2-Butene (d) (E)-But-2-enoic acid
Solution
CH3
4 4 | 1
2
(a) CH3 CH1 C == CHBr (1-Bromo-2-methylbut-1-ene)
Since CH2CH3 CH3 and Br H, the structure of (Z)-1-bromo-2-methylbut-1-ene

is that in which the higher-ranked groups (CH2CH3 and Br) are on the same side
of the double bond. Hence, the configuration is:
CH3CH2 Br
C C
H3C H
(Z)-1-Bromo-2-methylbut-1-ene
Cl
5 4 3 | 1
(b) CH3 CH2 CH22C = CHNO2 (2-Chloro-1-nitropent-1-ene)
Since Cl CH2CH2CH3 and NO2 H, the structure of (E)-diastereoisomer of this

compound is that in which the higher-priority groups (Cl and NO2) are on the
opposite sides of the double bond. Hence, the configuration is:
Cl H
C C
CH3CH2CH2 NO2
(E)-2-Chloro-1-nitropent-1-ene
4 3 2 1
(c) CH3 CH=CHCH3 (2-Butene)
Since CH3 H, the structure of (Z)-2-butene is that in which the higher priority
groups (-CH3) are on the same side of the double bond. Hence the configuration is:
H H
C C
H3 C CH3
( )-2-Butene
4 3 2 1
(d) CH3 CH=CHCOOH (but-2-enoic acid)
Since CH3 H and COOH H, the structure of (E)-diastereoisomer of this
compound is that in which the higher-ranked groups (CH3 and COOH) are on the
opposite sides of the double bond. Hence, the configuration is:
H3C H
C C
H COOH
( )-But-2-enoic acid
9. Designate each of the following structures with erythro or threo prefix:
COOH OH
(a) H Cl (b) H3C H
H Br H
F
COOH
CH3
CHO H
H
H Br
Cl
(c) (d) Br C C
H NO2 CH3
C2H5
CHO
Solution (a) erythro-isomer (b) threo-isomer (c) erythro-isomer (d) threo-isomer

10. What will be the change in conductivity of boric acid if (R, R)-2-,
3-butanediol, (R, S)-2, 3-butanediol and (S, S)-2, 3-butanediol are separately
treated with boric acid?
Solution The borate complex of a chiral isomer [(R, R) or (S, S)] of 2, 3-butanediol is
more stable than that of the achiral (meso) (R, S)-isomer because in the former complex
the bulky methyl groups are anti to each other (sterically more favoured), whereas in
the latter complex the methyl groups are gauch to each other (sterically less favoured).
Therefore, the conversion of the chiral diol in to the borate complex is relatively more
favourable than for the achiral diol. The conductivity of boric acid increases with increase
in concentration of the complex. Thus, chiral 2, 3-butanediol [(R, R), or (S, S)-enantiomers]
enhances the conductivity of boric acid more than does the achiral isomer.
anti-conformer
CH3 CH3 CH3
H OH H O 3 O OH
2 + H3BO3 m B + 3H2O
H OH H O H≈ O OH
CH3 CH3 CH3
(S, S)-2,3-Butanediol Borate complex

(chiral) (more stable)
anti-conformer
CH3 CH3 CH3
H OH H O 3 O H
2 + H3BO3 m B + 3H2O
H H O H≈ O
OH H
CH3 CH3 CH3
(R, R)-2,3-Butanediol Borate complex
(achiral) (more stable)
H gauche-conformer H H
H OH H O 3 O H
2 + H3BO3 m B + 3H2O
H3 C OH H3C O H≈ O
CH3
CH3 CH3 CH3
steric steric
(R, S)-2,3-Butanediol strain Borate complex
strain
(achiral) (less stable)
11. Isomers of CH3CH2CH==CHCH2CH3 differ widely in chemical properties
but those of CH3CH2CH==C==CHCH2CH3 do not – Why?
Solution Isomers of 3-hexene (CH3CH2CH==CHCH2CH3) are cis-trans isomers, i.e., they
are diastereoisomers and because of this they differ widely in chemical properties. On the
other hand, isomers of 3, 4-heptadiene (CH3CH2CH==C==CHCH2CH3) are enantiomers
and because of this, they do not differ in chemical properties (except reaction with a chiral
reagent involving diastereoisomeric transition states having unequal energies).
cis trans
12. Assign R/S descriptions to the asterisked chiral centre (*) in each of the
following compounds:
NO2
OCH3 CH2OCH2CH3
* CH2CH2OCH2CH3
(a) H C* (b)
Br NO2 O
H3CO
C2H5
(c) H Cl
H Cl
H * Cl
Cl H
H Cl
C2H5
Solution
OCH3 OCH3
NO2
(a) Since —NO2 is close to OCH3 in , it gets priority over
NO2
. Therefore, the priority order of the ligands is:
NO2
Br H
NO2
OCH3 OCH3
NO
OCH
2 The arrow from 1 Æ 2 Æ 3 traces
H C a clockwise path suggesting the
Br NO R configuration.
1 3
CH O
(b) For the determination of absolute configuration, the other ring is to be written in
the disconnection form (as given below), where the chiral carbon itself serves as
a duplicated atom at the end of the expanded chains. The duplicated atoms are
written within parentheses. Phantom atoms (designated by subscript 0) are used
to bring the valency of duplicated atoms up to four.
Æ Æ
(c) In this compound, the chiral centres C-2, C-3, C-5 and C-6 have absolute
configurations, S, S, S and R, respectively. According to the sequence rules, (S, S)
gets priority over (S, R). The absolute configuration of the asterisked chiral centre
is determined as follows:
C2H5 2
S (S, S) 2
H Cl
S
H Cl Allowed
* 4 H Cl 1 1 3
H Cl interchange
S
Cl H
R (S, R) 4
H Cl
3
The arrow from 1 Æ 2 Æ 3
C2H5
traces a clockwise path,
suggesting the R configuration.
13. Identify all the asymmetric carbon atom(s) in each of the following
structures:
(a) CH CH CH CH (b) H3C—CH—COOH

CH CH NH2
CH3 3
(c) (d)
(e) CH3 CH2CH CHCH3 (f)

O CH3
Cl Br
CH3 CH3
(g) (h)
H 3C H 3C
O
Solution
* (* denotes an asymmetric carbon atom)
(a) CH3CH2CH2CH
CHCH3
*
(b) H3C—CH—COOH
NH2
(c) No asymmetric carbon
CH3
(d) OH
* *
* *
(e) CH3CH3 CHClCHBrCH3
(f)
3
(g) No asymmetric carbon
* CH3
(h)
H 3C *
O
14. State whether each of the following molecules is achiral or chiral
Br H H
≈
H C I (c) H3C N CH2CH3 Cl
–
(a) (b) H C H
F Cl CH2CH2CH3
H
14 Cl
(d) 2 (e) (f)
12
2 O 3 3
H
H
Cl
(g) (h)
Cl
Solution (a) chiral (b) achiral (c) chiral (d) chiral (e) achiral (f) achiral (g) achiral
(h) achiral
15. Locate the indicated centres as stereogenic/nonstereogenic or chirotopic/
achirotopic:
2 H
(a) (b) Cl CH3
2 Cl CH3
H
Br Cl CH3
(c) C (d) C (e) H Cl
He H Cl Br
Cl H
Et I Cl H
CH3
CH3
CO2Et CH3
H
(f) H OH (g) H C H (h) H3C
H Cl C == C
Cl H H
H OH CH3
CO2Et
H H H
CH3 Cl C
(i) C (j) H3C C C == C (k) CH3
CH2CH3
H H
Et H
Solution
H C==CH H achirotopic
C==C and stereogenic
(a) H C=CH H (b) Cl CH3
Cl CH3
achirotopic and
stereogenic H
Br Cl achirotopic
chirotopic
and nonstereogenic
(c) C and stereogenic (d) C I
HC
H Cl
Et Br
chirotopic chirotopic
CH3 and stereogenic CO2Et and stereogenic
chirotopic chirotopic
(e) H Cl (f) H OH
and nonstereogenic and nonstereogenic
Cl H chirotopic H Cl chirotopic
and stereogenic and stereogenic
Cl H H OH
CH3 CO2Et
chirotopic and CH3

CH3 stereogenic
H
(g) H C H chirotopic and (h) H3C
C == C stereogenic
Cl H H
CH3 achirotopic and
stereogenic
chirotopic and
stereogenic
3 chirotopic
(i) (j) and stereogenic
2 3
H H
H3C C C == C
H H
Et
chirotopic and
(k) nonstereogenic
3
16. Label the indicated (Æ) centres of the following compounds as stereogenic/
nonstereogenic or chirotopic/achirotopic. Give your reasoning.
Br H H H
(a) F (b) Cl Cl F Cl
H Br F
H
H
(c)
Solution
(a) The molecule is chiral, therefore, the indicated centre is chirotopic. However, this
centre is not stereogenic because interchange of positions of F and H does not
produce a new stereoisomer.
Br H Br H
interchange of the
F positions of F and H H
H Br H Br
H F
chirotopic identical molecules

and nonstereogenic because one on upside
down produces the other
(b) Due to the same reason, the indicated centre is chirotopic and nonstereogenic.
(c) The indicated centre is achirotopic because a s-plane passes through it. It is
stereogenic because interchange of positions of F and H produces a new stereoisomer.
achirotopic
and stereogenic
Br F Br H
interchange of
Br positions of F and H Br
H H H F
H two different H
stereoisomers
17. Indicate whether each of the following compounds is chiral. Identify the
asymmetric carbons and stereocentres or stereogenic centres (if any) in
each.
Br
H ≈
(a) H5C2 CH—CH2CH2CH3 (b) C N
C C H3C
H5C2 CH3
Br
Cl
Cl
(c) (d)
H
3 H
Solution
(a) The compound is chiral because it contains one asymmetric carbon or chiral carbon.
Br
H 5 C2 *CH—CH2CH2CH3
C C
H5C2 CH3 Stereocentre
(b) The compound is chiral because it contains one asymmetric carbon.
(c) The compound is achiral because it has a s-plane. It contains no asymmetric

carbon. However, it has four stereocentres.
stereocentres
CD3
(d) This molecule is chiral due to presence of a chiral axis. It contains no asymmetric
carbon. However, it has three stereogenic centres.
Cl
Cl
H
H
stereocentres
18. Identify Ha and Hb in each of the following structures as homotopic,

enantiotopic or diastereotopic ligands:
Ha COOH Ha Hb
(a) H Br (b) Ha OH (c) C
Br Hb OH
C
H Hb COOH H Br
H
H Cl Cl Ha F
(d) Cl (e) (f) Cl
Ha Hb Ha Hb Hb H
Ha Br Hb H
(g) C Cl (h) Ha Hb (i) C C C
Hb H Br CH3
CH3 Ha
CH3
COOH COOH Ha
(j) H OH (k) Ha H (l) HO2C CHOHCH3
Ha Hb H Hb Hb
HO H
COOH
COOH
Solution (a) Homotopic, (b) Enantiotopic, (c) Diastereotopic, (d) Enantiotopic,

(e) Enantiotopic, (f) Diastereotopic, (g) Enantiotopic, (h) Diastereotopic, (i) Enantiotopic,
(j) Homotopic, (k) Homotopic, (l) Diastereotopic.
19. Complete the equations and label each product with D or L. Comment.
COOH
NaOH NaN3 H2
III low [OH–] H Br I Ni II
CH3
Solution
COO3 COOH COOH COOH

NaOH NaN3 H2
H OH low [OH–] H Br N3 H H 2N H
Ni
CH3 CH3 CH3 CH3
III (D–) D– I (L–) II (L–)
D-2-Bromopropanoic acid reacts with NaN3 (an SN2 reaction) with inversion of
configuration. For this reason, the configuration of I becomes L. When I is reduced to II,
no bond to the chiral centre is broken and retention of configuration occurs. As a result,
the configuration of II becomes L. When D-2 bromopropanoic acid is treated with NaOH
solution having low [OH–], a neighbouring group participation occurs and the hydrolysis
proceeds with retention of configuration (a result of double inversion). Therefore, the
configuration of III is also D.
Mechanism:
HO O
COOH COOH C
OH
–
SN2
; H
–
N3 + H Br N3 H + Br
–
Br
CH3 CH3 CH3
D L D
O
O
–
O O
–
O
C C SN2 C
OH
–
S N2
H Br 1st inversion O H 2nd inversion H OH
CH3 CH3 CH3
D
20. Find out the configuration of the final product when a hydride ion from a
chiral source is allowed to attack on the re-face of acetophenone.
Solution Since 1 Æ 2 Æ 3 traces a clockwise path, the re-face of acetophenone is as given
below. Now attack of H@ on the re-face leads to the formation of an alkoxide having S
configuration.
re-face 4
H H
3 H C
–
C==O 1 H C C – ∫ 3 H C Ph 2
O
2 Ph Ph O
–
S configuration 1
21. Draw all the possible stereoisomers of EtCH == CH—CHBr—CH == CHEt.

Mention their configurations in terms of R/S and E/Z and predict whether
they are chiral or achiral.
Solution Four stereoisomers are possible. Two of them are achiral and two of them are
chiral (exists as enantiomers). These are as follows:
Et H H Et H H H H
C C C C C C C C
H E C E H Et Z C Z Et
Br H Br H
(E, E)-isomer diastereo- (Z, Z)-isomer
(achiral) isomers (achiral)
H H H Et Et H H H
C C C C C C C C
Et Z R C E H H E SC Z Et
Br H Br H
enantiomers
(E, R, Z)-isomer (E, S, Z)-isomer
(chiral) (chiral)
22. Explain why camphor exists only as a pair of enantiomers, even though
it has two asymmetric carbons. Designate the asymmetric carbons as R
or S.
H3 C CH3
H
CH3 O
Camphor
Solution Since camphor contains two dissimilar asymmetric carbon atoms, it might be
expected to have 22 or 4 stereoisomers. However, it exists only as a pair of enantiomers
because the diastereoisomers having a trans-oriented (CH3)2C bridge is structurally
not possible for this rigid structure.
H3 C 7 CH3 Mirror H3 C 7 CH3
plane RH
H
5 4 3 4
S 3 5
R 2 1
6 1 2 S 6
CH3 O CH3
O
I II
Enantiomers of camphor
Configuration of C-1 of I:
H3C CH3
2 2
H 4 2 2
3
2 120° rotation
1 C 1 4
1 3 1 3
3
1 1 3
CH3 O 4
4 The arrow from 1 Æ 2 Æ 3
traces a clockwise path,
suggesting the R
configuration.
Configuration of C-4 of I:
2 2 The arrow from 1 Æ 2 Æ 3
4 traces a counterclockwise path,
3 1 3 1 suggesting the configuration.
The absolute configurations of C-1 and C-4 of II will be S and R, respectively.

23. Tri-sec-butylmethane has four optically active stereoisomers. Write down
the isomers in terms of R/S designation.
Solution Each sec-butyl group [CH3CH2(CH3)CH–] has a chiral C that can be R or S.
Since all three groups are equivalent, the order of groups is immaterial. RSR is identical
with SRR or RRS. The four isomers are
1. Give the R/S designation of each of the stereoisomers of 1, 3-dibromo-2-

methylbutane.
2. Write down all the possible stereoisomers of MeCH==CH–CH==N–OH and specify
their configuration as E/Z.
3. Give an example of a molecule with enantiomeric groups along with pro-R and
pro-S hydrogen atoms on a propseudoasymmetric centre.
[Hint: COOH
R
H OH
pro-S H H pro-R
H S OH
COOH ]
4. Predict configurations of the products when
(a) Pro-R-hydrogen of propanal is replaced by Br.
(b) CN@ is allowed to attack on the Re-face of propanal.
5. Designate the marked (*) centres of the following compounds as stereogenic/

nonstereogenic and chirotopic/achirotopic. Give reasons.
H Br
CH3 Me Me
H3C * *
(a) * * (b) C C
Et Et
Br H
COOH H Cl
*
C C C *
(c) H Cl (d)
Cl * H CH H
Cl H
COOH
6. Assign the following compounds with R/S descriptor.
Ph
H H H Me
(a) D3 C C CH2CH3 (b) C C C C
Me H
C
C∫CCH3
OH
Ph H
N∫C CH(OCH3)2
OHC CH2OH
(c) (d) C C C
H H H3C—C CHO
Ph
O
H 3C CN COOH
(e) C (f) H3C NH2

Ph C∫CH
Ph
OHC O
CH2OCMe3
(g) C (h) P CH2Ph
D3CH2 CD2CH3 Me Ph
CD2CH3 CH3
Cl
Cl
(i) H3C C OH (j) Cl CH3
Cl
CH3 H
CH2CH2CH3
H Br CH2OH
(k) (l) C CD2OH

H2 N CH2SH
Cl
H CH3
(m) C F (n)
I
Br
H Br
C
(o)
Cl
7. List the substituents in each of the following sets in order of decreasing priority:
(a) –Br, –OH, –SH, –H
(b) –CH2Br, –CH2Cl, –CH2OH, –CH3
(c) –CH=CH2, –CH2CH3, –Ph, –CH3
(d) –CH2CH2Cl, –CH(CH3)2, –Cl, –CH2CH2CH2Br
(e) –CH(CH3)2, –C(CH3)3, –H, –CH==CH2
(f) –OH, –OPO3H2, –CHO, –H
(g) –OCH3, –CH3, –NEt2, –H
(h) –D, –H, –OH, –CHO
(i) –NH2, –F, –CH3, –OCH3
8. Shown below are Newman projection formulas for (R, R)-, (S, S)- and (R, S)-4,
5-dibromooctane. (a) Which is which? (b) Which formula represents an achiral
meso compound?
CH2CH2CH3 CH2CH2CH3 CH2CH2CH3
H Br H Br Br H
H Br Br H Br H
CH2CH2CH3 CH2CH2CH3 CH2CH2CH3
I II III
9. For the following molecule, draw its enantiomer as well as one of its diastereoisomers.
Assign the R or S configuration at each chirality centre.
H
Cl
F
H5C2
H
H
10. Draw one chiral (active) and one achiral (meso) isomer of EtO2C(CHOH)3CO2Et in
Fischer projection formula. Will the interchange of H and OH at C-3 of the chiral
stereoisomer, you have drawn, lead to another stereoisomer? What will happen if
H and OH are interchanged at C-3 of the achiral (meso) isomer you have drawn?
Explain stating whether C-3 is a stereogenic centre in each case.
11. There are four diethylcyclopropane isomers. Write 3D-formulas for these isomers.
Which of the isomers are chiral?
≈
12. PhCH2CH N H3COO@ has two diastereotopic ligands. Label them as pro-R and pro-S.
13. Identify the enantiotopic, diastereotopic and homotopic hydrogens (HA, HB) in each
of the following compounds:
CO2H
(a) A B (b) HA HB (c)
HB OH
3
CH3
CH3 COOH
HA HB 2 2
(d) C == C == C (e) HB H H HA (f)
CH3
H3C A B
COOH CH3
HA
Me
3
ClHB O
` (g) A (h) A (i)
Cl
B Me
B
3
H A
(j) H C (k)
2 B
H
2
14. Locate the stereogenic centre(s) in each of the following compounds. A molecule
may have no stereogenic centre.
(a) CH3CH2CH2CHClCH2CH2CH3
(b) CH3CH2CH2OCH(CH3)CH2CH3
(c) CH3CH2CHDCH2CH2CH3
(d) (CH3)2CHCH2CH(CH3)CH2CHClCH2CHBrCH3
(e) (f)
O
3
Br
(g) CH2 (h)

Et
3
15. Identify the missing substituents (A and B) in each of the following compounds:
A A
(a) C C2H5 (b) HO B

B CH3
(S)-2-Bromobutane COOH
(R)-2-Hydroxypropanoic acid
16. When (S)-2-methylbutan-1-ol is allowed to react with concentrated HBr, (S)-1-
bromo-2-methylbutane is obtained. Explain with reasons whether the reaction
proceeds with inversion or retention of configuration.
17. Redraw each of the following molecules as a Fischer projection and then assign R
or S designations to each stereocentre:
COOH
3
OHC C2H5
(a) (b)
HO C2H5
3
OH
H2N OH H3 C H
H Cl
(c) C C (d) H C C
D3 C
H COOH F CH3
18. Indicate whether the hydrogens marked Ha and Hb in each of the following
compounds are homotopic, enantiotopic or diastereotopic:
Cl
Ha
Ha
(a) (b) Ph
Hb
Hb Ph
Hb H H Ha
Ha H H
Hb
C C C C
(c) (d)
CH3 C H H 3C C CH3
Br F
H H
H H Hb
H
Ha Ha
Hb H
C C C
(e) C (f)
H5C2 D3 C C CD3
C C2H5
Cl H Br
H
H Hb
3
(g) 3 (h) H5C2 C C
a C C C2H5
b Ha H
Ha Hb
(i) (j) Ha–CH2 S

O
Hb–CH2
O
Ha H CMe3 Hb
Hb
(k) H (l)
CMe3
H Ha CMe3
H
OH b
a
O– CH3
≈
(m) S—C Hb (n)
Ha
Me
19. Write down the flying wedge projections of each of the following compounds having
noted configuration and ligands bonded to the chirality centre:
14
(a) —Ph, — C ∫ CH, —H, —C ∫ CH (with R configuration)
(b) —I, —CH2NH2, —CH2CH3, —CH2Cl (with S configuration)
(c) —OD, —18OD, —OMe, —OH (with R configuration)
(d) Me, —OH, —D, —CN (with S configuration)
20. Predict the number of diastereoisomers for the compound with molecular formula
C2BrClFI. Write down their structures and label each of them as (E) or (Z).
[Hint: Six diastereoisomers are possible; these are:

Br Cl , F Cl , F Br , I Br , I Cl Br Cl
and
F I Br I I Cl F Cl Br F I F
21. Give examples of the following:
(a) A chirotopic as well as stereogenic centre
(b) A chirotopic but nonstereogenic centre
(c) A stereogenic but achirotopic centre
22. A triol [CH3CH2(CHOH)3CH2CH3] has three dissimilar asymmetric carbon atoms.
List the isomers in terms of R/S designation.
È R S R S R S R S ˘
ÍHint: R S ; R S ; S R ; S R ; four pairs of enantiomers ˙
Í ˙
ÍÎ R S S R R S S R ˙˚
23. Draw all possible stereoisomers of EtCH(OH)CH==CHEt and designate them by
R/S and E/Z notations.
[Hint: (3R, 4E), (3S, 4E), (3R, 4Z), (3S, 4Z).]
24. How many stereoisomers of the following compound would you expect
(mathematically speaking) to have? Only two of them are actually isolated – Why?
Predict the number of stereogenic centre in the compound.
3
25. Which of the following structures represents the naturally occurring amino acid
(2S, 3R)-threonine?
H H H
H3C ≈ ≈
C C NH3 H3N C C OH
HO
COO
– –
OOC CH3
H
I II
≈
H CH3 HO NH3
≈
H3N C C H C C COO –
H
H3C
–
OOC OH H
III IV
2.6 OPTICAL ACTIVITY OF CHIRAL COMPOUNDS

Enantiomers share many of the same properties, including the same melting points, the
same boiling points, the same indices of refraction, same solubilities in common solvents,
same infrared spectra, same rate of reaction with achiral reagents, etc. In fact, all the
physical and chemical properties of enantiomers are the same except those that arise due
to different spatial arrangement of groups bonded to the asymmetric or chiral centre. One
property that enantiomers do not share is the way they interact with the plane-polarized
light resulting in its rotation in opposite directions.
2.6.1 Plane-polarized Light

Light is an electromagnetic phenomenon. A beam of normal or ordinary light (composed
of rays of different wavelengths) consists of an electric field and a magnetic field which
are oscillating at right angles to each other in every plane perpendicular to the direction
of propagation. The same is true for monochromatic light, i.e., light of single wavelength.
These oscillations of the electric field can be made to occur in a single plane when a beam
of monochromatic light is allowed to pass through a polarizer (a Nicol prism) that interacts
with the electric field. Such light is called plane-polarized light.
2.6.2 Optical activity

In 1815, the physicist Jean-Baptiste Biot discovered that some naturally occurring organic
compounds rotate the plane of polarization of plane-polarized light. Some of them rotate
it clockwise and some rotate it counterclockwise. It has been proposed that the rotation of
the plane of plane-polarized light is due to some asymmetry in the molecule.
When the plane-polarized light is allowed to pass through an achiral compound in pure
liquid form or through its solution in a suitable solvent, the light emerges from the
solution with its plane of polarization unchanged, i.e., there occurs no rotation of the
plane of polarization. However, when the plane-polarized light is allowed to pass through
a solution of a chiral compound or through the pure liquid chiral compound, the light
emerges with its plane of polarization rotated either clockwise or counterclockwise. If one
enantiomer of the compound rotates it clockwise, its mirror image will rotate it in exactly
the same amount counterclockwise.
A compound that rotates the plane of polarization of plane-polarized light is said to be
optically active. It thus follows that chiral compounds are optically active, while achiral
compounds are optically inactive. If an optically active compound rotates the plane of
polarization clockwise, then the compound is called dextrorotatory (Latin dexter, meaning
right), which can be indicated in the name of the compound by the prefix (+). If it rotates
the plane of polarization counterclockwise, then it is called levorotatory (latin laevus,
meaning “left”), which can be indicated by (–). Sometimes d and l are used instead of (+)
and (–).
The (+) and (–) signs are not to be confused with R and S. In fact, there is no correlation
between R and S configuration of enantiomers and the direction [(+) or (–)] in which they
rotate the plane of plane-polarized light. The (+) and (–) symbols indicate the direction in
which an optically active compound rotates the plane of plane-polarized light, whereas R
and S indicate the spatial arrangement of the groups about a chiral centre. Compounds
with the R configuration may be (+) or (–). Similarly, compounds with S configuration may
be (+) or (–). For example, (R)-(+)-2-methyl-1-butanol and (R)-(–)-1-chloro-2-methylbutane
have the same configuration, that is, they have the same general arrangement of their
groups about the chiral centre. However, they have an opposite effect on the direction of
rotation of the plane of plane-polarized light. The first one is dextrorotatory (+) and the
second one is levorotatory (–).
2 5 2 5
2 3 2 3
R R
The configuration (R or S) of a compound can be ascertained by looking at its structure,

but the only way to know whether a compound is dextrorotatory (+) or levorotatory (–) is
to put the compound in a polarimeter, an instrument that measures the direction as well
as the amount the plane of polarization of plane-polarized light is rotated.
2.6.3 Polarimeter
A polarimeter is an instrument by which the rotation of the plane of a plane-polarized light
by an optically active compound is detected and measured. The main working parts of a
polarimeter are: (i) a light source, (ii) a polarizer, (iii) a cell or tube for holding the optically
active compound or its solution in the light beam, (iv) an analyser, and (v) a scale for
measuring the angle (in degrees) that the plane of plane-polarized light has been rotated.
A polarimeter uses a monochromatic (single-wavelength) light source (usually a light from
a sodium arc which is called the sodium D-line; wavelength = 589 nm) which produces
unpolarized light. This light passes through the polarizer to generate the plane-polarized
light. The plane-polarized light then passes through the cell or tube and then through
the analyzer and finally reaches the eye of an observer. Before taken to the compounds
in the cell, the polarizer and the analyser are so adjusted that the light passes through
them without loss of intensity, i.e., the maximum amount of light reaches our eye. The
compound in pure liquid form or as a solution in a suitable solvent like water, ethanol, etc.,
is then poured into the cell. If the intensity of light is found to be reduced (due to rotation
of the plane of the polarized light), the compound is said to be optically active. In that case,
the analyser is rotated in either clockwise or counterclockwise direction to bring back the
original intensity of light. The angle in degrees that the analyser needs to be rotated is
called the observed rotation (a) of the optically active compound. The clockwise rotation of
the plane of plane-polarized light is taken as positive (+), while counterclockwise rotation
of the plane is taken as negative (–). Two enantiomers rotate the plane of plane-polarized
light by exactly the same amount, i.e., to an equal degree but in opposite directions. The
observed rotation depends on the number of optically active molecules that the light
encounters in the sample, which in turn depends on the concentration of the sample and
the length of the polarimeter tube. The observed rotation also depends on the temperature
and the wavelength of the light used.
A schematic drawing of a polarimeter is as follows:
Direction of light propagation
Light Ordinary Polarizer Plane-

source light polarized
(Sodium-D-line) light
Observed
Direction of light propagation a rotation
Viewer
Polarimeter tube Plane- Analyser
containing a chiral polarized
compound light
(after clockwise
rotation)
2.6.4 Specific Rotation

The observed rotation (a) is found to be proportional to the concentration of the optically
active compound and the length of the cell through which the plane-polarized light passes.
a
a = [a ] ◊ l ◊ c or [a ] =
l◊c
where, [a] = the proportionality constant and is called the specific rotation
a = the observed rotation
c = the concentration of the solution in grams per millilitre of solution
(or density in gm/L for pure liquids)
l = the length of the cell in decimetre (1 dm = 10 cm).
Thus, the specific rotation, [a], of a compound may be defined as the rotation in degrees
brought about by a pure liquid or solution containing 1 g of optically active substance
per mL of solution, placed in a 1 decimetre polarimeter tube. Since the specific rotation
is independent of c and l, it is used as a standard measure of optical activity. The specific
rotation also depends on the temperature and the wavelength of light that is employed
and for this, [a] is conventionally reported with a subscript that indicates the wavelength
of light, used and a superscript that indicates the temperature in °C. Therefore, a specific
rotation reported as [a]D25 indicates that it has been determined at 25°C and the wavelength
of light used is a yellow emission line in the spectrum of Na, called the sodium D line. The
specific rotation of (+)-2-methyl-1-butanol, for example, might be given as follows:
[a ]25
D = + 5.756∞
Although the unit of specific rotation is degree cm2g–1, it is usually expressed in degree
only. When the specific rotation is measured in solvents other than water, the name of
the solvent must be mentioned. Under a given set of conditions the specific rotation of
any compound is constant and the value may be used as a criterion for identification of
optically active compounds. Specific rotation is a physical property of a compound just as
mp, bp, density, etc.
t t M t
Optical rotation is also expressed by Molecular rotation, [M]D , such that [M]D = [a ]D ¥
100
where M indicates the molecular mass of the chiral compound. Thus, optical rotation of 1
mole of compound in a sample tube of 1 dm length is called its molecular rotation.
A single experiment of measurement of rotation of any chiral compound cannot be used
to conclude that the rotation is either dextro (+) or levo (–). For example, if the observed
rotation of a compound is reported to be +50° one can conclude that it could be –130°
(concentration is 1 M). One can find out the correct direction of rotation by changing
the concentration of the solution. If the concentration is reduced by ten times, then the
decrease in dextrorotation should be ten times of +50° and consequently the new value
would be +5°. Similarly, the change in levorotation should be ten times less than that of
–130° and the changed value would be –13°. Therefore, if the second reading is +5°, the
rotation must be dextro (+) and if the second reading is –13°, the rotation must be levo (–).
Thus, the reading taken at two different concentrations almost always determine the sign
of a unequivocally.
Calculation of specific rotation The observed rotation of a chiral compound, for example,
(–) glyceraldehyde at 25°C using sodium D line has been found to be –1.74° when its
solution containing 2.0 g/10 mL is placed in a 1 dm polarimeter tube. Then its specific
rotation:
a -1.74
[a ]25
D = = = - 8.7∞
cl (1)(0.2)
Doubling the concentration doubles the observed rotation because the number of molecules
the plane of polarized light encounters doubles. Therefore, the observed rotation, a =
–1.74° ¥ 2 = –3.48°. Doubling the length of the sample tube also doubles the observed
rotation due to the same basic reason. However, the specific rotation, [a ]25
D , is a constant
and is independent on concentration and the length of the cell because the value of a/c or
a/r remains unchanged.
2.6.5 The Necessary and Sufficient Condition

(the ultimate criterion) for Optical Activity
The same chirality (nonsuperimposability on the mirror image) that gives rise to
enantiomerism also is responsible for optical activity. The relationship between optical
activity and chirality is absolute. Many thousands of cases have been found in accord with
it. The ultimate criterion, then, for optical activity is chirality. This is both a necessary
and a sufficient condition.
Optical activity: A close look All optically active compounds are chiral, but the reverse is
not always true. The reasons are as follows:
(i) Some chiral molecules are configurationally unstable. Due to rapid pyramidal
inversion (umbrella effect) such molecules undergo racemization (formation
of an equimolar mixture of two enantiomers) and becomes optically inactive
[the (+)-enantiomer would rotate the plane of polarized light clockwise and the
(–)-enantiomer would rotate the plane counterclockwise by exactly the same
amount]. For example, ethylisopropylamine (EtNHCHMe2) is found to be optically
inactive although the nitrogen atom has four different groups around it (a hydrogen,
an ethyl group, an isopropyl group and an electron pair) and the geometry of the
molecule is essentially tetrahedral, i.e., it is a chiral molecule.
enantiomers
Et Et Et
N N N
H H
Me2HC H CHMe2 CHMe2
(Chiral) Transition state (Chiral)
(achiral)
A racemic mixture
of ethylisopropylamine
As the inversion takes place, the large lobe of the electron pair appears to push
through the nitrogen to the other side. As this occurs, the three other groups move
first into a plane containing the nitrogen, then to the other side.
(ii) Because of extremely low differences in polarizabilities among the alkyl groups,
some chiral molecules have a vanishingly small rotation and are found to be
optically inactive. For example, the specific rotation of the chiral alkane ethyl-
n-propyl-n-butyl-n-hexylmethane is too low and actually far below the limits of
deflection by any existing polarimeter (calculated specific rotation (0.00001°) and
so, each of its enantiomer is found to be optically inactive.
CH2CH3
C
CH3
CH3(CH2)3
(CH2)5CH3
(R)-Ethyl-n-propyl-n-butyl-n
hexylmethane
(chiral but exhibits no
optical activity)
(iii) A 50:50 mixture of enantiomers is optically inactive. If we are to observe optical
activity, the material we are dealing with must contain an excess of one enantiomer
so that the net optical rotation can be detected by the particular polarimeter at the
hand.
2.6.6 Racemic Modification

An equimolar mixture of two enantiomers is called a racemic modification or a racemic
mixture or a racemate. A racemic mixture is symbolized by the prefix (±) to the name of
the compound. For example, racemic 2-butanol would be symbolized by (±)-2-butanol. A
racemic modification is optically inactive because the rotation caused by a molecule of one
isomer is exactly cancelled by an equal and opposite rotation caused by a molecule of its
enantiomer.
Mirror
CH3 plane CH3
C C
H5C2 H H C2H5
OH OH
(±)-2-Butanol
(optically inactive)
2.6.7 Enantiomeric Excess(ee) or Optical Purity (op)

Whether a particular sample of a compound consists of a single enantiomer, a racemic
mixture, or a mixture of two enantiomers in unequal amounts can be determined by its
observed specific rotation, which is the specific rotation of the sample. For example, if
a sample of (S)-(+)-2-butanol is enantiomerically pure or has an enantiomeric excess of
100% (meaning only one enantiomer is present), it will have an observed specific rotation
of +13.52° because its specific rotation is +13.52. If, however, the sample of 2-butanol is
a racemic mixture, it will have an observed specific rotation of 0°. If the observed specific
rotation is positive but less than +13.52°, the sample must be a mixture of two enantiomers
containing more of the S enantiomer than the R enantiomer, because the S enantiomer is
dextrorotatory.
The enantiomeric excess (ee), also known as optical purity (op), can be defined as the
specific rotation of a mixture of two enantiomers, expressed as a percentage of the specific
rotation of a pure enantiomer.
Observed specific rotation
Therefore, Enantiomeric excess = ¥ 100%
Specific rotation of the pure enantiomer
For example, if a sample of 2-butanol has an observed rotation of +6.76°, then the
enantiomeric excess is 50%. In other words, the excess of one of the enantiomers comprises
50% of the mixture.
+6.76∞
Enantiomeric excess = ¥ 100 = 50%
+13.52∞
When we say that the enantiomeric excess of this mixture is 50%, we mean that 50% of the
mixture is excess S enantiomer and 50% is a racemic mixture. Half of the racemic mixture
plus the amount of excess S enantiomer equals the amount of the S enantiomer present
in the mixture. It thus follows that 75% of the mixture is the dextrorotatory S enantiomer
[(50 ¥ 1/2) + 50] and 25% is the levorotatory R enantiomer.
The enantiomeric excess may also be defined as the excess amount of one enantiomer
in a mixture of enantiomers expressed as a percentage of the entire mixture and can be
expressed as follows:
moles of one enantiomer - moles of other enantiomer
Enantiomeric excess = ¥ 100
total moles of both enantiomers
For example, the enantiomeric excess of a mixture containing 10 g of (S)–(+)-2-butanol
and 6 g of (R)-(–)-2-butanol is 25%.
10 6
-
4
Enantiomeric excess = 74 74 ¥ 100 = ¥ 100 = 25%
16 16
74
2.6.8 Racemization
Racemization is the process of producing a racemic modification starting from either of
the pure enantiomer. A racemic modification is an equimolar mixture of two enantiomers,
i.e., a mixture of two different molecular species. Since mixing causes increase in disorder,
DS of racemization is a positive quantity and therefore, DG in the expression DG° = DH°

– TDS° is negative (assuming DH° constant). Because of this, racemization is a process
which is thermodynamically favourable. For example, at 27°C (300 K), DG° change is –1.8
kJ/mol. Driving force for racemization is, in fact, entirely entropic.
Various methods used for racemization are discussed below:
(a) Thermal racemization In this process of racemization, one of the four bonds to the
chiral carbon is broken temporarily by applying heat. An achiral free radical is obtained
by homolytic bond cleavage. The carbon radical then combines with the breakaway
radical from either faces of the sp2 (planar) carbon to form an equimolar mixture of two
enantiomers (a racemic modification). For example, (+)- or (–)- a-chloroethylbenzene
undergoes racemization when distilled under normal pressure.
Cl
D
C Cl
H3C H
Ph C
( )-(+)-a-Chloro- H3 C H
ethylbenzene Ph
H3C C H + Cl Mirror
or plane H3C
Ph
Ph H
Cl An achiral free C
D radical
C Cl
Ph H
H3 C (±)-a-Chloroethylbenzene
( )-(–)-a-Chloro-
ethylbenzene
(b) (i) Acid-catalyzed racemization via an enol This process of racemization involves
temporary separation of an acidic hydrogen atom bonded to a chiral centre. Chiral carbonyl
*
compounds of the type RR ¢ CHCOR in which the chiral centre containing hydrogen is
adjacent to the C=O group undergoes racemization in the presence of acid through the
formation of a planar and achiral enol (intermediate). Recombination of the proton with
the achiral enol molecule then occurs from either side of the double bond with equal facility
to produce a racemic modification. For example, 1-phenyl-2-methyl-1-butanone undergoes
racemization when dissolved in ethanol containing HCl. The solution gradually loses its
optical activity.
HO
O OH H C C(CH3)CH2CH3
* H OH OH Ph
Ph C CHCH2CH3 Ph C C CH2CH3
CH3 CH3 HO –
(+) or (–)-1-Phenyl-2-methyl- C C(CH3)CH2CH3

1-butanone Ph
Enol form
(kclo form)
(achiral)
(a)
2 3
2 3
2 3
3 H O H
– 2 3
3
(b)
2 3
2 3 3
3 2
(ii) Base-catalyzed racemization via an enolate anion This process of racemization involves
the formation of a planar (achiral) enolate anion formed by expulsion of a proton from the
chirality centre by base. The enolate anion then recombines with a proton from water.
This process occurs from either side of the enolate with equal facility to give a racemic
modification. For example, 1-phenyl-2-methyl-1-butanone undergoes racemization when
dissolved in ethanol containing NaOH.
O H –
O
–
O –
Ph C C CH CH C C(CH )CH CH C C(CH )CH CH

Ph Ph
CH
Enolate anion (achiral)
(+)- or (–)-1-Phenyl-2-
methyl-1-butanone
H
Ph C (a)
C CH2CH3
CH3 (a)
O
HO H O
Mirror CH2CH3
(±)-1-Phenyl-2-
–
(–OH – ) C C
methyl-2-butanone plane Ph
CH3
O
CH2CH3 (b) (b)
C CH3
Ph C
H
(c) By cation formation This process involves the formation of a carbocation by separating
a group from the chirality centre with its bonding electrons leaving behind a planar
(achiral) carbocation. The anion then attacks the intermediate carbocation from either
side with equal facility to form both the enantiomers in equal amounts. This process
actually occurs when a substrate which can form a stable carbocation (benzylic, allylic
or tertiary) is treated with a Lewis acid (e.g., SbCl5, AlCl3, BF3 and ZnCl2). For example,
when (+)- or (–)-a-chloroethylbenzene is heated with antimony pentachloride (SbCl5),
(±)-a-chloroethylbenzene is obtained.
3 3
SbCl –
–
5 3 6
Cl
C 5
(a)
Ph CH3
H
Mirror CH3
(±)-a-Chloro- Ph C
plane
ethylbenzene H
(b)
CH3
Ph 5
C H
Cl
A mineral acid may also cause racemization. For example, when optically active (2R, 4S)-
or (2S, 4R)-camphene is treated with acid, racemization occurs through the formation of a
carbocation which undergoes a methyl shift followed by loss of a proton.
CH3 1,2-methyl–shift CH3

CH3
CH3 1,2-methyl–shift
CH3
CH3
+H –H
+H –H
CH3
CH3
CH2
180° rotation
Mirror
plane
* H2C *
CH2
* H3C *
CH3
H3C
(2R, 4S) CH3 (2S, 4R)
(±)-Camphene
2.6.9 Resolution of Racemic Modification

Resolution is the method of separation of a racemic modification into its pure enantiomers.
The two enantiomers in a racemic mixture have identical chemical and physical
properties except towards optically active reagents. So, they cannot be separated by usual
techniques like fractional distillation, fractional crystallization (unless the solvents are
optically active), chromatography (unless the absorbents are optically active), etc. The
chemical method, which involves the formation of diastereoisomers, is the best method
of resolution of a racemic modification. Since diastereoisomers, unlike enantiomers, have
different physical properties like solubility, boiling point, absorption coefficient, etc., they
can easily be separated by fractional crystallization (if they are solids) or by fractional
distillation (if they are liquids). The separated diastereoisomers are then individually
treated with a suitable reagent to regenerate the pure enantiomer. Resolution of three
types of compounds are discussed below.
1. Resolution of a racemic carboxylic acid The racemic modification of a carboxylic acid

*
(RCOOH) is first allowed to react with an optically active base (the resolving agent, e.g.,
methylamine, 1-phenylethylamine and alkaloids like brucine, strychnine, ephedrine,
morphine, etc.) to form two diastereoisomeric salts. These salts are then separated by
fractional crystallization. Each of the separated diastereoisomer is then treated with
dilute mineral acid (e.g., dil. HCl) to regenerate the enantiomer. Finally, the resolving
agent and the pure enantiomer of the acid are isolated.
A schematic representation of the process is as follows:
O
C H ( ) N
*
( ) R O
+ +
O
* C H ( ) N
( ) R O
A racemic mixture of An optically

a carboxylic acid active amine (base)
(the resolving agent)
An acid-base reaction
O O
– –
* C ( ) NH * C ( ) NH
( ) R O + ( ) R O
Two diastereoisomeric salts
having different physical properties
separated by fractional
crystallization
* *
( ) RCOO ◊ ( ) RCOO ◊ ( )
–
NH ( ) NH
Single diastereoisomer Single diastereoisomer
dil. HCl dil. HCl
* *
( ) RCOOH + ( ) NH Cl ( ) RCOOH + ( ) NH Cl
– –
Pure enantiomer Hydrochloride Pure enantiomer Hydrochloride

(organic layer) salt of amine (organic layer) salt of amine
(aqueous layer) (aqueous layer)
Example: Resolution of racemic mandelic acid may be shown as follows:
– –
– –
– –
The organic layer is separated and from that pure enantiomer is isolated by removing the
solvent.
*
2. Resolution of a racemic base The racemic modification of an organic base (RNH2 ) is
converted into two diastereoisomeric salts by the reaction of an optically active carboxylic
acid (the resolving agent, e.g., camphoric acid, camphor-10-sulphonic acid, glutamic
acid, tartaric acid, etc.). The salts are then separated by fractional crystallization. Each
diastereoisomeric salts is then treated with dilute alkali to regenerate the enantiomeric
amines, leaving the acid as its salt.
A schematic representation of the process is as follows:
(R) RNH
+ + 2(S) RCOOH
(S) RNH An optically active
A racemic mixture carboxylic acid
of an amine (the resolving agent)
[(R) RNH (S) RCOO + (S) RNH (S) RCOO ]

– –
Two diastereoisomeric salts
(R) RNH (S) RCOO (S) RNH (S) RCOO

– –
(R) RNH + (S) RCOONa + H O (S) RNH + (S) RCOONa + H O

Pure enantiomer Sodium salt Pure enantiomer Sodium salt
(organic layer) of the acid (organic layer) of the acid
(aqueous layer) (aqueous layer)
Example: Resolution of racemic 1-phenylpropylamine by using (S)-(+)-mandelic acid as

the resolving agent may be shown as follows:
2 2
R S S
An acid-base
reaction
– –
3 3
R S S S
Separated by fractional
crystallization
– –
3 3
R S S S
dil. NaOH followed by dil. NaOH followed by
extraction with an extraction with an
organic solvent organic solvent
– –
2 2
2 2
R S
S S
The organic layer is separated and from that the pure enantiomer is isolated by removing
the solvent.
3. Resolution of a racemic alcohol (±)-Alcohols are usually resolved by converting them

into their half-ester by treating them with phthalic or succinic anhydride. These half-
esters are then resolved as typical active acids. Pure enantiomeric alcohols are finally
regenerated either by hydrolysis with hot aqueous sodium hydroxide or by reduction with
LiAlH4. Reduction is preferred when there is possibility of racemization by base.
Example: Resolution of (±)-2-methyl-1-butanol may be shown as follows:
C H C H O
C
HOH C CH + H C CH OH + O
C
H H
O
(R)-2-Methyl-1- (S)-2-Methyl-1- Phthalic anhydride
butanol butanol
A racemic mixture
COOH C H C H HOOC
C O H C CH + H C CH O
C
O H H
O
(R)-Half-ester (S)-Half-ester
A racemic mixture
C H C H HB OOC
– –
COO BH
C O H C CH + H C CH O
C
O H H
O
[(R), (–)]-salt [(S), (–)]-salt
A mixture of two diastereoisomeric salts

C H HB OOC
–
COO BH
–
C H
CH H C CH O
C O H C C
H H
O O
[(R), (–)]-salt [(S), (–)]-salt
COOH C H C H HOOC
– –
CH + BHCl BHCl + H C CH O
C O CH C
(aqueous (aqueous H
O H layer) O
layer)
(R)-Half-ester (S)-Half-ester
(organic layer) (organic layer)
The half-esters are separated and then treated individually as follows:

2 5
aq.NaOH/D 2 3
2 5
R
2 3
2 5
LiAlH4 2
R 2 3
2
R
2 5
aq.NaOH/D 3 2
S
2 5
3 2 2 5
LiAlH4 2
3 2
S 2
S
1. Give an example of each of the following compounds:

(a) an optically inactive compound containing chiral carbons,
(b) an optically active compound containing no chiral carbon.
Solution
(a) Meso-tartaric acid containing two chiral carbons is optically inactive.
COOH
H OH
H OH
COOH
meso-Tartaric acid
(b) Pent-2-3-diene (CH3CH=C=CHCH3) is an optically active compound containing no
chiral carbon.
2. Write structural formulas for the simplest optically active (a) alkene, (b)
alkyne, (c) alcohol, (d) aldehyde, (e) ketone, (f) carboxylic acid with no
isotopic atom.
Solution
* *
(a) CH2=CH– C H(CH3)CH2CH3 (3-methylbut-1-ene), (b) CH∫∫C– C H(CH3)CH2CH3
* *
(3-methylbut-1-yne), (c) CH3 C H(OH)CH2CH3 (2-butanol), (d) CH3CH2 C H(CH3)
*
CHO (2-methylbutanal), (e) CH3CO C H(CH3)CH2CH3 (3-methylpent-2-one), (f)
*
CH3CH2 C H(CH3)COOH (2-methylbutanoic acid).
3. In which of the following ways should enantiomers differ?
(a) melting point, (b) boiling point, (c) sign of specific rotation, magnitude
of specific rotation, (e) absolute configuration, (f) solubility in water,
(g) solubility in chloroform, (h) solubility in benzene, (i) solubility in
(+)-2-chloropentane, (j) solubility in (–)-2-chloropentane, (k) solubility
in (±)-2-chloropentane, (l) refractive index, (m) infrared spectrum, (n)
NMR spectrum, (o) rate of reaction with an optically inactive reagent, (p)
rate of reaction with an optically active reagent, (q) mass spectrum, (r)
colour, (s) interaction with left circularly polarized light, and (t) toxicity
to human beings.
Solution Enantiomers differ in (c), (e) (i), (j), (p), (s) and (t).
4. Can a tertiary amine of the type RR ¢R ¢¢ N or a carbanion of the type
RR ¢R ¢¢ C: exist as a pair of enantiomer?
Solution Theoretically it can, but not in practice because a rapid pyramidal inversion
(umbrella effect) converts either enantiomer to a racemic mixture. The energy required
for this inversion is very low at room temperature, thus racemization is unavoidable. The
inversion process is so fast that the enantiomers cannot be separated or isolated.
enantiomers enantiomers
R≤ R≤
R≤ R≤ R≤ – – –
R≤
N N N C C C
R¢ R¢ ; R¢ R¢
R R¢ R R R R¢ R R
(chiral) T.S. (chiral) (chiral) T.S. (chiral)
(achiral) (achiral)
A racemic mixture A racemic mixture
5. An optically active alkene A has the molecular formula C5H9Cl. On

hydrogenation A is converted to the optically inactive compound B.
Predict the structure of A and B.
Solution Since on hydrogenation the compound A becomes optically inactive, i.e.,
achiral, the unsaturated group and the other alkyl group attached to the chiral centre
must contain the same number of carbon atoms and become identical after hydrogenation.
Therefore, the unsaturated group must be –CH==CH2 and the saturated group which may
be obtained on its hydrogenation is –CH2CH3. Hence, the compound A is 3-chloropent-1-
ene (any enantiomer) and the compound B is 3-chloropentane.
Cl Cl Cl
H2/Ni
H3CH2C CH CH2 or H2C CH CH2CH3 CH3CH2 CH2CH3
H H H
R S s-plane
A B
3-Chloropent-1-ene 3-Chloropentane
(optically active) (optically inactive)
∞
6. Esterification of (+)-lactic acid, [a ]25
D = + 4∞ , with methanol give (–)-methyl
∞
lactate, [a ]25
D = –8°. Has the configuration changed?
Solution When lactic acid (CH3CHOHCOOH) undergoes esterification by reacting with

MeOH in the presence of acid, none of the bonds to the chiral centre is broken. Therefore,
although the sign of rotation is changed, the configuration remains unchanged.
COOH COOMe
MeOH/H
HO CH3 HO CH3
H H
(+)-Lactic acid (–)-Methyl actate
Configuration retained
7. No necessary correlation exists between the R and S designations and the

direction of rotation of plane-polarized light. Explain.
Solution The rotation of a compound, (+) or (–), is something that we measure in the
polarimeter, depending on how the molecule interacts with light. On the other hand, the
R and S designation is our own artificial way of describing how the atoms or groups are
arranged in space around a chiral centre. In the laboratory, we can measure a rotation and
see whether a particular substance is (+) or (–). On paper, we can determine whether a
particular drawing is designated as R or S. But it is difficult to predict whether a structure
with R-configuration will rotate the plane of polarized light clockwise or counterclockwise.
Similarly, it is difficult to predict whether a (+) substance in a bottle has the R or S
configuration.
8. (–)-Epinephrine is biologically active (stimulating effect is notified when
given to a patient) but (+)-epinephrine is not. Why?
Solution Enzymes in living systems are chiral, and they are capable of distinguishing
between enantiomers. Usually, only one enantiomer of a pair fits properly into the chiral
active site of an enzyme. Only the (–)-enantiomer of epinephrine fits into the active site of
an enzyme and for this reason, (–)-epinephrine is biologically active.
9. (S)-Lactic acid has the physical characteristics listed under the structure.
COOH
C
H
H3C OH
(S)-Lactic acid
25°
[a]D = + 3.8°
mp = 53°C
(a) What is the melting point of (R)-lactic acid? (b) How does the melting
point of a racemic mixture of (R)- and (S)-lactic acid compare to the
melting point of (S)-lactic acid? (c) What is the specific rotation of (R)-
lactic acid, recorded under the same conditions as that of (S)-lactic acid?
(d) What is the optical rotation of a racemic mixture of (R)- and (S)-lactic
acid? (e) Label each of the following as optically active or inactive: (i) a
solution of pure (S)-lactic acid, (ii) a solution of an equimolar mixture

of (R)- and (S)-lactic acid and (iii) a solution that contains 70%(S)- and
30%(R)-lactic acid.
Solution (a) 53°C. (b) The melting point of a racemic mixture of (R)- and (S)-lactic acid
will be different from that of (S)-lactic acid. (c) The specific rotation of (R)-lactic acid is
–3.8°. (d) The optical rotation of a racemic mixture of (R)- and (S)-lactic acid is zero. (e) (i)
optically active, (ii) optically inactive, (iii) optically active (dextrorotatory).
10. Calculate the specific rotation [a ]20
D of (S)-2-butanol if a sample of (S)-2-
butanol has an observed rotation of +1.03°. The measurement was made
with a 1.0 M solution of (S)-2-butanol in a polarimeter tube that is 10 cm
long.
Solution To calculate the specific rotation, the sample concentration in g/mL must be
determined. Since the molecular mass of 2-butanol is 74.12, the solution contains 47.1 g/L
or 0.0741 g/mL of 2-butanol. This is the value of C. The value of l is 10 cm or 1 dm.
a +1.03
Therefore, [a ]20
D = = = + 13.9∞
cl (0.0741)(1)
11. A sample of 2-methyl-1-butanol has an observed specific rotation, [a ]25
D ,
equal to +1.151. (a) Calculate the enantiomeric excess of the sample.
(b) What is the actual stereoisomeric composition of the mixture? (The
specific rotation of the pure enantiomer is + 5.756°).
Solution
(a) Enantiomeric excess = ¥ 100%
Specific rotation of pure enantiomer
+1.151
= ¥ 100 = 20.00%
+5.756
(b) Of the total mixture, (100 – 20) or 80% consists of the racemic form, which contains
an equal number of the two enantiomers. Therefore, half of this 80%, or 40%, is
the (+) enantiomer and 40% is the (–) enantiomer. The other 20% of the mixture
(the excess) is also the (+) enantiomer. Consequently, the mixture contains 60% (+)
enantiomer and 40% (–) enantiomer.
12. A solution obtained by mixing 20 mL of a 0.1 M solution of the R enantiomer
of a compound and 60 mL of a 0.1 M solution of the S enantiomer was
found to have an observed specific rotation of +4.8°. What is the specific
rotation of each of the enantiomers?
Solution (20 mL ¥ 0.1 M), i.e., 2 mmol of the R-enantiomer is mixed with (60 mL ¥
0.1 M), i.e., 6 mmol of the S enantiomer. Now, 2 mmol of the R enantiomer and 2 mmol
of the S enantiomer will form 4 mmol of a racemic mixture. So, there will be 4 mmol of
the S enantiomer as left-over. Since 4 out of 8 mmol is excess S enantiomer, therefore
the solution has 50% enantiomeric excess. Now, from the values of enantiomeric excess
and the observed specific rotation, the specific rotation of the pure enantiomer can be
calculated.
Enantiomeric excess = ¥ 100%
+ 4.8∞
50% = ¥ 100%
+ 4.8∞
Therefore, specific rotation of the pure enantiomer = ¥ 100 = + 9.6∞
50
Hence, the S enantiomer has a specific rotation of +9.6° and the R enantiomer has a
specific rotation of –9.6°.
13. What is the enantiomeric excess for each of the following mixtures of
enantiomers X and Y?
(a) 90% X and 10% Y (b) 75% X and 25% Y
Solution Enantiomeric excess tells how much one enantiomer is present in excess of the
racemic mixture.
(a) Since the mixture contains 90% of X enantiomer and 10% of Y enantiomer, the
enantiomeric excess is 90% – 10% = 80%. There is an 80% excess of X enantiomer
over the racemic mixture. (b) Similarly, the enantiomeric excess in this mixture
is (75% – 25%) = 50%. There is a 50% excess of X enantiomer over the racemic
mixture.
14. Which of the following are optically active?
Et Et Et
(a) (b) (c)
Et Et Et
Et Et Br Et Br
(d) (e) (f)
Et Et Br Et
Et Br Cl
(g) (h)
Br Et Et
Solution The compounds (c), (e) and (h) are chiral (nonsuperimposable on mirror image)
and no Sn axis is. So, these are optically active. The compounds (a), (b), (d), (f) and (g)
have a plane of symmetry and they are superimposable on their mirror images, i.e., these
compounds are achiral and optically inactive.
15. The following conformer (I) of perfluorobutane is chiral, then why is

perfluorobutane optically active?
CF3
F CF3
F F
F
I
Solution Perfluorobutane has another conformation (II) which is a nonsuperimposable
mirror image of the conformer (I). The two enantiomers may interconvert simply by
rotation about the C-2—C-3 bond and the rate of interconversion is much higher at room
temperature because rotational barriers are generally very small. Since there exists an
equimolar mixture of two conformational enantiomers I and II, all the rotations produced
by individual molecules are cancelled and prefluorobutane is found to be optically inactive.
In fact, these two exist as a nonresolvable dl-pair (a racemic mixture).
CF3 CF3
F CF3 120° or 240° rotation F3C F
of the back carbon
F F F F
F F
I II
nonresolvable
enantiomers
16. Calculate the concentration of a solution of the alkaloid coniine
([a ]25
D = + 16∞ ) with an observed rotation of +4.0° in 1.0 dm polarimeter tube.
Observed rotation
Solution Concentration of the solution =
Specific rotation ¥ length of tube
+ 4.0∞
= = 0.25 g/mL
+16∞ ¥ 1.0
17. Draw all the stereoisomers of (a) 1, 2-dimethylcyclopropane, (b) 1,
2-dimethylcyclobutane and (c) 1,3-dimethylcyclobutane and indicate the
optically active and inactive isomers.
Solution
s-plane
(a) H3 C CH3 H3C H H CH3

,
H H H CH3 CH3 H
cis (meso) trans trans

(optically active
individually)
s-plane
(b) H3 C CH3 H3C H H H3C
H H H CH3 CH3 H
cis (meso) trans trans
(optically active
individually)
Centre of
CH3 symmetry H
(c)
H3 C s-plane H3C
H CH3
H H
cis trans
(optically inactive) (optically inactive)
18. The ketone (+)-(CH3)2CHCOCH(OH)CH2CH3 racemizes on treatment with

alkali, whereas the isomeric ketone (+)-CH3CH2COCH2CH(OH)CH2CH3
does not. Explain this observation.
Solution For racemization to take place the hydrogen attached to the chiral carbon must
be acidic and for this to happen the C==O group must be adjacent to the chiral carbon.
The carbonyl function is attached to the chiral carbon in the ketone (CH3)2CHCOCH(OH)
CH2CH3 but not in the isomeric ketone CH3CH2COCH2CH(OH)CH2CH3. Because of this,
the former ketone racemizes on treatment with alkali, whereas the latter ketone does not.
19. (+)-MeO— —CH(CH3)C2H5 racemizes in the presence of AlCl3. Explain.

Solution AlCl3 (a Lewis acid) abstracts a hydride ion from the chiral carbon of (+)
*
-MeO CH(CH3)C2H5 to form a planar (achiral) carbocation. The process
is facilitated by the resonance stabilization of the resulting carbocation by the
p-methoxylphenyl group. A hydride ion from a second molecule of the compound
then attacks the carbocation from either side with equal facility to form racemic or (±)
-MeO CH(CH3)C2H5.
H
–
H – Shift –
MeO C—C2H5 + AlCl5 MeO C—C2H5 + AlHCl3
CH3 CH3
A resonance-stabilized
carbocation
(achiral)
OMe
(a)
C2H5
MeO C + H—C—C2H5
CH3
CH3
(b)
(±)MeO—
H
(a)
MeO— —C(CH3)C2H5 + C
C2H5
CH3
—CH(CH3)C2H5
MeO
–
H – Shift Mirror
plane
e
OM
C2H5
(b) CH3
MeO— —C(CH3)C2H5 + C
H
20. Explain whether the following compounds are resolvable or not:

O
H3CHC==C==CHCH3; PhCH2P(Me)Ph; PhN(Me)Et

I II III
Solution The allene I and the phosphine oxide II are resolvable because they are chiral
and configurationally stable. The amine III is not resolvable because this chiral amine is
not configurationally stable due to rapid pyramidal inversion (umbrella effect).
21. On heating two moles of (±)-lactic acid loses two moles of water to give
two diastereoisomeric products of which one is resolvable and the other
is not. Draw the structures of the diastereoisomeric products and indicate
which is resolvable and why?
Solution When heated (±)-lactic acid (two molecules) reacts intermolecularly to give cis-
and trans-products (two diastereoisomers). The cis-isomer is optically active (it has no
Sn axis) and resolvable, whereas the trans-isomer is optically inactive (it has a centre of
symmetry).
D
—2H2O
3 3
3 3
cis
3 3 3 3
D
cis
—2H2O
cis
3 3
D
—2H2O
3 3
trans
meso
22. Which of the following amines could in principle be used as a resolving

agent for a racemic carboxylic acid?
(±) – Ph—CH—NH (–) – Ph—CH—NH Ph(CH ) C—NH
CH CH III
I II
Solution A resolving agent must be a pure enantiomer. Therefore, II could in principle

be used as a resolving agent for a racemic carboxylic acid.
23. Enantiomers have identical reactivities with achiral reagents — Why?
Solution Enantiomers have identical free energies. The enantiomeric transition states
have also identical free energies, as do the enantiomeric starting materials. Because
relative reactivity is determined by the difference in free energies of transition state and
starting material, and because of this difference in same for both enantiomers, enantiomers
react at identical rates.
24. A compound is found to be optically inactive. Predict the nature of the
compound.
Solution It may be (i) an achiral compound with no stereogenic centre or with two or
more stereogenic centres (meso), (ii) a compound whose molecules are chiral but not
configurationally stable, (iii) a compound whose molecules are chiral but each enantiomer
has immeasurably small optical activity, or (iv) a racemic mixture—an equal amount of
two enantiomers.
1. Explain why an optically inactive product is obtained when (–)-3-methyl-1-pentene

is subjected to catalytic hydrogenation.
2. What is the necessary and sufficient condition for a molecule to show optical
activity?
3. The presence or absence of a chirality centre is no criterion for optical activity.
Explain.
4. Explain why the compound I can be resolved into enantiomers, but the compound
II cannot.
2 5 2 5
–
2 2 2 2
5. When an optically active compound A (molecular formula C9H16) is allowed to

react with hydrogen in the presence of a catalyst, an optically inactive product is
obtained. Which of the following structures for A is (are) consistent with all the
data?
Et Et Et
(a) (b) (c)
Et Et
Et
Et
Et
(d) (e) Et
Et Et
6. Which of the following compounds can be resolved into enantiomers at room

temperature? Explain.
CH2CH2CH3
N
(a) (b) C2H5
2 5 2 3 3
C2H5
2 3 2
Et CH2CH2CH3
Et
N
(c) N (d) CH3
H
N Et
Et
7. Of the nine possible stereoisomers of 1, 2, 3, 4, 5, 6-hexachlorocyclohexane, only

two can be isolated in optically active form under ordinary conditions. Write down
the structures of these enantiomers.
8. Explain why compound A has two optically active stereoisomers, but B and C exist
as optically inactive single compounds.
≈ ≈
CH3CH2CH2 N H(CH3 )CH = CH2Cl@ ; (CH3 )2 N HCH = CH2Cl@ CH3NHCHCH2CH = CH2
A B C
9. The observed rotation of a solution of compound (0.187 g/mL) is found to be –6.52°

when placed in a polarimeter tube. What is the percent of each enantiomer in the
solution if [a ]25
D of the compound is –39.0° and the length of the tube is 1 dm?
10. A sample of (R)-(–)-lactic acid was found to have an enantiomeric excess of 74%.
How much R enantiomer is present in the sample?
11. Are the following compounds chiral and exhibit optical activity? Give your
reasoning.
(a) (CH3)2C==C==C(CH3)2 (b) CH3CH==C==CHCH3
(c) CH3CH==C==C==CHCH3 (d) (CH3)2C==C==C==C(CH3)2

H O
F Br F Br N
(e) (f) (g)
Br F Br F NH
O
12. Draw all possible stereoisomers of the following compound and indicate which are
optically active.
HOCH2CH CH CHCH2OH
OH OH OH
13. Pure (R)-mandelic acid has a specific rotation of –154°. If a sample contains 60% of
the R and 40% of the S enantiomer, what is the observed rotation of this solution.
14. When butane is allowed to react with Br2/hn, a bromoalkane is obtained which can
be resolved. Explain this observation.
15. Predict the products of the following reaction and indicate whether each product is
optically active or not.
CH2
H /Ni
CH2
16. A chiral sample gives a rotation that is close to 180°. How can one tell whether this
rotation is +180° or –180°?
17. A solution of 4.0 g of (+)-glyceraldehyde, HOCH2CHOHCHO, in 20 mL of water was
placed in a 100-mm polarimeter tube. Using the sodium D line, a rotation of +1.74°
∞
was found at 25°C. Determine the specific rotation [a ]25 D of (+)-glyceraldehyde.
18. Calculate the enantiomeric excess and the specific rotation of a mixture containing
6 g of (+)-2-methyl-1-butanol and 4 g of (–)-2-methyl-1-butanol.
19. When (+)-2-iodoactane is reflexed with KI in acetone, it loses optical activity.
[Hint: Racemization occurs by a reversible SN2 reaction in which the nucleophile
and the leaving group are identical.]
20. Predict the stereochemical outcome of the following reactions:
Br Br
(a) cis-2-Butene æææ 2
Æ (b) trans-2-Butene æææ 2
Æ
21. How can an amino acid be resolved?
[Hint: A racemic amino acid can be resolved by initial conversion to its formyl
derivative (–NH2 Æ NHCHO) in which the basic character of the –NH2 group
is eliminated owing to the –I and –R effects of the C==O group. This derivative
then behaves as an acid and resolved by using an optically active base. The basic
resolving agent and the formyl group are finally removed by acid hydrolysis.]
22. In a given solution, a compound shows optical rotation of +300°. How will you
prove that it is dextrorotatory?
23. Explain why (S)-3-phenylbutanone loses its optical activity in alkali medium.
24. Predict the specific rotation of a mixture of 30% (–)-2-bromobutane and 70%
∞
(+)-2-bromobutane. [a ]25
D of the pure enantiomer is –23.13°.
25. The product of the first reaction can be resolved but the product of the second
reaction cannot be resolved into two enantiomers. Explain.
(i) PhCHO + HCN Æ; (ii) Me2CO + HCN Æ
26. The diketone CH3COCH(CH3)COCH3 is allowed to react with HCN (2 moles)
in the presence of a basic catalyst. Identify the products and give their Fischer
projections. Discuss their behaviour towards plane-polarized light.
27. Which one of the following two reactions gives a product which is optically inactive?
28. Tetra-sec-butylmethane has four optically active and one optically inactive
stereoisomers. List them in terms of R/S designation.
2.7 CONFORMATION OF ACYCLIC ORGANIC MOLECULES

Stereoisomers that can be interconverted through rotation around single bonds (or twisting
of bonds) are known as conformational isomers or rotational isomer or simply conformers
or rotamers and the phenomenon is known as conformational isomerism. A molecule could
have an infinite number of confomations. The rotation around a C—C single bond is not
actually completely free. It is hindered by very small energy barrier of 16–25 kJ/mol due
to very weak repulsive interaction between the electron clouds of eclipsed s-bonds. Since
the energy barrier to this rotation is actually very small, the different conformations
are rapidly interconvertible and cannot be isolated, except in a few cases at very low
temperatures, or when they possess unusual structures like ortho-substituted biphenyls.
A particular conformer having a considerable amount of internal strain (bond angle
strain, torsional strain and steric strain) is less stable and hence less populated than that
without having strain. The more stable conformer is more populated and therefore, it
represents the properties of a molecule at ordinary temperature. At higher temperatures,
the population of the less stable conformers increases and the properties of molecules are
then represented by those isomers. The study of physical as well as chemical properties of
a compound with respect to the relative stability as well as the population of its conformers
is known as conformational analysis.
Unlike readily interconvertible conformers of molecules like ethane, propane, etc., the
conformers of ortho-substituted biphenyls are stable enough at a particular geometry
and do not interconvert. For example, 6, 6¢-dinitro-diphenyl-2,2¢-dicarboxylic acid is
not planar since this geometry is highly destabilized due to strong steric interaction
between the substituents at ortho positions of the ring. In order to avoid this strain, the
compound exists in two isomeric nonplanar forms (I) and (II) in which the two rings are
orthogonal. These are actually conformers since they are related to the rotation around
the C—C s-bond connecting the two aromatic rings. They do not interconvert at ordinary
temperature (energy barrier 80–100 kJ/mol) as interconversion involves highly unstable
planar structure as an intermediate. These conformational isomers are nonplanar and
nonsymmetric and hence chiral. In fact, they are enantiomers and therefore, it is a case of
conformational enantiomerism, commonly known as atropisomerism. Their interconversion
may, however, be effected by applying heat and this results in racemization.
2 Bond rotation requires 2 2

higher energy
2
¢ ¢
Dihedral angle: The dihedral angle (q), which is usually symbolized by the Greek letter
theta, is the angle between the X–C–C plane and the C–C–Y plane of X–C–C–Y unit in a
molecule. In the Newman projection formula, it is the angle between the C–X bond on the
front carbon atom and the C–Y bond on the back carbon atom.
dihedral angle
60°
q=
Torsion angle is synonymous with the term dihedral angle. However, in contrast to dihedral
angle, torsion angle has been given directional sense. It is positive (+) when measured in
a clockwise direction and negative (–) when measured in a counterclockwise direction,
starting from the front substituent and ending at the rear substituent. Therefore, torsion
angle has a range of value from 0° to ±180°.
2.7.1 Conformations of Ethane (CH3—CH3)

A molecule of ethane contains a carbon–carbon single bond (s-bond) and each carbon atom
is attached to three hydrogen atoms. The two –CH3 groups can rotate freely around the
C—C bond axis. Rotation of one carbon keeping the other fixed results into an infinite
number of spatial arrangements of hydrogen atoms attached to the rotating carbon
atom with respect to the hydrogen atoms attached to the fixed carbon atom. These are
conformational isomers or conformers or conformations of ethane. Thus, there are an
infinite number of conformations of ethane. However, there are two extreme cases. The
conformation in which the hydrogen atoms attached to two carbons are as close together
as possible, i.e., in which the dihedral angle between two nearest C—H bonds of two –CH3
groups is zero is called the eclipsed conformation. This is so-called because the Newman
projection shows the hydrogen atoms on the back carbon atom to be hidden (eclipsed) by
those on the front carbon atom. The other extreme conformation in which the hydrogen
atoms are as far apart as possible, i.e., the dihedral angle between two C—H bonds is 60°
called the staggered conformation. Any other intermediate conformation is called a skew
conformation.
q = 0 – 60°
°
H 60 H
H H q = 0° H q=H H
H H
H H
H H H H
H H H
H
Eclipsed Staggered Skew
conformation conformation conformation
In the eclipsed conformation, the electron clouds in the eclipsed C—H bonds on adjacent
carbon atoms repel each other. The repulsive destabilization caused due to eclipsing of
bonds on adjacent carbons is called torsional strain. On the other hand, in the staggered
conformation, the electron clouds in the C—H bonds are separated as much as possible and
so, the staggered conformation is free from torsional strain. Because of this, the eclipsed
conformation is less stable (about 2.86 kcal/mol or 12 kJ/mol higher in energy) than the
staggered conformation. Since a molecule must rotate from one staggered conformation to
another through the more energetic eclipsed conformation, the rotation about the C—C s
bond in ethane is somewhat restricted, i.e., not completely free. There is an energy barrier
of about 2.86 kcal/mol or 12 kJ/mol. This is called torsional energy.
How the potential energy of ethane changes with dihedral angle or torsion angle as one
CH3 group rotates relative to the other can be represented graphically in a potential
energy diagram or torsional curve as shown below:
60° 60° 60° 60° 60° 60°

I II III IV V VI I
rotation rotation rotation rotation rotation rotation
H H H H
H H H H
eclipsed
H H H H H H H H conformations
H H H H H H H H (energy maximum)
I III V I
12 kJ mol–1 (2.86 kcal mol–1)
All three eclipsed
Potential energy
conformations (I, III and V)

are identical and the
same is true for all three
staggered conformations
(II, IV & VI).
H H H Staggered
H H H H H H conformations
(energy minimum)
H H H H H H
H H H
II IV VI
0° 60° 120° 180° 240° 300° 360° = 0°

Dihedral angle
The staggered conformation is the most stable arrangement, so it is at an energy minimum.

As the C—H bonds on one carbon are rotated relative to the C—H bonds on the other
carbon, the energy increases as the C—H bonds come close to each other until a maximum
is reached after 60° rotation to the eclipsed conformation. As rotation continues, the energy
decreases until after 60° rotation, when the staggered conformation is reached once again.
This energy barrier is not large enough to prevent rotation at room temperature as collision
between the molecules supply sufficient kinetic energy to overcome this energy barrier.
Therefore, at any given moment, unless the temperature is extremely low (–250°C),
most ethane molecules will have enough energy to undergo rapid bond rotation from
one staggered conformation to another, i.e., at any given moment, all ethane molecules
do not exist in the more stable staggered conformation; rather a higher percentage of
molecules is present in the staggered conformation than any other possible conformation.
The steric contribution to torsional strain or Pitzer strain due to non bonded interaction
between the vicinal H atoms is negligible since the internuclear distance (0.23 nm) is
almost equal to twice the value of van der Waals radius of hydrogen (0.12 nm).
2.7.2 Conformations of Propane (CH3CH2CH3)

The torsional curve, i.e., the potential energy diagram for propane is very similar to that
of ethane because both C-1—C-2 and C-2—C-3 bonds are equivalent. However, the energy
barrier between the eclipsed and staggered forms is slightly higher (14.2 kJ mol–1). This
small difference in potential energy barrier between propane and ethane (14.2 – 12 =
2.2 kJ/mol) further shows that steric interaction has no significant role (at least in this
case) in the origin of energy barrier. If it would be, the eclipsed propane would have much
higher potential energy due to CH3/H interaction in place of H/H interaction.
Potential energy changes of propane as function of dihedral angle or torsion angle may be
shown by the following energy diagram:
60° 60° 60° 60° 60° 60°
I II III IV V VI I
rotation rotation rotation rotation rotation rotation
H H H H
H3C H H H3C
eclipsed
H H H H H H H H conformations
H H H CH3 H3C H H H (energy maximum)
I III V I
Potential energy
All three eclipsed

14.2 kJ mol–1
conformations (I, III and V)

are identical and the
same is true for all three
staggered conformations
(II, IV & VI).
H H H Staggered
H CH3 H H H3C H conformations
(energy minimum)
H H H H H H
H CH3 H
II IV VI
0° 60° 120° 180° 240° 300° 360° = 0°

Dihedral angle
2.7.3 Conformations of Butane (CH3CH2CH2CH3)

Butane contains two kinds of C—C bonds. So, conformations likely to be generated depend
on that the particular C—C bond around which C-atoms are made to rotate.
C-2 C-3 C-3 C-4
bond bond
1 2 3 4
CH3 CH2 CH2 CH3
C-1 C-2
bond
(i) Rotation about C-3—C-4 bond Rotations around C-3—C-4 and C-1—C-2 bonds generate
homomeric conformations. The case is very similar to the case of propane (methyl group
is replaced by ethyl group) and so, the torsional curve or the potential energy diagram is
very similar to that of propane.
(ii) Rotation about the C-2—C-3 bond Clockwise rotation of C-2 (the front carbon in the
following figure) of butane in 60° increments (until the original conformation comes back)
results in six possible conformations:
4
CH3 H CH3
CH3
H H H3 C H
60° rotation 60° rotation
2 H H
H3C H
H 1
H H H
CH3 H
Anti I Eclipsed II Gauche III
60° rotation 60° rotation
H CH3 H3C
CH3 CH3
60° rotation H CH3 60° rotation
H H H H
H CH3 H H
H H
H
Eclipsed VI Gauche V Totally eclipsed
IV
The order of the increasing stability of the conformations of butane is as follows:
IV < II=VI < III = V < I
Potential energy changes of butane as a function of dihedral angle or torsion angle may be
shown by the following energy diagram:
The staggered conformations (I, III and V) of butane are at energy minima and are the
stable conformations of butane. The different staggered conformations have been given
special names. The conformations with a dihedral angle of ±60° (or in Fig. 60° and
300°) between the two C—CH3 bonds are called gauche conformations (III and V) and
the conformation in which the dihedral angle is 180° is called the anti-conformation (I)
[gauche is French meaning “to turn aside” and anti is Greek for “opposite off”]. The anti-
conformation (I) does not have torsional strain because the groups are staggered. Again,
the two —CH3 groups are located farthest from each other and so also no steric stain
operates. The methyl groups in the gauche conformations (III and V) are close enough
to each other and for this, the gauche conformations are destabilized by van der Waals
repulsions between nonbonded hydrogens on the two —CH3 groups. This repulsion causes
the gauche conformation to have approximately 0.9 kcal/mol (3.8 kJ/mol) more energy than
anti-conformation in which such van der Waals repulsions are absent. Therefore, the anti-
conformation is more stable than the gauche conformation. The eclipsed conformations
of butane are unstable for the same reason that the eclipsed conformations of ethane
are unstable. The eclipsed conformations (II, IV and VI) represent energy maxima in the
potential energy diagram. The eclipsed conformations II and VI not only have torsional
strain, but also have van der Waals repulsions arising from the eclipsed methyl groups
and hydrogen atoms. The eclipsed conformation IV has the greatest energy (least stable)
of all because, in addition to torsional strain, there is a large van der Waals repulsive
interaction because the hydrogens in the two eclipsed –CH3 groups are even closer than
they are in the gauche conformation.
A particular structure of gauche-form without having any symmetry property is chiral
but the compound as a whole is not resolvable (and optically inactive) due to rapid
interconversion between them (III and V). The same is true for the enantiomeric
conformations II and VI.
The most stable anti-conformation is the predominant conformation of butane. At room
temperature, there are about twice as many molecules of butane in the anti-conformation as
there are in the gauche conformation (66% anti and 34% gauche). These two conformations
interconvert rapidly at room temperature. Because the eclipsed conformations are
unstable, they do not exist to any measurable extent.
Butane-gauche interaction: The steric interaction between the two methyl groups in
the gauche conformations of butane (q = 60°) makes them less stable (0.9 kcal/mol higher in
energy) than the anti conformation (q = 180°). This fundamental interaction is encountered
in many organic molecules and is specifically called butane–gauche interaction.
CH3 butane-gauche
H CH3 interaction
(0.9 kcal/mol)
H H
H
A gauche conformation
of butane
2.7.4 Conformations of Chloroethane (CH3CH2Cl)

This case is very similar to the case of propane (methyl group is replaced by Cl atom) and
so, the torsional curve or the potential energy diagram is very similar to that of propane.
The barrier to rotation is about 15 kJ/mol.
2.7.5 Conformations of 1,2-dichloroethane (ClCH2CH2Cl)

Clockwise rotation of C-1 (the front carbon in the following figure) of 1,2-dichloroethane
in 60° increments (until the original conformation comes back) results in six possible
conformations:
The order of increasing stability of the conformations of 1,2-dichloroethane is as follows: I

III = V II = IV = VI.
The potential energy changes of 1,2-dichloroethane (in liquid state) as a function of
dihedral angle may be shown by the following energy diagram:
Potential energy
I I
III V
II IV VI
0° 60° 120° 180° 240° 300° 360°= 0°
Dihedral angle
The difference in energy between the anti- and gauche-conformations of 1,2-dichloroethane

in the liquid state is about zero (in the vapour phase, the anti-conformation is more stable
by 1.2 kcal/mol.). Two opposing forces are to be considered to account for this observation,
and these are the dipole–dipole repulsion and the van der Waals attraction between the
two Cl atoms. In the liquid phase, the dipole–dipole repulsion becomes equal to the van
der Waals attraction (in the vapour phase the dipole–dipole repulsion is stronger than van
der Waals attraction). Due to the presence of a number of 1,2-dichloroethane molecules in
the neighbourhood of a particular molecule, the intermolecular dipole–dipole interaction
causes reduction of the intramolecular dipole–dipole interaction to such an extent that the
van der Waals attraction just cancels the intramolecular dipolar interaction (repulsion) in
the gauche conformation. As a consequence, the energy difference between the anti- and
gauche-conformations is zero.
The anti-conformation as well as the gauche conformations are the most stable insofar
as the torsional strain or Pitzer strain and dipolar repulsions are concerned. The eclipsed
conformations are less stable than the anti and gauche conformations mainly because
of torsional strain and a very small H/Cl steric interaction. The fully eclipsed form is
the least stable one because of torsional strain and large electrostatic repulsion between
the two negatively polarized Cl atoms which are close to each other (q = 0). Also, there
operates Cl/Cl steric interaction of about same size as that operates in butane (the van der
Waals radius of Cl atom is about the same as that of a –CH3 group).
2.7.6 Conformations of Some Typical Acyclic Molecules

The gauche conformation of butane is less stable than the anti-conformation. However,
molecules like HOCH2CH2X (where, X = halogen, OH, NH2, OR, NHR, NR2, etc.) are
found to be more stable in their gauche conformations due to intramolecular hydrogen
bonding as shown below:
2.7.7 Invertomerism
Two conformers which interconvert by inversion at an atom having an unshared pair of
electrons are known as invertomers and this phenomenon is known as invertomerism. (R)-
and (S)-ethylmethylamine are two invertomers which cannot be resolved.
p orbital H sp
H H
N N N
Me Me
Et Me Et Et
R-(Ethylmethylamine) (S)-Ethylmethylamine
T.S.
invertomers
If the nitrogen atom is contained in a small ring, for example, it is prevented from attaining
the 120° bond angles that facilitate inversion. Such a compound has a higher activation
energy (Eact) for inversion and for this reason, the enantiomers can be resolved. The
following chiral aziridine derivative, for example, can be resolved into two enantiomers
because it is configurationally stable.
enantiomers
H3C CH3
H3C N N CH3
CH3 H3C
(R)-1,2,2-Trimethylaziridine (S)-1,2,2-Trimethylaziridine
Unlike N and C (second period elements), the activation energy for pyramidal inversion at
S or P (third period elements) is much higher. For this reason, a sulfonium salt with three
nonidentical substituents on S, a sulfoxide with two nonidentical substituents on S and a
phosphine with three nonidentical substituents on P do not undergo pyramidal inversion
at room temperature and can be resolved into enantiomers.
R1 R1 O O
S S S S
R2 R2 R1 R1
R3 X
–
X
–
R3 R2 R2
enantiomeric sulfoxides
enantiomeric sulfonium salts
(configurationally stable
(configurationally stable
and resolvable)
and resolvable)
R1 R1
P P
R2 R2
R3 R3
enantiomeric phosphines
(configurationally stable and
resolvable)
1. Explain why the following two conformations (I and II) of 3,4-di

(1-adamantyl)-2,2,5,5-tetramethylhexane can be isolated.
CMe3 CMe3
Me3C R R H
R = (adamantyl
group)
R H R H
H CMe3
I II
Solution Due to severe steric interaction between two bulky alkyl groups (–CMe3/–R) in
the eclipsed conformation, the rotational energy barrier is much higher in this case and
for this reason, these two conformations of 3,4-di(1-adamantyl)-2,2,5,5-tetramethylhexane
can be isolated.
2. Write the most stable conformation of n-pentane (in Newman projection

formula).
Solution The anti-conformation is the most stable conformation for rotation around the
C-2—C-3 bond because this conformation is free from torsional and steric strain (the –
C2H5 and –CH3 groups are anti to each other)
CH3
H H
H H
C2H5
Anti conformation
of n-pentane
3. The potential energy barrier for the direct interconversion of the two
gauche conformations of butane is 15 kJ mol–1 or 3.6 kcal mol–1, while the
potential energy barrier for the direct interconversion of the two gauche
conformations of 1,2-dichloroethane, ClCH2CH2Cl, is 38.9 kJ mol–1 or 9.3
kcal mol–1 even though the van der Waals radius of a covalently bound
Cl atom is about the same as that of a methyl group (–CH3). Explain why
the energy barrier in 1,2-dichloroethane is so much higher than that in
butane.
Solution Due to very strong electrostatic repulsion between the two very close negatively
polarized Cl atoms in the fully eclipsed conformation, the potential energy barrier for
the direct interconversion of the two gauche conformations of 1,2-dichloroethane is much
higher (38.9 – 15 = 13.9 kJ/mol) than butane.
electrostatic
repulsion
d–
d–
d– Cl d–
d– Cl Cl Cl
d–
Cl H H Cl
H H
H H H H H H
H H
(unstable)
4. Use a Newman projection, about the indicated bond, to draw the most
stable conformation for each compound:
(a) 3-methylpentane about the C-2—C-3 bond and
(b) 3,3-dimethylhexane about the C-3—C-4 bond.
Solution
4 3
2 3
3
1 2 3 4 5 2
(a) 3 2 2 3
1 3
3
(b)
5. The torsional energy in propane (CH3CH2CH3) is 14 kJ/mol. If each H/H

eclipsing interaction is worth 4.0 kJ/mol of destabilization, then how
much is one H/CH3 eclipsing interaction worth in destabilization?
Solution Two H/H interactions + One H/CH3 interaction = 14 kJ/mol
Therefore 2 ¥ 4.0 kJ/mol + One H/CH3 interaction = 14 kJ/mol
Therefore One H/CH3 interaction = (14 – 8) or 6 kJ/mol
6. Haloethanes (CH3CH2X, X = Cl, Br, I) have similar barriers to rotation
(13.4 – 15.5 kJ/mol) despite the fact that the size of the halogen increases
on going from Cl to I. Suggest an explanation.
Solution Since the size of halogen atom increases in the order: Cl < Br < I, the H/X
eclipsing interaction is expected to increase. However, due to increase in size of halogen,
the C—X bond length gradually increases and as a result, the distance between H and
X increases which causes decrease in eclipsing interaction. For this reason, haloethanes
have similar barriers to rotation despite the fact that the size of the halogen increases on
going from Cl to I.
7. Draw the two conformations of 1,3-butadiene. Which one of them is more
stable? Give your reasoning.
Solution The two conformations of 1,3-butadiene (CH2 = CH – CH = CH2) resulting from
rotation about the C – C single bond are s-cis and s-trans. The prefix s stands for single
bond.
H H
H C H C
C H rotation C H
180°
H
H C C
C H H C
H H
s-trans conformation s-cis conformation
(more stable) (less stable)
1,3-Butadiene
Because of steric interactions between two terminal H atoms, the s-cis conformation is
thermodynamically less stable than the s-trans conformation.
8. The intramolecular H-bonding in active butane-2,3-diol is relatively
stronger than that in meso butan-2,3-diol. Explain.
Solution
stronger
H-bond
H
CH3 CH3
H OH H OH H O
H OH HO HO H
H
HO H H CH3 H O
CH3 CH3 CH3
CH3 H3 C
Active Butane-2, 3-diol
weaker
H-bond
H
CH3 O
H OH H OH H O
H OH H H OH
H
H OH OH CH3 H CH3
CH3 CH3 CH3
CH3 H3C
meso-Butane-2,3-diol
butane–gauche
interaction
Because of butane–gauche interaction, the intramolecular H-bonding in the meso form

of butane-2,3-diol is much weaker. On the other hand, in the H-bonded staggered
conformation of active butane-2,3-diol, the two –CH3 groups are placed anti to each other.
So, unlike the meso form, it does not suffer from any steric strain and for this reason, the
H-bonding in this case is relatively much stronger.
9. Considering rotation around the indicated C—C bond in each compound.
Draw Newman projections for the most stable and least stable
conformations.
1 2 3 4 5 6 1 2 3 4 5 6
(a) CH 3 — CH 2CH 2CH 2CH 2CH 3 (b) CH 3CH 2CH 2 — CH 2CH 2CH 3
≠ ≠
Solution
(CH2)4CH3 H
(CH2)4CH3
H H
(a) H H
;
H H H H
H
Most stable Least stable
conformation conformation
C2H5
C2H5 H5C2
(b) H H
; H H
H H H H
C2H5
Least stable
Most stable
conformation
conformation
10. Arrange the staggered conformations of 2,3-dimethylbutane in order of
decreasing energy.
Solution
11. Which in between butane (CH3CH2CH2CH3) and 2-methylbutane

[(CH3)2CHCH2CH3] has more torsional strain or Pitzer strain?
Solution 2-methylbutane has more torsional strain than butane. It has two staggered
conformations with two adjacent methyl groups and one methyl group at a distance. The
third staggered conformation has all the three methyl groups adjacent to each other, and
has the highest energy. Only two methyl groups are to be considered in the case of butane.
It has two staggered conformations, one with two methyl groups adjacent and the other
with two methyl groups anti (the lowest energy conformation). The energy difference
between the lowest and highest energy conformations (the barrier to rotation) is, therefore,
greater in 2-methylbutane than in butane, i.e., 2-methylbutane has more torsional strain.
12. Explain why: (a) the gauche conformation of 1-chloropropane is found to
be more stable than its anti-form and (b) the eclipsed form of CH3CH2CHO
predominates over the anti-form.
Solution
(a) Due to electrostatic forces of attraction between the two electronically opposite
groups (Cl and –CH3), the gauche conformation of 1-chloropropane is found to be
more stable than its anti-form.
d+
Me Me d–
H H H Cl
anti form gauche form
H H H H
Cl H
(less stable) (more stable)
(b) Due to electrostatic forces of attraction between the electronegative oxygen atom,
and the electron-deficient –CH3 group (an electron-releasing group), the eclipsed
from of CH3CH2CHO predominates over the anti-form.
d+
d–
Me
O O
H H
anti form eclipsed form
H H
Me
H H
13. Explain why 2,3-di-tert-butyl-1,3-butadiene exists nearly exclusively in

s-trans conformation.
Solution Because of severe steric stain involved in the s-cis conformation due to close
proximity of very bulky tert-butyl groups (–CMe3), 2,3-di-tert-butyl-1,3-butadiene exists
nearly exclusively in the more stable s-trans configuration.
severe steric
interaction
Me3C CH2 Me3C CH2
C C
C C
H2C CMe3 Me3C CH2
s-trans conformation s-cis conformation

(more stable) (much less stable)
1. Which conformation in each pair is more stable?

Calculate the difference in energy between the two conformations (butane–gauche
interaction = 0.9 kcal/mol and H/CH3 eclipsing interaction = 1.4 kcal/mol).
(a)
(b)
2. The eclipsed conformation of ethyl chloride (CH3CH2Cl) is 15 kJ/mol less stable

than the staggered conformation. How much is the H/Cl eclipsing interaction
worth in destabilization?
3. Label the sites of torsional and steric strain in each of the following conformations:
(a) (b) (c)
4. The gauche conformation of ethylene glycol is more stable than the anti-
conformation. Offer an explanation.
5. Draw Newman projections of all staggered and eclipsed conformations that result
from rotation around the C-2—C-3 bond in (a) pentane and (b) 3-ethylpentane.
Also, draw a graph of potential energy versus dihedral angle for rotation around
this bond for each of these two molecules.
6. Considering rotation around the indicated bond in each of the following compounds,
draw Newman projections for the most stable and least stable conformations:
(a) CD3 CD2CD2CD2CD3 (b) CF3CF2CF2 CF2CF2CF3
≠ ≠
7. The anti-conformation of butane is more stable than the gauche conformation
by about 0.9 kcal/mol. (a) Calculate the equilibrium constant and (b) the relative
amounts of the two conformations at 25°C.
[Hint: DH = –900 cal @ DG = –RT Keq. \ ln Keq = 1.52 and Keq = 4.57 for gauche m
anti.
The ratio of anti/gauche is 4.6/1. This indicates that about 82% of the molecules
exist in anti-conformation and 18% in the gauche conformation at any moment.]
8. Draw the eclipsed conformations of 2,3-dimethylbutane (rotation around the
C-2—C3 bond) in order of increasing energy.
9. Explain why the rotation about the C—C s-bond in ethane (CH3—CH3) is not
completely free?
10. What do you mean by the term conformation? How many conformations of ethane
are possible? Name and draw the conformations with the highest and lowest
energies at room temperature. Mention the term used for low energy conformations.
[Hint: the low energy conformations are usually called ‘conformers’.]
11. Explain how the diagram of potential energy versus angle of rotation around the
C-2—C-3 bond of 2,2,3,3-tetramethylbutane to differ from that for ethane, if at all.
12. Considering only rotation about the bond shown, draw a potential energy versus
dihedral angle graph for:
(a) Me2CH – CHMe2 (b) Me2CH – CH2Me (c) Me3C – CMe3
13. Both calculations and experimental evidence indicate that the dihedral angle
between the two –CH3 groups in the gauche conformation of butane is actually
somewhat larger than 60°. How would you account for this observation?
14. Which of the following structures represent conformers of (a)?
(a) (b) (c)
(d) (e)
15. In between the two conformations (I and II) of acetylcholine which one is more
stable and why?
[Hint: II is more stable due to electrostatic attraction between the positively

charged quaternary ammonium nitrogen atom and the slightly negative oxygen
atoms in the acetyl group.]
16. Draw the preferred conformation(s) of the following compounds:
(a) ClCH2CH2Br ≈ (b) FCH2CH2OH
(c) CH3COOCH2CH2 S (CH3)2
(d) (E) – CH2==CH—CH==CH—CH==CH2 (e) CH3CH2CHO
(f) CH3COOCH3 (g) OHCCH2F (h) OHCNMe2
Explain your choice of conformations.
17. Predict the optical nature of Tröger’s base.
18. The following conformation of hexane is chiral, then why hexane is found to be
optically inactive?
19. At room temperature, the anti-conformer of butane is more stable than the gauche
conformer by about 0.9 kcal/mol of energy; this results in an equilibrium constant
(Keq) of about 4.6 favouring the anti form. Would you expect any change to the value
of the equilibrium constant as the temperature is increased? Give you reasoning.
20. Draw the most stable conformer for each of the following using Newman projections:
(a) 3-Methylhexane (rotation around C-3—C-4 bond)
(b) 3-Methylpentane (rotation around C-2—C-3 bond)
(c) 3,3-Dimethylhexane (rotation around C-3—C-4 bond)
Hint: (a) (b) (c)
21. Which of the following conformations of 1-bromo-2-methylpropane is the most

stable?
22. For rotation around the C-3—C-4 bond of 2-methylhexane:

(a) Draw the Newman projection of the least stable conformation.
(b) Draw the Newman projection of the most stable conformation.
(c) How many of the C—C bonds in the compound have staggered conformation
that are all equally stable.
23. The potential energy diagram for the rotation about the C-2—C-3 bond in (R)-
2-iodobutane is given below. Assume that the energies of the conformations are
determined by steric effect alone. Draw Newman projections of the conformations
corresponding to position I – IV on the figure. Why is the plot of potential energy
against dihedral angle (q) unsymmetrical?
3
CHAPTER
NUCLEOPHILIC
SUBSTITUTION REACTIONS
AT SATURATED CARBON
ATOM
Chapter Outline
Introduction 3.2.9 Summary of reactivity of Alkyl
3.1 The SN2 Reaction Halides in Nucleophilic Substitution
Reaction
3.1.1 Example of SN2 Reaction
3.2.10 Factors Favouring SN1 and SN2
3.1.2 Kinetics of SN2 Reaction
Reactions
3.1.3 Mechanism of SN2 Reaction
3.2.11 SNi and SNi¢ Mechanisms
3.1.3 Stereochemistry of SN2 Reaction
3.2.12 SN1¢ Mechanism
3.1.5 Evidence in Favour of SN2
3.2.13 Isotope Effects and Salt Effects
Mechanism
(Methods used to Distinguish
3.1.6 Factors Influencing SN2 Reaction Between SN1 and SN2 Type
Rate or SN2 Reactivity Reactions)
3.2 The SN1 Reaction 3.3 Neighbouring Group Participation
3.2.1 Example of SN1 Reaction (NGP)
3.2.2 Kinetics of SN1 Reaction 3.3.1 Definition;
3.2.3 Mechanism of SN1 Reaction 3.3.2 Mechanism of NGP;
3.2.4 Stereochemistry of SN1 Reaction 3.3.3 Example of NGP;
3.2.5 Evidence in Favour of SN1 3.3.4 Anchimeric Assistance;
Mechanism 3.3.5 Evidence for Participation by a
3.2.6 Factor Influencing SN1 Reaction Neighbouring Group
Rate or SN1 Reactivity 3.3.6 Various Cases of Neighbouring
3.2.7 Carbocation Rearrangements in Group Participation
SN1 Reactions
3.2.8 Comparison of the SN2 and SN1
Reactions
INTRODUCTION
Substitution reactions involving displacement of a nucleophile by another nucleophile
are called nucleophilic substitution reactions. A nucleophilic substitution reaction may be
depicted in the following way:
*
Nu: + R — LG Æ R - Nu +
ææ LG*:
Nuelophile Substrate Product Leaving group
(LG is bonded
to an sp3 - hybridized C
atom)
*
In such a reaction, a nucleophile (Nu) displaces a leaving group (LG *:) in the molecule
that undergoes substitution (known as substrate). The nucleophile is always a Lewis base
and it may be negatively charged or neutral. The leaving group (also known as nucleofuge)
is always a species that takes a pair of electrons (the bonding electrons) with it when it
departs. The substrate is often an alkyl halide (R — X) and the leaving group is a halide
�� :* . For example:
anion : X
��
�� * + CH Cl ææ
:OH Æ CH 3OH + :Cl�� *:
3
��
Nucleophilic substitution reactions are among the most important and fundamental types
of organic reactions because they make it possible to convert readily available alkyl halides
into a wide variety of other compounds.
–
OH
R—OH (alcohols)
–
OR
R—OR¢ (ethers)
–
CN
R—CN (nitriles)
–
R—X – X R¢COO
R—OCOR¢ (esters)
Alkyl halide –
SH
–
R—SH (mercaptans)
SR¢
R—SR¢ (sulphides)
NH3
R—NH2 (amines)
d+ d-
The polar C — X bond causes the alkyl halides to undergo substitution reactions. Since
substitution at an sp3 hybridized carbon involves breaking of the s bond to the leaving
group and formation of the s bond to the nucleophile, we must know the timing of these
two events, i.e., whether bond breaking and bond making occur at the same time (one-step
mechanism) or bond breaking occurs before bond making (two-step mechanism).
Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom 3.3
One-step mechanism:
Two-step mechanism:
The mechanism that predominates depends on (i) the structure of the alkyl halide, (ii) the
strength and the structure of the nucleophile, (iii) the concentration of the nucleophile and
(iv) the solvent in which the reaction is carried out.
3.1 THE SN2 REACTION

Example, kinetics, mechanism, stereochemistry, evidence in favour of the mechanism and
factors influencing SN2 reactivity, i.e., the SN2 reaction rate.

The alkaline hydrolysis of methyl chloride is a typical example of SN2 reaction.
�� : * + CH Cl æææ H2O �� :*
HO 3 60∞C
Æ CH 3OH + : Cl
�� Methyl chloride Methyl alcohol
��

The rate of this reaction depends on the concentrations of the reagents. For example,
doubling the concentration of either the nucleophile, i.e., OH① or the substrate, i.e., the
alkyl halide doubles the rate. If the concentrations of both the reactants are doubled, the
rate of the reaction quadruples. From this relationship between the rate of the reaction and
the concentration of the reactants, we can write the rate law for the reaction as follows:
rate = k [CH3Cl] [OH①]
Because the rate of this reaction depends on the concentration of both the reactants, it is
a second-order reaction. Therefore, the reaction is bimolecular, i.e., both methyl chloride
and the hydroxide ion are involved in the transition state of the rate-determining step.

The most straight forward explanation for the observed second-order kinetics is a concerted
reaction, i.e., bond breaking and bond making occur simultaneously. The mechanism of
this reaction is designated as SN2 (Substitution, Nucleophilic, bimolecular) because the
two reactants (the substrate and the nucleophile) are involved in the transition state of
the rate-determining step (the only step).
In this substitution reaction, the hydroxide ion (OH①), the nucleophile, attacks the sp3
hybridized C atom in methyl chloride from the opposite side of bromine. This approach
of the nucleophile requires lowest energy because this avoids electrostatic repulsions
between the negatively charged entering and leaving groups. The backside attack is
also more feasible sterically. Since it is a one-step (concerted) reaction, it involves no
intermediate. The formation of the C — O bond and the cleavage of the C — Cl bond take
place at the same time. The reaction proceeds through a single transition state in which the
carbon atom is half-bonded to the three hydrogens. In the transition state, the hydroxide
possesses a diminished negative charge because it starts sharing its electrons with carbon
and chlorine hand also acquires a partial negative charge as it tends to depart with the
bonding electrons. The carbon and the three attached hydrogens become coplanar (every
bond angle is 120°). In the transition state, the carbon is pentacoordinated (fully bonded to
three atoms and partially bonded to two) rather than tetrahedral and it is sp2 hybridized.
The energy required to break the C — Cl bond is supplied by the formation of the C — O
bond. The process can be represented as follows:
The energy profile diagram for the reaction may be shown as follows:

When an SN2 reaction carried out with a chiral substrate, a product having configuration
opposite to that of the substrate is obtained. For example, when (S)-(+)-2-chlorobutane is
subjected to alkaline hydrolysis, (R)-(–)-2-butanol is obtained.
Therefore, an SN2 reaction results in inversion of configuration and since inversion occurs,
it must be a case of backside displacement. The nucleophile attacks the sp3 hybridized
carbon of the tetrahedral substrate from the rear and forces the carbon to achieve a planar
(Csp2 ) configuration in the transition state. The plain containing the three groups attached
to the central carbon is perpendicular to the remaining p orbital, which by means of its two
*
lobes overlaps with the orbitals of the nucleophile and the leaving group (L G). When the
leaving group is completely detached, the central carbon becomes once again tetrahedral,
but the configuration is opposite to that of the substrate. Therefore, the backside attack
causes the arrangement of the three nonreacting groups on the central carbon at which
the substitution occurs to be turned inside out like an umbrella in a high wind. The process
may be visualized as follows:
Frontier Molecular orbital theory offers a good explanation for the backside attack in an
SN2 reaction. In fact, the inversion of configuration can easily be explained by looking at
the frontier orbitals, which will be the HOMO of the nucleophile and the LUMO of the
electrophile (the C — X bond of the substrate molecule). The overlap is bonding when the
nucleophile approaches the electrophile from the rear leading to inversion of configuration
(i), but the approach from the front leading to retention configuration (ii), is both bonding
and antibonding. Therefore, the former approach, i.e., the backside attack, is clearly
preferred.
3.1.5 Evidence in Favour of SN2 Mechanism

(a) Stereochemical evidence The SN2 reaction involves inversion of configuration which
can be established by an experiment, carried out by Hughes and his co-workers. In the
experiment the optically active 2-iodooctane is allowed to react with radioactive iodide ion
(128 I* ) in dry acetone solvent. The alkyl iodide is found to lose its optical activity and gets
to exchange its ordinary iodine for labelled iodine. The reaction follows a second-order
kinetics (first order in the concentration of each reactant) and, therefore, it is an SN2
one.
I 128
I
| |
C6 H13CHCH 3 + 128I* ææ Æ C6 H13 CHCH 3 + I*
rate = k [C6 H13CHICH 3 ][128 I* ]

The rate of substitution, i.e., the rate of radioactive exchange, is determined by measuring
the radioactivity of the alkyl iodide at certain intervals. If each nucleophilic attack of
the iodide ion leads to inversion of configuration, a (+)-2-iodooctane molecule will be
converted to a molecule of its mirror image, i.e., its enantiomer (–)-2-iodooctane, which
will cancel the rotation of a second (+)-2-iodooctane molecule by forming a recemic
mixture. The reaction of one molecule of alkyl iodide, i.e., each inversion, will thus result
in cancellation of optical rotation of two molecules of alkyl halide. As a consequence, the
rate of racemization is expected to be twice the rate of inversion. The rate of recemization
is determined polarimatrically and it is found to be twice the rate of iodine exchange. The
rate of inversion and of iodine exchange are, therefore, identical and it thus follows that
every act of substitution much proceed with inversion of configuration.
(b) Kinetic evidence Usually SN2 reaction involving nucleophiles other than solvents
exhibits the first-order kinetics with respect to both the substrate and the nuclephile.
Therefore, SN2 reactions usually follow the second-order kinetics and hence these are
bimolecular (involve rate-determining attack of the nuclephile on the substrate). However,
the kinetic evidence is not enough for establishing the mechanistic pathway actually
involved because several pathways may be suggested by considering these data.
3.1.6 Factors Influencing SN2 Reaction Rate or SN2 Reactivity

3.1.6.1 The effect of substrate structure on rate
As the number of R groups on the carbon with the leaving group increases, the rate of SN2
reaction decreases.
CH3—X R CH2—X R2CH—X R3C—X

Methyl 1° 2° 3°
SN2 reaction rate decreases
The SN2 reaction is very much less affected by the electrical effects of the groups at the
reaction centre and this is because the central carbon does not become considerably
more negative or positive in the transition state as compared to its initial state. Steric
effects have profound influence on the rates of SN2 reactions. As a small H atom is
replaced by larger alkyl groups, hindrance caused by bulky R groups make the attack by
the nuceleophile from the back side more difficult, slowing the reaction rate. This may
also be explained in terms of the energy of the transition state. In an SN2 pathway, the
tetrahedral central carbon becomes surrounded by five substituents in the transition state
(carbon is pentacoordinated). Thus, on going from the substrate to the transition state the
steric crowding around the central carbon increase. As consequence, the SN2 transition
state becomes progressively more crowded with increase in the number of alkyl groups
at the reaction centre. As crowding increases, the transition state energy increases and
thus, the reaction rate decreases. Hence, due to steric reason, the SN2 reactivity decrease
progressively on going from methyl to tert-butyl halide. In fact, the steric crowding in the
transition state of a tertiary alkyl halide is so large, i.e., the transition state is so unstable
that it is unable to undergo the SN2 reaction.
The following observations demonstrate how various structural effects influence the rate
(reactivity) of SN2 reactions:
(1) Increased branching at the b-carbon retards an SN2 reaction. For example, the rate of
SN2 reaction of the following bromides with I① in acetone decreases as the following series
is traversed:
acetone
R — Br + :��I : * ææææ �� : *
Æ R — I + : Br
��
CH 3CH 2Br > CH 3CH 2CH 2Br > (CH 3 )2CHCH 2Br > (CH 3 )3CCH 2Br
Ethyl bromide n–Propyl bromide Isobutyl bromide Neopentyl bromide
relative rate: 1 0.082 0.036 0.000012
The differences in electron-releasing inductive effect (+I) of the methyl groups at the
b-carbon through the two saturated carbon atoms are very small to show any detectable
difference in the SN2 reaction rate. It is the steric effect which is responsible for the
rate differences. As the number of CH3-groups at the b-position increases, it becomes
progressively more different sterically for the nucleophile I① to attack on C 3 bearing
sp
bromine form the backside and consequently, the SN2 reaction rate decreases on going
from ethyl to neopentyl bromide. Conformational analysis gives us a clear picture. In
conformation I, the backside of n-propyl carbon is seriously blocked, whereas in other two
conformations (II and III) the situation is no worse than ethyl. As a consequence, n-propyl
bromide undergoes SN2 reaction only slightly less readily than does ethyl bromide.
In the case of isobutyl bromide, there is only one conformation in which the nucleophile
can avoid a blocking methyl group. Also, that conformation is highly congested and
therefore, has relatively high energy. Consequently, isobutyl bromide is much less reactive
than either ethyl or n-propyl bromides.
In neopentyl bromide, there is no conformation in which the attacking nucleophile can avoid
a blocking methyl group. Thus, except under very drastic condition neopentyl bromide is
practically unreactive in SN2 reactions. Its unreactive nature is due partially to the fact
that the leaving group experiences severe van der Waals repulsions with hydrogens of the
methyl branches.
Hence, the order of relative SN2 rate of these alkyl bromides is
CH3CHBr > CH3CH2CH2Br > (CH3)2CHCH2Br > (CH3)3CCH2Br

(2) Small bicyclic systems containing leaving group at the bridgehead carbon do not undergo
SN2 reaction: For example, 1-bromobicyclo [2.2.1] heptane is extremely unreactive in
an SN2 reaction. This is because the backside attack on the carbon bearing bromine is
prevented sterically by its cage-like structure and also by the impossibility of assuming the
required planar distribution of bonds to the bridgehead carbon in the transition state.
The matter can be well demonstreated by the fact that the following ether (I) undergoes
cleavage by HBr to give the corresponding bicyclic alcohol and CH3Br.
The reaction proceeds through the SN2 pathway involving nucleophilic attack by Br① on
the methyl carbon of the protonated ether.
Because of the formation of an unstable bridgehead carbocation, the reaction does not
proceed through the SN1 pathway to give methanol and 1-bromobicyclo[2.2.1]heptane.
(3) A double bond at the b-position helps to increase the rate of an SN2 reaction. For
example, allyl bromide, CH2 == CH — CH2 Br, undergoes SN2 reaction slightly more
readily than does ethyl bromide. This is because the unhybridized p orbital involved in the
transition state interacts with the p orbital system of the allyl group and as a result, the
transition state becomes much stable and the activation energy of the reaction becomes
relatively low.
(4) A carbonyl group at the b-position helps to increase the rate of an SN2 reaction. For
example, chloroacetone (CH3COCH2Cl) undergoes SN2 reaction at a rate faster than does
n-propyl chloride (CH3CH2CH2Cl). This is because the b-carbonyl group accepts some of
the negative charge of the nucleophile and thereby facilitates the nucleophilic attack on
carbon bearing the leaving group by stabilizing the transition state.
(5) Cyclopropyl substrates does not undergo SN2 reaction. For example, halocyclopropanes
are unreactive towards SN2 reactions. This is due to increased angle strain in the system.
In the transition state of an SN2 reaction, the central carbon becomes sp2 hybridized
in which the normal bond angle is 120°. In a cyclopropyl system, the bond angle strain
increases (109.5° – 60° = 49.5° to 120° – 60° = 60°) with change in hybridization (sp3 to sp2)
of the central carbon atom. As a result, the energy of the transition state becomes too high
to make halocyclopropanes unreactive towards SN2 reactions.
(6) Halocyclohexanes react slowly by the SN2 mechanism. The SN2 reaction of
halocyclohexanes take place through the relatively stable equatorial conformation. Ring
strain does not appear to be an important factor in this case. However, the axial hydrogen
at C-3 and C-5 interfere sterically with attacking nucleophile and as a result, the transition
state becomes relatively unstable. For this reason, halocyclohexanes react by the SN2
mechanism rater slowly.
(7) In cyclohexane systems, the conformer with axial leaving group reacts more rapidly than
the conformer with equatorial leaving group. For example, the cis-isomer of 4-tert-butyl-
* ≈
cyclohexyl bromide reacts with sodium thiophenoxide (Ph S Na) in aqueous ethanol at a
much faster rate than its trans-isomer. This is because the axial approach of the nucleophile
(specially bulky) for an equatorial leaving group is sterically more hindered due to the
axial hydrogens at C-3 and C-5, relative to the carbon (C-1) undergoing substitution. Also,
the ground state energy of the axial isomer is higher than the equatorial isomer, i.e., the
energy of activation for the axial isomer is relatively low. In this case, the bulky tert-butyl
group locks the cis-isomer of 4-tert-butylcyclohexyl bromide in that conformation which
places bromine in the axial position and so, this isomer reacts with the bulky nucleophile
thiophenoxide ion (Ph S①) by the SN2 mechanism at a much faster rate.
On the other hand, the bulky tert-butyl group locks the trans-isomer in that particular
conformation which places bromine in the equatorial position and so, due to steric
hindrance, displacement of bromine occurs at a much slower rate in this case.
If the leaving and entering groups are the same in such a reaction, any difference in
reaction rate must due to difference in the energy of starting materials. For example,
axial cyclohexyl iodide reacts with radioactive iodide ion, 128I① (an SN2 reaction) at a rate
faster than equatorial cyclohexyl iodide. Since the equatorial conformer of cyclohexyl
iodide is more stable than the axial conformer, the equatorial conformer must have the
lower ground state energy. Since the energy of activation reflects the difference in energy
between the transition state and the starting material, the reaction involving the axial
conformer must have lower Ea. For this reason, the axial conformer reacts at a rate faster
than the equatorial conformer.
128I d – 128I
H H
S N2 d– H 128
128 – I –
I + I + I
equatorial cyclohexyl Transition state

iodide
d–
I I H H
S N2 –
H +128 I –
128I d – 128I + I
axial cyclohexyl Transition state

iodide
(8) Vinylic and aryl halides are unreactive towards SN2 reactions. Displacement of an aryl
or vinyl halogen by an SN2 mechanism is prevented because the p electron cloud of the
benzene ring of an aryl halide or the double bond of a vinylic halide repeals the approach of a
nucleophile from the backside. Again, the C — X bond of aryl or vinylic halides is unusually
strong because the bond has considerable double bond character due to delocation of the
unshared electron pair on halogen with the p electrons of the ring or the double bond.
Consequently, bond breaking requires more energy. It is for these reasons, vinylic and aryl
halides are unreactive towards SN2 reactions.
The unreactive nature of phenyl halides can be demonstrated by the fact that ethyl phenyl
ether reacts with HBr to give ethyl bromide and phenol instead of bromobenzene
and ethyl alcohol. The conjugate acid of ethyl phenyl ether undergoes SN2 attack by Br①
not on the ring carbon (the p electron cloud of the ring repeals the backside approach of the
nucleophile) to give bromobenzene and ethanol but on the methylene carbon (a favourable
site for SN2 attack) to yield ethyl bromide and phenol.
(9) Intramolecular SN2 reaction is possible if geometry allows backside attack. For example,
when treated with alkali, trans-2-bromocyclopentanol produces cyclopentene oxide but its
cis-isomer produces trans-cyclopentane-1,2-diol.
Since in trans-2-bromocyclopentanol the hydroxyl group is placed stereochemically anti

with respect to bromine, intramolecular backside displacement of Br① by oxyanion occurs
to yield cyclopentene oxide.
In the cis-isomer, on the other hand, the hydroxyl group and bromine are on the same
side of the ring. Therefore, backside displacement of bromine by the oxyanion does not
take place. An ordinary SN2 reaction with inversion of configuration occurs to give trans-
cyclopentane-1,2-diol.
(10) Williamson synthesis of ethers involves an SN2 reaction between an alkyl halide and
an alkoxide. Therefore, if there are two possible ways for the synthesis of a particular
ether, that alkyl halide should be used which causes less steric hindrance during during
backside attack. For example, tert-butyl ether can be expected to be prepared either by
treating ethyl chloride with potassium tert-butoxide or by treating tert-butyl chloride with
sodium ethoxide.
SN 2
CH 3CH 2Cl + (CH 3 )3COK æææ Æ (CH 3 )3C — O — C2H5 + KCl
Ethyl chloride Potassium tert- Butyl ethyl ether
tert-butoxide
S 2
or (CH 3 )3CCl + CH 3CH 2ONa æææ
N
Æ (CH 3 )3C — O — CH 2CH 3 + NaCl
tert- Butyl chloride sodium ethoxide tert-Butyl ethyl ether
In the latter combination of reactants, a tertiary alkyl halide is used and because of this,
an SN2 reaction leading to the formation of tert-butyl ethyl ether does not take place
due to steric reason. However, in the presence of a very strong base like EtO①, the alkyl
halide undergoes facile b-elimination reaction to yield 2-methylpropene as the major
product. Such as elimination reaction does not take place when ethyl chloride is allowed to
react with potassium tert-butoxide. Being a primary substrate, eithyl chloride undergoes
ready substitution by the SN2 pathway to form tert-butyl ethyl ether predominantly. This
combination of reactants is, therefore, more preferred than the other.
(11) SN2¢ mechanism: When steric factors prevent normal SN2 attack at an allylic carbon,
the nucleophil attacks at the g-carbon, i.e., an allylic rearrangement occurs under SN2
conditions. This is what is called SN2¢ (SN2-prime) mechanism (bimolecular nucleophilic
substitution with allylic rearrangement). For example, when a, a-dimethylallyl chloride
is allowed to react with sodium thiphenoxide (a bulky is nucleophile) in ethanol, the
rearranged substitution product is obtained is good yield.
3.1.6.2 The effect of solvent on rate

In general, SN2 reactions are very slightly affected by the polarity of the solvent. In a most
common type of SN2 reaction, an anionic nucleophile attacks a neutral substrate:
– d– d– –
Nu + R—X Nu R X Nu—R + X
Reactants Transition state Products
(concentrated charge: (dispersed charge:
more stabilized by less stabilized by
solvation) solvation)
Since the charge is dispersed in the transition state, therefore, stabilization by the solvent
is somewhat greater for the reactants than for the transition state. As a consequence,
a small decrease in reaction rate is observed in increasing solvent polarity. A marked
decrease in reaction rate occurs by the change from an aprotic to a protic solvent. An aprotic
polar solvent [e.g., N, N-dimethyl formamide (DMF) Me2NCHO, dimethyl sulphoxide
(DMSO) Me2S=0, dimethylacetamide (DMA), Me2NCOMe, hexamethylphosphoramide
(HMPA) (Me2N)3PO, N-methylpyrrolidone (NMP), N, N-dimethylpropyleneurea (DMPU),

etc] preferentially solvates the cations. They perform this by orienting their negative ends
around the cation and by donating an unshared electron pair to the vacant orbitals of the
cations. However, because they cannot form hydrogen bonds and their positive centres are
well shielded by steric effects from any interaction with anions, aprotic polar solvents do
not solvate anions to any appreciable extent. Consequently, the anions become relatively
free and highly reactive as a nucleophile.
SMe2
==
O
Me2S == O O== SMe2
Sodium cyanide +
in DMSO: Na :CN
Me2S == O O== SMe2

O (A relatively free
or ‘naked’ cyanide ion)
==
SMe2
(A sodium ion solvated by

molecules of the aprotic polar
solvent DMSO)
On the other hand, a protic solvent solvates the anions effectively through hydrogen
bonding and thereby reduces its reactivity abnormally. The cyanide ion (CN①) in sodium
cyanide (NaCN), for example, is far more reactive as a nucleophile in DMSO than in the
protic solvent aqueous methanol.
Therefore, an aprotic polar solvent is much more effective for an SN2 reaction which
particularly depends on the reactivity of the nucleophile.
Protic and aprotic solvents A protic solvent is one that consists of molecules having acidic
protons (usually attached with oxygen or nitrogen atom), i.e., protic solvent has O — H
or N — H group. These groups form hydrogen bonds to negatively charged nucleophiles.
Water, alcohols (ROH) and carboxylic acids (RCOOH) are examples of protic solvents.
Solvent that cannot form hydrogen bonds are called aprotic solvents. Acetone (CH3COCH3),
ether (ROR) and benzene (C6H6) are examples of aprotic solvents.
Polar and apolar solvents A solvent which has a high dielectric constant (e), i.e., which
effectively separates ions of opposite charges from one another, is called a polar solvent
and a solvent which has a low dielectric constant is called an apolar solvent. In fact, if a
solvent has a dielectric constant of about 15 or greater, it is considered to be polar. Water
(e = 79), methanol (e = 33) and formic acid (e = 59) are examples of polar solvents, whereas
hexane (e = 1.9), ether (e = 4.3) and acetic acid (e = 6) are examples of apolar solvents.
[N.B. In organic chemistry, the word polar has a double usage. When we say that a
molecule is polar, we simply mean that it has a significant dipole moment (m). However,
when we say that solvent is polar, we mean that it has a high dielectric constant (e).
Actually, molecular polarity or dipole moment is a property of individual molecules, but
solvent polarity or dielectric constant is a property of many molecules acting together. The
contrast between acetic acid and formic acid is particularly striking:
These two compounds contain identical function groups and have very similar structure
and dipole moments. Both are polar solvent. Yet there is substantial difference in their
dielectric constants and they differ substantially in their solvent properties. Formic acid
is a polar solvent, but acetic acid is not. Although it is true that all polar solvent consists
of polar molecules, the converse is not true.]
Dielectric Constants (Œ) of Some Solvent at 25°C
Solvent Dielectric constant
Cyclohexane (C6H12) 2.01

p-Dioxane ( ) 2.20
Carbon tetrachloride (CCl4) 2.23
Benzene (C6H6) 2.27
Carbon disulphide (CS2) 2.60
Diethyl ether (C2H5OC2H5) 4.24
Chloroform (CHCl3) 4.8
Chlorobenzene (C6H5Cl) 5.63
Ethyl acetate (CH3COOC2C2H5) 6.09
Acetic acid (CH3COOH) 6.1 (20°C)
Tetrahydrofuran ( ) 7.6
1,2-Dichloroethane (ClCH2CH2Cl) 10.36
Increasing solvent polarity
Sulphur dioxide (SO2) 14.1

n-Propyl alcohol (CH3CH2CH2OH) 20.1
Acetone (CH3COCH3) 20.7
Ethanol (C2H5OH) 24.33
Methanol (CH3OH) 32.63
Nitromethane (CH3NO2) 35.9
Ê O ˆ
�
Dimethylformamide H - C - N Me2 ˜
Á 36.7
Á ˜
ÁË ˜¯
Acetonitrile (CH3CN) 37.0

Sulfone ( ) 42.0
Dimethyl sulfoxide (Me2S=O) 46.7

Formic acid (HCOOH) 59.0
Water (H2O) 79.0
That marked increase in reaction rate occurs by the change from a protic solvent to an
aprotic solvent can be shown by carrying out the following reaction in methanol and in
dimethyl sulphoxide (Me2SO).
* SN 2
N3 + CH3 — I æææ Æ N3 — CH3 + : ��I : *
��
The rate of the reaction increases very much (109-fold) on transfer from methanol to
*
DMSO. The nucleophile N3 remains highly solvated through hydrogen bonding in the
protic polar solvent methanol. However, in the aprotic polar solvent DMSO (Me2SO), it
remains weakly solvated (only through ion-dipole interactions). Also, its counterion (the
cationic part) is solvated by this solvent making the nucleophile much freer to react. Thus,
*
N3 reacts as a very powerful nucleophile in DMSO than when it reacts in methanol. For
this reaction, the rate of this reaction increases very much on transfer from methanol
(e = 32.63) to DMSO (e = 46.7).
3.1.6.3 The effect of the nucleophile on rate

The nature of the nucleophile plays a vital role in determining the rate of SN2 reaction
because the nucleophile is involved in the transition state of the rate-determing step (the
only step here). The SN2 reaction rate increases with increasing the concentration and
the nucleophilic power (strength) of the nucleophile. A good nucleophile is one that reacts
rapidly in an SN2 reaction with a given substrate and a poor nucleophile is one that reacts
slowly in an SN2 reaction with the same substrate under comparable reaction conditions.
Every nucleophile has basic character and in general, nucleophilicity and basicity run in
parallel. For reagents having the same attacking atom, a parallelism between them is
observed. For example, in the following anions (the attacking atom is oxygen) decreasing
nucleophilicity parallels decreasing basicity:
�� * > HO:
CH 3 O: �� * > PhO:
�� * > CH COO: �� * > HCOO:�� *
3
��
These two also run in parallel if the attacking atoms are elements with the same period
but belonging to different groups of the periodic table. For instance, in the following series,
the nucleophilicity decreases with decreasing basicity:
* *
��
CH : ��
3 > NH2 > OH
: �� * > : F
�� :* (Period-2)
Group-14 Group-15 ��
Group-16 ��
Group-17
*
: PH
�� 2 > : SH
�� * > :Cl
�� :* (Period-3)
Group-15 Group-16 ��
Group-17
The parallelism between nucleophilicity and basicity, however, is not observed when the
attacking atoms are of elements belonging to the same group of the periodic table. For
instance, the sequence of nucleophilic reactivity for the halides: I① > Br① > Cl① > F① (in
protic solvent) is opposite to the sequence of basicity: F① > Cl① > Br① > I①. This order of
nucleophilicity tells us that the nucleophilic reactivity of halides increases with increasing
the size of the halogen atom. In fact, polarizability and solvation are involved in determining
the nucleophilic power of a nucleophile. The outermost electrons of larger and less
electronegative atoms are less tightly held by the nucleus. Thus, they are more polarizable
and readily available for making bonds to C 3 in the SN2 transition state. Therefore,
sp
the polarizability and hence nucleophilicity increases on going from the smaller F① ion to
the larger I① ion. Again, in a protic solvent, a nucleophile remains solveted by hydrogen
bonding and to participate in the reaction it must be released from the solvation cage.
Solvation energies of ions increase with increasing charge to size ratio. Hence, desolvation
of a smaller ion, in which the charge is more concentrated, requires more energy compared
to a larger one bearing the same charge. The larger ions are, therefore, better nucleophiles
in a protic solvent. In aprotic polar solvents, the cations remain well slovated but the anions
remain less solvated and relatively free. Under such conditions, a small anion having its
negative charge spread out over a comparatively small volume is more energetic than a
large anion and tends to be a better nucleophile. Nucleophilicity and basicity, therefore,
run in parallel. For example, in dimethyl sulphoxide (Me2S=0), the order of nucleophilicity
for the halide ions is the same as their relative basicity: F① > Cl① > Br① > I①.
A negatively charged nucleophile is always a more reactive nucleophile than its conjugate
acid. For example, OH① is a better nucleophile than H2O and MeO① is better than
MeOH.
The nucleophilicity of a reagent may also be influenced by the steric effect. For example,
the bulky tert-butoxide ion is unable to attack the central carbon of R — X (except
R == Me) smoothly, even though it is more basic than the ethoxide ion which is a better
nucleophile.
An interesting example of steric effect is as follows. Quinuclidine is a better nucleophile

than tirethylamine. Since the carbon atoms in quinuclidine are held back by its cage-
like structure, therefore, the unshared electron pair on nitrogen is more exposed and its
approach towards the C 3 atom from the backside is not sterically hindered. On the other
sp
hand, the ethyl groups in triethylamine hinder sterically its close approach towards carbon
in an SN2 reaction. Quinuclidine is, therefore, a better nucleophile than triethylamine.
Like basicity, nucleophilicity depends on the availability of the unshared electron pair on
the attacking atom. For example, trimethylamine, (CH3 )3 N, �� is a good nucleophile while
tris(trifluoromethyl)amine, (CF ) N, �� is completely nonnucleophilic. In trimethylamine, the
3 3
three electron-releasing (+I) methyl groups increase the electron density on nitrogen and
thereby increase the availability of its unshared electron pair. For this reason, (CH3 )3 N
��
is a good nucleophile. In tri(trifluoromethyl)amine, however, the powerfully electron-
withdrawing (–I) — CF3 groups decrease the electron density on nitrogen considerably and
thereby decrease the availability of the unshared pair makedly. For this reason, (CF3 )3 N
��
is completely nonnucleophilic.
H3C
+
S N2 –
N + R X (CH3)3N—RX
H3C
CH3
F3C
S N2
N + R X No reaction
F3C
CF3
Basic difference between nucleophilicity and basicity
Nucleophilicity Basictiy
1. The nucleophilicity of a regent is its tendency 1. The basicity of a reagent is its tendency to form
to form a bond with a carbon atom. a bond with a proton.
2. Nucleophilicity is a measure of how rapidly a 2 Basicity is measure of the position of equilibrium
reagent attacks the substrate in a substitution in an acid-base reaction.
reaction 3 Basicity is determined by the equilibrium
3. Nucleophilicity is determined by the reaction constant which depends on the overall energy
rate which depends on the free energy of change (DG°) of the reaction.
activation (DG=) of the reaction.
|
Some relevant topics are discussed below:
(a) The order of basicity of halide ions The order of basicity of a halide ions is opposite
to the order of their nucleophilicity. Basicity is a measure of the position of equilibrium in
the following reaction and it depends on the overall energy change (DG°) of the reaction:
The equilibrium constant (Keq) of the reaction is related DG° by the following expression:
DG° = – 2.303 RT log Keq
Basicity increases as the value of Keq increases, i.e., the value of DG° increases. Again,
the value of DG° decreases as the value of DH° decreases (DG° = DH° – DS°), i.e., the value
of H — X bond strength increases. Hence, basicities of halides increase with increasing
H — X bond strength. Proton is a hard acid. Base hardness of the halide ions increases
in the order: I① < Br① < Cl① < F①. A hard base prefers to combine with a hard acid and
this result in formation of a stronger bond. Therefore, the bond strength of halogen acids
decreases in the order: H — F > H — Cl > H — Br H — I and so, basicity increases in the
order: I① < Br① < Cl① < F①.
(b) Gas phase nucleophilicity of halide ions In the gas phase, the order of SN2 reactivity,
i.e., nucleophilicity of the halide ions towards CH3Br is F① > Cl① > Br① > I①. In the gas
phase, where solvation effects are absent, the smaller anion having its negative charge
spread over a comparatively small volume is more energetic than the larger anion and
tends to be a better nucleophile. Therefore, in the gas phase, the order of SN2 reactivity,
i.e., the order of nucleophilicity of halide ions towards CH3Br is F① > Cl① > Br①> I①.
(c) Role of ion pairing in determining nucleophilicity In the following reaction, the order
of nucleophilicity for various halides is I① > Br① > Cl① when LiX is used, but the order is
≈ *
Cl① > Br①> I① when Bu 4 N X is used.
acetone
�� :* ææææ
CH3 CH2CH2CH2OBs + : X SN2
�� :*
Æ CH3CH2CH2 X + BsO
��
In the weakly polar solvent acetone (CH3COCH3), there is considerable ion pairing and the
electrostatic attraction between the concentrated charges on two smaller ions is strongest.
Therefore, the small Li≈ ion from tight ion pair with a small halide ion makes it less
reactive as a nucleophile. Again, as the size of the halide ion increases, the freeness of
the halide ion increases. Therefore, in the presence of Li≈ ion, the order of nucleophilicity
is the same as that observed in protic solvents, i.e., I① > Br① > Cl① and for much the
same reason. The large quaternary ammonium ion, with its charge shielded by the bulky
n-butyl groups, forms only a very loose ion pairs. The halide ions are then relatively free
and the small ion is more energetic and more nucleophilic, i.e., their reactivity order is
same as their basicity order, i.e., Cl① > Br① > I①.
(d) Hard and soft nucleophiles and electrophiles Pearson’s HSAB (Hard and Soft
Acids and Bases) theory has also been introduced in the domain of nucleophiles and
electrophiles. Hard bases are hard nucleophiles in which the attacking atom possesses
high electronegativity, low polarizability and low reducing power. Soft bases, on the other
hand, are soft nucleophiles in which the attacking atom has low electronegativity, high
polarizability and high reducing power. So, I① ion is a soft nucleophile and OH① ion is a
hard nucleophile. Proton and all proton donors are hard electrophiles while a saturated
carbon (C 3 ) is a soft electrophile. A few hard and soft electrophiles and nucleophiles are
sp
given in the following table.
Hard and so electrophiles and nucleophiles

Electrophiles (Acids)
Hard Soft
≈ ≈ ≈ ≈ 2≈ 3≈ 3+ ≈ ≈ ≈ ≈ ≈
H , Li , Na , K , Mg , Cr , Fe , BF3, AlCl3, Saturated carbon, Cu , Ag , Hg , I , Br , I2, Br2
≈ metal atoms, carbenes, trinitrobenzene, quinones.
AlH3, CO, RCO , SO3
≈ ≈
Borderline: Fe2+ , Cu2+ , Zn2+ , B(CH ) , Pb2+ , NO, Me C, C H≈
3 3 3 6 5
Nucleophiles (Bases)
Hard Soft
NH 3 , RNH 2 , N2H 4 , H 2O, OH * , O* , ROH, H* , R * , C2 H4 , C6 H6 , CN* , SCN* , R 3 P,

2*
3 , F , Cl , SO 4 , R 2 S, RSH, RS
RO* , R 2O, CH 3COO* , NO* * * *
PO34* , ClO*
4
Borderline: PhNH2, N3① N NO2 , Br

① ①
There is an empirical rule which governs acid–base and nucleophile–electrophile

interactions. The rule sates that hard acids prefer to bind with hard bases and soft acids
with soft bases. This means that hard–hard as well as soft–soft interactions are strong,
i.e., these lead to the formation of strong bonds but hard–soft interactions are weak and
these lead to weak bonds. For example, H≈ (a hard acid) reacts faster with OH① (a hard
base) than with I≈ (soft base). Along a period from left to right, the harness of nucleophiles
increases and the softness decrease and down a group, the softness of nucleophiles
increases and the harness decreases. Thus, the order of increasing hardness amongst the
following site of nuceleophiles is
@ @
(i) : H@ < : CN < : OH < : F
�� :@
��
(ii) I < Br < Cl < F
@ @ @ @
(iii) C@ < N@ < O@ < F@
A nucleophile with negative charge on the attacking atom is harder than its conjugate
@
acid. Therefore, :N H2 is harder than NH �� * is harder than H O
�� , :OH �� @: is harder
�� : and HS
3 2
�� . ��
than H2 :S
Furthermore, a positively charged carbon is a harder centre (electrophile) than a partially
positively charged carbon, while the latter is a harder centre than a less partially positively
charged carbon.
(e) Ambident nucleophiles An ambident nucleophile is a species which has two (or
more) different types of mucleophilic centres through which it can attack the electrophilic
centres. Cyanide, nitrite and enolate ions are the examples of ambident nuceleophiles
(ambi in Latin for ‘both’; dent in Latin ‘teeth’)
The nucleophilic centres in a cyanide ion are C and N atoms since both bear a lone pair
of electrons. The same is the case with nitrite ion where the two O atom and the N atom
are nucleophilic centres. In an enolate ion, anionic O and C atoms are the nucleophilic
centres.
The following sets of reaction illustrate the behaviour of ambident nucleophiles in

distribution of products in substitution reactions:
≈ *
(a) attack by
CH 3Br + KCN ææææææ Æ CH 3 — C ∫∫ N: + CH 3 — N ∫∫ C:
CN* through
C atom Ethanenitrile Ethaneisonitrile
(major product) (minor product)
≈ *
attack by
CH3 Br + AgCN ææææææ Æ CH3 — N ∫∫ C: + CH3 — C ∫∫ N
CN* through
N atom (major product) (minor product)
(b)
These observations can be explained in terms of hard–soft nucleophile–electrophile

interaction as discussed before. In each case. The reaction involves either a hard–hard
interaction or a soft–soft interaction. The hard and soft centres of each nucleophile are
labelled here following the fact that the hardness increases along a period from left to
right.
In the presence of Na≈ or K≈ ions, the C atoms of CH3Br, CH3I and CH3CH2I are the
@
soft centres. So, : CN : and NO@2
attack the electrophilic carbons through soft centres
(less electronegative and more polarizable) C and N, respectively, and we get methyl
cyanide and nitromethane as major products. In the presence of Ag≈, which is generated
from the reagent AgCN or AgNO2, alkyl halides become more carbocation-like in which
the central carbons are hard centres (being appreciably positively polarized). So, the
nucleophiles attack the positively polarized (or cationic) carbons through their hard centres
*
(N of : CN : and O of NO2*) and naturally the major products are alkyl isocyanides and
alkyl nitrites. In the presence of Na≈ or K≈ ion, the reactions are SN2 but in the presence
Ag≈ ion, the reactions are SN2 with considerable SN1 character or pure SN1.
This can be explained simply as follows. In the presence of AgCN, the reaction proceeds
by the SN2 mechanism with considerable SN1 character or by pure SN1 mechanism (in
the case of 2° or 3° substrates) and this is because Ag≈ ion coordinates with iodine and
causes the development of considerable positive charge on central carbon (or promotes
fromation of a carbocation by precipitating AgI). Nucleophilic attack on carbon then takes
place through the more electronegative and more electron dense nitrogen atom and as a
result, RNC (an alkyl isocyanide) is obtained predominatly. However, in the presence of
NaCN, the reaction proceeds by the SN2 mechanism because unlike silver ion, Na≈ ion
cannot make the central carbon positively polarized (or promote ionization of the C — I
bond). Since bond formation occurs in the transition state of the rate-determining step, the
cyanide ion attacks R — I through the more polarizable and less electronegative, i.e., more
nucleophilic carbon atom, and thus, leads to the formation of R — CN (an alkyl cyanide)
predominantly.
(f) Nucleophilic catalysis Highly polarizable nucleophile like I① (a good entering group
and a good leaving group) can often be used as a catalyst to promote an otherwise slow
displacement reaction. This is known as nucleophlic catalysis. For example, methyl
chloride undergoes hydrolysis at a much faster rate in the presence of sodium iodide. The
hydrolysis of CH3Cl in an aqueous medium take place at a much slower rate because H2O
is a weak nucleophile (because the molecule is neutral and the attacking atom is highly
electronegative and less polarizable) and Cl① is not very good as a leaving group (because
it is not a very weak base and the C — Cl bond is not very weak).
SN2 (i)
H2O: + CH3—Cl very slow HO—CH3 + HCl
Methyl Methyl
chloride alcohol
The iodide ion (I ) catalyzes the hydrolysis of methyl chloride. Due to large size and
①
low electronegativity, the iodide ion is highly polarizable. Also, its solvation energy is
realatively small because the charge to size ratio is small. Furthermore, it is a very good
leaving group because it is a weaker base and the C — I bond is relatively weak. For
this dual ability to attack and depart readily, I① is an effective catalyst in substitution
reactions. In the presence of NaI, the hydrolysis of CH3Cl takes place by the following two
steps.
– SN2 –
I + CH3—Cl I—CH3 + Cl (ii)
fast
SN2
H2O + CH3—I HO—CH3 + HI (iii)
fast
Each of the reaction (ii) and (iii) takes place at a rate faster than the reaction (i) because I①
is a better nucleophile than H2O and is a better leaving group than Cl①. Thus, the iodide
ion increases the rate of formation of ethanol by changing a relatively slow SN2 reaction
into two relatively fast reactions. The overall hydrolysis, therefore, occurs rapidly.
(g) Phase-transfer catalysts A compound that catalyzes a reaction by transferring a
reagent (usually an inorganic ion) into the phase in which it is needed is called a phase-
transfer catalyst. In the following reaction, tetrabutylammonium hydrogen sulphate acts
as a phase transfer catalyst.
≈
Bu 4 N HSO4*
CH3 (CH2 )5 Br (in decane) ææææææææ
aqueous NaCN/105∞ C
Æ CH3 (CH2 )5 CN
1-Bromohexane Heptanenitrile
≈
A quaternary ammonium salt like Bu4 N HSO@
4 ,
because of its nonpolar alkyl groups, is
soluble in a nonpolar solvent (like dissolves like), but it is also soluble in water because of
its charge. Therefore, it can act as a mediator between the two immiscible solvents. When
≈
Bu4 N HSO4 ion passes into the nonpolar layer, it must carry a counter ion with it. The
@
counter ion may be either HSO4* or CN* . Because there is more CN① ion than HSO4*
ion in the aqueous layer, the former will be the main accompanying ion. Once it is in the
nonpolar decane layer, the CN① ion can react with the alkyl halide. The catalysis effected
by this technique is called phase-transfer catalysis. Some HSO* 4
ion is also transported
into the organic layer, but it remains unreactive because it is a very poor nucleophile. The
≈
Bu 4 N ion then passes back into the aqueous layer carrying with it either HSO*
4 or Br .
*
The reaction continues with the phase–transfer catalyst shuttling back and forth between
the organic and aqueous phases. Phase transfer catalysis has been used successfully in a
wide variety of organic reactions.
Two other quaternary ammonium slats used commonly as phase transfer catalysts are as
follows:
≈ * + –
CH 3 (CH 3 )14 CH 2 N(CH 3 )3 HSO4 — CH2N (C2H5)3 HSO4
Hexadecyltrimethylammonium
hydrogen sulphate Benzyltriethylammonium
hydrogen sulphate
Crown ethers: A group of large-ring polyethers (cyclic polymers of ethylene glycol),
because of their three-dimensional crown shape, are called crown ethers. They are named
as x-crown-y, where x is the total number of atoms in the ring and y is the number of oxygen
atoms. For example, 18-crown-6, 15-crown-5, 12-crown-4, dicyclohexano-18-crown-6, etc.
Crown ethers act as phase-transfer catalysts. They are able to bind various metal ions
through ion-dipole interactions involving the unshared pair of electrons on the oxygen
atom. A crown ether coordinates effectively with a particular cation which fits well into
the cavity of the crown. For example, 18-crown-6 (hole diameter 2.7 Å) coordinates very
effectively with K≈ ions (diameter 2.66 Å) because the cavity size is correct and because
the six oxygen atoms are ideally situated to donate their unshared electron pairs to the
central ion in a Lewis acid-base complex.
The resulting complex is lipophilic on the outside, and has a positive charge burried
within the molecule. Because of lipophilicity (hydrocarbon like properties), it is soluble
in organic solvent of low polarity. When it passes into organic solvent, it gets the anion
with it. The anion is shielded from the positive charge on K≈ by the bulky size of the
crown, thus forming only loose ion pairs and is highly reactive. In this way, the crown
ether permits an inorganic salt like KCN to be dissolved in nonpolar or less polar solvents
like benzene, acetonitrile, etc. and thereby makes some SN2 reactions successful by
acting as a phase-transfer catalyst. KCN (insoluble in benzene), for example, reacts with
chloroethane in benzene containing a catalytic amount of 18-crown-6 to form CH3CH2CN
almost quantitatively.
18- crown -6
K ≈ CN* + CH3CH2 Cl æææææ
Benzene
Æ CH3CH2CN + K ≈ Cl*
Chloroethane (SN 2) Ethanenitrile
Alkyl fluorides (R–F) cannot be prepared easily. 18-Crown-6 provides a good method for
the synthesis of alkyl fluorides. Potassium fluoride (KF) is an ionic compound which is
particularly insoluble in the aprotic nonpolar solvent benzene. The crown ether 18-crown-6
makes this salt soluble in benzene. Every K≈ ion that enters into the benzene layer (as
a Lewis acid-base complex with the crown ether) brings a fluoride ion (F①) with it. These
relatively unsolvated fluoride ions with a greatly enhance reactivity displace halide ions
from alkyl halides quantitatively. For example:
KF, benzene 90∞ C
CH 3 (CH 2 )4 Br æææææææ
18-crown-6
Æ CH 3(CH 2 )7 F
1-Bromoctane 1-Flurooctane (92%)
The relationship between a crown ether and the ion it binds is called a host–guest
relationship.
3.1.6.4 The effect of the leaving group on rate

The rates of SN2 reactions are influenced by the nature of the leaving group because the
process of expulsion of the leaving group occurs in the rate-determining step (the only
step here). In most of the SN2 reactions, the leaving group begins to acquire a negative
charge as the transition state is reached. Stabilization of this developing negative charge
at the leaving group stabilizes the transition state, i.e., lowers its energy. This, in fact,
lowers the free energy of activation (DG=| ) and thereby increases the rate of the reaction.
Because weak base stabilizes a negative charge effectively, these are good leaving groups.
For instance, in the following series, the leaving-group ability (fugicity) increases with
decreasing basicity:
@ @
: NH
�� :@ < PhO
2 < CH3 O
�� :@ < NH
�� : : �� :@ < : C
3 < CN < F
�� l :@ < : Br
�� : @ < : ��I : @
��
Also, an effective leaving group should be linked to the central C 3 atom with a relatively
sp
weak bond. The C — X bond strength, for example, increases in the following order:
C — I < C — Br < C — Cl < C — F
Bond dissociation
energy (kcal/mol) : 53 69 81 109
Therefore, the leaving ability of halide ions decreases in the order I① > Br① > Cl① > F①.
The following sulfonate ions are good leaving groups like halides because these are all
conjugate bases of strong acids:
The following fluorinated alkanesulfonate ions are very good leaving group (‘super’ leaving
groups) because these are conjugate bases of very strong acids (pKa ≈ < –10):
:O: :O: :O:
� � �
F3C — S — O �� :* CF3 (CF2 )3 — S — O�� :* CF2CH 2 — S — O �� :*
� ��
:O: :O: :O:
(OTf *)
Trifluoromethane- Nonafluorobutane- 2,2,2-Trifluoroethane
sulfonate ion sulfonate ion sulfonate ion
(Triflate ion) (Nonaflate ion) (Tresylate ion)
Strongly basic ions (e.g., OH①) rarely act as leaving groups and very powerful bases such
as hydride ion (H①) and alkanide ions (:R①) virtually never acts as leaving groups.
An SN2 reaction proceeds in the direction that allows the stronger base (the nucleophile)
to displace the weaker base (the leaving group). Therefore, reactions such as the following
do not take place:
– –
Nu: + CH3—H CH3—Nu + H:
– –
Nu: + CH3—CH3 CH3—Nu + :CH3
(a) SN2 reactions are usually irreversible For example, the hydrolysis of CH3Cl with
aqueous NaOH is essentially irreversible.
HO�� :@ + CH — Cl ææ Æ HO — CH 3 + Cl��:@
3 ¨æ¥æ
��
We can correctly predict the direction of equilibrium by comparing the basicity of the
nucleophile and the leaving group. Equilibrium favours the products of an SN2 reaction
when the leaving group is a weaker base than the nucleophile. In this reaction, the basicity
of the nucleophile (OH①) and the leaving group (Cl①) may be compared by comparing the
pKa values of their conjugate acids. The stronger the conjugate acid, the weaker the base
and the better the leaving group.
Nucleophile OH
① Leaving group Cl–
(Stronger base) (Water base)
Conjugate H2O HCl
acid (Weaker acid) (Stronger acid)
pKa 15.7 –7
Because Cl①, the leaving group, is a weaker base then OH①, the nucleophile, the reaction
favours the product, i.e., the reaction is essentially irreversible.
If the difference between the basicities of the nucleophile and the leaving group is not very
large, the reaction will be reversible. In the reaction of ethyl bromide with iodide ion, Br①
is the leaving group in the forward direction and I① is the leaving group in the backward
direction. Because the pKa values of the conjugate acids of the two leaving groups are
similar (pKa of HBr = –9; pKa of HI = –10), the reaction is reversible.
CH3CH2 Br + K ≈ I* � CH3CH2I + K ≈ Br *
A completely reversible reaction is as given below:
When (+)-2-iodobutane is allowed to react with iodide ion (I①) in acetone, instead of only
(–)-2-iodobutane (expected to be obtained by inversion of configuration) an optically inactive
equimolar mixtrue of (+)- and (–)-2-iodobutane is obtained. This observation suggests
that the reaction is completely reversible. Each SN2 attack by I① on (+) -2-iodobutane
leads to the formation of (–)-2-iodobutane, a product with opposite relative configuration.
However, complete conversion of the (+)-enantiomer into the (–)-enantiomer does not take
place because the (–)-enantiomer also undergoes SN2 attack by I① with equal facility.
In fact, as the concentration of the (–)-enantiomer in the reaction mixture increases, its
tendency to react with I① increases. A time comes when the mixture contains an equimolar
amount of (+)- and (–)-2-iodobutane and since both the enantomers then react with I① at
the same rate (identical Ea), the composition remains unchanged. Hence, (+)-2-iodobutane
reacts with I① in acetone to yield optically inactive racemic or(±)-2-iodobutane instead
only (–)-2-iodobutane.
The energy diagram for this reversible reaction is as follows:
T.S.
Free energy of activation
is the same for the forward
and the backward reaction
Free energy
DG
(+)-2-Iodobutane (–)-2-Iodobutane
+ I– + I–
(b) A poor leaving group converts to a good one by coordination with an electrophilic
species Hydroxide ion, alkoxide ions and other strong bases are poor leaving groups for
SN2 reactions. For example, the — OH group of an alcohol is a poor leaving group because
it would have to leave as hydroxide ion (OH ).
①
– –
Br + CH3—OH Br—CH3 + OH (strong base)
Ions that are strong bases are poor leaving groups:
*
: OH
�� * : OR
�� * :N
�� H2
��
Hydroxide ��
alkoxide ��
amide
Some neutral molecules like H2O, ROH, R2NH, etc. are good leaving groups. A neutral
molecule often severs as the leaving group from a positively charged electrophile.
For example, if an alcohol is placed in an acidic solution, the hydroxyl group becomes
protonated. Water then serves as the leaving group. It is to be noted that the need to
protonate the alcohol (requiring acid) limits the choice of nucleophiles to those few that
are weak bases, such as bromide and iodide. A strongly basic nucleophile would become
protonated in an acidic medium.
SN2
There are two main reasons for smooth SN2 reaction: (a) Br① is now attacking a positively
charged, an opposed to neutral species, and (b) the very weakly basic H2O is a very much
better leaving group than the strongly basic OH① ion.
The well-known use of HI is to cleave ethers. I①, generated in a strongly acidic solution,
is a very good nucleophile. The strong acid initiates the reaction by protonation of the
—O�� — group. For example, the cleavage of anisole by HI is as follows:
��
SN2
1. Does the rate of SN2 reaction depend on the amount of d+ on the attacked
Csp3 of R–X? Explain.
Solution The rate of an SN2 reaction does not depend an the amount of d+ on the attacked
Csp3 of R — X. The amount of d+ on the Csp3 decreases with decreasing electronegativity

of X in the order F > Cl > Br > I. Therefore, RF is expected to be the most reactive and RI
is the least. However, the reactivity order is RI > RBr > RCl > RF. The reactivity actually
depends on leaving ability of X①.
2. tert-Butylamine cannot be prepared by Gabriel synthesis — Why?
Solution The Gabriel synthesis of an 1° amine (usually with 1° R) may be given as follows:
O O O O
C C C C NH
KOH R—X NH2NH2
NH NK SN2
NR EtOH/D
H2O NH
C C C C
O O O O
Phthalimide Potassium N-Alkylphthalimide + RNH2
phthalimide Phthalimide
hydrazide
To prepare tert-butylamine (Me3C — NH2) we have to use tert-butyl halide (Me3C — X)

in step II. A tertiary substrate cannot be used in this step because it is an SN2 reaction
which, due to steric hindrance, does not take place. It is for this reason, tert-butylamine
cannot be prepared by Gabriel synthesis.
O
C
+ SN2
NK C—X No reaction
C
O
Potassium A tert-butyl halide
phthalimide
3. Explain the following observation
1. TsCl/Py
2. LiBr / acetone
CH2OH CH2Br
Solution In the first step, the bicyclic alcohol undergoes tosylation by TsCl in the presence
of pyridine and as a result, the — OH group becomes a good leaving group. Although the
intermediate tosylate is a neopentyl type of substrate, displacement of OTs① by Br① occurs
under SN2 conditions in the second step to form the corresponding bromide. This is because
the groups on the 3° bridgehead carbon are tied back and do not interfere sterically with
the incoming nucleophile Br①.
4. Cleavage of ordinary ethers does not take place in the presence of base.
However, basic regents may cause ready opening of epoxides. Explain
these observations.
Solution An ordinary ether (R — O — R) does not undergo cleavage under basic conditions
because that SN2 reaction would result in expulsion of strongly basic alkoxide ion (RO①)
which, in fact, is a very poor leaving group.
HO + R—OR R—OH + RO
A nucleophile An ether An alcohol An alkoxide
(a strong base and a
poor leaving group)
Epoxides, on the other hand, suffers from significant ring strain (about 25 kcal/mol) and
the strain is released when the ring opens up. For this reason, epoxides undergo ring-
opening reactions (SN2) under basic conditions although the leaving group in this case also
an alkoxide ion which actually remains within the same molecule. The strain is enough
to compensate for the poor alkoxide leaving group. In fact, the ring opening has a lower
energy of activation because an epoxide is 25 kcal/mol higher in energy than an ordinary
ether.
O O
SN2
HO + C——C —C——C—
OH
A nucleophile An epoxide An alkoxide ion
(suffers from (a strong base)
angle strain)
5. Are SN2 reactions stereospecific and/or stereoselective?

Solution SN2 reactions are stereospecific because stereoisomeric reagents give stereo-
chemically different products. Also, they are stereoselective because they form exclusively
only one of a possible pair of enantiomers or one of the possible diastereoisomers.
6. Predict the product and suggest a mechanism of the following reaction:
14 NaOMe (1 mole)
H2C——CH——CH2Cl
MeOH
O
Solution The epoxide first undergoes SN2 attack on the less crowded labelled carbon of
the ring. The resulting alkoxide then undergoes intramolecular SN2 reaction to form again
an epoxide in which the labelled carbon is not a part of the ring. This labelling experiment
proves that the reaction is not a simple SN2 displacement of Cl① by OMe①.
Cl
14 SN2 14 SN2 14
Me O + CH2—CH—CH2Cl MeO—CH2—CH—CH2 MeO—CH2—CH—CH2
O O O
7. Ethers can be easily cleaved by acids like HBr and HI. However, HCl
cannot be used for this purpose. Explain these observations.
Solution In the presence of strong acids like HBr or HI, ether undergoes protonation and
as a result, the very poor leaving group RO① becomes converted into a very good neutral
leaving group ROH. The protonated ether then undergoes ready SN2 attack at the less
crowded alkyl group by the good nucleophiles Br① or I① to yield an alcohol and an alkyl
halide. For example:
HCl cannot be used for ether cleavage because it is not as strong an acid such as HI or HBr
to protonate ether sufficiently and also Cl① is a very poor nucleophile.
8. Ethers can be used safely as a solvent during Grignard synthesis — Why?
Solution Being a very strong base the leaving group of ether (R—O—R), i.e., RO① is
a very poor leaving group. Consequently, ethers are unreactive towards nucleophilic
substitution and so, they cannot be cleaved by base, i.e., the carbanion R① (from RMgX)
does not cause cleavage of ROR.
R¢—MgX + R—O—R R¢—O—R + RO MgX
Grignard reagent
(source of :R¢ )
9. Give SN1 and SN2 mechanisms for the cleavage of ethers with HI.
Solution
SN1 mechanism: If the departure of the leaving group (an alcohol) creates a relatively
stable carbocation (e.g., a 3° carbocation), i.e., if R in ROR¢ is 3°, an SN1 reaction occurs.
The steps involved are as follows:
Step 1: Protonation of ether oxygen by HI.
R—O—R¢ + H—I R—O —R¢ + I

Step 2: Departure of the leaving group
R—O —R¢ R + R¢OH

(3°)
Step 3: Nuclephilic attack by I① on the carbocation
R + I R—I
SN2 mechanism: If departure of the leaving group would create an unstable carbocation
(e.g., methyl cation or a primary carbocation), the leaving group cannot depart. It has to
be displaced by the halide ion by an SN2 process. The halide ion preferentially attack that
alkyl group which is less sterically hindered. The steps involved are as follows:
Step 1: Protonation of ether oxygen by HI
R—O—R¢ + H—I R—O —R¢ + I
Step 2:
SN2
I + R—O —R¢ R—I + R¢OH
(R = methyl or 1°)
10. Predict the product(s) and suggest a mechanism for each of the following
reactions of ether cleavage:
(a) + HBr (excess)

O
(b) PhOCH2CH2CH(CH3 )CH2OCH2CH3 + HBr (excess) ææÆ
1. BBr
(c) R — O — R ¢ ææææ 3
2. H O
Æ
2
(R == CH 3 or 1∞)
Solution
(a) H—Br SN2

—Br
H—Br
Br Br + — Br
O O OH OH2
SN2
Br — — Br
—H2O
(b)
(c)
11. Explain why methyl propyl ether reacts with HI to yield 1-propanol and
methyl iodide while methyl tert-pentyl ether is cleaved by HI to yield
methanol and tert-pentyl iodide.
Solution If the presence of HI, the conjugate acid of a dialkyl ether (R — O — R) undergoes
nucleophilic attack by I① on the less crowded alkyl group. It thus follows that the conjugate
acid of methyl propyl ether undergoes SN2 attack by I① on the methyl carbon to give
methyl iodide and 1-propanal.
Step 1: CH3—O—CH2CH2CH3 + H—I CH3—O—CH2CH2CH3 + I

H
Methyl propyl ether
CH2CH3
Step 2: I + C —O—C SN2 CH3—I + CH3CH2CH2OH

Methyl iodide 1-Propanol
less crowded more crowded

centre centre
If, however, the protonated ether can cleave to give a stable carbocation, the reaction takes
place by the SN1 mechanism. The conjugate acid of methyl tert-pentyl ether, therefore,
cleaves to give methanol and stable tert-pentyl cation which subsequently reacts with I①
to give tert-pentyl iodide.
Step 1: CH3—O—C(CH3)2CH2CH3 + H—I CH3—O—C(CH3)2CH2CH3 + I

H
SN1
Step 2: CH3—O:—C(CH3)2CH2CH3 CH3OH + C(CH3)2CH2CH3
H
Methanol A 3° carbocation
(stable)
CH3
Step 3: I + C(CH3)2CH2CH3 CH2CH2C—I
CH3
tert-pentyl iodide
12. Arrange the following alkyl bromides in order or decreasing reactivity as
a substrate in an SN2 reaction and explain the order.
Br Br
Br
Br
I II III IV
Solution To assess the reactivity of these alkyl bromides, we have to examine the
steric hindrance to an SN2 reaction at the carbon bearing the leaving group. In I, it is
3° and, therefore, three groups would hinder the approach of a nucleophile towards the
carbon bearing bromine. So, this alkyl bromide would react so slowly that it seems to be
unreactive. In II, the carbon bearing the leaving group is 2°, therefore, two groups would
hinder the approach of the nucleophile. In III and IV, the carbon bearing bromine is 1° and
so, only one group would hinder the nucleophile’s approach. Therefore, II is expected to
react at a rate faster than I but slower than III and IV. Now, III has two methyl groups on
the b-carbon (the carbon adjacent to the one bearing bromine), which provide more steric
hindrance to the approaching nucleophile, while IV has only one alkyl substituent on the
b-carbon which provides less steric hindrance to the approaching nucleophile. Hence, IV
reacts at a rate faster than III. The order of decreasing reactivity, therefore, is IV > III >
II > I.
13. Explain why each of the following ethers cannot be prepared by the
Williamson synthesis:
(a) Me3C — O — CMe3 (b) Me3C — O — CHMe2
(c) Et2CH — O — CHEt2 (d) Ph — O — Ph
(e) MeCH == CH — O — CH == CHMe, and (f) Me3CCH2OCH2CMe3
Solution The Williamson ether synthesis consists of an SN2 reaction of a sodium alkoxide
with an alkyl halide, alkyl sulfonate or alkyl sulfate. Because of steric reason, methyl
and primary substrates are the best. Secondary and tertiary substrates are not suitable
because steric hindrance causes them to undergo E2 elimination rather than substitution.
Neopentyl type substrates are also inert due to steric reason. Aryl and vinyl halides are
inert because backside attack does not take place due to repulsive interaction caused
by the p electron cloud, and the C — LG bond in them is shorter and stronger because it
possesses considerable double bond character due to electron delocalization.
Therefore, the given ethers cannot be prepared by the Williamson synthesis because in
(a) both alkyl groups are tertiary, in (b) one alkyl group is tertiary and the other is
secondary, in (c) both alkyl groups are secondary, in (d) both groups are aryl, in (e) both
groups are vinyl and in (f) both groups are neopentyl.
14. Predict the product and suggest a mechanism for the following reaction:
Solution The reaction of cis-1-chloro-3-methylcyclopentane (in which the carbon that

bears the leaving group is chiral) with hydroxide ion (OH①) is an SN2 reaction which proceeds
with complete inversion of configuration. The product is trans-3-methylcyclopentanol. The
reaction occurs as follows:
Cl
H3 C Cl SN2 H3C H3C H
+ OH H
H H H H OH
cis-1-Chloro-3-methyl OH
cyclopentane
Transition state trans-3-Methylcyclopentanol
15. A cyclic ether can be prepared by an intramolecular Williamson synthesis.

Give the mechanism of the following reaction and explain why it is carried
out at high dilution.
CH2 CH2 CH3 Na
Cl CH2 C high dilution CH3
CH3
OH O CH3
Solution The mechanism of the reaction is as follows:
The reaction is carried out at high dilution because under such conditions the intermolecular
distance becomes greater than the intramolecular reactive centres and as a consequence,
the cyclic compound is formed instead of polymeric product (a likely result in concentrated
solution) as given below.
Cl SN2 O
Cl O Na Cl ONa
O Na
etc.
16. 18–Crown–6 may be prepared by treating a mixture of triethylene
glycol and the corresponding dichloride with aqueous KOH. Suggest a
mechanism for the reaction.
Solution The reaction proceeds through the SN2 mechanism and involves two successive
displacements with chloride ion being the leaving group. Even though, a large ring is being
formed, the reaction needs not be carried out at high dilution. After the initial alkylation,
i.e., formation of ether linkage at one end, the K≈ ion apparently acts as a ‘template’ to
bring the two reacting ends of the long chain close together for rapid reaction.
The mechanism of the reaction is as follows:
17. Predict the product in each of the following SN2 reactions and give your
reasoning:
1 3 5
PhS Na
(a) Br 2
4
Br
EtOH
1 3 5
PhS Na
(b) I 2
4
I
EtOH
1 3 5
PhS Na
(c) Br 2
4
I
EtOH
Solution
(a) The dibromide undergoes ready SN2 attack by the nucleophile PhS① at the
sterically less crowded primary site C-1. Nucleophilic attack does not take place
at the sterically more crowded tertiary site C-4. Hence, the halogen at C-1 is more
reactive than the halogen at C-4 in an SN2 reaction and the product corresponding
to such displacement obtained.
PhS /EtOH
Br PhS + Br
SN2
Br Br
(b) Both C-1 and C-5 are primary sites and both are expected to be equally favourable
for SN2 attack from steric point of view. However, the SN2 attack at C-1 is more
favourable because the corresponding transition state is relatively more stable
(stabilized by the adjacent double bond by p orbital overlap). Therefore, the iodine
atom at C-1 is more reactive than the iodine atom at C-5 in an SN2 reaction and the
product corresponding to that of the reactive centre is obtained predominantly.
PhS / EtOH
I I PhS I + I
SN2
(c) Both of the carbons bearing halogen (i.e., C-1 and C-5) are primary sites and
hence, they are equally favourable sites for SN2 attack from steric point of view.
However, since I① is a better leaving group than Br①, the halogen at C-5 is more
reactive. Nucleophilic attack, therefore, takes place preferably at C-5 to yield the
corresponding product predominantly.
PhS / EtOH
Br I Br S Ph + I
SN2
18. Since SN2 reactions proceed with inversion of configuration, an (R)–

compound always produce and (S)–compound and an (S)–compound
always produce an (R)–compound. Comment on this statement and give
example in favour of your answer.
Solution This statement is not correct. In an SN2 reaction of a chiral substrate, the
relative configuration changes but the absolute configuration may or may not change
since the absolute configuration depends on the priority of the groups or atoms present
on the chiral carbon. For example, (S)-1-chloro-1-iodoethane reacts with Br① to give (R)-
1-bromo-1-chloroethane, while (S)-1-bromo-1-iodoethane reacts with Cl① to give (S)-1-
bromo-1-chloroethane.
19. Comment on the stereochemical course involved in each step of the

following reaction sequences and give three-dimensional structures (with
R or S designation) for the compounds I — XIX:
CH3
NaN H2 CH I
(a) C ææææ 3
Acetone
Æ I æææ
Pt
Æ II æææ
3
Æ III
Ph
Br
H
(S)-1-Bromo-1-phenylethane
CH3
CH CH O Na HI
(b) C ææææææ
3 2
CH CH OH
Æ IV æææ ÆV
3 2
H
Br
CH2CH2CH3
(R)-2-Bromopentane
TsCl CH COOK KOH/ H O

(c) ææææææ
Pyridine, 25∞ C
Æ VI æææææ
3
Acetone
Æ VII æææææ
D
2
Æ VIII
PBr NaCN H SO 1. LiAlH

(d) C ææææ 3
Pyridine
Æ IX ææææ
DMSO
Æ X ææææ
2
H O
4
Æ XI æææææ
≈ Æ XII
4
CH3 2 2. H3O
HO
C2H5
(R)-2-Butanol
OH
C
(e) CH3(CH2)4 H
CH3
(S)-2-Heptanol
TsCl NaN H / Pt NaNO / ACOH
ææææ
Pyridine
Æ XIII ææææ 3
Acetone
Æ XIV ææææ
2
Æ XV ææææææ
2
0 - 5∞ C
Æ XVI + XVII
C Dioxan
(f) H3C OBs æææææ
H O/ NaN
Æ XVIII + XIX
2 3
(CH2)5CH3
(S)-2-Octyl brosylate
Solution (a) Step 1: Being a secondary substrate, (S)-1-chloro-1-phenylethane reacts

with the powerful nucleophile N3 (azide ion) in the aprotic polar solvent acetone by the
①
SN2 mechanism to yield (R)-1-azido-1-phenylethane (I) with inversion of configuration.

CH3 CH3
N3 NaN3 / acetone
C C
Ph Br (inversion of configuration) Ph + Br
:N==N==N
H H
(S)-1-Bromo-1-phenylethane I
(R)-1-Azido-1-phenylethane
Step 2: The azide I on catalytic hydrogenation produces the corresponding amine II.
Since the bond between the chiral carbon and nitrogen is not cleaved in this reaction, the
configuration of the chiral centre is retained.
CH3 CH3
H2 / Pt
C C
Ph (retention of configuration) Ph + N ≠
:N==N==N H2 N 2
H H
I II
(R)-1-Amino-1-phenylethane
Step 3: Since methylation of the amine II does not involve reaction at the chiral carbon,
no change in configuration takes place in this step.
CH3 CH3
CH3—I
C C
Ph SN2 Ph + HI
H2N CH3NH
(retention of configuration)
H H
II III
(R)-N-Methyl-1-phenyl-1-ethanamine
(b) Step 1: When (R)-2-bromopentane is treated with ethoxide ion in ethanol, it undergoes
an SN2 reaction to yield (S)-2-ethoxypentane with inversion of configuration.
CH3 CH3
C2H5O C2H5O /C2H5OH
C C
H SN2 H + Br
Br C2H5O
(inversion of configuration)
H CH2CH2CH3
IV
(R)-2-Bromopentane (S)-2-Ethoxypentane
Step 2: In this step, the ether IV undergoes cleavage by HI. The conjugate acid of the
ether undergoes SN2 attack at the less crowded methylene carbon of the ethyl group. In
the resulting alcohol V, the configuration is retained.
(c) Step 1: Since tosylation of this chiral alcohol does not involve reaction at the
asymmetric carbon no change in configuration occurs in this step.
Step 2: Being a primary substrate the tosylate VI undergoes SN2 displacement of OTs①
(a good leaving group) by CH3COO① in acetone to yield (R)-1,2-dideuterioethyl acetate
(VII) with inversion of configuration.
Step 3: The alkaline hydrolysis of the acetate VII is known not to involve cleavage of
the bond joining the acetate group to the chirality centre. Therefore, the alcohol VIII must
have the same configuration as the acetate VII.
(d) Step 1: (R)-2-Butanol reacts with phosphorous tribromide (PBr3) to yield a phosphite
ester with retention of configuration. The phosphite ester being a secondary substrate
with a good leaving group (phosphoryl group, P == O) undergoes SN2 attack by Br① (a good
nucleophile to yield (S)-2-bromobutane (IX) having configuration opposite to that of the
phosphite ester.
H H H
N –
C + PBr2 C C + Br
H5C2 H5C2 + – H5C2 + N
OH O P Br2 Br O PBr2 H
Br
CH3 CH3 CH3
H
Step 2: The bromide IX (a secondary substrate) reacts with CN① (a very powerful
nucleophile) in the aprotic polar solvent DMSO by an SN2 mechanism to give
(R)-1-methylbutanenitrile (X) with inversion of configuration.
Step 3: Hydrolysis of the nitrile X produce (R)-2-methylbutanoic acid (XI). Since the
bond joining the — CN group to the chiral carbon is not cleaved during hydrolysis, the
configuration of the chirality centre is preserved.
Step 4: The reduction of the carboxylic acid XI by LiAlH4 to yield the corresponding
1° alcohol does not involve reaction at the chiral carbon. Hence, the configuration of the
chirality centre remains unchanged in this case also.
(e) Step 1: Since tosylation of the alcohol does not involve reaction at the chiral carbon,
the tosylate XIII has the same configuration as the original alcohol.
Step 2: The tosylate XIII undergoes SN2 attack by the powerful nucleophile N3① (azide
ion) in the aprotic polar solvent acetone to yield (R)-2-azidoheptane (XIV) with inversion
of configuration.
H H
N3
O
C NaN3 / acetone C
CH3 + OTs
H3C O—S— — CH3 :N∫∫N—N
SN2
(CH2)4CH3 O (inversion of configuration) (CH2)4CH3
Cl
XIII XIV
(R)-2-Aziodoheptane
Step 3: The azide XIV on catalytic hydrogenation produces (R)-2-Heptanamine (XV).
Since hydrogenation does not involve cleavage of the C — N bond, the configuration
remains unchanged.
H H
H2 / Pt
C C
:N∫∫N—N CH3 (retention of H2N CH3
configuration)
(CH2)4CH3 (CH2)4CH3
XIV XV
(R)-2-Heptanamine
Step 4: When the aliphatic amine XV is treated with nitrous acid (obtained from NaNO2
plus HCl), an unstable diazonium salt is obtained which readily dissociates to form a
carbocation. The carbocation undergoes nucleophilic attack by water from either side with
equal facility to give, after proton loss, an equimolar mixture of (R)- and (S)-2-heptanol.
(f) (S)-2-Octyl brosylate undergoes SN2 attack by the more nucleophilic solvent dioxane
to give an intermediate oxonium ion. The unstable intermediate reacts readily with water
in another SN2 process to from (S)-2-octanol (XVIII) with the same configuration as the
starting brosylate. The azide ion reacts similarly with the oxonium ion to yield (S)-2-
azidooctane (XIX). In both the cases, the retention of configuration has been brought about
by two successive inversions.
20. Arrange the following compounds in order of decreasing reactivity

�� :* in an S 2 reaction carried out in methanol:
towards CH 3 O N
��
CH3Cl, CH3I, CH3OH, CH3F, 14CH3OSO2CF3, CH3Br
Solution The leaving groups of the substrates can be arranged in order of increasing
leaving ability as follows:
: OH
�� * �� :* < : Br:
�� :* < : Cl
<:F �� * < : ��I :* < CF SO O
3 2
�� :*
(most �� basic) ��
(least basic)
A better leaving group makes the substrate more reactive is an SN2 reaction. Therefore,
the given compound can be arranged in order of decreasing SN2 reactivity as follows:
14
CH3OSO2CF3 > CH3I > CH3 Br > CH3Cl > CH3 F > CH3OH
(most reactive) (least reactive)
21. Explain the observed effects of increasing solvent polarity (solvent

ionizing power) on the rates of the following SN2 reactions:
�� * Æ CH OH + : Cl
(a) CH3Cl + : OH �� :* (a small decrease in the reaction rate)
3
��
≈ ≈
(b) CH 3 — N(CH 3 )3 + H 2S: Æ CH 3 — SH 2 + N(CH
��
3 )3
(a large increase in the reaction
��
rate)
≈
:��: (a large increase in the reaction rate)
�� *
(c) CH 3I + N(CH 3 )3 Æ CH 3 — N(CH 3 )3 + I
��
≈
(d) CH 3 — N(NH 3 )3 + :OH Æ CH 3OH + N(CH 3 )3 (a large decrease in the reaction
�� * ��
��
rate)
Draw a comparative energy profile diagram to show the effect of solvent
on the rate of attack by OH① ion on CH3Cl.
Solution
(a) This SN2 reaction proceeds through the transition state as follows:
HO + CH3—Cl HO CH3 Cl HO—CH3 + Cl

T.S.
In this reaction, an initial charge is dispersed in the transition state. Therefore,
increasing solvent polarity stabilizes the reactants (specifically the negatively
charged nucleophile) more than the transition state. Because of this differential
stabilization, the difference in energy between the transition state and the
nucleophile increases, i.e., the energy of activation (Ea) increases. The net result is
that there is a slight decrease in the reaction rate.
(b) This SN2 reaction proceeds through the transition state as follows:
In this reaction also, an initial charge is dispersed in the transition state. Therefore,
increasing solvent polarity stabilizes the reactants (specifically the positively
charged substrate) more than the transition state. As a result of this, the difference
in energy between the transition state and the substrate increases, i.e., the energy
of activation (Ea) increases. The net result is that there is slight decrease in the
reaction rate.
(c) The reaction proceeds through the transition state as follows:
N(CH3)3 + CH3—I (CH3)3N CH3 I (CH3)4 N + I

T.S.
As both the substrate and the nucleophile are neutral, a charge is expected to be
developed and concentrated in the transition state. Solvation is, therefore, much
greater for the transition state than for the reactants. Increasing solvent polarity
stabilizes the transition state much more than the reactants and thereby decreases
the energy of activation (Ea) considerably. The net result is a large increase in the
reaction rate.
(d) The reaction proceeds through the transition state as follows:
Both of the reactants are charged and the initial charges are decreased in the
transition state. Increasing solvent polarity, therefore, stabilizes the reactant
much more than transition state and thereby increases the energy of activation
(Ea) considerably. The net result is a large decrease in the reaction rate.
A comparative energy diagram showing the effect of solvent on the rate of the
alkaline hydrolysis of CH3Cl is as follows:
22. Which one would you expect to be the stronger nucleophile in ethanol
and why?
(a) �� * or NH
:NH �� (b) n-C4 H9 O �� :* or tert-C H O �� :*
2 3 4 9
��
(c) �� or Ph P
Ph3N �� (d) ��
CH 3O : *
or CH 3S �� : *
3
��
≈
�� :*
p-CH3OC6 H4 O or p-NO2C6 H4 O �� :*
(e) CH 3NH 2 or CH 3 NH 3
(f)
��
(g) —O or —CH2O
Solution
(a) Anions are always stronger nucleophiles than their conjugate acids. The anion
��
�� * is, therefore, more nucleophilic than its conjugate acid NH
:NH 2 3.
(b) Because of steric hindrance, tert-butoxide ion is a weaker nucleophile than

n-butoxide ion, even though the former ion is a stronger base. The carbon adjacent
to the negative oxygen atom is relatively less crowded in n-butoxide than in tert-
butoxide. Consequently, the approach of n-butoxide towards C 3 in an SN2 reaction
sp
is more easier than the approach of the very bulky tert-butoxide and therefore,
the activation energy of the reaction involving n-butoxide is relatively low. Hence,
n-butoxide ion is a stronger nucleophile than tert-butoxide ion.
CH3CH2CH2 H3C
C——O C——O
H H3C
H CH3
n-Butoxide ion tert-Butoxide ion

(stronger nucleophile) (weaker nucleophile)
(c) Phosphorous is less electronegative and larger in size than nitrogen; so, it is
�� is relatively less solvated than Ph N
relatively more polarizable. Also, Ph 3 P �� in the
3
protic solvent. For these reasons, Ph3P (triphenylphosphine) is more nucleophilic
than Ph3N (triphenylamine).
(d) Because of low electronegativity and larger size, sulphur is more polarizable than
oxygen. Furthermore, a protic solvent solvates CH3O① (through H-bonding) better
than CH3S①. The more polarizable and weakly solvated CH3S① is, therefore, a
better nucleophile than CH3O①.
(e) Due to the presence of an unshared pair of electrons on nitrogen, CH3 NH ��
3
can act
≈
as a nucleophile. However, CH3 NH3 is nonnucleophilic because the unshared pair
of electron on nitrogen is involved in the formation of a coordinate bond and is no
longer available to exhibit nucleophilic activity.
CH3 NH2 + H CH3N H or CH3NH3
(f) The stronger base is expected to be the better nucleophile because the attacking
atom is the same in these two anions. In p-CH3OC6H4O①, the CH3O-group
increases the electron density of the phenoxy oxygen by its +R effect (by inhibiting
the interaction of its negative charge with the delocalized p electrons of the ring).
However, in p-NO2C6H4O①, the nitro group reduces the electron density of the
negative oxygen by its — R and — I effects. The more stable anion p-NO2C6H4O①
is, therefore, less basic and hence, less nucleophilic than p-CH3OC6H4O①.
(g) In phenoxide ion (Ph — O①), the negative charge on oxygen is delocalized with the
ring p electron. However, in the alkoxide of benzyl alcohol, i.e., in PhCH2O①, the
negative charge cannot be delocalized with the ring p electrons due to the presence
of saturated carbon between oxygen and the ring. Therefore, PhO① is less basic
than PhCH2O① and hence, PhO① is less nucleophilic than PhCH2O①.
23. Arrange the following ions/molecules in decreasing order of nucleophilicity
in protic solvents. Give reasons.
*
�� :* , : OH
�� * , C H O �� *
: CH* (b) CH3 CH2COO 2 5 :
: �� * : �� :
3 , OH , N H2 , H2 O
(a)
��
(c) �� HOO
H2O, �� :* , : OH
�� * �� : , : OH
(d) CH3OH, PhO * �� * �� :* , CH COO
, CH3S �� :*
3
��
Solution
(a) Since the atoms across a period of the periodic table have approximately the same
size, the basicity and nucleophilicity run in parallel (stronger bases are better
nucleophiles). Since the basicity of the given anions (the attacking atoms are
*
elements of Period 2) decreases in the order :CH * : �� : �� *
3 > NH 2 > OH , the nucleophilicity
��
decreases in the same order. Again, a negatively charged nucleophile is always a
more reactive nucleophile than its conjugate acid. Thus, OH① is a better nucleophile
than H2O. Hence, the decreasing order of nucleophilicity of the given species is
@
:CH * : �� : �� *
3 > NH 2 > OH > H 2O .
�� :
��
(b) For reagents having the same attacking atom, nucleophilicity and basicity run in
parallel, i.e., stronger bases are better nucleophiles. A stable anion (the conjugate
base of a strong acid) is less basic and is, therefore, less nucleophilic. In C2 H5 O�� :*
��
ion, the negative charge on oxygen is intensified by the +I effect — C2H5 group;
this ion is, therefore, less stable and more basic than :OH�� * ion. Because of charge
��
�� : ion is much
delocalization by resonance, the C2 H5COO *
more stable than :OH
�� *
��
and so, it is less basic and less nucleophilic than :OH
�� *
. Hence, the nucleophilicity
��
of these ions decreases in the order C2 H5 O �� :* > :OH

�� * > C H COO
2 5
�� :*
��
(c) Although the attacking atom is the same, nucleophilicity and basicity do not run
in parallel in this series. Bases are better nucleophiles than their conjugate acids.
Therefore, :OH
�� * ion is more nucleophilic than H O
2
�� : . Again, when the attacking
��
atom is bonded to an atom containing an unshared electron pair, the nucleophilicity
of the species increases (a-effect). Therefore, HOO �� :* is more nucleophilic than
�� * , even though it is less basic. Hence, the ��nucleophilicity
:OH
��
of these species
��
decreases in the order HOO : > HO : > H 2O:.
�� * �� * ��
��
(d) Since the sulphur atom is larger and more polariable than the oxygen atom, CH3S �� :*
is the strongest nuclephile among these five species. CH3OH, containing neutral ��
oxygen atom, is the poorest nucleophile. The decreasing order of basicity of the three
other reagents having the same attacking atom is OH �� :* > PhO �� :* > CH COO
3
�� :* (a
��
carboxylic acid is more acidic than phenol which in turn is more acidic than water). ��
Therefore, the decreasing order of nucleophilicity is OH �� :* > PhO �� :* > CH COO
3
�� :*.
��
Thus, the given species can be listed in the order of decreasing nucleophilicity in a
protic solvent as follows:
CH3S�� :* �� :* > PhO

> HO �� :* > CH COO
�� :* > ��
CH3OH
3
��
(most nuclephilic) ��
(least nucleophilic)
24. Imidazole increases the rate of hydrolysis of phenyl acetate. Explain this
observation.
O O
C C
CH3 OPh + H2O CH3 OH + PhOH (slow reaction)
O O
C :N NH
C
CH3 OPh + H2O CH3 OH + PhOH (fast reaction)
Imidazole
Phenyl acetate Acetic acid Phenol
Solution Imidazole is a better nucleophile than water, so it reacts at a rate faster

than water with the ester phenyl acetate. The resulting acyl imidazole is particularly
reactive because the positively charged nitrogen makes imidazole a very good leaving
group. Therefore, it undergoes hydrolysis much more rapidly than the ester. Because
the formation of acyl imidazole and its subsequent hydrolysis are both faster than the
hydrolysis of ester, imidazole increases the rate of the overall reaction. The mechanism of
the reaction is as follows:
25. The rate of SN2 reaction CH3CH2Cl + OH① CH3CH2OH + Cl① is given by the
following equation:
rate = k [CH3CH2Cl] [OH①]
What happens to the rate of the reaction under each of the following
conditions?
(a) [CH3CH2Cl] is halved and [OH①] is doubled.
(b) [CH3CH2Cl] is tripled and [OH①] remains the same.
(c) [CH3CH2Cl] is halved and [OH①] stays the same.
(d) Both [CH3CH2Cl] and [OH①] are doubled.
Solution
(a) The rate expression for the reaction involving half the concentration of CH3CH2Cl
and double the concentration of hydroxide ion is
[CH 3CH 2Cl]
rate = k ¥ 2 [OH * ]
2
= k[CH 3CH 2Cl][OH * ]
Hence, there will be no change of the reaction rate.
(b) The rate expression for the reaction involving triple the concentration of CH3CH2Cl
is
rate = k. 3[CH3CH2Cl] [OH①] = 3 k[CH3CH2Cl] [OH①]
Thus, the rate increases by a factor of three.
(c) The rate expression for the reaction involving half the concentration CH3CH2Cl is
[CH 3CH 2Cl]
rate = k ◊ ◊ [OH * ]
2
The rate thus becomes half of the original rate.

(d) The rate expression for the reaction involving double the concentration CH3CH2Cl
as well as OH① is
rate = k ∙ 2 [CH3CH2Cl] ∙ 2[OH①] = 4 k CH3CH2Cl] [OH①]
Thus, doubling the concentration of both reactants increases the rate by a factor of
four.
26. The nucleophilicity of F① in a weakly polar solvent like acetone increases
in the order: LiF < KF < CsF — Why?
Solution There is considerable ion pairing in a weakly polar solvent. As the alkali
metal cation gets smaller, ion-pairing to F① gets stronger and a result, its nucleophilicity
decreases. Since Cs≈ ion is the largest and Li≈ ion is the smallest cation, F① in CsF is the
most nucleophilic and F① in LiF is the least nucleophilic, i.e., the nucleophilicity of F①
increases in the order: LiF < KF < CsF.
27. List the following ions in decreasing order of leaving ability. Give your
reasoning.
O
H3C— —S—O , —O , O2N— —O
Solution The better leaving groups are those that become either a relatively stable
anion or a neutral molecule (i.e., an weaker base) when they depart. p-Toluensulfonate
ion (OTs①) is more stabilized by resonance than p-nitrophenoxide ion, which in turn is
more resonance–stabilized than the phenoxide ion. Therefore, the leaving ability of these
anions decreases in the order: p-CH3C6H4SO3① > p-NO2C6H4O① > C6H5O①.
28. Methylation of propaonate ion (CH3CH2COO①) can be carried out smoothly
by treating it not with methyl iodide, CH3I, but with dimethyl sulfate,
Me2SO4. Explain these observations.
Solution Methyl sulfate ion, CH3OSO3①, being a very weak base, acts as a much better
leaving group than I①. Because of this, methylation of propanoate ion occurs smoothly
when treated with Me2SO4 than that with CH3I.
29. Suggest a mechanism for each of the following reactions (methylation by

diazomethane):
(a) ether
CH 3CH 2COOH + CH 2N2 æææ
0∞ C
Æ CH 3CH 2COOCH 2 + N2
Propanoic acid Diazomethane Methyl propanoate
OH OCH3
(b) ether
+ CH2N2 + N2
0°C
Phenol Diazomethane Anisole
Al(OEt)3
(c) CH3CH2OH + CH2 N2 ææææææ
(a Lewis acid)
Æ CH3CH2OCH3 + N2
Ethanol Ethyl methyl ether
Explain why methylation of the alcohol requires a Lewis acid.
Solution
(a)
O
O C
ether
CH3CH2—C—O—H + :CH2—N∫∫N CH3CH2 O + CH3—N∫∫N
0°C
Propanoic acid —N2 SN2
O
C
CH3CH2 OCH3
Methyl propanoate
Diazomethane methylates is a compound only in the presence of a proton source.

≈
CH3 — N ∫∫ N is a very reactive methylating agent because of having an excellent leaving
group, gaseous N2.
(b)
(c)
Because of low acidity, alcohol cannot supply a proton to diazomethane. However, when
alcohol forms a complex with the Lewis acid Al(OEt)3, its hydroxyl proton becomes
considerably acidic and then methylation occurs smoothly.
30. For each of the following pairs of SN2 reaction indicates which takes place
with the larger rate constant:
(a) (CH3 )2CHCH2 Br + OH* or CH3CH2 CH Br + OH*
|
CH3
(b) CH3CH2 Br + EtOH �� or CH3CH2 Br + EtO �� :*
��
(c) CH3CH2Cl + CH3S �� :* or CH CH Cl + CH O* (both in ethanol)
3 2 3
��
CH3CH2Cl + : ��I : or CH3 Br + : ��I :
* *
(d)
��
(e) I CH2CH2CH2 NH ��
2 or I CH 2 CH ��
2 CH2 CH2 CH2 NH2
(f) �� :* or PhBr + MeO

Ph CH2 Br + MeO �� :*
��
(g) CH3CH == CBrCH3 + OH or CH3CH == CHCH BrCH3 + : OH
: �� * �� *
��
(h) CH3CH2CH2 Br (1.0 M) + MeO* (1.0 M) or CH3CH2CH2 Br (1.0 M) + MeO* (2.0 M)
Solution
(a) (CH3 )2CHCH2 Br + OH* (Because the substrate is 1∞ and there is only two
substituents at the b-position, i.e., the reaction is strerically more feasible.)
(b) CH3CH2 Br + EtO �� :* (The anion is more nucleophilic than its conjugate acid.)
��
(c) CH3CH2Cl + CH3S �� :* [The more polarizable (less electronegative and larger in
��
size) sulphur atom is more nucleophilic than less the polarizable oxygen.]
(d) CH3 Br + I* (CH3Br is a relatively less crowded substrate and Br is a better
①
leaving group compared to Cl①.)

(e) ��
I(CH2 )5 NH 2 (The intramolecular SN2 reaction leads to the formation of relatively
strain free six-membered ring.)
(f) PhCH2 Br + MeO* (Since the substrate is a benzylic halide, the unhybridized p
orbital involved in the transition state interacts with the p orbital system of the
ring and as a result, the transition state is stabilized. Bromobenzene does not
undergo backside attack due to repulsive interaction.)
Br
|
(g) CH3CH == CH — CH — CH3 + OH* (Since the substrate is an allylic halide, the
unhybridized p orbital involved in the transition state interacts with the p orbital
system of the double bond and as a result, the transition state is stabilized.
CH3CH == CBrCH3 being a vinylic halide does not undergo SN2 attack from the
backside due to repulsive interaction.)
(h) �� :* (2.0 M) (The concentration of the nucleophile is
CH3CH2CH2 Br(1.0 M) + MeO
��
higher.)
31. Explain why each of the following SN2 reaction is unsuccessful:
(a) EtS �� :* + CH — O — CH �� :*
CH3 — S — Et + CH3O
3 3
��
(b) : CN* + CH3CH2 C(CH3 )2 Br �� *
CH3CH2 C(CH3 )2 CN + : Br:
��
��
(c) :OH + CH3CH2CH3
*
CH3CH2CH2OH + : H *
��
(d) : ��I : + CH3 — CH3 CH3I + : CH3*
*
��
*
(e) : OH
�� * + ��
HOCH2CH2CH2 CH 2
��
(f) ��
PhS: + CH3CH2CH == CH — Br
* �� *
CH3CH2CH == CH — SPh + : Br:
��
≈ ≈
(g) ��
CH3 NH 2 + CH3 CH2 OH2
CH3 NH2CH2CH3 + H2O
Solution
(a) Since the very strong base CH3O① is a very poor leaving group, it cannot be displaced
by EtS�� :* .
(b) �� of steric hindrance, this tertiary substrate does not undergo backside
Because
attack required in an SN2 reaction.
(c) The reaction does not take place because the hydride ion, :H① (a very strong base),
is a very poor leaving group.
*
(d) Since methyl anion, :CH3 , is a very strong base and hence, a very poor leaving
group, it cannot be displaced by : ��I : .
*
��
(e) The reaction does not a take place because a carbanion (a very powerful base) is a
very poor leaving group.
(f) This vinylic substrate does not undergo SN2 attack by PhS
�� :* because the p electron
��
cloud of the double bond repels the nucleophile and inhibits its approach from the
backside of the leaving group.
�� ≈
(g) Since CH3 NH 2 is more basic than CH3CH2OH, it will take up H from the
*
conjugate acid of ethanol, i.e., CH3CH2 OH2 , and instead of an SN2 reaction, an
acid-base reaction will take place.
CH3NH2 + H—O—CH2CH3 CH3NH3 + CH3CH2OH

H
32. Which SN2 reaction of each pair is expected to take place at a faster rate?
≈
(a) �� ææÆ CH CH CH NPh + : Br
CH3CH2CH2 Br + Ph 3 N �� :*
3 2 2 3
≈
��
�� ææÆ CH CH CH PPh + : Br
CH3CH2CH2 Br + Ph 3 P �� :*
3 2 2 3
��
EtOH/ H2O
�� :* æææææ �� :*
(b) CH3CH2Cl + SCN Æ CH3CH2SCN + : Cl
EtOH/ H2O
��
�� :* æææææ
CH3CH2Cl + SCN �� :*
Æ CH3CH2 NCS + : Cl
��
H2 O
(c) CH3I + NaOH æææ Æ CH3OH + NaI
H O
CH3I + CH3COONa æææ
2
Æ CH3 COOCH3 + NaI
Br SPh
(d) + PhS + Br
Br SPh
+ PhS + Br
CH3OH �� :*
(e) CH3CH2CH2CH2 Br + N3* ææææ Æ CH3CH2CH2CH3 N3 + : Br
CH3CN
��
CH3CH2CH2CH2 Br + N3* ææææ �� :*
Æ CH3CH2CH2CH3 N3 + : Br
��
acetone �� :*
(f) CH3CH2 Br + CN* ææææ Æ CH3CH2CN + : Br
acetone
��
CH3CH2 Br + CN* ææææ �� :*
Æ CH3CH2 NC + : Br
��
Solution
(a) The second reaction is expected to take place more rapidly because
triphenylphosphine, Ph 3 P �� , is a stronger nucleophile than triethylamine, Ph N
3
��
(larger and less electronegative phosphorus atom is more polarizable than nitrogen
atom).
(b) SCN① is an ambident nucleophile in which sulphur and nitrogen are two nucleophilic
sites. Its less electronegative and larger sulphur atom is more polarizable (soft
base) while its more electronegative and smaller nitrogen atom is less polarizable
(hard base). Attack by SCN① on the carbon of CH3CH2Cl (soft acid) thus occurs
at a faster rate through sulphur (favourable soft–soft interaction) than through
nitrogen (unfavourable hard–soft interaction). Therefore, the first reaction will
take place at a faster rate.
(c) Since OH① is a stronger base than CH3COO①, it is a better nucleophile and
therefore, the first reaction is expected to take place at a faster rate than the second
reaction.
(d) Since the SN2 transition state involving cyclopentyl bromide is stable (less angle
strain) compared to that involving cyclobutyl bromide (more angle strain), the
second reaction is expected to take place at a faster rate than the first reaction.
(e) The polar aprotic solvent CH3CN does not solvate N3① ion through hydrogen
bonding while it solvates the cation and thereby makes the nucleophile much freer
to react. The protic polar solvent CH3OH solvates N3① by forming hydrogen bonds
and thereby lowers its reactivity. So, the second reaction is expected to take place
at a faster rate.
(g) :CN① is an ambident nucleophile. Its more polarizable carbon atom is the soft
centre while its less polarizable nitrogen atom is the hard centre. The C atom of
— CH2Br is a soft centre. Since the soft–soft interaction takes place rapidly than
the hard–soft interaction, the first reaction is expected to take place at a faster rate.
33. 1–Bromobutane (CH3CH2CH2CH2Br) can be prepared by heating 1–butanol
(CH3CH2CH2CH2OH) with a mixture of NaBr and H2SO4. However, the
reaction fails in the absence of H2SO4. Explain.
Solution The very bad leaving group OH① (a stronger base) cannot be displaced directly
by Br①. So, 1-bromobutane cannot be prepared from 1-butanol by treating it with NaBr.
However, in the presence of H2SO4, the — OH group undergoes protonation and becomes
≈
converted to a — OH 2 group. Since H2O is a very much better leaving group (as it is a very
weak base), it can be displaced readily by Br① to yield 1-bromobutane.
34. Apply Corey–House method to synthesize compound A using two suitable

substrates having four carbon and three carbon units respectively. Give
the possible routes. Which route will you choose and why?
(CH 3 )3 CCH 2 CH 2 CH 3
(A)
Solution The two possible routes for the synthesis of the compound A by Corey–House
method are as follows:
Li CuI
Route 1: CH3CH2CH2I CH2CH2CH2Li (CH3CH2CH2)2CuLi
(CH3)3CCl
SN2 (a 3° substrate)
(CH3)3CCH2CH2CH3
Li CuI
Route 2: (CH3)3CI (CH3)3CLi (CH3)3C 2 CuLi
CH3CH2CH2Br
SN2 (a 1° substrate)
(CH3)3CCH2CH2CH3
A
The SN2 reaction in step III of route 1 involves a tertiary substrate and so, because of
steric hindrance, the reaction does not take place. That is, no desired product A will be
obtained through this route. On the other hand, the SN2 reaction in step III of route 2
involves a primary substrate and so, because of small steric hindrance, the reaction takes
place smoothly to give the desired compound A in good yield.
35. Predict the products A, B and C obtained in the following two routes:
NaNH (CH ) CHBr
Route 1: CH 3C ∫∫ CH ææææ
liq. NH
2
Æ A ææææææ
3 2
liq. NH
ÆB
3 3
NaNH2 CH Br
Route 2: (CH 3 )2 CHC ∫∫ CH ææææ
liq. NH
Æ C ææææ 3
liq. NH
ÆB
3 3
Are both the routes equally efficient for the preparation of the compound
B? Explain your answer.
Solution The three compound A, B and C are as follows:
* *
≈
�� Na ≈
�� Na
A = CH 3C=C ; B =CH 3C ∫∫ C— CH(CH 3 )3 ; C= (CH 3 )2CHC ∫∫ C
Both the routes are not equally efficient for the preparation of B. Route 2 is more efficient
because its 2nd step (an SN2 reaction which is very susceptible to steric effect) involves
a relatively unhindered substrate methyl bromide (CH3Br) while the 2nd step of Route 1
involves a sterically hindered secondary substrate isopropyl bromide (Me2CHBr).
36. Complete the equations (products with structures) and assuming all but
one of the steps is SN2, label each of the product with D or L.
COOH
NaOH NaN3 H2
? H Br ? ?
Ni
CH3
Solution
O
1. ether
37. Complete the following: MeCH——CH2 + Me2CH MgBr
2. H3O
Solution SN2 attack by the carbanion (from the Grignard reagent) takes place at the less
crowded epoxy carbon (the methylene carbon) to form the corresponding halomagnesium
alkoxide. This is subsequently converted to the corresponding alcohol by the addition of
dilute mineral acid.
OMgBr
O
SN2 attack
MeCH——CH2 + Me2CH—MgBr MeCH—CH2CH2Me2
at the less hindered
carbon of the epoxide
H 3O
OH
MeCHCH2CHMe2
4-Methyl-2-pentanol
38. Give methods for the following conversions:
(R)-2-Ethoxypentane
(R)-2-Pentanol —
(S)-2-Ethoxypentane
Solution
39. Compare the yields of the following two routes:
(a) (CH3)2CH ONa + PhCH2Cl

(CH3)2CH—O—CH2Ph
(b) (CH3)2CHCl + PhCH2ONa Benzyl isopropyl ether
Solution The route (a) gives better yield of benzyl isopropyl ether because this SN2
reaction (Williamson synthesis) involving a less crowded primary alkyl halide is relatively
more facile from steric point of view.
40. Explain why 2,2,5,5-tetramethyl-3-hexyne cannot be prepared by using an
acetylide anion.
Solution The alkyl halide and the acetylide anion required for the synthesis of 2,2,5,5-
*
≈
tetramethyl-3-hexyne (Me3C — C ∫∫C — CMe3) are Me3CBr and Me3CC ∫∫ C
�� Na. However,
an SN2 reaction involving a tertiary alkyl halide does not take place because of steric
hindrance caused by three alkyl groups.
41. Predict the product of the following reaction and explain:
Solution The epoxide undergoes SN2 attack by the acetylide ion at the less crowded ring
carbon to give an alkoxide. The reaction occurs with inversion of configuration around
the stereogenic centre. The alkoxide on subsequent protonation by water produces the
corresponding alcohol.
SN2 O H OH H + OH
H—OH
O + :C∫∫ CH (inversion of
configuration) —— ————
CH3 H CH3 C∫∫ CH CH3 C∫∫ CH
42. How can the following transformation be carried out using a 2,4,6-
triphenylpyrylium salt?
H H
C C
D D
NH2 I
CH3 CH3
Solution The amine is converted to a pyridinium salt by treating with 2,4,6–
tiphenylpyrylium iodide and as a result of this, — NH2 is converted into a good leaving
group. When the salt is heated, the counterion I① acts as a nucleophile to give the desired
compound with inversion of configuration.

CH2CH2Cl
OH
OH O
EtOH
I A
CH2CH2OH CH2CH2OH CH2CH2OH

OH
Cl +
EtOH
II B C
Solution The compound I is first converted into the corresponding alkoxide which then
undergoes an intramolecular SN2 reaction (the carbon bearing chlorine is primary) to
yield the spirocyclic compound A.
Similar intramolecular SN2 reaction leading to the formation of the same spirocyclic
compound does not take place in the case of the compound II because the carbon bearing
chlorine is tertiary and the backside attack on it is not sterically feasible. As a consequence,
the compound II undergoes elimination (E2) reaction to yield two isomeric alkenes B and C.
44. Predict the product of each of the following reactions:

1. CH3O
1. CH3O
(a) Me3C (b) Me3C 2. H2O
2. H2O
O O
Solution Epoxide ring fused with six-membered ring opens in a trans diaxial fashion.
Thus, the nucleophile attacks from the axial direction to give a leaving group that also
axial. Because of this, only one product is formed in each of the given reactions, even
though both ends of the epoxide are equally substituted.
(a)
(b)
45. Which of the following two isomeric halocyclohexane derivatives (I and

II) would you expect to be more reactive in an SN2 reaction and why?
CH3 CH3
CH3 H CH3 Br
H H
H Br H H
I II
Solution The two relatively bulky methyl groups lock the isomer I in that conformation
which places bromine in the equatorial position. Since the axial approach of a nucleophile
for an equatorial leaving group is sterically hindered due to the axial hydrogens at C-3
and C-5, relative to the carbon (C-1) undergoing substitution, displacement of bromine
occurs very slowly.
On the other hand, the two relatively bulky methyl groups lock the isomer II in that
conformation which places bromine in the axial position. In this case, the approach of the
nucleophile is not sterically hindered and also, the ground state energy of this isomer is
higher (i.e., Ea is lower) than the isomer I. For these reasons, this isomer reacts with the
nucleophile by the SN2 mechanism readily. The cyclohexane derivative II is, therefore,
more reactive than I in an SN2 reaction.
46. Potassium permanganate (KMnO4), a common oxidizing agent, is insoluble

in a nonpolar solvent because it is ionic. How, then, can KMnO4, oxidize
compounds that are soluble only in nonpolar solvents?
Solution KMnO4 can be dissolved in a nonpolar solvent like benzene if 18-crown-6 is
added to the reaction mixture. The crown ether binds K≈ in its cavilty and the crown
ether-potassium complex with a nonpolar hydrocarbon-like exterior dissolves in benzene.
In order to maintain electrical neutrality, the permanganate ion (MnO4①) accompanies
the complexed K≈ ion into the benzene layer, permitting MnO4① to act as an oxidizing
agent in the nonpolar solvent.
47. Predict the product of each of the following reactions and explain the
stereochemical course involved:
Br
(a) MeO + MeOH
D
Br
D
H
Me
(b) CH(CO Et) +
2 2
O—C— — NO2
H O
Solution
(a) This SN2¢ reaction proceeds with syn stereochemistry as given below:
The C — Br bond is attacked by the p bond from the back and the p bond is
attacked from the back by the nucleophile. The result is an overall syn attack of
the nucleophile MeO① with respect to the leaving group.
(b) This SN2¢ reaction also proceeds with syn stereochemistry.

H
H
Me
Me
CH (CO2Et)2 + SN2¢
O—C— — NO2 (EtO2C)2CH
H O H O O
C
NO2

(a) A small amount of NaI catalyzes the following reaction:
R¢ONa + RCl Æ R¢OR + NaCl
(b) Neopentyl halides are notoriously slow in nucleophilic substitution reactions,
whatever the experimental conditions.
(c) The decreasing order of reactivity of Me3CCH2Br(I), ClCH2CH ==CH2(II),
ClCH2CH2CH3 (III) and BrCH2CH2CH3 (IV) in the Williamson synthesis of
ethers is II > IV > III > I.
2. Indicate the product of each of the following reactions with plausible mechanism:
CH2OH COOH
CH2N2 CH2N2
(a) (b)
ether ether
OH OH
[Hint: The more acidic functional group will undergo methylation by diazomethane.]
3. (a) 2,6–Di–tert–butylpyridine acts as a base but not as a nucleophile – Why?
Ê �� * ˆ
(b) Azide (N3①) ion is weaker base than amide Á : N H2 ˜ ion – Why?
Ë ¯
(c) Many nucleophiles are anions, but some anions are not nucleophiles. Explain
why BF4① is not a nucleophile.
4. How will you convert (S)-1-phenyl-2-propanol to (R)-1-phenyl-2-propanol?
5. Rank the following molecules/ions in the order of decreasing nucleophilicity and
give your reasoning:
(a) CH3COO①, CH3CH2O①, CH3CH2S① in MeOH
(b) I①, Br①, Cl① and F① in DMSO and in ethanol
(c) : NH
�� *, NH
2
�� , NH
3
�� NH
2
�� , PhNH
2
��
2 in MeOH
(d) �� H O , HOO
H 2O, 2 2
�� :* , H O
�� :* in MeOH
��
6. When PhCH2Br is added to a suspension of KF in benzene, no reaction occurs.
However, when a catalytic amount of 18-crown-6 is added, PhCH2F can be isolated
in good yields. If, however, LiF is substituted for KF, there is no reaction even in
the presence of crown ether.
[Hint: Li≈ ion cannot be trapped by 18-crown-6 because its cavity size is larger
than the size of Li≈.]
7. Which compound in each pair exhibits greater SN2 reactivity. Explain your
reasoning.
(a) Br or CH2Br
Cl
(b) or
Cl
(c) I or Cl
(d) Cl or
Cl
8. Treatment of substituted anisoles with HI produces CH3I and phenols rather than
CH3OH and iodobenzenes. Explain.
[Hint: SN2 attack by I① ion cannot take place at the ring carbon of the protonated
ether]
9. A concentrated aqueous solution of HBr reacts with CH3CH2OH to give CH3CH2Br,
but a concentrated aqueous solution of NaBr does not. Explain.
10. Acids of the type R3CCOOH cannot be prepared by the cyanide synthesis using a
tertiary alkyl halide — Why?
[Hint: R3C — CN cannot be prepared by treating R3C — Cl with NaCN in DMSO
(an SN2 reaction) because of steric reason.]
11. A three-membered ring formation reaction (by and intramolecular SN2) occurs
102–103 times faster than a comparable noncyclic reaction. Explain.
[Hint: Ready formation of a three-membered ring is the consequence of a more
favourable entropy contribution to the free energy of the reaction.]
12. What is the main difference is using RS① and RO① as nucleophiles in SN2
reaction?
[Hint: Elimination with RS① (a much weaker base) is not a serious problem as
with RO① (a much stronger base).]
13. What do you mean by necleofugality? Explain with suitable examples.
[Hint: A leaving group that carries away an electron pair is known as a nucleofuge.
Hence, nucleofugality is the leaving ability of a group that comes away with the
electron pair. Nucleofugality increases with decreasing basicity. For examples,

CN①, Br①, I①, SH①, etc. are good nucleofuges and HO①, RO①, NH2①, etc. are very
bad nucleofuges.]
14. Compare the nucleophilicities and basicities of tBuO① and EtO①.
15. Predict the product of the following reaction and suggest a plausible mechanism:
HI
(S)- Me(Et)CH — O — CH 3 ææ æÆ
[Hint: (S)-Me(Et)CHOH; the configuration is retained because no bond to the
asymmetric carbon is broken.]
16. Convert (S)-2–pentanol to (R)-2–pentanol.
≈
17. Cl① ion of Bu 4 NCl* in acetone is a better nucleophile than that of LiCl in the
same solvent. Explain.
18. Predict the major product obtained from each of the following reactions and account
for its formation:
(a) HI(exces) (b) HI(exces)

O O
CH3
(c) HBr (d) HI

CF3 — O — CH 3 æææ Æ O CH2CH3
CH3
(e) HI
CH 3CH == CH — O — CH 2CH 3 ææ æÆ (f) HI
O
CH3
19. An increase in solvent polarity will increase the rate of the SN2 reaction between
≈
an amine, R 3 N,
�� and an alkyl halide, RX, to form an ammonium salt, R NX * .
4
Explain.
20. Write the appropriate combination of alkyl halide and alkoxide for the preparation
of the following ethers by Williamson ether synthesis:
(a) CH2CH3 O (b) CH2 == CH — CH2 — O — CH(CH3)2
21. For the SN2 reaction: KCN + R — X Æ R — CN + KX, which solvent among methanol,
acetone and DMSO would be most appropriate and why?
22. Which would you expect to be the stronger nucleophile in protic solvent?
(a) CH3O① or CH3COO① (b) H2O or H2S (c) Me3P or Me3N
23. Arrange CH3COO , PhO and
① ①
PhSO3① anions as leaving groups in the decreasing
order if the pKa values of their conjugate acids are 4.5, 10 and 2.6, respectively.
24. Mention the solvent characters of the following:

H2O, CCl4, CH3OCH3, DMSO
25. Which of the following statements are correct?
(a) RO �� :* is a stronger nucleophile than : OH
�� * .
��
(b) RCOO �� :* is a stronger nucleophile than : OH �� * .
��
(c) RCOO �� :* is a stronger nucleophile than ROH. ��
��
(d) RO �� :* is a weaker nucleophile than H O
2
�� :.
��
26. In which of the following substrates chlorine is least reactive and why?
(a) Methyl chloride (b) Allyl chloride
(c) Ethyl chloride (d) Vinyl chloride
27. Account for the following reaction sequences and give three-dimensional structures
for A — K (with R and S designation):
NaH CH I
(a) ( R)-2- Butanol æææ Æ A æææ
3
ÆI
T sC l CH COOK
KOH/ H2O
(b) ( R)-1- Deuterioethanol ææææææ
Pyridine, 25∞ C
Æ C æææææ
3
acetone
Æ D æææææ
D
ÆE
Na CH I
(c) (S)-sec-Butyl alcohol æææ Æ F æææ
3
ÆG
NaSH CH I NaN 1. LiAH
(d) (S)-1-Chloro-1-phenylethane ææææ
acetone
Æ H æææ
3
Æ I ææææ 3
acetone
Æ J ææææÆ
2. H O
4
2
28. (a) Which one of the following compounds will react faster in an SN2 reaction and
why?
Br Br
Me Me Me Me
I II
[Hint: I will react at a rate faster than II.]

(b) Is an SN2 reaction exothermic or endothermic. Give your reasoning.
29. Nucelophilicity and basicity order of NH3, NH2NH2 and NH2OH are reverse to
each other. Explain.
[Hint: The basicity order is NH
��
3 > NH2 NH2 > NH2 OH.
In the case of NH2NH2,the
basicity is partly decreased due to — I effect of the other N atom. Since oxygen is
more electronegative than nitrogen, basicity of NH2OH is less than that of NH2NH2.
In this case, the nucleophilicity does not run parallel with basicity. In both NH2NH2
and NH2OH, the tendency of nucleophilic attack by the lone pair on nitrogen is
enhanced by the lone pair of electrons on the adjacent atoms. The phenomenon is
called ‘a–effect’. This effect is more pronounced in the case of NH2NH2 compared to
NH2OH and this is because the lone pair on less electronegative nitrogen is more
polarizable than the lone pair on more electronegative oxygen. Therefore, the order
of nucleophilicity is NH
�� NH
2
��
2 > NH 2OH > NH 3 .]
��
30. F① in CsF is more nucleophilic than F① in LiF in an aprotic solvent like acetone.
Give an explanation.
31. Which SN2 reaction in each pair is faster and why?
–
MeOH
(a) (b) I + OMe
Me2 CHCH 2 Cl + N3* ææ
Æ
–
Me3 CCH 2 Cl + N3* ææ
Æ DMSO
I + OMe
–
(c) CH3 Br + NaOH æææ
2
Æ
H O
(d) Cl + EtO
H O
CH3 Br + CH3COONa æææ
2
Æ Cl + EtOH
EtOH
(e) Br + Ph3N
EtOH
Br + Ph3P
32. Which of the following SN2 reactions will take place and which will not? Give
reasons.
NH2 –
I –
(a) + I + NH2
(b) CH2CH2CH2I + CH3O* ææÆ CH3CH2CH2OCH3 + : ��I :

*
��
OH F +
(c) + F
–
OH
–
(d) Br + CN
–
CN + Br
–
(e)
– –
(f) CN + I I + CN
33. Rank the alkyl bromides in each group in the order of increasing SN2 reactivity and
give your reasoning:
Br
(a) Br , ,
Br
Br
(b)
Br , Br ,
Br Br
(c) Br
, ,
Br H H
Me
(d) H , Br , Br
H
H H H
34. (a) Phloroglucinol can behave as an ambident nucleophile. Explain with
reactions.
(b) Two isomeric SN2 products are obtained when sodium thiosulfate (Na2S2O3)
is allowed to react with one equivalent of CH3I in methanol.
[Hint: (b) Thiosulfate (S2O32- ) ion is an ambident nucleophile.]
35. Arrange the following carbanions in the order of increasing nucleophilicity:
≈ * * * *
:CH 3 , :C ∫∫ N, CH 2 == CH — CH
�� , (CH ) C :, HC ∫∫ C :
2 3 3
[Hint: The increasing acidity order of the conjugate acids is

(CH 3 )3CH CH 4 CH 2 == CH — CH 3 HC ∫∫ CH HCN]
36. Predict the major product in each of the following reactions and indicate the
mechanistic pathway involved therein:
(a) CH3CHONa + CH3CH2Br Æ ; (b) (CH3)2CHBr + CH3CH2ONa
[Hint:
(a) Since the substrate is a primary halide, substitution (SN2) is highly favoured
over elimination (E2).
CH 3CH 2Br + CH 3CH 2ONa ææ
Æ CH 3CH 2OCH 2CH 3 + CH 2 = CH 2
(90%) SN2 (10%) E2
(b) With this secondary halide, the elimination is favoured over substitution
because of steric hindrance.
(CH 3 )2 CHBr + CH 3CH 2ONa ææ
Æ (CH 3 )2 CHOCH 2CH 3 + CH 2 = CH CH 3
(21%)SN2 (79%) E2]
37. Alcohols (ROH)

�� are weaker acids than phenols (ArOH)
�� but are stronger
��
nucleophiles. Explain.
[Hint: The conjugate base of a phenol (ArO①) is relatively more stable (stabilized
by resonance) than the conjugate base of an alcohol (RO①). For this reason, phenols
are more acidic than alcohols. On the other hand, phenols are weaker nucleophiles
than alcohols because the unshared pair of electrons on oxygen is delocalized with
the ring p electrons in the case of phenols and therefore, less available compared to
that on oxygen of alcohols.]
38. Predict the product and explain their formations:
+
H
O
==
–
HO S O
Bisulfite ion PhCHO

[Hint: Hard–hard and soft–soft interactions are to be considered.]
39. Explain the order of reactivity when each of the following compounds reacts with
methyl iodide (CH3I) to yield a methyl pyridinium iodide salt.
≈
Nu : + CH 3 — I ææ
Æ Nu — CH 3 I*
CH3
Nu:
N N N N N C(CH3)3
CH3 CH3 CH3
Relative rate 2.3 1 0.5 0.04 0.0002
40. The trans-isomer of 2–chlorocyclohexanol can be converted to cyclohexene oxide by

base, but the cis-isomer cannot. Explain.
41. The optical rotation of a solution of NaBr and (+)–2–bromopentane in acetone goes
slowly to 0°. Explain this observation.
42. Give the mechanism of the following reaction:
What is the advantage of using 2,6–lutidine instead of OH①?

43. The rate of SN2 reaction decreases in the order CH3Cl > CHCl2 > CHCl3 > CCl4.
Explain.
44. Write a detailed mechanism for the following reaction:
HBr(excess) 2
O D Br
–
Hint: + H—Br + + Br
O O
H
SN2
–
SN2
Br + +
Br—H
Br OH2 OH + Br
45. Suggest a mechanism for each of the following reactions:
OH Br
PBr3
(a)
SOCl2
(b) Me OH Me Cl
PBr3
(c) OH Br
46. Predict the major product from each of the following competition experiments:
(a) Cl NaI(1 mole)
Cl
DMF
H F NaI(1 mole)
(b)
acetone
Br H
NaSPh(1 mole)
(c)
I I EtOH
47. Predict the product obtained when ethyl amine is allowed to react with excess of
methyl iodide in a basic solution of K2CO3.
[Hint: The final product of the reaction is a quaternary ammonium iodide. Its
formation may be shown as follows:
SN2
+ – K2CO3 Me—I + –
Et—NH2 + Me—I Et NH2Me I Et NH Me Et NHMe2I
SN2
K2CO3
+ – I—Me
EtNMe3 I EtNMe2
SN2
48. Ammonia reacts with an alkyl halide (R — X) to give a low yield of 1° amine. A
much better yield of 1° amine is obtained from the reaction of an alkyl halide (1°)
with azide ion (N3①) followed by catalytic hydrogenation. Explain.
[Hint: Unlike an alkyl amine (R — NH2) an alkyl azide is not nucleophilic.]
49. To maximize the amount of alkyl iodide, the reaction of an alkyl chloride with KI
is generally carried out in acetone (MeCOMe). What is the role of the solvent in
increasing the yield of alkyl iodide?
[Hint: KI is soluble in acetone but KCl is not. As a result, the backward reaction
cannot take place.]
50. How will the rate of the following SN2 reactions change if the polarity of the solvent
(protic polar) is increased?
(a) CH3CH2CH2 Br + HO* ææÆ CH3CH2CH2OH + Br *
≈
(b) CH 3 S CH 2CH 3 + C2H5O* ææ
Æ C2H5OCH 3 + CH 3SCH 2C H 3
|
CH3
≈
(c) CH3CH2CH2I + NH3 ææÆ CH3CH2CH2 NH3I*
51. Would you expect CH3COO① ion to be more reactive as a nucleophile in an SN2
reaction carried out in MeOH or DMSO?
52. Starting with (R)-2-bromobutane, outline the synthesis of each of the following
compounds:
SH
|
(a) (S)-CH 3CHCH 2CH 3 (b) (S)-C H 3 CHOCOCH 3
|
C2H5 C2H5
|
(c) (S)-CH 3CHOC2 H5
53. 1–Bromobicyclo[2.2.1]heptane is extremely unreactive in an SN2 reaction.

Explain.
54. Explain why halocyclohexanes react slowly by the SN2 mechanism.
55. The trans-isomer of 4-tert-butylcyclohexyl bromide reacts with sodium thiophenoxide
* ≈
(Ph SNa) in aqueous ethanol at a much slower rate than its cis-isomer. Explain
this observation.
56. What are the possible combination of reactants required for the preparation of
isopropyl methyl ether by Williamson synthesis? Which combination of reactants
is preferable and why?
57. Explain nucleophilic catalysis with a suitable example.
58. CH3Br reacts with AgCN to give CH3 — NC (an isocyanide), whereas it reacts with
NaCN to give CH3 — CN (a cyanide). Explain.
59. (a) Predict the major product from each of the following reactions:
1. t-BuO *
AgNO2
(i) CH 3COCH 2COCH 3 æææææ
2. CH I
Æ (ii) BrCH 2CO2 Me ææææ Æ
3
CH N NaNO2
CH 3COCH 2COCH 3 ææææ
2 2
Æ BrCH 2CO2 Me ææææ Æ
(b) Account for the following reaction, making clear the role played by tosyl
chloride (TsCl):
soft base soft base

H O CH3
– –
t-BuO CH3—I –
Hint: (a)(i) CH3COCHCOCH3 CH3—C—CH—C—CH3 CH3COCHCOCH3 + I
–
´
O O
CH3—C ==CH—C—CH3
– hard base hard acid
O—H – + O
CH2—N2 +
CH3COCH2COCH3 CH3C ==CHCOCH3 CH3—C ==CHCOCH3 + CH3—N2

Keto Enol OCH3
CH3—C ==CHCOCH3
60. Give the products of the following SN2 reactions:

(a) (3S, 4R)-3-Bromo-4-methylhexane + CH3S① Æ
(b) (3S, 4S)-3-Bromo-4-methylhexane + CH3S① Æ
(c) (3R, 4R)-3-Bromo-4-methylhexane + CH3S① Æ
(d) (3R, 4S)-3-Bromo-4-methylhexane + CH3S① Æ
61. For each pair of the following compounds, state which one is the better SN2
substrate.
(a) 2–Methyl-1-iodopropane or tert-butyl iodide
(b) Cyclohexyl bromide or 1-bromo-1-methylcyclohexane
(c) 2,2–Dimethyl-1-chlorobutane or 2-chlorobutane
(d) Isopropyl iodide or 1-iodo-2,2-dimethylpropane
62. Draw a perspective structure of a Fisher projection for the products of the following
SN2 reactions:
(a) (S)-2-Bromopentane + KCN Æ
(b) trans-1-Chloro-3-methylcyclopentane +KOH Æ
CH3
(c) Br H acetone
+ NaI
H CD3
CH2CH3
Br
(d) C + NaSH
F CH2CH3
H
63. Elucidate the structures of compounds I–IV:
NaNH2 I(CH ) Cl
1-Bromo-8-Fluorooctane ææ Æ I(C10 H17 F) ææææ Æ II (C10H16FNa) æææææ
2 7
Æ
+ sodium acetylide
NaCN
II (C17 H 30Cl F) ææææ Æ IV(C18 H 30 NF)
[Hint:
64. Predict the product obtained when 1–bromopropane is treated separately with
methanolic solution of KSCN and AgSCN, respectively. Give your reasoning.
65. Give the major product in each of the following reactions and account for its
formation:
O
� *
Na salt + CH3I
(a) CH 3 — C — N �� — C H ææææææ
2 5 Æ
��
O
� *
Ag salt + CH3I
(b) CH 3 — C — N �� — C H ææææææ
2 5 Æ
��
66. When methyl bromide is dissolved in methanol and an equimolar amount of NaI is
added, the concentration of iodide ion quickly decreases, and then slowly returns
to its original value. Explain these observations.
67. Give the structures, including stereochemistry, of compounds I and II in the
following sequence of reactions:
68. The reaction of cyclopentyl bromide with sodium cyanide to give cyclopentyl cyanide
proceeds faster if a small amount of NaI is added to the reaction mixture.
H NaCN H
Br ethanol-water CN
Can you suggest a reasonable mechanism to explain the catalytic function of

sodium iodide?
69. Select the combination of alkyl bromide and potassium alkoxide that would be the
most effective in the synthesis of the following ethers:
(a) OCH3 (b) CH3OC(CH3)3 (c) (CH3)3CCH2OCH2CH3
70. Suggest a mechanism for the following reaction:

Br
NH
N + N
I II
[Hint: I is an SN2¢ product.]
3.2 THE SN1 REACTION

Example, kinetics, mechanism, stereochemistry, evidence in favour of the mechanism and
the factors influencing SN1 reactivity, i.e., the SN1 reaction rate.

The alkaline hydrolysis of tert-butyl chloride is a typical example of SN1 reaction.
�� * + H2O �� *
:OH (CH 3 )3 CCl ææææ
acetone
Æ (CH 3 )3 COH + : Br:
�� tert-Butyl chloride tert-Butyl alcohol ��

The rate of the reaction is found to depend on the concentration of only the alkyl halide.
For example, doubling the concentration of only the alkyl halide doubles the rate of the
reaction and changing the concentration of the nucleophile has no effect on the rate of the
reaction. From this relationship between the rate of the reaction and the concentration of
the alkyl halide, we can write the rate law for the reaction as follows:
rate = k [(CH3)3CCl]
Because the rate of the reaction depends on the concentration of only one reactant, it is a
first-order reaction. Therefore, the reaction is unimolecular, i.e., only tert-butyl chloride is
involved in the transition state of the rate determining step.

The most straightforward explanation for the observed first-order kinetics is a two-step
reaction in which bond breaking occurs before bond making. The mechanism of the reaction
is designated as SN1 (Substitution, Nucleophilic, unimolecular) because only one reactant
(the substrate) is involved in the transition state of the rate-determing step. The two steps
through which the reaction proceeds may be shown as follows:
CH3 CH3
slow
Step 1: CH3—C—Cl C + Cl
CH3 H3C CH3

CH3
H3C OH
Step 2: C—CH3 CH3—C—OH
fast
H3C CH3
In the first step, tert-butyl chloride undergoes heterolysis of the C — Cl bond to yield tert-
butyl cation and a chloride ion (the leaving group). This is the rate-determining (slow) step
of the reaction. The cleavage of the C — Cl bond is an endothermic process. In the second
step, nucleophilic attack of the hydroxide ion on the carbocation leads to the formation of
tert-butyl alcohol. This is a Lewis acid-base reaction. The formation of C — OH bond is an
exothermic process and hence, a fast process.
The energy profile diagram for the reaction may be shown as follows:

In the first step in an SN1 reaction, the tetrahedral substrate undergoes slow dissociation
of C — LG bond to form a trigonal carbocation. The hybridization of the central carbon
changes from sp3 to sp2. The three s bonds connected to the positive carbon lie in a single
plane which is perpendicular to the vacant p orbital of C 2 . The nucleophile may attack
sp
the planar carbocation from either side or face with equal facility. Thus, if the reaction is
carried out with a chiral substrate (the leaving group is bonded to the chirality centre), a
mixture of equal number of molecules having configuration identical and opposite to the
substrate will be obtained, i.e., the product will be mixture of exactly equal amounts of
two enantiomers. Hence, complete racemization is the expected stereochemical result of
an SN1 reaction. The process may be shown as follows:
However, in SN1 reactions, complete racemization is not frequently observed. The product
of inverted configuration generally exceeds its enantiomer, i.e., the usual stereochemical
result of an SN1 reaction is racemization plus a few percent of inversion. This becomes clear
if we consider the process of ionization of the substrate in the first step. When the alkyl
halide ionizes, an intimate or tight ion pair is formed first. The ion pair then undergoes
complete separation to give a pair of separately solvated free ions. A free carbocation
obtained in a solvolytic reaction finally becomes symmetrically solvated by the solvent
molecules. Nucleophilic attack on the carbocation by the solvent molecules then occurs
from either side with equal readiness and results in formation of a racemic product. Attack
on the ion pair, on the other hand, does not take place at either face of the carbocation
with equal ease and this is because the leaving group remaining in close proximity shields
the front side from nucleophilic attack. As a result, nucleophilic attack by the solvent
molecule occurs preferably from the backside of the unsymmetrically solvated carbocation
and substitutions product of inverted configuration is obtained predominantly. The process
may be shown as follows:
In fact, the degree of racemization depends on the stability (lifetime) of the intermediate
carbocation. A more stable carbocation that gets enough time to become symmetrically
solvated reacts to form racemic product predominantly. However, if the carbocation
is not much stable (short lived), the nucleophilic attack occurs mainly through the
unsymmetrically solvated carbocation to produce the inverted product predominantly.
An example of complete racemization is as follows. When optically active (S)-3-bromo-

3-methylhexane is heated with aqueous acetone, 3-methyl-3-hexanol is obtained as a
mixture of 50% (R) and 50% (S).
3.2.5 Evidence in Favour of SN1 Mechanism

(a) Kinetic evidence The rate law, rate = k[RX], shows that the rate of SN1 reaction
depends only on the concentration of the alkyl halide. This means that we must be
observing a reaction whose rate-determining transition state involves only the alkyl
halide, RX. The rate-determining transition state, therefore, is unimolecular because it
involves only one molecule. In fact, a rate law similar to the experimentally observed one
can be derived by applying the assumption of steady state and considering the first step to
be rate-determining. Therefore, the two-step mechanism involving the rate-determining
cleavage of the C — X bond is the most plausible one.
(b) Stereochemical evidence The SN1 reaction of an alkyl halide in which the halogen
atom is bonded to a chirality centre from two stereoisomers: one with the same relative
configuration as the starting alkyl halide, and the other with the inverted configuration.
These two enantiomers are often obtained in 1:1 ratio. Formation of two enantiomers
suggests that these reactions proceed through planar carbocation which undergoes
nucleophilic attack at either face to give product molecules with inversion and retention
of configuration.
(c) Change of rate with structure When the methyl groups of tert-butyl chloride are
successively replaced by hydrogens, the rate of the SN1 reaction decreases progressively.
The stability of carbocations decrease progressively for 3° to 1°. This observation,
therefore, suggests that the rate-determining step of an SN1 reaction involves formation
of a carbocation.
3.2.6 Factors Influencing SN1 Reaction Rate or SN1 Reactivity

3.2.6.1 The effect of substrate structure on rate
As the number of R groups on the carbon with leaving group increases, the rate of an SN1
reaction increases.
CH 3 — X RCH 2 — X R 2CH — X R 3C — X
methyl 1∞ 2∞ 3∞
An electrical effect is particularly important for SN1 reactions. An SN1 reaction involves
rate-determining formation of a carbocation. Since the transition state for the heterolysis
of C — X bond closely resembles the intermediate carbocation in energy, any effect that
stabilizes the carbocation will stabilize the transition state and thereby causes rate
enhancement. An alkyl group stabilizes a carbocation by releasing electrons through
inductive (+I) and hyperconjugation effects. Thus, the stability of carbocation increases
with increase in the number of alkyl groups on the central carbon and therefore, the
stability decreases in the following order:
A tertiary carbocation is more stable and is, therefore, easier to form than a secondary
carbocation, which in turn is more stable and easier to form than a primary carbocation.
Therefore, a tertiary alkyl halide is more reactive than a secondary alkyl halide, which
in turn is more reactive than a primary alkyl halide. Since methyl cation is least stable,
therefore, a methyl halide is least reactive. Thus, the reactivity order agrees with the
observation that the SN1 reaction rate increases as the hydrogen of methyl halide are
successively replaced by alkyl groups. It is to be noted that primary carbocation and
methyl cation are so unstable that primary alkyl halides and methyl halides do not at all
undergo SN1 reactions. The very slow reactions reported for primary and methyl halides
are actually SN2 reactions.
Rela ve rates of SN1 reac ons (hydrolysis) of several alkyl bromides
Class of alkyl bromide Example relative rate

Tertiary (3°) Me3C — Br 1,200,000
Secondary (2°) Me2CH — Br 11.6
Primary (1°) MeCH2 — Br 1.00*
Methyl CH3 — Br 1.05*
*SN1 rates are actually zero. These small rates are observed as a result of SN2 reactions.
Steric effects are less important in an SN1 reaction because the angular distance between
the alkyl groups increases on going from the substrate to carbocation. However, for SN1
reactions of alkyl halides having large alkyl groups, an enhancement of reaction rate
(steric acceleration) in observed. The rate enhancement is due to some relief of steric
compression between the R groups (B strain) on going from the tetrahedral substrate
(bond angle 109.5°) to the planar trigonal carbocation (bond angle 120°). The solvolysis (a
nucleophilic substitution in which the nucleophile is a molecule of the solvent) of the alkyl
chloride A in aqueous ethanol, for instance, proceeds about 600 times faster than the alkyl
chloride B.
H3C H3C
C—Cl C—Cl
Me3CCH2 H3C
CH2CMe3 CH3
A B
(much steric strain) (less steric strain)
The following observations demonstrate how various structural effects influence the rate
(reactivity) of SN1 reactions:
(1) The rate of hydrolysis (solvolysis in water) of alkyl halide increases as the
number of the phenyl groups on the carbon bonded to the leaving group
increases. For example, the SN1 reaction rate of the three phenyl substituted
chloromethanes increases in the order: PhCH2Cl < Ph2CHCl < Ph3CCl. Because
≈
of charge delocalization by resonance, benzyl cation (Ph CH2 ) derived from benzyl
chloride (PhCH2Cl) is resonably stable. For this reason, benzyl chloride undergoes
heterolysis of the C — Cl bond readily in water (a protic polar solvent), i.e., it
undergoes ready hydrolysis by the SN1 mechanism.
CH2—Cl CH2 CH2 CH2 CH2 CH2

+ Cl
´ ´ ´ ´
Benzyl chloride Benzyl cation

(resonance-stabilized)
As the number of phenyl groups on the carbon bonded to the leaving group
increases, resonance stabilization of the resulting carbocation increases in the
≈ ≈ ≈
order: Ph CH2 < Ph 2 CH < Ph 3 C and consequently, the rate of hydrolysis increases
in the order: PhCH2Cl < Ph2CHCl < Ph3CCl.
(2) Small bicyclic systems containing leaving group at the bridgehead carbon
does not undergo SN1 reaction. For example, 1-bromobicycl[2.2.1]heptane is
exceedingly unreactive towards nucleeophilic substitution by the SN1 mechanism.
Because of extremely rigid cage-like structure, the bridgehead carbocation
expected to be formed cannot assume its usual planar trigonal orientation of
bonds and the angle between bonds is somewhat less than the sp2 bond angle of
120°. So, the carbocation suffers from angle strain. Again, it will not be stabilized
by hyperconjugation because the formation of a double bond at the bridgehead
position is not possible according to Bredt’s rule. For these reasons, ionization of
1-bromobicyclo[2.2.1]heptane leading to the formation of an unstable bridgehead
carbocation does not take place, i.e., the compound is unreactive towards SN1
reaction.
(3) A halide producing an aromatic carbocation reacts at rate faster than a

halide producing a nonaromatic carbocation which in turn reacts at a
rate faster than a halide producing an antiaromatic carbocation. The SN1
reactivity of the following bromo compounds is II < I < III.
Br Br Br
O
I II III
The compound III on ionization produces a carbocation which is aromatic

[a(4n + 2)p electron system, where n = 1]. The compound I on ionization produces
a carbocation which is nonaromatic and the compound II on ionization produces a
carbocation which is antiaromatic (a 4np electron system, where n = 1).
The aromatic carbocation IIIa is more stable than the nonaromatic carbocation Ia
which in turn is more stable than the antiaromatic carbocation IIa. Because the
SN1 reactivity depends on the stability of the intermediate carbocation obtained
in the rate-determing step, the order of increasing SN1 reactivity of these bromo
compounds is II < I < III. In fact, the compound II does not undergo SN1 reaction.
(4) SN1 reactivity depends on the angle strain in the cyclic carbocation
obtained from a cyclic substrate. The more it will be unstable and
less readily it will be formed. The order of SN1 reactivity of the following
chlorocycloalkanes is I < II < III.
Cl
—Cl —Cl
I II III
An SN1 reaction proceeds through the rate-determining formation of a trigonal
planar carbocation in which the angle between any two bonds is 120°. The normal
bond angles of three-, four- and five-membered rings are of the order of 60°, 90° and
108°, respectively. Thus, there are deviations of 60°, 30° and 12°, respectively from
the trigonal angle. The bond-angle strain in the resulting cyclic carbocation thus
increases in the order: cyclopentyl cation
( ) < cyclobutyl cation
( ) <
cyclopropyl cation
( ) . Also, in a cyclopentane system, change of hybridization
from sp3 to sp2 involves relief of four bond oppositions. Therefore, the increasing
order of stability of these carbocations is cyclopropyl cation < cyclobutyl cation
< cyclopentyl cation. Because the SN1 reactivity depends on the stability of the
intermediate carbocation obtained in the rate-determining step, the reactivity
order of these chlorocyloalkanes is I < II < III.
(5) Vinylic and aryl halides are unreactive towards SN1 reactions. The positively
charged carbons of vinyl or aryl cations are sp hybridized and because sp carbon
are more electronegative than sp2 carbons that carry the positive charge of alkyl
carbocations, sp carbons are more resistant to becoming positively charged, i.e., sp
hybridized positive carbons are relatively very unstable. The formation of vinylic
and aryl cations by the heterolysis of the C — X bond is, therefore, disfavoured, i.e.,
an SN1 mechanism is prevented from operating.
Again, the C — X bond, the aryl and vinylic halides are unusually short and strong
because in bond is formed by sp2– p overlap and the bond has a considerable double
bond character due to delocalization of the unshared electron pair on halogen with
the ring p electrons or with the p electrons of the double bond. Consequently, bond
breaking requires more energy. For these reasons, aryl and vinylic halides are also
unreactive towards SN1 reactions.
(6) A primary substrate may even undergo SN1 reaction readity if the carbon
going to a carbocation is bonded to a heteroatom containing unshared
electron pair. For example, chloromethyl ethyl ether, ClCH2OC2H5, undergoes
ready SN1 solvolysis reactions, even though it is a primary substrate.
�� — CH — Cl ææææCH3OH �� — CH OCH
C2 H 5 — O 2 25∞ C
Æ C2 H 5 — O 2 3
��
The carbocation obtained from C2H5OCH2Cl on heterolysis of the C — Cl bond is
stabilized by resonance. It is for this reason chloromethyl ethyl ether undergoes
ready SN1 solvolysis.
SN1
C2H5—O—CH2—Cl C2H5—O—CH2 C2H5—O==CH2 + Cl
A 1° carbocation
(resonance-stabilized)
–H CH3OH
C2H5OCH2OCH3 C2H5—O—CH2—O—CH3
H
(7) Competition experiments are those in which two reactants at the same
concentration (or one reactant with two reactive sites) compete for a
reagent. When there are two halogen atoms in a substrate, which will
take part in an SN1 reaction depends on the carbocation expected to be
formed by C — X bond cleavage. For example, the circled halogen atom in each
of the following compound is more reactive than the other in an SN1 reaction.
Br
I I
Br Br
Br
CH2CH3
I II III
Cl CH2— Cl
CH3O
Cl
CH2CH2CHClCH3
IV V
Heterolysis of the C — Br bond in compound I does not actually take place because
this leads to the formation of a very unstable 1° carbocation. On the other hand,
heterolysis of C— Br bond leads to the formation of a very much stable 3°
carbocation. The circled bromine atom is, therefore, more reactive than the other.
The C — Br bond in compound II does not actually undergo heterolysis because this
leads to the formation of a very unstable 3° carbocation (a nonplanar bridgehead
carbocation which is not stabilized by hyperconjugation). Heterolysis of the C— Br
bond, on the other hand, leads to the formation for a very much stable 3° carbocation.
Therefore, the circled halogen atom is more reactive than the other one.
SN 1 SN 1
Br Br ; + Br ; Br Br + Br
Br Br Br
CH2CH3 CH2CH3 CH2CH3 CH2CH3
II A bridgehead carbocation II A 3° carbocation

(very unstable) (very stable)
Heterolysis of the C — I bond in compound III leads to the formation of a less stable
2° carbocation while heterolysis of the C — I bond leads to the formation of a very
stable allylic cation (resonance-stabilized). Therefore, the circled iodine atom is
relatively more reactive than the other one is an SN1 reaction.
Heterolysis of the C— Cl bond in compound IV leads to the formation of a very

stable (resonance-stabilized) 2° carbocation while heterolysis of the C — Cl bond
leads to the formation of a relatively less stable simple 2° carbocation. Therefore,
the circled chlorine atom is relatively more reactive than the other one.
Cl
SN1
+ Cl
CH3O CH3O CH3O
Cl Cl
Cl
IV A 2° carbocation
(more stable)
Cl Cl
SN1
CH3O + Cl
CH3O
Cl
IV A 2° carbocation
(less stable)
Heterolysis of the C— Cl bond in compound V leads to the formation of a

very stable (resonance-stabilized) benzylic cation while heterolysis of the
C — Cl bond leads to the formation of a relatively less stable 2° carbocation.
3.2.6.2 The Effect of Solvent on Rate

The rate of an SN1 reaction increases markedly with increase in the ionizing power of the
solvent. In the rate-determining step of the reaction a carbocation and an anion are formed
from a neutral substrate. Since stretching of the C — X bond starts before its cleavage,
therefore, the transition state is relatively more polar than the substrate.
The ionizing power of a solvent actually depends on two factors: (i) its dielectric constant
(e) and (ii) its ability to solvate ions. As the dielectric constant of the solvent or solvent
polarity increases, the electrostatic force of attraction between the two incipient ions
decreases. This facilitates the reaction by lowering the energy of the polar transition state.
As the molecular polarity, i.e., the dipole moment (m) of the solvent increases, solvation
(through dipole–dipole interactions) and consequent stabilization of both the substrate and
the transition state increases. However, such stabilization is far greater for the more polar
transition state than the less polar substrate. Because of this differential stabilization, the
energy of activation for ionization of the substrate is considerably reduced and the rate is
increased. This rate enhancement, however, depends on whether the solvent is protic or
aprotic. Since the transition state closely resembles the ions in energy and geometry, any
factor that stabilizes the ions also stabilizes the transition state. Therefore, to understand
the role played by a particular solvent, its ability to solvate the separated ions must be
considered. Protic polar solvents (e.g., methanol, ethanol, water, etc.), which effectively
solvate both cations and anions, are found to be very much effective for SN1 reactions. In
such a solvent, cation and anion solvation occur through donation of the unshared electron
pairs to the vacant orbital of the cation and through the formation of ion-dipole bonds or
hydrogen bonding, respectively.
d–
d+
H O H
d–
H2O OH2 O d+ d+d–
H H H O H
H2O R OH2 d– X
O d+ d+ d–
H H H O
H
H2O OH2
H O d– Hd+
A solvated cation A solvated anion
Polar aprotic solvents like CH3CN, Me2SO, Me2NCHO, etc. processing high dielectric
constants are not suitable for SN1 reactions because they unable to solvate anions (a very
important factor for ionization) through ion-dipole bonds or hydrogen bonding.
Water is the most effective solvent for promoting ionization because its dielectric constant
is much higher and it can stabilize the anion through hydrogen bonding. However, most
organic compounds do not dissolve appreciably in water. They usually dissolve in alcohol,
and quite often in mixed solvents. Methanol-water and ethanol-water are the two common
mixed solvents for nucleophilic substitution reactions.
The rate of an SN1 reaction increases when the percentage of more polar component of
a mixed solvent increases. For example, the rate of solvolysis of tert-butyl bromide in
methanol-water mixed solvent increases with increasing the percentage of water. As the
percentage of the more polar solvent water increases, the polarity of the mixed solvent
increases. Increasing the polarity of the mixed solvent increases the rate of solvolysis
because the more polar transition state is relatively more stabilized than the less polar
substrate resulting in a decrease of activation energy (Ea).
SN1 d+ d–
Me3C—Br Me3C Br Me3C + Br
Substrate Transition state

(less polar) (more polar)
The effect of the polarity of the solvent on the rate of reac on of tert-butyl bromide in an SN1 reac on
Solvent relative rate
100% H2O 1200

80% H2O + 20% EtOH 400
50% H2O + 50% EtOH 60
20% H2O + 80% EtOH 10
100% EtOH 1
If, however, the compound undergoing SN1 reaction is charged, increasing the polarity
of the solvent will decrease the rate of the reaction because the more polar solvent will
stabilize the full charge on the reactant to greater extent than it will stabilize the dispersed
charge on the transition state.
3.2.6.3 The effect of the nucleophile on rate

The rate of an SN1 reaction is independent of the nature of the nucleophile because the
nucleophile is not involved in the rate-limiting (slow) step of this two-step process. However,
if a better nucleophile is added to an SN1 reaction mixture, the rate remains unchanged
but the product containing the added nucleophile is obtained predominantly. In some SN1
reactions, the solvent is the nucleophile. These are called solvolysis. For example, when
water reacts with alkyl halides it serves as both the nucleophile and the solvent.
3.2.6.4 The effect of the leaving group on rate

The rate of an SN1 reaction is influenced by the nature of the leaving group because
expulsion of the leaving group from the substrate occurs in the rate-determing step of the
reaction. The leaving group in an SN1 reaction begins to acquire a negative charge as the
transition state is reached. Stabilization of the developing negative charge at the leaving
group stabilizes the transition state, i.e., lowers its energy. This in fact, lowers the free
energy of activation (DG=| ) and thereby increases the rate of the reaction. Because the weak
bases stabilize a negative charge effectively, i.e., stabilize the transition state effectively,
these are good leaving groups. For instance, an alkyl iodide in the most reactive and alkyl
fluoride is the least reactive of the alkyl halides in the SN1 reaction and this is because
their basicity increases in the order: I① < Br① < Cl① < F①. This order of leaving ability is
also due to the fact that C — X bond strength increases gradually from C — I to C — F.
Relative reactivity of alkyl halides is an SN1 reaction:
(a) Electrophilic catalysis The expulsion of a halogen atom from an alkyl halide is facilitated
by the presence of Ag≈ ion in the reaction mixture. This is called electrophilic catalysis of
an SN1 reaction. For example, when optically active 1-bromo-1-phenylethane is treated
with acetic acid in the presence of CH3COOAg, a racemic mixture of two enantiomeric
acetates is obtained.
CH3 OCOCH3 OCOCH3

CH3COO Ag
C C + C
H CH3COOH H H
Br CH3 H3C
Ph Ph Ph
Optically active
Racemic mixture
The mechanism of the reaction is as follows:
(b) Alcohols and ethers can act as SN1 substrates in the presence of acid The hydroxide ion
is a strong base and thus reaction like the following do not take place.
R——OH R + OH
A 2° or 3° alcohol A very bad
leaving group
However, when an alcohol is dissolved in a strong acid, it can undergo C — O bond cleavage
to form a carbocation. Because the acid protonates the — OH group of the alcohol, the
leaving group no longer needs to be a hydroxide ion; it is now a molecule of H2O which is
@
a much weaker base than an OH ion and a good leaving group.
H
R——OH R——OH2 R + H2O
A good leaving
group
When a chiral alcohol like (R)-2-butanol is allowed to stand in aqueous acid, it is found to
have lost its optical activity.
H H H
H3O
C C + C
C2H5 C2H5 C2H5
OH OH HO
CH3 CH3 CH3
(R)-2-Butanol (R)-2-Butanol (S)-2-Butanol
Racemic mixture
In the presence of aqueous acid, (R)-2-butanol undergoes protonation. The conjugate
acid thus obtained undergoes heterolysis of the C — O bond (an SN1 process) to form a 2°
carbocation by loss of water (a good leaving group). Recombination of the carbocation with
water, which occurs with equal readiness at either face of the trigonal planar ion, leads to
the formation of an equimolar mixture of two enantiomeric 2-butnaol (a racemic mixture)
which is optically inactive.
An ether containing a 3° alkyl group (e.g., Me3C — O — Me) can easily be cleaved by HBr.
Protonation makes–OMe a good leaving group (MeOH) and the ether undergoes ready
≈
cleavage of Me3C — O bond leading to the formation of a stable 3° carbocation (Me3 C).
3.2.7 Carbocation Rearrangements in SN1 Reactions

The SN1 reaction involves a carbocation intermediate. This intermediate can rearrange,
usually by a hydride (H①) shift or an alkyl (R①) shift, to give a more stable carbocation.
For example, the product obtained when 2-bromo-3-methylbutane undergoes an SN1
reaction is different from the product obtained when it undergoes an SN2 reaction. When
the reaction is carried out under condition favouring an SN1 reaction, the initially formed
secondary (2°) carbocation undergoes a 1,2–hydride shift, rearranging to a relatively more
stable tertiary (3°) carbocation.
H
SN1 1,2-hydride shift H2O
(CH3)2C—CHCH3 (CH3)2CCH2CH3
A secondary carbocation A tertiary
(CH3)2CHCHCH3 (less stable) carbocation
Br (more stable)
OH2
2-Bromo-3- SN2
methylbutane (CH3)2CHCHCH3 (CH3)2C CH2CH3
OH
—H
OH
3-Methyl-2-butanol OH
(CH3)2CHCH2CH3
2-Methyl-2-butanol
The product obtained when 3-bromo-2,2-dimethylbutane is allowed to react with a

nucleophile also depends on the conditions under which the reaction is carried out. The 2°
carbocation formed initially under SN1 conditions undergoes rearrangement involving a
1,2–methyl shift to give a relatively more stable 3° carbocation. Since a carbocation is not
formed under SN2 conditions, therefore, the carbon skeleton remains unchanged.
When neopentyl bromide is boiled with ethanol, it gives only a rearranged substitution
product. The product results from a methyl shift (migration of a methyl group together
with its pair of electrons). Without rearrangement, ionization of neopentyl bromide would
form a very unstable primary (1°) carbocation.
(CH3)3C—CH2—Br (CH3)2C CH2 + Br

Neopentyl bromide A 1° carbocation
(very much less stable
and not formed)
The methyl shift occurs while bromide ion is leaving, so that only more stable 3° carbocation
is formed. Attack by ethanol on the 3° carbocation gives the product.
Similarly, neopentyl chloride cannot be prepared from neopentyl alcohol by treating with
HCl. tert-Amyl chloride is obtained as the only product in this reaction.
Neopentyl alcohol first undergoes protonation in the presence of acid to form its conjugate
≈
acid. The — OH group is converted into — OH 2 which acts as a very good neutral leaving
group (weak base). However, expulsion of water does not lead to the formation of an
unstable 1° carbocation. It undergoes rearrangement at the same time by a 1,2–methyl
shift to form a stable 3° carbocation. Nucleophilic attack by Cl① on this carbocation leads
to the formation of tert-amyl chloride.
[It is to be noted that although neopentyl is a primary (1°) alkyl group, because of steric
hindrance, neopentyl alcohol reacts with HCl extremely slowly to give neopentyl chloride
by an SN2 mechanism.
The following bicyclic compound (a 1° alcohol) undergoes neopentyl type rearrangements

in the presence of HBr/ZnCl2 to a bromide.
HBr / ZnCl2
+ H2O
SN1
CH2OH Br
Under this SN1 condition, the bridge methylene group migrates to bond with the incipient
1° carbocation that would be formed by the loss of water. The resulting 3° bridgehead
carbocation then undergoes attack by the nucleophile Br① to yield a bromo compound.
2 HBr
ZnBr2 ZnBr 24 1,2-bond shift
2 2 2
– H2O
CH2OH CH2—OH2 2
Br—ZnBr2—Br
2 + ZnBr2
Br
3.2.8 Comparison of the SN2 and SN1 Reactions

SN2 reaction SN1 reaction
1. A one-step reaction 1. A two-step reaction
2. A bimolecular rate-determining transition 2. A unimolecular rate-determining transition
state state
3. There is no carbocation rearrangements. 3. Carbocation rearrangements may take place.
4. The product has inverted configuration compared 4. The products have both identical and inverted
with the reactant. configuration compared with the reactant.
5. Reactivity order: methyl >1° > 2° > 3°. In fact, 3° 5. Reactivity order: 3° > 2° > 1°> methyl. In fact,
substrates do not undergo SN2 reaction because 1° and methyl substrates do not undergo SN1
of severe steric hindrance. reaction because the corresponding carbocations
are extremely unstable.
3.2.9 Summary of Reactivity of Alkyl Halide in Necleophilic Substitu-

tion Reactions
Alkyl halide Reaction they undergo
methyl and 1° SN2 only
2° SN1 and SN2

3° SN1 only
vinylic and aryl neither SN1 nor SN2
1° and 2° benzylic and 1° and 2° allylic SN1 and SN2
3° benzylic and 3° allylic SN1 only
3.2.10 Factors Favouring SN1 and SN2 Reactions

Factor SN1 SN2
Substrate R3CX (3°) > R2CHX (2°) CH3X > RCH2X (1°) R2CHX (2°)
(The reaction involves a relatively (The reaction requires unhindered substrates)
stable carbocation)
Nucleophile The reaction is favoured by weak The reaction is favoured by strong nucleophiles
nuclepophiles (usually neutral (usually bearing a net negative charge). The rate
solvent molecules increases by the high concentration of the nu-
cleophile. In protic solvents, within a group of the
periodic table, rate μ polarizability of the attacking
atom of the nucleophile and for the same attacking
atom, rate μ basicity of the nucleophile.
In aprotic solvents, rate μ basicity of the nucleophile.
Solvent Polar protic (e.g., alcohol, water or a Polar aprotic (e.g., DMF, DMSO, acetone, etc.)
mixture of alcohol and water).
Leaving group The weaker the base after the group departs the better the leaving group for both SN1
and SN2 reactions.
Depending on the reaction conditions a 2° substrate R2CHX may react by the SN1 or SN2
mechanism. Strong nucleophiles favour the SN2 mechanism over the SN1 mechanism.
For example, RO@ favour SN2 while ROH favours SN1 mechanism. Protic polar solvents
favour the SN1 mechanism while aprotic polar solvents favour the SN2 mechanism. For
example, water and CH3OH favour the SN1 mechanism while acetone and DMSO favour
the SN2 mechanism.
3.2.11 SNi and SNi¢ Mechanisms

3.2.11.1 SNi mechanism
Some nucleophilic substitution reactions occurs with retention of configuration in contrast
to SN1 or SN2, even though there is no neighbouring group participation (see article 3.3).
The mechanism of such reactions is termed as SNi mechanism (Substitution Nuclephilc
internal). For example, (R)-2-octanol reacts with thionyl chloride (SOCl2) by the SNi
mechanism to yield (R)-2-chlorooctane with the same configuration as the alcohol.
Mechanism: The mechanism of the reaction involves the steps as follows:

Step 1: When (R)-2-octanol is allowed to react with thionyl chloride, an alkyl chlorosulphite
with retained configuration is obtained and this is because the bond between the chiral
carbon and oxygen is not cleaved during this reaction.
Step 2: The intermediate chlorosulphite ester undergoes slow dissociation to give an

intimate ion pair.
n-C6H13 C6H13-n –
Cl
+
O
slow
C—O— S ==O C S ==O
H3C D
Cl
H CH3 H
Intimate ion pair in solvent cage

Step 3: The ion pair rapidly collapses to yield (R)-2-chlorooctane.
@
The nucleophile :Cl

�� @ : ion originated from :O
�� : SOCl ion attacks the carbocation from
��
the front side because it is unable to get to ��
the rear. As a result, the configuration is
retained.
The overall reaction between the chiral alcohol and thionyl chloride, therefore, takes place
with retention of configuration.
Evidence in favour of ion pair The reaction involves the formation of an ion pair that can
be proved by the fact that 3-methyl-2-butanol, when allowed to react with SOCl2, produces
3-chloro-2-methylbutane. This is possible only if a carbocation forms and subsequently
rearranges to the more stable one.
CH3 CH3 O
(CH3)2CH—CH—OH + SOCl2 Æ (CH3)2CH—CH—O—S—Cl

3-Methyl-2-butanol slow
– H
+ O – + –
(CH3)2C—CH2CH3 S ==O H-shift (CH3)2C—CH—CH3 O SOCl

Cl
Cl
fast
(CH3)2C CH2CH3
–SO2
2-Chloro-2-methylbutane
Reaction of (R)-2-octanol with SOCl2 in the presence of pyridine The reaction proceeds
with inversion of configuration when carried out in the presence of pyridine. The base
pyridine reacts with HCl generated during the formation of the chlorosulphite ester to
yield pyridinium chloride salt. The alkyl chlorosulphite then undergoes nucleophilic attack
by the Cl@ ion from the back side (SN2) and results in formation of the corresponding alkyl
chloride with inverted configuration.
The reaction may also take place as follows:

3.2.11.2 SNi¢ mechanism

When an allylic substrate undergoes SNi¢ reaction, the nucleophilic attack occurs at the
g-carbon rather than at the a-carbon. This is what is called SNi¢ mechanism. When pent-
2-en-1-ol, for example, is treated with thionyl chloride, 3-chloropent-1-ene is obtained
exclusively.
Cl
|
ether
CH3CH2CH == CHCH2OH + SOCl2 æææÆ CH3CH2 CHCH == CH2
Pent-2-en-1-ol 3-Chloropent-1-ene
(100%)
The reaction takes place as follows:
3.2.12 SN1¢ Mechanism

Under SN1 conditions, allylic substrates undergo nucleophilc substitutions to give a
rearranged product in addition to the normal product. For example, CH3CH==CHCH2Cl on
solvolysis with 0.8(N) NaOH at 25°C yields 60% of CH3CH==CH CH2OH (normal product)
and 40% of CH3CHOHCH== CH2 (rearranged product).
0.8 (N) NaOH
CH 3 CH == CH CH 2Cl ææææææ25∞C
Æ CH 3CH == CHCH 2OH + CH 3CH OH CH == CH 2
1-Chlorobut-2-ene But-2-en-1-ol (60%) But-1-en-3-ol (40%)
(normal prroduct) (rearranged product)
The mechanistic pathway which results in formation of the rearranged product is known
as the SN1¢ (SN1-prime) mechanism (unimolecular nucleophilic substitution with allylic
rearrangement).
The reaction proceeds through the formation of a delocalized allylic carbocation as
follows:
+ + –
CH3CH ==CHCH2—Cl [CH3CH ==CH—CH2 ´ CH3CH —CH ==CH2 ] + Cl
Allylic
1-chlorobut-2-ene carbocation
– –
OH OH
CH3CH ==CH—CH2OH CH3CHOH CH ==CH2

But-2-en-1-ol But-1-en-3-ol
(SN1 product) (SN1¢ product)
3.2.13 Isotope Effects and Salt Effects (Methods Used to Distinguish

between SN1 and SN2 Type Reactions)
Isotope effects a and b secondary isotope effects have been employed as useful probes to
distinguish between SN1 and SN2 type reactions. In b secondary isotope effect, substitution
of D for H b to the position of bond cleavage slows the reaction. The effects are greatest
when the T.S. possess considerable carbocation character. A large value of kH/kD, therefore,
suggests that an SN1 reaction takes place. The kH/kD value for the solvolysis of tert-butyl
chloride-d9 [(CD3)3CCl] in 50% acetone at 25°C, for example, is 2.38. This observation
indicates that the compound undergoes solvolysis by the SN1 mechanism.
a Secondary isotope effect results from a replacement of H by D at the carbon bearing the
leaving group. These effects are also correlated with carbocation character. Substitution
reactions not involving carbocation intermediates, i.e., SN2 reactions, have an isotope effect
near unity. However, reactions involving carbocation, i.e., SN1 reactions, have higher a
isotope effects (which depend on the nature of the leaving group). For instance, the kH/
kD value for the solvolysis of Ph2CDCl in 80% aqueous acetone is 1.16. This observation
suggests that the compound undergoes solvolysis by the SN1 mechanism. The kD/kD value
for the solvolysis of CH3(CH2)5CD(CH3)OTs in ethanol is near unity. This suggests that
the compound undergoes solvolysis by the SN2 mechanism.
Salt effects SN1 reaction rates increase when non-common ion salts (which increase the
ionic strengh of the medium) are added to the reaction mixture. This is what is called
salt effect. SN1 reaction rates are sufficiently increased when there are ions present
that specifically help in pulling the leaving group from the substrate. Some especially
important ions are Ag ! , Hg 2! and Hg 2! . H ! helps to pull off F from alkyl fluorides by
forming H-bond).
The non-common ion such as the azide ion (N3@ ) can be used as a means of distinguishing
between SN1 and SN2 mechanisms. This is because the azide ion markedly accelerates the
rate of reaction as it is a strong nucleophile and provides an additional mode of attack in
the alkyl derivative. A large enhancement in the rate of reaction on addition of an azide
salt (e.g., NaN3) has been considered as an indication of SN2 mechanism. A slight increase,
on the other hand, points to an SN1 mechanism.
1. Rank the following compounds in order of increasing reactivity:
Solution The compound II on ionization produces a carbocation which is aromatic

[a(4n + 2)p electron system, where n = 1], the compound III on ionization produces a
carbocation which is nonaromatic and compound I on ionization is expected to produce a
carbocation which is antiaromatic (a 4np electron system, where n = 1)
An aromatic system is more stable than a nonaromatic system, which in turn is more
stable than an antiaromatic system. Hence, an increasing order of stability of the resulting
carbocations is: (least stable) Ia < IIIa < IIa (most stable). Because the SN1 reactivity
of a compound depends on the stability of the intermediate carbocation obtained in the
rate-determining step of the reaction, the order of SN1 reactivity of these compounds is
I < III <II. In fact, the compound I does not undergo SN1 reaction because the resulting
carbocation is very unstable.
2. Arrange the following compounds in order of increasing reactivity
towards SN1 and SN2 reactions:
(a) 1-Bromo-3-methylbutane (b) 2-Bromo-2-methylbutane
(c) 2-Bromo-3-methylbutane
Solution Because the rate-determining step of an SN1 reaction involves formation of a
carbocation and because stabilities of carbocations increase on going from 1° to 3°, 2-bromo-
2-methylbutane (a tertiary substrate) is more reactive than 2-bromo-3-methylbutane
(a secondary substrate), which in turn is more reactive than 1-bromo-3-methylbutane
(a primary substrate) towards SN1 reaction. In fact, the SN1 reactivity of the primary
substrate is zero.
Because an SN2 reaction is very susceptible to steric hindrance at the site of substitution
and because steric crowding increases on going from primary to tertiary substrate, 1-bromo-
3-methylbutane (a primary substrate) reacts by the SN2 mechanism at a faster rate than
2-bromo-3-methylbutane (a secondary substrate), which in turn reacts at a much faster
rate than 2-bromo-2-methylbutane (a tertiary substrate). In fact, the SN2 reactivity of the
tertiary substrate is zero.
3. Solvolysis of the alkyl chloride I in aqueous ethanol proceeds about 600
times faster than the alkyl chloride II. Explain.
CH3
|
(Me3CCH 2 )2 CCl Me3 CCl
I II
Solution Compressed reactants try to avoid their steric strain. If the strain is relieved
in attaining the intermediate of the rate-determining step or the transition state leading
to the formation of the intermediate, that reaction will be speeded up. This phenomenon
is termed as steric acceleration and because of this, an alkyl halide containing bulky
substituents undergoes solvolysis (SN1) at a faster rate than that expected on the basis of
stabilization of the carbocation through inductive and hyperconjugative electron release.
In the alkyl chloride I, two out of three alkyl groups are very bulky neopentyl groups
(— CH2CMe3), while in the alkyl chloride II, all the three groups are less bulky methyl
groups. Steric acceleration is, therefore, more important in the case of I than in the case
of II and so, the solvolysis of the alkyl chloride I in aqueous ethanol proceeds at a much
faster rate (600 times) than the alkyl chloride II.
4. In the investigation of the mechanism of solvolysis (in 80% EtOH) of

2-chloro-2-methylbutane, the following results were obtained for the
solvolysis of: (a) Me2CClCD2Me (kH/kD = 1.41), (b) (CD3)2 CClCH2Me (kH/kD =
1.78). Explain the observations.
Solution From the kH/kD values it becomes clear that the substitution of deuterium for
six b hydrogens [in (b)] causes much retardation of reaction (solvolysis) rate than the
substitution of deuterium for two b hydrogens [in(a)] and this is because the carbocation
! !
(CD3 )2C CH 2Me is less stable than the carbocation Me2 C CD2Me (the C — D bonds
are somewhat stronger than the C — H bonds and thus result in less stabilization by
hyperconjugation).
5. Draw the conformational structures of the product(s) expected to be
obtained from each of the following reactions:
H Br Cl –
H2O H I
(a) ; (b)
CH3OH CH3OH
D CH3 H
D
Solution
(a) This tertiary substrate undergoes solvoysis (SN1) in methanol-water mixed solvent.
Either nucleophile (H2O or CH3OH) can attack the intermediate carbocation from
either the top or bottom side to give four substitution products (two with retention
and two with inversion of configuration).
H Br
OCH3
D CH3 CH3
(a) (a) (retention)
H
∫∫
Br CH3
H2O D
– CH3OH
CH3 Br + CH3
CH3OH (–H + )
H H (b) OCH3
(b)
D D H (inversion)
∫∫
D OH
CH3
(a) (a) (retention)
H
CH3 D
+ H2O
+
(–H ) CH3
H (b) OH
(b)
D H (inversion)
(b) This secondary substrate undergoes SN2 reaction with the strong nucleophile I@ to
yield the corresponding iodo compound. The reaction, therefore, proceeds only with
inversion of configuration around the stereogenic centre.
6. Formation of ethers from alcohols by intermolecular dehydration

using concentrated H2SO4 may proceed either by the SN1 or by the SN2
mechanism. Explain.
Solution The formation of an ether by acid-catalyzed intermolecular dehydration of
an alcohol is, in fact, a nucleophilic substitution reaction in which the conjugate acid
of alcohol is the substrate, H2O is the leaving group and a second molecule of alcohol
is the nucleophile. Because of steric reason, methanol and primary alcohols react by
the SN2 mechanism while, because of electronic reason, secondary and tertiary alcohols
react by the SN1 mechanism. Both of these mechanistic pathways may be outlined as
follows taking ethanol (CH3CH2OH) as an example for the SN2 route and isopropyl alcohol
(CH3CHOHCH3) as an example for the SN1 route.
SN2 route:
Step 1:
+ fast +
CH3CH2 OH + H CH3CH2 OH2
Step 2:
Step 3:
SN1 route:
fast
Step 1: (CH3)2CH—OH + H (CH3)2CH—OH2
slow
Step 2: (CH3)2CH—OH2 (CH3)2CH + H2O
SN1
slow
Step 3: (CH3)2CH—OH + (CH3)2CH (CH3)2CH—O—CH(CH3)2
H
fast
Step 4: (CH3)2CH—O—CH(CH3)2 + H2O (CH3)2CH—O—CH(CH3)2 + H3O:
H
7. 2-Ethyl-2-methyloxirane reacts with acidic methanol to yield the 1° alcohol
A as the major product while it reacts with basic methanol to yield the 3°
alcohol B as the major product:
Explain these observations regarding the regiochemistry of epoxide ring

opening in acidic and basic medium.
Solution In the presence of acid, 2-ethyl-2-methyloxirane undergoes ring cleavage by a
pathway that is partially SN1 and SN2. It is not a pure SN1 reaction because a carbocation
intermediate is not fully formed. It is not a pure SN2 reaction because the leaving group
begins to depart before the compound is attacked by the nucleophile. In acid, the epoxide
or the oxirane first undergoes protonation. In protonated epoxide, the positive charge is
borne not only by the oxygen but also by the two ring carbons. The sharing of the positive
charge can be represented by drawing the following three resonance structures:
The structure I is the conventional structure for the protonated epoxide while the structures
II and III show that the epoxied carbons share part of the positive charge.
The structure II (a very stable 3° carbocation) is more important (contributing) than the
structure III (a very unstable 1° carbocation). Therefore, the tertiary carbon bears a larger
part of the positive charge, and it is more strongly electrophilic. As a consequence, the more
substituted carbon–oxygen bond becomes longer and weaker than the less substituted
carbon–oxygen bond. Thus, the transition state for attack at the more substituted carbon
is lower in energy than that for attack at the less substituted carbon (even though attack
at this carbon is less hindered) and so, attack by weakly nucleophilic methanol (which
is sensitive to the strength of the electrophile) takes place preferentially at the more
electrophilic tertiary carbon to give predominantly the alcohol A.
Under basic conditions, the cleavage of the unprotonated epoxide ring occurs purely by
the SN2 mechanism and so, the nucleophile CH 3O�� :@ attacks the less substituted and less
��
sterically hindered methylene carbon of the epoxide preferentially to give the tertiary
alcohol B as the major product.
Solution In an anhydrous medium, the low polarity of the solvent (ether) favours the
SN2 mechanism and the nucleophile I@ attacks the sterically less hindered methyl carbon
of the protonated ether to yield methyl iodide and tert-butyl alcohol.
H—I SN2
(CH3)3C—O—CH3 (CH3)3C—O—CH3 + I (CH3)3COH + CH3I
tert-Butyl methyl ether tert-butyl Methyl
H alcohol iodide
In an aqueous medium, the high polarity of the solvent (H2O) favours the SN1 mechanism
and therefore, the protonated ether dissociates to give methanol and tert-butyl cation
which then undergoes nucleophilic attack by I@ to yield tert-butyl iodide.
H—I SN1
(CH3)3C—O—CH3 (CH3)3C—O—CH3 + I (CH3)3C + CH3OH + I
tert-Butyl methyl ether H
(CH3)3C—I + CH3OH
tert-Butyl Methyl
iodide alcohol
9. When the following ester of acetic acid is allowed to undergo acidic

hydrolysis, a racemic alcohol is obtained:
CH2CH (CH3)2 CH2CH (CH3)2 CH2CH (CH3)2
C H3O C C
H5C2 O—C—CH3 CH3COOH + H5C2 OH + HO C2H5
CH3 O CH3 CH3
(optically active) Racemic mixture

Solution Because heterolysis of the C — O bond results in the formation of a stable 3°
carbocation, acid-catalyzed hydrolysis of this ester occurs by alkyl-oxygen cleavage. The
reaction thus follows the SN1 mechanism and leads to the formation of acetic acid and a
racemic alcohol. The reaction takes place through the steps as follows:
Step 1:
CH2CH (CH3)2 CH2CH (CH3)2
O OH
C C
H5C2 O—C—CH3 + H—OH2 H5C2 O—C—CH3 + H2O
CH3 CH3
CH2CH (CH3)2 CH2CH (CH3)2
OH
C slow C + CH3 COOH
Step 2: H5C2 C
O CH3 SN1
CH3 H5C2 CH3
Step 3:
10. When the following alkyl bromides are subjected to hydrolysis in dilute
aqueous ethanolic sodium hydroxide solution, the rates of the reactions
showed the following order:
(CH3)3CBr > CH3Br > CH3CH2Br > (CH3)2CHBr
Provide an explanation for this order of reactivity.
Solution Two different mechanisms are involved in the hydrolysis of these alkyl bromides.
(CH3)3CBr undergoes hydrolysis by the SN1 mechanism and so, its hydrolysis takes place
most readily. The other three alkyl bromides undergo hydrolysis by the SN2 mechanism
(in the presence of strong nucleophile OH@, the 2° substrate also follow the SN2 pathway
predominantly). Since the activation energy of an SN2 reaction is normally much higher
compared to an SN1 reaction, the hydrolysis of these three alkyl bromides takes place
relatively slow. Since an SN2 reaction is very susceptible to steric hindrance, the order of
their SN2 reactivity is CH3Br > CH2CH3Br > (CH3)2CHBr.
11. One reaction in each pair takes place more readily. Identify the reaction
and explain your answer.
(a) Me3CCl + H 2O ææ
Æ Me3COH + HCl
Me3CBr + H 2O ææ
Æ Me3COH + HBr
Ph Ph
| |
(b) (CH 3 )2 C — Br + H 2O ææ
Æ (CH 3 )2 C — OH +HBr
(CH 3 )3C — Br + H 2O ææ
Æ (CH 3 )C — OH + HBr
(c) (CH3 )3CCl + H2O ææÆ (CH3 )3 COH + HCl

(CH3 )3CCl + CH3OH ææÆ (CH3 )COCH3 + HCl
(d) Me3CBr + H 2O ææ
Æ Me3COH + HBr
H2O �� @
Me3CBr + OH @ æææ Æ Me3COH + :Br:
��
EtOH
(e) Me CCl (1.0M) + EtO@ (1.0M) ææææ �� @
3 Æ Me3COEt + Cl:
��
EtOH �� @
Me3CCl (1.0M) + EtO@ (2.0M) ææææ Æ Me3COEt + Cl:
��
(f) Me3CCl + H 2O ææ
Æ Me3COH + HCl
PhCl + H 2O ææ
Æ PhOH + HCl
Solution
(a) The second reaction will take place more rapidly because Br@ is a better leaving
group than Cl@.
!
(b) Since the 3° carbocation (CH 3 )2 C Ph is relatively more stable (resonance-stabilized)
!
than the 3° carbocation (CH 3 )3 C, the first reaction (solvolysis) is expected to take
place at a faster rate than the second.
(c) SN1 reactions take place faster in more polar solvents. Since water is a much polar
solvent (e = 79) than methanol (e = 32.63) therefore, the first reaction will take
place more rapidly.
(d) Since the nucleophile is not involved in the rate-determining step of an SN1
reaction, the rate is independent of the nature of the nucleophile. Therefore, the
two reactions are expected to take place at the same rate.
(e) SN1 reactions are independent of the concentration of the nucleophile because the
nucleophile is not involved in the rate-determining step of the reaction. Therefore,
the two reactions will take place at the same rate. However, the predominant
process in this pair of reactions would be E2.
(f) The first reaction will take place rapidly because the substrate is a tertiary halide.
Phenyl halides are unreactive towards SN1 reaction because, due to repulsive
interaction, backside attack is not possible.
12. Which of the following tosylates would undergo solvolysis in 80% ethanol
at a faster rate and why?
H OTs
OTs H
or
H H
I (trans) II (cis)
Solution Owing to steric interaction between the axial-OTs group and the axial hydrogen
atoms at C-3 and C-5, the cis-isomer (II) has a higher ground state energy than the trans-
isomer (I). Now, solvolysis of both of them involves a common carbocation intermediate.
So, as a first approximation, the transition state may also be assumed to be common.
Therefore, the activation energy (Ea) for the cis-isomer (higher ground state energy) will be
less than that for the trans-isomer and because of this, the cis-isomer undergoes solvolysis
at a rate faster than the trans-isomer.
13. Compare the reactivities of isopropyl chloride and benzyl chloride in

nucleophilic substitution reaction under SN1 and SN2 conditions.
Solution Benzyl chloride (PhCH2Cl) is more reactive than isopropyl chloride (Me2CHCl)
!
both in SN2 and SN1 reactions. An SN1 reaction is favoured because benzyl cation (Ph CH 2 )
!
is relatively more stable (resonance-stabilized) than isopropyl cation (MeCH) and an SN2
reaction is favoured because it is a primary substrate (Me2CHCl is a secondary substrate
and sterically more hindered) and also, the transition state is stabilized by p overlap.
14. Write the structure of the intermediate carbocation expected to be formed

in the following reaction:
solvolysis
Br
Solution
SN1
∫
+ Br
Br
Norbornyl cation
(a nonclassical carbocation)
15. In each of the following dihalides, determine which halogen atom is
relatively more reactive in SN1 reaction. Give your reasoning.
Br
O
(a) (b) (c)
CH2Br
CH3O Br
Br
Solution
(a) The bromine atom adjacent to the ring, i.e., the benzylic bromine atom exhibits
greater reactivity because the carbocation obtained on its expulsion (the benzylic
cation) is relatively more stable (resonance-stabilized) compared to the carbocation
obtained on expulsion of the other bromine atom (a 2° carbocation which is not
stabilized by resonance due to the presence of intervening saturated carbon).
Br Br
(a)
–Br
Br
(b) (a)
CH3O CH3O Br
(b)
A 2° carbocation
(less stable)
etc.
Br CH3O Br
OCH3
A benzylic carbocation
(more stable)
(b) The carbocation obtained on expulsion of the iodine atom attached to the ring is a
very unstable antiaromatic system ( 4np electron system, where n = 1) while the
carbocation obtained on expulsion of the iodine atom attached to the side chain is
a stable (resonance-stabilized) 3° allylic carbocation. Therefore, the latter iodine
atom exhibits greater SN1 reactivity. In fact, the former iodine atom is unreactive
towards SN1 reaction.
(c) The carbocation obtained on expulsion of the Br atom attached to the ring is a
very stable aromatic system [a(4n + 2)p electron system, where n = 1] while the
carbocation obtained on expulsion of the other Br atom is a relatively less stable
allylic cation. Therefore, the former Br atom exhibits greater SN1 reactivity.
16. Which one of the following two halides is more reactive in SN1 reaction
and why?
CMe2Br CMe2Br
O O
I II
Solution The halide I is more reactive than the halide II in SN1 reaction because the
intermediate 3° carbocation expected to be formed from the halide I in step 1 (the rate-
determining step) of an SN1 reaction is relatively more stable (stabilized by resonance)
than the 3° carbocation (not stabilized by resonance) expected to be formed from the
halide II.
–Br etc.
CMe2—Br
O O CMe2 O CMe2
I
A 3° carbocation
(more stable)
–Br
O CMe2—Br O CMe2
II A 3° carbocation
(less stable)
17. Predict the major product obtained from each of the following reactions
and account for its formation:
O O
H / H2O CH3O
(a) (b) CH3OH
Solution
(a) The reaction under acidic conditions proceeds to form the major product A as
follows:
In the protonated epoxide, one of the C — O bonds starts breaking in that direction
which places a partial positive charge on the more substituted carbon and this
is because the substituted carbocation is relatively more stable. Therefore, the
protonated epoxide undergoes nucleophilic attack at the more substituted carbon to
form the 1,2-diol A (with the — CH2OH group axial and the — OH group equatorial)
predominantly. The reaction in fact occurs by a pathway that is partially SN1 and
partially SN2.
(b) The reaction under basic conditions proceeds to give the major product B as follows:
O O OH
CH3O H—OCH3
CH2OCH3 CH2OCH3 + CH3O
SN2
B
When a nucleophile attacks an unprotected epoxide, the reaction is a pure SN2.

Therefore, in this case, the nucleophile CH 3O:
�� @ attacks the less substituted and
��
less satirically hindered ring carbon to yield the alcohol B (with the — OH group
axial and the — CH2OCH3 group equatorial) predominantly.
18. Propose a mechanism for each of the following reactions:
O OH
H3O
(a)
OH
OH
HBr CH2 CH==CHCH3
(b) CHCH3 Br CH2
H
AgNO3
(c) H2O / EtOH
CH2I OH
OH
CH3
H2O CH3
(d) CH—Br
(e)
Solution
d+ H OH OH OH
O d+
O
H—OH2
(a) H2O —H
OH2 OH
(b)
(c)
CH3 bond CH3 CH3

H2O shift H2O
(d) CH—Br Br + CHCH3
SN 1 –H OH
(e)
H 3C CH3
(a)
CH3 CH3 CH3 H
(b)
H2O
CHCH3 Br + C I
SN1
(a) H (b)
Br
CH3
CH3 II H
19. The relative rates of solvolysis of the following three tertiary halides in
80% ethanol at 25°C are as follows:
Br Br Br
I II III
relative rate: 1 ª10–6 ª10–14
Solution The bridgehead carbocation expected to be obtained from the [2.2.2] bicyclic
!
system II is relatively very unstable than the 3° carbocation, Me3 C expected to be
obtained from I because, being nonplanar, it suffers from angle strain and also it is not
stabilized by hyperconjugation (according to Bredt’s rule double bond cannot be formed at
the bridgehead position). Again, the bridgehead carbocation expected to be obtained from
the [2.2.1] bicyclic system III is relatively more unstable as compared to the carbocation
obtained from II and this is because it is less planar (as one of the bridges contains one
less carbon) and hence, suffers from greater angle strain. It is also not stabilized by
hyperconjugation. Because of such order of stability of the intermediate carbocations, the
halide I undergoes solvolysis very much faster than the halide II, which in turn undergoes
solvolysis very much faster than the halide III.
(from I) (from II) (from III)
Stability increases
20. Describe the Lucas test for distinguishing primary, secondary and
tertiary alcohols and discuss the reactions involved. How do you account
for the anomalous behaviour of allyl alcohol, CH2 ==CHCH2OH, and benzyl
alcohol, PhCH2OH, in this test?
Solution Whether an alcohol is primary, secondary or tertiary can be determined by
taking an advantage of the relative rates at which the three classes of alcohols react with
concentrated HCl plus ZnCl2 (the Lucas reagent). This is what is called Lucas test. When
the reagent is added to the alcohol, the mixture forms a single homogeneous phase and
this is because the concentrated HCl solution is very polar, and the polar alcohol-zinc
chloride complex dissolves in it.
When an alcohol reacts with the Lucas reagent, the relatively nonpolar and insoluble
alkyl chloride is obtained and so, the solution turns cloudy or the chloride separates into a
second phase. When the test is carried out at room temperature, the solution turns cloudy
immediately if the alcohol is tertiary, in about five minutes if the alcohol is secondary,
and remains clear if the alcohol is primary. ZnCl2, a strong Lewis acid, encourages an
SN1 pathway for the formation of the alkyl chlorides (R — Cl). Tertiary alcohols react
almost instantaneously because they form relatively stable 3° carbocations. Secondary
alcohols react in about 1–5 minutes because secondary carbocations are less stable than
the tertiary ones. Primary carbocations are too unstable to be formed and so, primary
alcohols do not react to form the corresponding chlorides at room temperature. However,
the turbidity appears when the mixture is heated or kept for a long period. In fact, the
primary alcohols then react by the SN2 mechanism which is slower than the SN1 reaction
of secondary and tertiary alcohols.
Reactions: Secondary and tertiary alcohols react with the Lucas reagent by the SN1
mechanism as follows:
Primary alcohols react with the Lucas reagent at higher temperature by the SN2 mechanism
as follows:
Because each of the allyl and benzyl alcohols produces a stable carbocation (resonance-
stabilized), they readily react with the Lucas reagent to make the solution turbid very
quickly even though they are primary alcohols.
21. When cis-2-methylcyclohexanol reacts with the Lucas reagent, the

major product is 1-chloro-1-methylcyclohexane. Propose a reasonable
mechanism to account for the formation of this product.
Solution This secondary alcohol reacts with the Lucas reagent by the SN1 mechanism
as follows:
The initially formed 2° carbocation rearranges by hydride shift to a more stable 3°

carbocation which then undergoes nucleophilic attack by Cl@ to give 1-chloro-1-methylcy-
clohexane.
22. Predict the products of the following reaction and explain their formations:
H
CH3
SOCl2
CH2 ==CH—C A+B
OH
Solution
The allyl cation obtained from the chiral chlorosulphite ester undergoes attack at the
g-carbon in two fashions to yield two geometric isomers A and B.
23. Predict the product and suggest a mechanism of the following reaction:
Solution The reaction takes place by an SNi mechanism i.e., the reaction proceeds with
complete retention of configuration to yield (R)-a-chlorethylbenzene. The steps involved
are as follows:
Ph Cl Ph O Ph –
O
–HCl slow +
C—O + C O C—O—C—Cl C C ==O
Me Me
H Cl
H Cl Me H
H
(R)-a-Phenylethanol Phosgene An alkyl chloroformate Intimate ion pair
(retention of configuration) in solvent cage
Ph
Me C——Cl + CO2
H
(R)-a-Chloroethylbenzene
(retention of configuration)
(R)-a-Phenylethanol reacts with phosgene to form an alkyl chloroformate with retained

configuration. This ester undergoes slow dissociation to give an intimate ion pair. The ion
pair rapidly collapses to form (R)-a-chloroethylbenzene. The nucleophile Cl@ ion originated
@
from OCOCl ion attacks the carbocation from the front side because it is unable to get to
the rear. This results in retention of configuration . It thus follows that the overall reaction
between the chiral alcohol and carbonyl chloride or phosgene occurs with retention of
configuration.
24. Predict the major product expected to be obtained when 1-methyl-1,2-
epoxycyclopentane is treated with (a) sodium ethoxide in ethanol and
(b) H2SO4 in ethanol.
Solution
(a) In this case, an SN2 reaction takes place. EtO@ (a strong nucleophile) attacks the
less hindered secondary carbon to give (E)-2-ethoxy-1-methylcyclopentanol.
+ – –
O Na OEt/EtOH O H H—OEt OH H –
+ EtO
SN 2
H3C H H3C OEt CH3 OEt
1-Methyl-1,2-epoxycyclopentane
(b) Under acidic conditions, the weakly nucleophilic alcohol attacks the more
electrophilic tertiary carbon atom of the protonated epoxide. In this case, therefore,
nucleophilic attack occurs from the backside (an SN2 characteristic) at the more
substituted carbon (an SN1 characteristic).
+ H
H2O
CHHO CHHO
H—OH2 d+ Et OH +
O O 3 3 + H3O
d+ SN2 + SN1
+
H3C H H3C H EtO H OEt H
H
1. Arrange the following compounds in the order of increasing reactivity towards

aqueous alkali and give your reasoning:
CH3(CH2)3Br, CH3CH2CH== CHBr, CH2== CHCH2CH2Br, CH3CH== CHCH2Br
(a) Triphenylmethyl bromide and 1-bromotryptecene are found to differ in their
rate of solvolysis under parallel conditions by a factor of 1:10–23.
CH3
|
(b) 2-Chloromethoxypropane (CH 3CHOCH 2Cl) undergoes solvolysis at a rate
faster than 1-chloro-2-ethoxyethane (CH3CH2OCH2CH2Cl).
(c) The hydrolysis of ethyl vinyl ether in dilute aqueous acid takes place 103
times faster than the hydrolysis of diethyl ether.
[Hint: Ethyl vinyl ether (C2H5OCH==CH2) undergoes hydration to form
initially a hemiacetal. The hemiacetal then undergoes hydrolysis
(by the SN1 mechanism) with greater facility because the resulting
carbocation is resonance-stabilized.]
(d)
MeOH (very fast reaction)
EtO CH2Cl EtO CH2OMe
Solvolysis
MeOH (very slow reaction)

O2N CH2Cl O2N CH2OMe
Solvolysis
*
(e) * Cl
CH3COOAg
* O COCH3 + O COCH3
14
(* ∫ C) (33.33%) (66.66%)
[Hint: Cyclopropenyl cation is a hybrid of three equivalent resonance
+
structures: *+ ´ *´ * having equal contribution (33.33%) to

+
the hybrid.]
(f) The relative rate of solvolysis of the following allylic chlorides in formic acid
containing a small amount of water is as follows:
Relative rate:
CH3 CH3 CH3
| | |
CH 2 == CH — CH 2Cl CH 2 == C — CH 2Cl CH == CH CH 2Cl CH 2 == CH — CHCl
I II III IV
1.0 0.5 3550 5670
3. Explain the following order of decreasing SN1 reactivity:

Cl—CH CH2CH3 Cl CH CH2CH3
Cl
CH CH2CH3
S N N CH CH2CH3
Cl
4. Which reaction in each of the following pairs will take place more rapidly and
why?
H2O
(a) Me3CCl æææ Æ Me3COH + HCl
H2O
(Me3C)3Cl æææ Æ (Me3C)3 COH + HCl
(b) OH@
CH 3CH 2 CH 2 CH 2 Cl æææ �� @
Æ CH 3CH 2CH 2CH 2OH + Cl:
H2O
��
OH@ ��
CH 3CH 2 OCH 2 Cl æææH2O
Æ CH 3CH 2OCH 2OH + Cl: @
��
(c)
5. Which mechanism, SN1 or SN2, is favourable for reactions with each of the following
substrates? Explain.
(a) MeO CH2Cl (b) O2N CH2Cl (c) MeOCH2Cl
[Hint: SN1 for (a) and (c); SN2 for (b)]

6. Which one of the following two bromo compounds would undergo silver ion-assisted
hydrolysis at a faster rate? Explain your answer:
7. On ozonolysis, a 14C labelled allyl chloride yields formaldehyde with 100 percent
14
C. Ozonolysis of the allyl acetate obtained by treating that allyl chloride with
CH3COOAg in acetic acid yields formaldehyde with 50 percent 14C. Explain these
observations.
[Hint:
14 CH COOAg
CH 2 == CHCH 2Cl ææææææ
3
AcOH
Æ 14CH 2 == CH — CH 2OCOCH 3 + CH 2 == CH 14CH 2 OCOCH 3 ]
SN 1
8. Explain why 2-chloro-2-methylpropane undergoes solvolysis more slowly in 90%

D2O–10% dioxane than in 90% H2O–10% dioxane solution.
(a) Alcohols (ROH) react with CH3I in the presence of Ag≈ ion to form methyl
ethers (ROCH3). However, the reaction fails in the absence of silver ion.
(b) When tert-butyl alcohol is treated with equimolar mixture of HCl and HBr,
tert-butyl bromide is obtained as the major product and tert-butyl chloride is
obtained as the minor product.
(c) tert-Butyl chloride undergoes solvolysis much faster than 2-chloro-1,l,1-
trifluoro-2-methylpropane.
(d) Tetrahydrofuran can solvate a positively charged species better than diethyl
ether can.
(e) The ether A cleaves much faster than the ether B with concentrated HBr:
HBr + CH3CH2OH ; HBr + CH3CH2Br
H5C2 O CH CH H5C2 OCH2CH3 OH

2 3 Br
A B
10. Write structures for the alcohols that you would expect from the following
reaction:
I
CH3
H2O
(CH3)3C
(a) Cl NaCN (0.01M) CN
EtOH
(Major product)
(b) NaCN (0.01M)

EtOH
Cl OEt
(Major product)
[Hint: SN1 reactions show only a slight tendency to discriminate between
the weak and strong nucleophiles, whereas SN2 reactions show a marked
tendency to discriminate.]
12. When 1-chloro-3-methylbut-2-ene or 3-chloro-3-methylbut-1-ene is treated with
Ag2O in water, the following mixture of the two alcohols is obtained along with
AgCl.
Suggest a mechanism that accounts for the formation of these products. What
might explain the relative proportions of the two alkenes that are formed?
13. Under which of the following reaction conditions (S)-2-chlorobutane would form
the most (S)-2-butanol: (i) OH@ in 50% H2O / 50% EtOH or (ii) OH@ in 100% EtOH.
14. Which of the following carbocation would you expect to rearrange and why?
+
!
(a) (CH3 )2CHCHCH3 (b) CH3 (c) +
C2H5
! + !
(d) CH3CH2 CHCH3 (e) CH2 (f) CH2 == CH — CH2 — CHMe3
15. Explain each of the following observations:

(a) SN1 type of reactions are extremely uncommon in the gas phase.
[Hint: solvation effect is absent.]
(b) Hexane is not a common solvent for either SN1 or SN2 reaction.
(c) (CH3)3CBr reacts at the same rate with F@ and H2O in substitution reaction,
even though F@ has a net negative change but H2O is neutral.
(d) When treated with silver perchlorate in propanoic acid, iodocyclopentane (I)
undergoes solvolysis, whereas 1-iodocyclopenta-2,4-diene (II) does not.
I I
I II
16. Give the mechanism and the stereochemical course involved in each of the following
reactions:
SOCl2
(a) trans-5-Methyl-2-cyclohexenol ææææ Æ
t-Bu
(b) Me SOCl2
C——C∫∫ C CH3
HO
[Hint: (a)
(b)
17. Account for the stereochemical course involved in each step of the following
reaction sequences and give three-dimensional structures (with R or S designation)
of the compounds A-G:
SOCl CH COOK KOH/H2O TsCl /Pyridine Na I
(a) (R)-1-Phenylethanol æææ
2
Æ A æææææ
3
acetone
Æ B ææææÆ
D
C ææææææ25°
Æ D ææææ
acetone
ÆE
ether
��
BsCl/Py Dioxane
(b) (S)-2-Octanol ææææÆ F G
H 2O
18. Predict the product(s) of the following reaction and explain the mechanism:
HCl/EtOH
(CH 3 )3CCH 2C(CH 3 )Cl æææææ Æ
19. Carry out the following transformation and give the mechanism involved:
Me OH Me Cl
20. When 1-chloromethyl-4-methyl-1,3-cyclopentadiene is solvolyzed in aqueous

acetone, three alcohols are obtained which are isomeric to each other. Give
structures of the alcohols and arrange them in increasing order of their yield. Give
your reasoning.
21. Explain why the following alkyl chloride does not undergo a substitution reaction
regardless of the conditions under which the reaction is carried out.
Cl
22. The compound I undergoes SN1 reaction very much slowly, but the isomeric
compound II reacts readily, even though it is a primary alkyl halide. Explain.
Me
CH2Cl
Cl
I II
23. Predict the major product and explain its formation in each of the following
reactions:
–
MeO MeOH
(a) O (b) O
MeOH H+
H Me H Me
24. State with reasons whether the following reactions will take place either by SN1 or
by the SN2 mechanism:
O
O Me H
+
O Me + MeOH
(a)
HO O OMe O
– –
Me N3 Me O + Br
(b) O
Br O N3 O
[Hint:
(a) The reaction proceeds by the SN1 mechanism because the carbocation
intermediate is highly stabilized by resonance involving two oxygen atoms.
(b) This is the only known example of SN2 reaction at a tertiary carbon and this
occurs because an SN1 reaction involving the formation of a carbocation next to
a carbonyl group cannot take place and also the carbonyl group accelerates the
SN2 reaction very much by stabilizing the transition state (the unhybridized p
orbital involved in the T.S interacts with the p orbital system of the carbonyl
group.). Furthermore, the azide ion is an excellent nucleophile which, being
sharp and marrow, is not affected much by steric hindrance.
25. In each of the following compounds, determine which of the two halogens will be
more reactive in the SN1 reaction. Explain.
Cl Cl
(a) (b) CH2 (c) Br— —CH2Br
O Cl Cl
26. Explain the observed effects of increasing solvent ionizing power on the rates of the
following reactions:
!
S 1
N Æ Ph C H + Br @ (a large increase in the reaction rate)
(a) Ph2CHBr æææ
! !
S 1
N Æ Ph C H + S(CH ) (a small decrease in the reaction
(b) Ph CH 2 S (CH 2 )3 æææ 2 3 2
rate)
Ph Ph Ph
C C + C
Cl Cl HO
H3C H3C H3C
H H H
(R)-(–)-1-Chloro-1- (R)-(–)-1-Phenyl-1- (S)-(–)-1-Phenyl-1-
phenylethane ethanol ethanol
H2O
æææ Æ 41% 59%
40% H2O
æææææ
60% acetone
Æ 47% 53%
20% H2O
æææææ
80% acetone
Æ 49% 51%
[Hint: The carbocation that is originally formed is not completely free to react
with the nucleophile water on both sides. One side of the carbocation is shielded
by the leaving group Cl@ for some time. Since the concentration of water molecules
in pure water is very high, some of the reactions take place while the carbocation
is still shielded by Cl@ ion in the front. As the percentage of water in the mixed
solvent decreases, an increasing number of carbocations can survive long enough
without reacting with water that the ions can diffuse apart. Reaction with water
can then take place on both sides of the carbocation to yield gradually an increased
amount of racemized product.]
28. tert-Alkyl fluorides (e.g., Me3C — F) are unreactive in solvolysis unless a strong
acid is present. Explain.
[Hint: :F:
�� @ ion is a poor leaving group (F@ ion is the strongest base among the
��
halide ions and the C — F bond is the strongest among the four carbon-halogen
bonds) and because of this, tert-alkyl fluorides are unreactive in SN1 solvolysis
reactions. However, in the presence of a strong acid, fluorine becomes strongly
hydrogen bonded to a proton because it is most electronegative. The leaving group
is then a stable H — F molecule rather than F@ ion and so, solvolysis takes place
smoothly. This is an example of electrophilic catalysis.]
29. Bridged alcohol as shown below cannot be chlorinated by SOCl2. Explain.
SOCl2
OH Cl
30. The acetolysis of compound I gives a mixture of two products II and III. Give
structures of II and III, and explain their formations.
31. Which of the two alkyl chlorides in each set will undergo hydrolysis at a faster rate
in an SN1 process? Give your reasoning.
(a) (Me3C)3CCl,Me3CCl (b)
32. Propose a mechanism for the following reaction:
CH3
CH3 CH3
CH3 H2SO4
CH3 CH3
HO
33. Unsymmetrical ethers are generally not prepared by heating two alcohols with
H2SO4 — Why? However, when tert-butyl alcohol is heated in methanol containing
H2SO4, a good yield of tert-butyl methyl ether results. Explain this observation by
means of reaction mechanism.
[Hint: Unsymmetrical ethers cannot be prepared by acid-catalyzed dehydration of
two different alcohols because the reaction leads to the formation of all the three
possible ethers (one unsymmetrical and two symmetrical ethers) which are not
easy to separate. However, when a 3° alcohol is used with an 1° (or 2°) alcohol, a
stable 3° carbocation generates readily and reacts with the 1° (or 2°) alcohol to give
the desired mixed ether. In fact, there is no scope of formation of symmetrical ether
corresponding to the 1° (or 2°) alcohol. Also, because of steric strain, the 3° alcohol
is much more reluctant to attack the 3° carbocation to form the corresponding
ether. It thus follows that a good yield of tert-butyl methyl ether results when tert-
butyl alcohol is heated in methanol containing H2SO4.
H —H2O CH3OH H2O

Me3C—OH Me3C—OH2 Me3C Me3C—O—CH3 Me3COCH3
H
+ H3O
34. Explain why 2-cyclopropyl-2-propanol reacts with HCl to give 2-chloro-2-
cyclopropylpropane instead of 1-chloro-2, 2-dimethylcyclobutane.
[Hint: The initially formed carbocation —CMe2 is stabilized by resonance
involving the cyclopyl group and so, it does not rearrange to give finally a cyclobutane
derivative.]
35. Generally nucleophilic substitution reaction is difficult at the bridgehead position,
but reaction given below proceeds — Why?
AgBF4
+ CO2 + AgCl + HF
PhCl
O O
C
Cl
Cl
[Hint: Certain nucleophilic substitution reactions that normally involve carbocation
(SN1) can take place at norbornyl bridgeheads if the leaving group used is of
the type that cannot function as a nucleophile and thus come back once it has
expelled. Since the expelled — OCOCl group (as CO2 and AgCl) cannot function as
a nucleophile, PhCl acts as a nucleophile to give the product.
36. Select the reaction conditions that would allow you to carry out each of the following
stereospecific transformations:
O
(R)-1,2-Propanediol
(a) H
CH3
O
(S)-1,2-Propanediol
(b) H
CH3
[Hint:
(a) The epoxide is to be cleaved by NaOH in ethanol-water.
An SN2 reaction will take place at the less substituted carbon to form (R)-1-2-
propanediol.
(b) The epoxide is to be cleaved in the presence of acid.

An SN2 reaction with considerable SN1 character will take place at the more
substituted carbon to form (S)-1,2-propanediol.
37. When (R)-(+)-2-phenyl-2-butanol is allowed to stand in methanol containing a few

drops of sulphuric acid, racemic 2-methoxy-2-phenylbutane is formed. Suggest a
reasonable mechanism for this reaction.
3.3 NEIGHBOURING GROUP PARTICIPATION (NGP)
3.3.1 Definition
A suitably located group (with a pair of electrons to offer) close to the site of the reaction in
a substrate is called a neighbouring group. In some substitution reactions, a neighbouring
group participates temporarily in the reaction to control both the rate and the stereo-
chemistry of the reaction. Such involvement of a group within the same molecule is known
as neighbouring group participation (NGP). The enhancement of the rate of a reaction
involving neighbouring group participation is called anchimeric assistance (Greek anchi +
meros, meaning neighbouring parts).
3.3.2 Mechanism of NGP

A reaction involving neighbouring group participation consists of two SN2 substitutions.
In the first step, the neighbouring group situated in the proper position for backside attack
pushes out the leaving group. Conformation plays an important role in such participation.
In the second step, the external nucleophile displaces the neighbouring group by another
backside attack and the neighbouring group goes back to its original position. The net
result of two consecutive SN2 substitutions involving inversion of configuration is retention
of configuration at the reacting carbon.
Step 1:
Step 2:
The intermediates in most neighbouring group mechanisms are not symmetrical and it
is therefore, not possible to get not a simple substitution product but a rearrangement
product. This will happen if Nu@ attacks not the carbon atom form which LG left, but the
one to which Z was originally attached.
3.3.3 Example of NGP

The displacement of chlorine form threo-3-chloro-2-butanol by the influence of dilute
MeONa is an example of a reaction that involves neighbouring group participation. The
product is formed with retention of configuration and the rate of the reaction does not
depend on the concentration of the nucleophile MeO@. The reaction is believed to proceed
through the steps as follows:
3.3.4 Anchimeric Assistance

When the reacting groups are in different molecules, they must find each other in solution
by random diffusion. It is relatively improbable that the two groups will diffuse together
and collide in just the right way for a reaction to take place. However, when the two
groups are present in the same molecule, they are already together; the only prerequisite
for the reaction is that the C — Z: bond bends towards the backside of the carbon in the
C — LG bond. Thus, the intramolecular attack of the neighbouring group on the reaction
centre has a greater probability of occurring. A higher probability of reaction results in
a lower standard free energy of activation. This means that this reaction occurring with
greater probability takes place at a larger rate. Again, a reaction between the substrate
and the external nucleophile :Nu@ involves a large decrease in entropy of activation
(DS==) because the reactants are far less free in the transition state than before. However,
|
the reaction of the neighbouring group -Z�� involves a much smaller loss of DS=|= because the
reactants are not very much free in the transition state than before.
For this reason, attack by the neighbouring group (rate-determining step) takes place
faster (lower DG== ) than by the external nucleophile and as a result, a large enhancement
|
of substitution rate occurs.

3.3.5 Evidence for Participation by a Neighbouring Group

The acceleration in the rate of a reaction is a good evidence for participation by a
neighbouring group. Other evidence may be obtained from the stereochemical results
of the reaction. Reactions involving neighbouring group participation are generally
stereospecific. Usually such reactions yield products with retention of configuration as
opposed to the normal racemization and inversion of the reaction centre.
3.3.6 Various Cases of Neighbouring Group Participation

A neighbouring group may be any group containing an atom like sulphur, nitrogen,
oxygen or bromine or a p bond or even a s bond. A successful NGP requires antiperiplaner
geometry of the leaving group and the neighbouring group so that the stereoelectronic
requirement for intramolecular SN2 reaction is fulfilled. Let us see different cases of
neighbouring group participation.
3.3.6.1 Heteroatoms as neighbouring groups

(1) There is a remarkable difference in the reaction rates of the following two
substitution reactions, which are superficially very similar:
100∞C
(i) CH3 (CH2 )4 CH2Cl ææææ
H2O
Æ CH3 (CH2 )4 CH2OH + HCl 1
1-Chlrohexane 1-Hexanol
100∞C
(ii) ��
CH3CH2 SCH 2 CH2 — Cl ææææ
H 2O
��
ÆCH3CH2 SCH 2 CH2 — OH + HCl 3200
��
1-Chloro-3-thiapentane ��
3-Thia-1-pentanol
(b -chloroethyl ethyl sulfide) (b -Hydroxydiethyl
sulfide)
At first sight, both reactions appear to be simple SN2 reaction in which chloride
ion is displaced as a leaving group by water. In fact, 1-chlorohexane reacts by the
following mechanism:
The presence of sulphur in the alkyl halide molecule should have little effect on
the rate of the SN2 reaction, because the SN2 reaction is not very sensitive to the
electronegativities of substituent groups. Yet the reaction of 1-chloro-3-thiapentane
is thousands of times faster than the reaction of 1-chlorohexane with water.
The rate of the second reaction is unusually large because a special mechanism
facilitates the reaction, a mechanism not available to 1-chlorohexane. In the
reaction of sulfide, it is the S atom (a powerful nucleophile that can participate
as a neighbouring group) that displaces the leaving chloride. This intramolecular

reaction is much faster than the intermolecular displacement of chlorine by water
either in the alkyl chloride or in the chlorosulphide.
CH2CH3 CH2CH3
S: :S
SN2 Cl
CH2——CH2 (r.d.step) CH2——CH2
Cl An episulfonium salt
The episulfonium ion that results from this internal nucleophilic substitution
reaction is very reactive because it contains a strained three-membered ring
and a good leaving group. This intermediate is opened up by water in a second,
intermolecular displacement step to give the product.
CH2CH3
:S H
SN2 Cl
CH2——CH2 + H2O CH3CH2 S CH2CH2—OH CH3CH2 S CH2CH2OH + HCl
3-Thia-1-pentanol
(2) When (S)-2-bromopropanoate ion is treated with concentrated solution of sodium
hydroxide, (R)-2-hydroxypropanoate is obtained. However, when the same reaction
is carried out with a low concentration of hydroxide ion in the presence of Ag2O
(where Ag ≈ ion acts as a Lewis acid), (S)-2-hydroxypropanoate ion is obtained.
conc.NaOH
CH 3CH Br COO@ æææææ
solution
Æ CH 3CH OH COO@
(S)-2-Bromopropanoate ion (R)-2-Hydroxypropanoate ion
dil.NaOH
CH 3CH Br COO@ æææææ
solution
Æ CH 3CH OH COO@
(S)-2-Bromopropanoate ion Ag2O (S)-2-Hydroxy propanoate ion
In the presence of concentrated alkali, (S)-2-bromopropanoate ion undergoes
normal SN2 reaction to yield (R)-2-hydroxypropanoate ion. The reaction proceeds
with inversion of configuration at the chiral carbon.
However, in the presence of Ag2O and a low concentration of OH@ ion, the same
reaction occurs with retention of configuration at the chiral carbon. In this case,
the reaction proceeds by the neighbouring group mechanism which consists of two
inversion steps. In the first step, an oxygen of the carboxylate group attacks the
stereocentre from the backside and displaces bromide ion to form an a-lactone
(a cyclic ester) with inversion of configuration. The silver ion here acts as an
electrophilic catalyst and aids the expulsion of bromine. In the second step, the
highly strained three-membered lactone ring undergoes SN2 attack by H2O from
the side as that of the expelled Br@ to produce (S)-2-hydroxypropanoate ion with
inversion of configuration. Therefore, the net result of two inversions in two
successive steps is an overall retention of configuration.
(3) The fact that acetolysis of both 4-methoxy-1-pentyl brosylate (I) and 5-methoxy-2-
pentyl brosylate (II) gave the same mixture of products (III and IV) is an evidence
for participation by a neighbouring group.
It is the ether oxygen that acts as a neighbouring group in the acetolysis of these
@
two isomeric brosylates. The sulfonate ion (OBs) is an excellent leaving group and
a five-membered ring is relatively unstained, so the intramolecular displacement
of p-bromobenzenesulfonate to give a cyclic oxonium ion takes place in both the
acetolysis reactions. The cyclic oxonium ion is opened by acetic acid in two ways to
give a substitution product (III) and a rearrangement product (IV).
Since the more substituted carbon of the oxonium ion is more positive, nucleophilic
attack by CH3COOH occurs preferably at that carbon to give the major product
(IV).
(4) Optically active threo-3-bromo-2-butanol reacts with HBr to give (±)-2,3-
dibromobutane. This observation suggests that the reaction proceeds not by the
ordinary SN2 or SN1 mechanism, but by the neighbouring-group mechanism in
which bromine acts as the neighbouring group. A meso-dibromide is expected to
be obtained if threo-3-bromo-2-butanel reacts with HBr by the SN2 mechanism.

However, if the reaction proceeds by the SN1 mechanism, a mixture of meso and
active products is expected.
Because the reaction produces only (±)-2,3-dibromobutane, it is not an SN2 or SN1

reaction. The formation of only racemic 2,3-dibromobutane can be explained by a
neighbouring group mechanism. The suitably oriented bromine atom on the adjacent
carbon acts as a neighbouring group and pushes out H2O (SN2) from the conjugate
acid of the alcohol to form a bromonium ion intermediate. This intermediate then
undergoes nucleophilic attack by :Br �� :@ on C-2 and C-3 equally well to produce
��
optically inactive racemic or (±)-2,3-dibromobutane.
(5) In a 1,2-disubstituted cyclohexane system, neighbouring group participation occurs

only when there is a trans diaxial arrangement of the departing and participating
groups and this is because antiperiplanar geometry is provided only in such
conformation. Also, such conformation is very suitable for backside attack required
for an internal SN2 .
:G
:G
H
H
G: H ∫
X
H X
X
trans diequatorial conformation trans diaxial conformation
(not a suitable geometry for (a suitable geometry for
backside attack) backside attack)
For example, the trans-isomer of 2-bromocyclohexanol, which can attain a diaxial

geometry necessary for backside attack, undergoes intramolecular SN2 reaction
(displacement of Br@ ion by the oxyanion) to yield cyclohexene oxide (an epoxide).
The cis-isomer, on the other hand, cannot attain the antiperiplanar geometry
necessary for backside attack and so, it undergoes a much slower reaction in which
migration of a hydride ion occurs to form cyclohexanone.
Among the following, two isomeric 1,2-bromohydrins I reacts with alkali to form an
epoxide but II does not.
H H
OH OH
Me Br Me Br
H H
I II
These are trans-decalin systems and so, these are locked in these two conformations.
i.e., they cannot flip to any other conformation.
The bromohydrin I will form an epoxide on treatment with alkali because the
nucleophile (—O@) and the leaving group (—Br) are suitably placed (antiperiplanar
arrangement) for a backside attack required for an internal SN2 reaction.
Since flipping is not possible, II cannot take up the conformation in which — OH and
— Br are diaxial. For this reason, II cannot react with alkali to form an epoxide.
(6) When active trans-2-acetoxyclohexyl tosylate (I) is treated with potassium acetate
(AcOK) in acetic acid (AcOH), (±)-trans-1,2-cyclohexanediol diacetate is obtained.
However, when active cis-2-acetoxycyclohexyl tosylate (II) is treated similarly,
active trans-1,2-cyclohexanediol diacetate is obtained.
An acetoxy group, containing oxygen with unshared electrons, can function as

a neighbouring group. In the diaxial form of the trans-isomer (I), the acetoxy
(— OCOCH3) group is suitably placed for backside attack (antiperiplanar
arrangement) and so, it pushes out the tosylate anion :OTs
�� @ to form an acetoxonium
��
ion (step 1) and, in doing this, inverts the configuration at the carbon under attack.
The acetoxonium ion undergoes ring flip to form its mirror image. These two
enantiomeric intermediates then undergoes SN2 attack by the acetate ion (step 2)
again with inversion of configuration at the carbon under attack and as a result,
racemic trans-1,2-cyclohexanediol diacetate is obtained.
This cis-isomer (II), on the other hand, cannot assume the diaxial conformation
required for backside attack by the acetoxy group and therefore, neighbouring group
participation is absent in this case. Thus, this isomer reacts by the normal SN2
mechanism to yield active trans-1,2-cyclohexanediol diacetate and the reaction is,
therefore, much slower than for the trans-isomer (I) to which anchimeric assistance
is available.
3.3.6.2 p bond as neighbouring group

(1) If an olefinic C == C bond at Cg is stereochemically anti to the leaving group, it
may act as a neighbouring group under SN1 conditions. Thus, acetolysis of anti-7-
norbornenyl tosylate (I) is 1011 times faster than that of its saturated analog (II)
and proceeds with retention of configuration.
H OTs H OTs
I II
Since the double bond in anti-7-norbornenyl tosylate (I) is situated in an especially
favourable position for backside attack on the carbon bearing the leaving group, it
acts as a very effective neighbouring group in the acetolysis of I and assists in the
departure of OTs@. For this reason, I undergoes acetolysis at a rate much faster than
II where there is no anchimeric assistance is available. The resulting nonclassical
or bridged carbocation then reacts with AcOH forming the side opposite to the
C == C group to from the corresponding acetate with retention of configuration.
(2) The C== C bond of an aryl ring at Cb with respect to the leaving group can act
as a neighbouring group. For example, optically active isomer of 3-phenyl-2-butyl
tosylate gives a racemic mixture of the corresponding acetates.
H 1 H H
3 2
CH3 3 2
CH3 2 3
CH3COOH H3C
C——C C——C + C——C H
H H
OTs OAc AcO CH3
H3C H3C
(2R, 3S) (2S, 3R)
Optically active threo
-3-phynyl-2-butyl tosylate Racemic mixture of 3-phynyl-2-butyl acetate
The formation of these two enantiomeric acetates can only be explained if the
phenyl group in the b position can function as a neighbouring group. Simple SN2 or
SN1 reaction cannot explain their formations. The reaction takes place as follows:
On the other hand, optically active erythro-3-phenyl-2-butyl tosylate undergoes

acetolysis to give optically active 3-phenyl-2-butyl acetate instead of racemic
mixture.
H H
CH3 CH3
CH3COOH
C——C C——C
H3C H3C
OTs OAc
H H
Optically active erythro-3-phenyl-2-butyl Optically active-3-phenyl-2-butyl
tosylate tosylate
(2R, 3R) (2R, 3R)
The reaction takes place as follows:
Since in the achiral or meso-phenonium ion the two methyl groups are cis while in
the chiral or active phenonium they are trans, therefore, the latter phenonium ion
is thermodynamically more stable (due to steric reason) than the former and for
this, the erythro-isomer undergoes acetolysis at a rate somewhat faster than the
threo-isomer.
(3) It is known that the properties of a cyclopropane ring are in some ways similar
to those of a double bond. Therefore, it is not surprising that a suitably placed
cyclopropyl ring can also be a neighbouring group. For example, acetolysis of endo-
anti-tricyclo[3.2.1.02,4]octan-8-yl-p-nitrobenzoate (A) is very much (~ 1014 times)
faster than that of the p-nitrobenzoate of 7-hydroxybicyclo[2.2.1]heptane (B).
O O
H O—C— —NO2 H O—C— —NO2
A B
In the compound A, the suitably placed cyclopropyl ring acts as a very effective
neighbouring group (even more effective than a double bond) and assists in the
departure of p-nitrobenzoate ion (p-NO2C6H4COO@). Therefore, this compound
undergoes acetolysis at a rate very much faster than that of the compound B where
no such anchimeric assistance is available. The resulting nonclassical carbocation
then undergoes nucleophilic attack by CH3COOH from the side of the expelled
leaving group to form the corresponding acetate with retention of configuration.
3.3.6.3 C — C s bond as a neighbouring group

Solvolysis in acetic acid of optically active exo-2-norbornyl brosylate produces a racemic
mixture of the two exo acetates.
AcOH
OBs OAc AcO

H H H
exo-2-Norbornyl brosylate (±)-exo-2-Norbornyl acetate

(active)
The brosylate undergoes s bond assisted ionization (the C-1 — C-6 s bond acts as a
neighbouring group) to give a nonclassical carbocation. This intermediate carbocation
then undergoes nucleophilic attack by CH3COOH with equal facility on the two equivalent
carbons to give equimolar mixture of two enantiomeric exo-2-norbornyl acetates.
Acetolysis of the exo-isomer is about 350 times faster than its endo-isomer and this because
the solvolysis of the endo-isomer is not assisted by the C-1 — C-6 bond because it is not in
a favourable position for the backside attack.
1. Account for the following observation of scrambling:

CF COOH
Ph14CH 2CH 2OTs æææææ
3
D
Æ PhCH 214CH 2OCOCF3 + Ph14CH 2CH 2OCOCF3
Solution A phenyl group at the b-position can function as a neighbouring group. The
reaction occurs to give a mixture containing equal amounts of 1–14C–2-phenylethyl
trifluoro acetate (I) and 2–14C–2-phenylethyltrifluoro acetate (II) as follows:
In the first step, the phenyl group pushes out the tosylate anion (an SN2) to form a bridged
ion, called a phenonium ion. In the second step, the phenonium ion undergoes nucleophilic
attack by fluoroacetic acid at either of two methylene carbons with equal facility to form
a mixture of I and II
2. Explain why b-(p-hydroxyphenyl)ethyl bromide reacts with MeO@/MeOH
about 106 times faster than b-(p-methoxyphenyl) ethyl bromide.
Solution In b-(p-methoxyphenyl)ethyl bromide, the — OMe group at the para-position
makes the benzene ring a better neighbouring group because it stabilizes the arenium ion
intermediate by its +R effect. However, the resulting intermediate is positively charged
and relatively less stable.
OMe OMe OMe

— Br MeO
slow MeOH
CH2—CH2—Br CH2 CH2 CH2CH2OMe
b-(p-Methoxyphenyl)ethyl An arenium ion
bromide intermediate
(less stable)
When b-(p-hydroxyphenyl)ethyl bromide is treated with MeO@/MeOH, it is first converted

into the corresponding phenoxide ion. The phenoxide ion is a much better neighbouring
group because the resulting bridged dienone intermediate is neutral and relatively
more stable. Because the neighbouring group effect or anchimeric assistance in b-(p-
hydroxyphenyl)ethyl bromide is much more pronounced than in b-(p-methoxyphenyl)
ethyl bromide, the former bromide reacts with MeO@/MeOH at a very much faster rate
than the latter bromide.
OAC O Me
EtOH
3. C
OBs OEt O
What is the significance of the reaction?
Solution Formation of such a product indicates that the — OAc group acts as a
neighbouring group in the displacement of OBs@.
4. Give mechanisms of the following reactions:
Cl
D
(a) H2O
N N
Et Et
O
CF3COOH
(b) CH3C∫∫ C CH2CH2OTs C—CH3 +
H2O
CH3 O
Solution
(a)
(b)
5. Both cyclopropylmethyl and cyclobutyl chlorides undergo hydrolysis

at abnormally high rate to give cyclopropylmethyl alcohol, cyclobutyl
alcohol and but-3-en-1-ol. Explain these observations.
Solution Cyclopropylmethyl and cyclobutyl chlorides undergo solvolysis very rapidly

due to participation of the C — C s bond of the ring. It has been suggested that a common
nonclassical intermediate is involved in both the processes and for this, both of these
substrates give the three common products.
6. When trans-3-bromocyclohexane carboxylic acid is treated with sodium

hydroxide solution, a g-lactone is obtained. However, when its cis-isomer
is similarly treated, no lactone is formed. Explain these observations.
Solution In conformation II of trans-3-bromocyclohexanecarboxylate anion, the — COO①
group is suitably placed to displace bromide ion by a backside attack and because of this,
an intramolecular SN2 reaction occurs to form a g–lactone.
–
O O
Ring H
C
O NaOH O flip
H C H2O C Br
H – +
O O Na
H
Br H – Br H II
H OH –
Br SN2
I
trans-3-Bromocyclohexane- O
carboxylic acid O C
H
H
A g -lactone
The backside attack is, however, not possible in either conformation of the cis-isomer
because the — COO① group is not properly placed to displace bromide.
7. Predict the product of the following reaction and explain the mechanistic
course involved:
HO H
–
CH3 OH
C——C
H2O
H3C
H Cl
(optically active)
Solution This optically active molecule of 3-chloro-2-butanol reacts with alkali to form
optically inactive meso-2, 3-butanediol. In the first step, the oxyanion formed from the
hydroxyl group pushes out chloride ion (an intramolecular SN2 to form an epoxide. In the
next step, the epoxide undergoes substitution by external hydroxide ion in an intermolecular
process. Ring opening occurs with equal probability at either carbon atom of the epoxide.
The optically inactive meso-2, 3-butanediol is obtained from two stereospecific inversion
steps.
8. Propose a mechanism for the following reaction:
OH OH Br
Me NaOH HBr Me + Me
Me Me Me
Cl Br OH
Solution The reaction proceeds through an intramolecular SN2 followed by an

intermolecular SN2 to give these two isomeric products.
OH
H H
Me Me –
–
OH O H
Me ∫∫ Me Cl O
–
Me Cl Me
HO
Me H Cl
H H Me
Cl
In the boat form, the
–
oxyanion is properly – Cl Intermolecular
placed to displace chloride SN2
by a backside attack.
H
+
O O
Br–H H H
H Me H Me
Me Me
– –
Br Intermolecular Intermolecular Br
SN2 SN2
H OH OH H
Me Me
Br H H Br
Me Me
OH H
Me Me
H
Me Me Br
H HO
Br H
∫∫
∫∫
Br OH
Me Me
Me Me
OH Br
9. Explain why the following two alcohols react with HCl to yield the same
alkyl chloride.
CH3 Cl CH3
| | |
HCl
�� — CH — CH — OH æææ HCl �� — CH — CH — OH + H O
EtS 2 Æ EtS — CH2 — CH — CH3 ¨ææ æ EtS 2 2
��
Solution Both of these two alcohols produce the same episulfonium salt by an
intramolecular SN2 reaction. In the episulfonium ion, the substituted carbon is more
positive and so, like a protonated epoxide, the more substituted carbon undergoes
nucleophilic attack by weak uncleophile Cl① to give the common product.
10. Explain the formation of exo-2-norbornyl acetate from the following reaction:
CH3COOH
OTs
OAc
H
Solution Displacement of the leaving group by the double bond of the five-membered
ring gives the 2-norbornyl cation.
Opening by acetic acid in an intermolecular SN2 reaction leads to the formation of the exo
acetate.
OH SN2 O—H +
+ O ==C + –H
+
CH3 O ==C
CH3 OAc
H
H
H OH
O +
H3O OH
(a)
(CH3)3COK
(b) HO— —(CH2)4OBs O
(CH3)3COH
Ph
C
== O
NHCOPh CH3 N
(c) H C3H5OH
C——C C——C
H KOAC H CH3
H3COOC OTs H3COOC H
Solution
(a)
+
+
O H—OH2 OH
SN2 OH
+
+
—H H2O
H OH
OH
∫ OH
HO
(b)
(c)
12. How do you account for the following relative rates of acetolysis of
p–substituted cyclohexyl brosylates? In which case there is evidence of
neighbouring group participation?
Z Relative rates
cis trans
Cl 1.6 5.9
Br 1.5 1250
I 2.2 ¥ 108
H 1.2 ¥ 104
Solution In solvolysis with SN1 character, halogen atom at b-position could exert two
opposing effects. By its electron-attracting inductive effect, it could tend to slow down
the reaction by intensifying the positive charge developing on the carbon atom. Through
anchimeric assistance, it could tend to speed up to reaction. Only in the trans isomers –X
and –OBs can take up the diaxial conformation required for anchimeric assistance.
:Z
H
H
OBs
From the given data it becomes clear that there is no anchimeric assistance by cis-Cl or
trans-Cl, or by cis-Br; there is only strong deactivation expected from the — I effect. The
trans-Br compound is much more reactive than the cis-Br compound. This indicates that
there is anchimeric assistance although not strong enough to offset the inductive effect
completely. The trans-I compound provides powerful anchimeric assistance, more than
offsetting any inductive (— I) effect and so, in this case solvolysis occurs at a rate 18000
times as fast as for the unsubstituted brosylate. Hence, the ability of halogens to give
nucleophilic assistance falls in the order I > Br > Cl.
EtO2C NaOEt
CO2Et
(a) CH(CH2)2Br
EtO2C EtOH
CO2Et
KOH
(b) Et 2N — CH 2 CHCH 3 æææ
H2O
Æ Et 2N — CH — CH 2OH
| |
Cl CH3
Solution
(a) The base OEt① abstracts the acidic hydrogen to form the corresponding conjugate
base. An intramolecular SN2 reaction then takes place to form a cyclopropane
derivative. Strain in the three-membered ring is offset by the relatively favourable
contribution of entropy to the activation energy.
H
OEt –Br
(EtO2C)2 C CH2CH2Br (EtO2C)2 C—CH2—CH2—Br
Intramolecular SN2
CH2
(EtO2C)2C
CH2
(b) Nitrogen pushes out chloride in an intramolecular SN2 process to form a three-
memebered ring of aziridinium ion. Attack by the strong nucleophile OH① then
takes place at the less hindered position and as a result, the ring is opened up to
give the product.

OCH3 Cl OCH3
Cl Cl
C C C
O Cl O OCH3 O Cl
I III II
Solution The reaction takes as follows:
Due to favourable geometry in the diaxial conformation of I, and nucleophilic substitution

on the carbonyl group occurs to form a lactone containing positive charge on oxygen. An
intermolecular SN2 reaction then takes place with inversion of configuration at the carbon
under attack.
The compound II does not react because in none of the conformations the — OCH3 group
is properly placed to displace Cl① from — COCl.
O Cl
C H
OCH3
O
∫ H :OCH3 No reaction
Cl C
H
H :OCH3
C
O Cl (e, a) (a, e)
II

* Ag2O *
S S S
Cl H2O OH + OH (* == 14C)
*
2. Predict the product of the following reactions and account for their formations:
OH *
OH *
(a) HO CH 2CH 2Br æææÆ
H O
(b) H 2N(CH 2 )4 Br æææÆ
H O
2 2
OH 18OH
OH
(c) H (d) H3PO4
H2O
OH
H
OH
Cl H2O OH
(e) (f) Br(CH2)3 COOH H2O
HS
Me
[Hint:
(e)
3. Account for the fact that the two isomeric bromoethers (A and B) undergo solvolysis
in acetic acid to give the same mixture of products (C and D).
Me
MeO Me MeO Me MeO
AcOH AcOH
Br OAc Br
A C(40%) B
+
Me
MeO
Br
D(60%)
HO—CEt2 HO—CEt2
OH
C——Cl C—OH + Cl
Me H2O Me
H H
[Hint: The retention of configuration is a result of two successive inversions
(SN2).]
5. Give the mechanism of the following reaction and account for the relative rate of
acetolysis of the following two tosylates (I and II):
TsO H TsO H
AcOH ;
I OAc II
rate = 104 rate = 1
[Hint: The C == C bond of the syn-tosylate I does not assist the ionization of the
substrate because it is not properly situated for participation as a neighbouring
group. The solvolysis occurs through the formation of a homoallylic carbocation
which rearranges to an allylic carbocation. The allylic carbocation then reacts with
AcOH to give an acetate. The high reactivity (104 times) of I compared to II is due
to participation of s electron of the two allylic s bonds (C-1 — C-6 and C-4 — C-5).
6. Explain why the compound I undergoes solvolysis at a rate much slower than the
compound II.
O O
CH3O— —S—O H CH3O— —S—O H
O O
CF3
CF3
I II
[Hint: The ability of C ==C to serve as a neighbouring group depends on its electron
density.]
7. Arrange the following compounds in order of increasing rate of solvolysis and
explain the order:
H OCOC6H4NO2-p H OCOC6H4NO2-p H OCOC6H4NO2-p
I II III
8. The brosylate I undergoes acetolysis ~ 140,000 times faster than its saturated
analog II. Explain.
H OBs H OBs
I II
[Hint: The double bond in I is geometrically fixed in an especially favourable
position for backside attack on the carbon bearing the leaving group and so, there
is a very large rate enchancement.]
9. When 5-chloro-2-hexyl tosylate is solvolyzed in acetic acid, there is little participation
by the Cl, but when the solvent is changed to trifluoroacetic acid, neighbouring-
group participation by the Cl becomes the major reaction pathway. Explain.
[Hint: Cl is a very weak neighbouring group and can be shown to act in this way
only when the solvent does not interfere.]
10. When G ==OCH3, about 93% of the products of solvolysis of I can be attributed to the
participation of the neighbouring phenyl group. But, when G==NO2, essentially none
of the products come from neighbouring phenyl group participation. Explain.
HCOOH
G
OTs
I
11. The compound I undergoes solvolysis in 80% aqueous ethanol 105 – 106 times faster
than its cis-isomer II. Explain.
[Hint: The high solvolysis rate of I is due to participation of the neighbouring

thiophenoxy group leading to a cyclic intermediate.]
12. The tosylate I solvolyzes 3 ¥ 105 times faster than the tosylate II. Explain.
TsO TsO
I II
13. Explain why ‘mustard gas’, S(CH2CH2Cl)2, a high boiling liquid used in World War
I as a combat gas in the form of aerosol, hydrolyses much more rapidly to give HCl
and a diol than its sulphur free analog, 1,5–dichloropentane.
[Hint: The reason for the high rate of hydrolysis of mustard gas is due to
participation of sulphur atom as the neighbouring group in an intramolecular SN2
reaction
14. In the nucleophilic substitution reaction of the following 14C labelled compound
with water, what labelling pattern is expected to be observed in the product (a) if
the neighbouring group participation does not take place and (b) if neighbouring
group participation does take place?
S *
Cl (* ==14C)
15. Although there is a substantial difference in the rate at which I and II solvolyze
(I undergoes solvolysis 4.4¥104 times faster than II is acetic acid), both compounds
give products of completely retained configuration.
Br H H Br
I II
Offer a mechanistic interpretation of the above observation.

[Hint:
17. Which reaction in each pair would be expected to be faster? Give your reasoning.
CH3 S CH3 S
(a) CF3— –C—O—C—Ph or CH3— –C—O—CPh (solvolysis in 100%
ethanol)
CH3 CH3
H CH2CH2ONoS H CH2CH2ONoS
(b) or (solvolysis in acetic acid)
(c) PhS(CH2 )3 Cl or PhS(CH2 )4 Cl (solvolysis in methanol)
(d) PhO2C— —CHBrPh or —CHBrPh (solvolysis in acetic acid)

CO2Ph
4
CHAPTER
ELIMINATION
REACTIONS
Chapter Outline
4.1 The E2 Reaction 4.2.8 Rearrangement of the Carbocation
4.1.1 Example of E2 Reaction Intermediate Involved in an E1
4.1.2 Kinetics of E2 Reaction Reaction
4.1.3 Mechanism of E2 Reaction 4.2.9 Acid-Catalyzed Dehydration of
Alcohols
4.1.4 Stereochemistry of E2 Reaction
4.2.10 Dehydration using POCl3 and
4.1.5 Evidence in Favour of the E2
Pyridine
Mechanism
4.2.11 Factors Influencing the Extent of
4.1.6 Factors Influencing E2 Reaction
E1 and E2 Reactions
Rate or E2 Reactivity
4.2.12 Factors that Govern the Proportions
4.1.7 Factors that Govern the Proportions
of E1 and SN1 Reactions
of E2 and SN2 Reactions
4.1.8 Regioselectivity in b-elimination 4.3 The E1CB Reaction
Reactions (Orientation of p Bond in 4.3.1 Example of E1cB Reaction
the Product Alkene) 4.3.2 Kinetics of E1cB Reaction
4.1.9 Hofmann Exhaustive Methylation 4.3.3 Mechanism of E1cB Reaction
or Hofmann Degradation 4.3.4 The Nature of the Substrate
4.1.10 Fragmentations 4.3.5 To Distinguish between E1cB and
4.1.11 Summary of the E2 Reaction E2 Mechanisms
4.2 The E1 Reaction 4.4 a- or 1,1-Elimination
4.2.1 Example of E1 Reaction 4.4.1 Example of a- or 1,1-elimination
4.2.2 Kinetics of E1 Reaction Reaction
4.2.3 Mechanism of E1 Reaction 4.4.2 Kinetics of a- or 1,1-elimination
4.2.4 Stereochemistry of E1 Reaction Reaction
4.2.5 Evidence in Favour of the E1 4.4.3 Mechanism of a- or 1,1-elimination
Mechanism Reaction
4.2.6 Factors Influencing E1 Reaction 4.4.4 Structure of the Substrate Involved
Rate or E1 Reactivity in a-elimination Reactions
4.2.7 Regioselectivity of E1 Reactions
INTRODUCTION
Elimination reactions belong to one of the major class of organic reactions involving the loss
of two atoms or groups from a substrate molecule. Thus, it is the reverse process of addition
reaction. Alkyl halides, in addition to the nucleophilic substitution reactions, also undergo
elimination reactions. The loss of one H atom and the Br atom (dehydrobromination)
from 2-methyl-2-bromopropane by the action of alcoholic alkali is a typical example of
elimination reaction.
alcohol
(CH 3 )3CBr + KOH ææææD
Æ (CH 3 )2 C == CH 2 + KBr + H 2O
2-Methyl-2-bromopropane 2-Methylpropene
The carbon atom that bears the leaving group (e.g., the halogen atom in the above reaction
is called an alpha (a) carbon atom and any carbon atom adjacent to it is called a beta (b)
carbon atom. A hydrogen atom attached to the b carbon atom is called a b hydrogen atom.
The carbon next to b is called a g carbon atom.
On the basis of the relative positions of the eliminating groups or atoms, the elimination
reaction may be classified as: (a) 1,1 or a-elimination reaction, (b) 1,2- or b-elimination
reactions and (c) 1,3 or g -elimination reactions.
(a) 1,1- or a-Elimination reactions: When the two atoms or groups are lost from
the same atom of a substrate molecule in an elimination reaction, the reaction is
said to be a 1,1- or a-elimination reaction. Such a reaction leads to the formation
of very reactive intermediate, carbene, rather than an alkene. For example, when
chloroform (CHCl3) is heated with alcoholic KOH solution, it undergoes a 1,1- or
a-elimination reaction to yield dichlorocarbene (:CCl2).
1(a )
alc. KOH
CHCl3 ææææÆ D
:CCl2 + H 2O + KCl
(b) 1,2 - or b-Elimination reaction: When the two atoms or groups are lost from the
two adjacent atoms of a substrate molecule in an elimination reaction, the reaction
is said to be 1,2-elimination reaction. In the older ab-terminology, a is commonly
omitted, and the reactions are referred to as b-eliminations.
In a 1,2 - or b-elimination reaction a p bond is formed.
The formation of cyclohexene from cyclohexyl chloride by the action of sodium

ethoxide in ethanol is an example of 1,2- or b-elimination reaction.
Elimina on Reac ons 4.3
(c) 1,3- or g-Elimination reaction: When in an elimination reaction the two atoms
or groups are lost from two different atoms which remain separated by another
atom, the reaction is called a 1,3- or g-elimination reaction. These less common
elimination reactions lead to the formation of three-membered rings. For example,
when the following quaternary ammonium iodide is treated with sodamide in liquid
ammonia it undergoes 1,3-elimination reaction to form a cyclopropane derivative.
Depending upon the mechanism involved 1,2- or b-elimination may be classified as

E1, E2 and E1cB. Most common b-eliminations follow E2 path.
• Competition between nucleophilic substitution and b-elimination reactions
Because the reactive part of both nucleophiles and bases is an unshared electron pair,
all nucleophiles are potential bases and all bases are potential nucleophiles. For this
reason, nucleophilic substitution reaction and elimination reaction often compete with
each other.
When the substrate is a primary halide and the base is strong and unhindered, like
ethoxide ion (OEt * ), substitution in highly favoured and this is because the base can
easily approach the carbon bearing the leaving group. For example:
* ≈
C H OH
CH 3CH 2 O Na + CH 3CH 2Br æææææ
2 5
55∞ C
Æ CH 3CH 2OCH 2CH 3 + CH 2 == CH 2
Primary SN 2 Product E2 Product
substrate Major(90%) Minor(10%)
With secondary substrates, however, a strong base favours elimination reaction because
substitution becomes more difficult due to steric hindrance in the substrate. For
example:
Br
* ≈ | C2H5OH
C2H5 ONa + CH 3 CH CH 3 æææææ
55∞ C
Æ CH 2 == CHCH 3 + (CH 3 )2 CH — O — C2H5
Secondary E2 Product SN 2 Product
substrate Major (79%) Minor (21%)
With tertiary halides, an SN2 reaction cannot take place because steric hindrance in the
substrate is severe and thus the elimination reaction is highly favoured, especially when
the reaction is carried out at higher temperature. Any substitution that occurs must take
place through an SN1 reaction. For example:
At room temperature
* ≈ C H OH
C2H5 O Na + (CH 3 )3 C — Br æææææ
2 5
25∞ C
Æ (CH 3 )2 C == CH 2 + (CH 3 )3 C — O — C2H5
Tertiary E2 Product SN1 Product
substrate Major (91%) Minor (9%)
At higher temperature
* ≈ C H OH
C2H5 O Na + (CH 3 )3 C — Br æææææ
2 5
55∞ C
Æ (CH 3 )2 C == CH 2
Tertiary OnlyE2 + E1 Product
substrate (100%)
4.1 THE E2 REACTION

Example, kinetics, mechanism, stereochemistry, evidence in favour of the
mechanism, the factors influencing E2 reactivity, i.e., the E2 reaction rate and
other relevant topics.
4.1.1 Example of E2 Reaction

Dehydrobromination of isopropyl bromide caused by the strong base sodium ethoxide
(NaOEt) to yield propene is a typical example of E2 reaction, i.e., a reaction which proceeds
through the E2 mechanism.
EtOH
EtO : * + (CH 3 )2 CHBr ææææ Æ CH 3 CH == CH 2 + EtOH + :Br :*
Isopropyl bromide Propene (79%)
4.1.2 Kinetics of E2 Reaction

The E2 reaction exhibits second-order kinetics (first order in the substrate and first order
in the base) and obeys the following rate law:
Rate = k ÎÈ(CH3 )2 CHBr ˚˘ [EtO :* ]
The rate of elimination thus depends on the concentration of both the substrate and the
base. Therefore, the reaction is bimolecular, i.e., the substrate (Me2CHBr) and the base
( : OEt * ) are involved in the transition state of the rate-determining step.
4.1.3 Mechanism of E2 Reaction

The most straight-forward explanation for the observed second-order kinetics is a concerted
reaction, i.e., bond breaking and bond making occur simultaneously. The mechanism of
this reaction is designated as E2 (E stands for elimination and 2 stands for bimolecular)
because the two reactants (the substrate and the base) are involved in the transition state
of the rate-determining step (the only step).
In the E2 reaction, the base (EtO :* ) abstracts a proton from the b-carbon of the isopropyl
bromide and simultaneously the leaving group ( :Br :* ) departs from the a-carbon with
the formation of a double bond. The electron pair in the b C — H bond forms the new
p bond and the leaving group Br① comes off with the electron pair in the C — Br bond.
Thus, it is a concerted elimination, i.e., a one-step process which passes through a single
transition state. In the transition state, there is a partially formed O — H bond, a partly
broken H — C bond, a partly formed C == C bond and a partly broken C — Br bond. The
process can be represented as follows:
Entropy favours the products of an E2 reaction because two molecules of starting material
are formed from three molecules of product.
The energy profile diagram for this E2 reaction is as follows:

There is a stereoelectronic requirement of an E2 process and the requirement is that
the groups to be eliminated should be anti-periplanar (i.e., conformationally trans to one
another with a dihedral angle of 180°) and not syn-periplanar (i.e., conformationally cis to
one another with a dihedral angle of 0°).
The reasons for such stereoelectronic requirement of anti-periplanarity are as follows:

(i) If the sigma orbitals of the departing groups become coplanar, there occurs
maximum overlap of the developing p orbitals. This causes the transition state to
have minimum energy and maximum stability.
(ii) Unlike syn geometry the anti geometry allows the b C — H bonding electrons to
displace the leaving group from the backside. Like SN2 this familiar backside
attack is energetically advantageous from the electronic point of view.
(iii) The attacking base and the leaving group remain far apart from each other in
anti geometry but not in syn geometry and, therefore, the steric interactions are
minimized.
(iv) Approach of the base towards the anti-periplanar conformation, but not towards the
syn-periplanar conformation, avoids electrostatic repulsion between the negatively
charged base and the leaving group.
The E2 is a stereospecific reaction, because different stereoisomers of the starting

material react to give different stereoisomers of the product. This sterospecificity
results from anti-coplanar transition state that is usually involved in the E2
reaction. The stereospecificity in E2 reaction may be observed in the following
examples.
(1) (1R, 2R)-1-Bromo-1,2-diphenylpropane undergoes base-induced dehydrobromination
to give cis-1,2-diphenylpropene but not trans-1,2-diphenylpropene. The diastereoi-
someric bromide (1R, 2S)-1-bromo-1,2-diphenylpropane, on the other hand, under-
goes base-induced dehydrobromination to give trans-1,2-diphenylpropene but not
cis-1,2-diphenylpropene. Also, the former reaction takes place at a much slower
rate than the latter.
In both of these two reactions, the anti-elimination product is obtained exclusively.

This stereospecificity is the proof of the existence of a stereoelectronic requirement or
stereoelectronic constraint upon the E2 reaction.
The molecule of 1-bromo-1,2-diphenylpropane starts to flatten with the extent of formation
of the double bond. In the transition state I, the bulky phenyl groups crowded together
on the same side of the molecule, whereas in the transition state II, they lie in roomier
positions on opposite sides of the molecule. Because of greater van der Waals strain, the
T.S. I is less stable than the T.S. II. Thus, the free energy of activation for the formation
of the cis-alkene is greater than that for the formation of the trans-alkene and because of
this, the (1R,2R)-isomer undergoes dehydrobromination at a much slower rate than the
(1R, 2S)-isomer does. (The ground states of these two diastereoisomeric bromides also
differ in stability, but the effect is not considerable and may be neglected in assessing
relative free energies of activation.)
(2) Iodide catalyzed debromination of (2R, 3S)-2,3-dibromo-2,3-dideuteriobutane pro-

duces trans-2,3-dideuterio-2-butene, but of (1R, 2S)-1,2-dibromo-1,2-dideuterio-
ethane produces cis-1,2-dideuterioethene.
In (2R, 3S)-2,3-dibromo-2,3-dideuteriobutane, both of the Br atoms are attached to
secondary carbon atoms and so, the compound undergoes E2 reaction (instead of
SN2) to give only trans-2,3-dideuterio-2-butene by an anti-elimination process. The
reaction proceeds as follows:
On the other hand, (1R, 2S)-1,2-dibromo-1, 2-dideuterioethane does not undergo

debromination directly. Since the Br atoms are attached to primary carbon atoms,
it undergoes first an SN2 reaction (displacement of Br① by I①) with inversion of
configuration and then an E2 reaction to yield cis-1,2-dideuterioethene by an anti-
elimination process.
(3) The following two diasteoisomeric bromides undergo E2 reaction to give different
products when treated with sodium methoxide in methanol and this is because of
anti stereospecificity of E2 reaction.
In the trans-isomer of the bromide, the H and Br atoms are anti-coplanar to each
other and it is the most suitable geometry for base-promoted E2 elimination. On
the other hand, in the cis-isomer, the vinylic hydrogen is cis to Br, but the allylic
hydrogens at C-3 can easily adopt the preferred anti conformation. For these
reasons, the trans-isomer undergoes base-promoted E2 elimination to yield oct-4-
yne (an alkyne), while the cis-isomer undergoes base-promoted E2 elimination to
yield octa-3,4-diene (an allene, i.e., a compound containing two double bonds that
share a single carbon atom).
(4) When the erythro and the threo isomers of sodium 2,3-dibromo-3-p-
nitrophenylpropanoate are heated, they undergo decarboxylative debromination
(E2) to give cis- and trans-1-bromo-2-p-nitrophenylethene, respectively. It thus
follows that these elimination reactions must have taken place in anti fashion.
(5) When (3R, 4R)-3,4-dibromohexane is heated with zinc in methanol, it undergoes

debromination (E2) to yield cis-3-hexene. However, when (3R, 4S)-3,4-
dibromohexane is similarly treated, debromination takes place to give trans-3-
hexene.
(6) In cyclohexane derivatives, the stereoelectronic requirement of E2 reaction, i.e.,

the anti-periplanar orientation of the departing groups is satisfied only when the
groups are in diaxial (a, a) positions but not in diequatorial (e, e) or equatorial-axial
(e, a) positions. In fact, when they are equatorial-axial, they are syn to each other
and also not coplanar. When they are diequatorial, they are anti to each other
but not coplanar. They are anti as well as coplanar only when they are diaxial.
Therefore, the diaxial conformer of cyclohexane derivatives is very susceptible to
E2 reaction as the stereoelectronic effect is helpful for this concerted process.
(a) When the cis-isomer of 4-tert-butylcylohexyl tosylate is heated with NaOEt in

EtOH at 75°C, it undergoes a facile E2 reaction to yield 4-tert-butylcyclohexene.
However, the trans-isomer reacts slowly with NaOEt to give the same product
when similarly treated.
The bulky tert-butyl group ( —< ) in both cis- and trans-4-tert-butylcyclohexyl

tosylate always assumes an equatorial position, holding the ring in a single
conformation. The cis-isomer exists nearly exclusively in the axial-tosylate
conformation, while the trans-isomer exists nearly exclusively in the
equatorial-tosylate conformation. In the thermodynamically stable axial-
tosylate conformation of the cis-isomer, the -OTs group and a b-H atom (either
at C-2 or at C-6) are anti-periplanar. Because of such arrangement of the
departing groups in which the stereoelectronic requirement is satisfied, this
isomer undergoes a facile E2 reaction, i.e., the loss of TsOH occurs at a faster
rate.
On the other hand, in the stable equatorial tosylate conformation of the trans-
isomer, the -OTs group cannot become coplanar with a b-H atom (either at
C-2 or at C-6), i.e., in this conformation the stereoelectronic requirement
is not fulfilled. Although the stereoelectronic requirement is satisfied in
the energetically unfavourable axial-tosylate conformation, flipping to this
conformation, which suffers from strong 1,3-diaxial interactions, does not
practically take place. For these reasons, the trans-isomer does not undergo
E2 reaction under the same reaction conditions. Slow elimination does take
place, but by a unimolecular (E1) mechanism.
(b) When neomenthyl chloride is treated with sodium ethoxide in ethanol, it

undergoes E2 elimination of HCl at a much faster rate (about 200 times
faster) than does menthyl chloride under similar reaction conditions. Again,
neomenthyl chloride produces a mixture of 75 percent 3-menthene and
25 percent 1-menthene, while menthyl chloride produces 2-menthene as the
only product.
In the thermodynamically more stable conformation (the two relatively bulky

alkyl groups are in equatorial positions) of neomenthyl chloride, the chlorine
atom is axial and there are axial hydrogens at both C-2 and C-4. This substrate,
therefore, reacts readily with ethoxide ion (EtO①) through this energetically
favourable conformation in which the stereoelectronic requirement for E2
reaction is satisfied and gives a mixture of 2- and 3-menthenes. 3-Menthene
predominates in the product mixture because it is more substituted and
thermodynamically more stable than 2-menthene. The reaction is governed
by the Saytzeff rule.
On the other hand, the thermodynamically more stable conformation of

menthyl chloride has all three groups (including the chlorine) equatorial.
However, this conformation is not suitable for facile E2 elimination of HCl
because chlorine is equatorial. For the chlorine to become axial, menthyl
chloride has to assume a conformation in which the large isopropyl group
(—CHMe2) and the methyl group are also axial. This conformation is of much
higher energy, and the free energy of activation for the reaction is large
because it includes the energy required for the conformational change. For
this reason, menthyl chloride undergoes E2 dehydrochlorination very slowly.
In this conformation, the only axial hydrogen b to the axial chlorine is at
C-2. The reaction, therefore, leads to the formation of only 2-menthene (the
Hofmann product).
(c) Both the trans- and the cis-isomer of 1,2-dibromocyclohexane produces

cyclohexene when treated with KI in ethanol.
The trans-isomer of 1,2-dibromocyclohexane undergoes debromination by the

iodide ion (I①) through the more stable diaxial conformation (in which the
stereoelectronic requirement for E2 is satisfied) to yield cyclohexene.
The cis-isomer of 1,2-dibromocyclohexane, on the other hand, does not undergo

E2 debromination, since the two Br atoms are not anti-periplanar. In fact,
it yields the same product by a merged substitution-elimination (SN2 – E2)
reaction. As a nucleophile I① displaces the axial Br atom by an SN2 reaction.

The resultant bromo-iodo compound then undergoes anti-elimination (E2) of
IBr to yield cyclohexene.
(7) The lactone I undergoes dehydrobromination to give the lactone II when treated
with NaOMe in MeOH, but the lactone III undergoes dehydrobromination to give
the lactone IV when treated similarly.
In the lactone I, the bromine atom and the b-H atom at the ring junction are anti-
periplanar. Therefore, it undergoes E2 elimination of HBr smoothly to yield the
lactone II (the Saytzeff product) when treated with NaOMe/MeOH. On the other
hand, in the lactone III, the Br atom is syn to the junction b-H atom and also they
are not coplanar. However, it can be anti-periplanar with a methyl b-hydrogen.
Therefore, this lactone undergoes E2 dehydrobromination involving a methyl
hydrogen smoothly to give the lactone IV (the Hofmann product).
(8) That E2 elimination is stereospecifically trans (anti) can be well demonstrated by

the following example. The trans-decaline derivative I in which Cl atom is axial
undergoes dehydrochlorination to give two isomeric alkenes (Cl is anti-periplanar
with two axial b–H atoms) when treated with sodium ethoxide in ethanol.
On the other hand, the trans-decaline derivative II in which Cl atom is equatorial

does not undergo E2 dehydrochlorination because Cl is not anti-periplanar with
any b-H atom. Furthermore, a trans-decaline system cannot flip and so, chlorine
cannot become axial for a smooth E2 elimination.
H H
H NaOEt/EtOH
No elimination reaction
Cl
H
H
II
(9) Due to anti stereochemical disposition of H and Br, trans-2-bromobut-2-ene
undergoes E2 dehydrobromination at 50 times faster rate than the cis-isomer
during treatment with NaNH2 in liquid NH3. Because of less favourable syn-
elimination, the latter reaction is slow.
• Syn-elimination
There is experimental evidence that compounds in which the departing groups are coplanar
on the same side of the molecule (i.e., eclipsed) may also undergo concerted elimination.
The groups are said to be syn-periplanar, and the process is called syn-elimination. Syn-
eliminations take place only under forcing conditions and only when the departing groups
cannot achieve anti-periplanarity.
(1) syn-elimination most often occur in certain rigid cyclic systems. In the following
bicyclic compound, for example, the quaternary ammonium group and the b-H
atom cannot become coplanar (the dihedral angle is about 120°). The syn deuterium
atom is, however, coplanar with the ammonium group (the dihedral angle is about
0°) and so, it is lost in the elimination reaction (syn-elimination) that takes place
when this hydroxide is heated.
Similarly, the following deuterated norbornyl bromide undergoes base-induced

syn-elimination of HBr to yield the corresponding deuterated alkene.
Stereoelectronic considerations suggest that syn-periplanarity of the departing

groups allows a reasonable conversion of the reactant molecular orbitals to the p
orbital of the resulting alkene. Coplanarity of the departing groups, whether anti or
syn, is the most important factor for reactions that proceed by the E2 mechanism.
However, the syn-elimination taking place from the higher energy ‘eclipsed’
conformation is considerably less favourable than anti-elimination taking place
from the lower energy ‘staggered’ conformation.
That coplanarity of the departing groups, whether syn or anti, is most important
for reaction that proceed by the E2 mechanism can be established by the following
observations.
The trans-isomer of 11,12-dechloro-9,10-dihydro-9,10-ethanoanthracene undergoes
base-induced dehydrochlorination at a rate faster than its cis-isomer. In the cis-
isomer, the dihedral angle between the departing groups is about 120°. Because of
regidity of the molecule, the departing groups cannot achieve a dihedral angle of
180°, i.e., the departing groups cannot become anti-periplanar. It thus undergoes
anti-elimination with a dihedral angle of about 120°. In the trans-isomer, on the
other hand, the dihedral angle between the departing groups is about 0°, i.e., the
groups are syn-periplanar. This isomer thus undergoes base-induced syn-elimination
of HCl. Since coplanarity is a very important requirement for E2, therefore, syn-
elimination with a dihedral angle of about 0° is energetically more favourable
than anti-elimination with a dihedral angle of about 120°. Dehydrochlorination,
therefore, takes place more rapidly from the trans-isomer than from the cis-
isomer.
(2) Syn-elimination may also take place in acyclic molecules when anti-periplanar
transition state is highly destabilized by the steric strain caused due to eclipsing of
the two very bulky alkyl groups.
For example, both (4S, 5R)- and (4R, 5R)-4-bromo-3,3,5,6,6-pentamethyloctane
undergo base-induced E2 dehydrobromination to give the same alkene.
In the (4S, 5R)-diastereoisomer of this bromoalkene, the b-H atom and the Br
atom can be anti-periplanar with the very bulky tert-pentyl groups anti to each
other. Thus, the transition state for anti-elimination is considerably stable
(unencumbered by any steric strain from the tert-pentyl groups) and so, this
diastereoisomer undergoes smooth E2 dehydrobromination through the favoured

staggred conformation to yield (E)-3,3,4,6,6-pentamethyl-4-octene (the alkene with
trans tert-pentyl groups). On the other hand, the transition state for anti-elimination
of HBr from the (4R, 5R)-diastereoisomer is very much destabilized because the
bulky tert-pentyl groups are crowded together on the same side of the molecule and
become involved in server steric interaction. However, syn-coplanarity of Br and
H atom can be attained with the bulky tert-pentyl groups anti to each other. This
diasteroisomer, therefore, undergoes syn-elimination of HBr involving a relatively
stable transition state to form the same diasterioisomeric alkene.
(3) There are a number of organic compounds (e.g., esters like acetates and xanthaltes,
amine oxides, etc.) that undergo syn-elimination through a cyclic transition state
when heated with no other reagent present. The formation of cyclic transition states
in these reactions is evidenced by a large decrease of entropy. These reactions,
which are run either in inert solvents or in the gas phase, are referred to as Ei
eliminations (elimination, intramolecular). The degree of syn stereoselectivity also
reflects the exent to which they proceed via cyclic transition states.
(a) Pyrolysis of acetates
The pyrolytic elimination of an acetate containing at least one b-H atom proceeds
concertedly through a six-membered cyclic transition state. The cyclic pathway required
a syn-coplanar arrangement of the departing groups and so, the reaction is a highly
stereoselective syn-elimination. The stereochemical course and the mechanism of this
intramolecular E2 reaction may be shown as follows:
When there are two possibilities of such intramolecular elimination, the major product is
that which is obtained through a more stable transition state. The following acetate, for
example, undergoes pyrolytic syn-elimination to yield the alkene I as the major product
and the alkene II as the minor product.
The six-membered cyclic transition state leading to the formation of I is planar and
geometrically as well as thermodynamically more favourable, while the six-membered
transition state leading to II is not planar (C — O and b C — H bonds are not coplanar) and
thermodynamically less favourable.
When there are two diastereotopic b-hydrogen in an acetate, that hydrogen will be
eliminated which leads to the formation of a more stable transition state to give the major
product. For example (1S, 2R)-2-deuterio-1,2-diphenylethyl acetate undergoes pyrolytic
elimination to give trans-stilbene predominantly, while the (1S, 2S)-isomer undergoes
pyrolytic elimination to give trans-1-deuteriostilbene predominantly.
Since these pyrolytic syn-eliminations involve expulsion of the acetoxy group with a cis
b-deuterium, therefore, the (1S, 2R)-isomer of 2-deuterio-1,2-diphenylethyl acetate is
expected to give trans-stilbene as well as cis-1-deuteriostilbene. However, because of steric
interaction between two bulky and eclipsing phenyl groups, the transition state leading
to the formation of cis -1-deuteriostilbene is less stable than that leading to the formation
of trans-stilbene. For this reason, trans-stilbene is formed readily and predominantly.
Due to the same basic reason, trans-1-deuteriostilbene instead of cis-stilbene is obtained
predominantly form the (1S, 2S)-isomer.
Mechanism:
(b) Pyrolysis of xanthates

Pyrolysis of xanthates to alkenes is known as the Chugaev reaction. The following
reaction sequence involving the Chugaev reaction is an indirect method of dehydration of
alcohols.
In the reaction sequence, the alcohol is first converted to the corresponding alkoxide ion
by the action of alkali. The alkoxide then reacts with carbon disulphide to yield O-alkyl
sodium xanthate. The O-alkyl sodium xanthate subsequently reacts with methyl iodide by
an SN2 mechanism to form a xanthate ester. The xanthate ester then undergoes pyrolytic
elimination (Ei elimination), a concerted reaction that proceeds through a six-membered
cyclic transition state to yield an alkene. The reaction exhibits higher degree of syn
stereoselectivity.
Mechanism:
(c) Pyrolysis of tertiary amine oxides

When tertiary amine oxides containing at least one b-H atom are heated, they undergo
decomposition to yield an alkene and a dialkylhydroxylamine. This is called Cope
elimination reaction. For example:
≈
D
PhCH 2CH 2 N(CH 3 )2 ææææ
ª 150∞ C
Æ Ph CH == CH 2 + (CH 3 )2 NOH
| Styrene Dimethylhydroxyl
:O* - amine
A tertiary amine oxide
In this Ei reaction, the negatively charged oxygen of the amine oxide is the base that
removes a proton from the b-carbon. The reaction proceeds through a planar five-membered
cyclic transition state. The mechanism of the reaction requires the departing groups to be
syn-coplanar and in fact, the reaction exhibits the greatest degree of syn stereoselctivity
of any of the Ei reactions.
The Cope elimination reaction is very useful for the preparation of many olefins because
at relatively low reaction temperature, possible isomerization of the resulting alkene is
minimized and also because of mild conditions, the side reactions are few.
The five-membered cyclic transition state involved in the Cope elimination reaction must
be completely planar can be well demonstrated by the following reactions.
Since it is relatively easier to force an axial and an equatorial bond into a place than two
equatorial bonds, therefore, an equatorial and an axial substituent (i.e., cis to each other)
become involved in pyrolytic elimination more easily (low activation energy) than two
equatorial substituents (i.e., trans to each other).
The cis-amine oxide undergoes unfavourable anti-elimination of the equatorial
— NMe2 Æ O group and the equatorial b-H at C-2 to yield 1-phenycyclohexene as the minor
product. However, it undergoes favourable syn-elimination of the equatorial — NMe2 Æ O
group and the axial b-H at C-6 to yield 3-phenylcyclohexene as the major product (almost
exclusively).
The pyrolytic elimination of trans-amine oxide leading to the formation of both 1- and
3-phenylcyclohexene involve a favourable syn-elimination of an equatorial — NMe2 Æ O
group and an axial b-H (at C-2 or C-6). An equal amount of both the alkenes is expected to
be obtained from this isomer. In fact, 1-phenylcyclohexene is obtained predominantly and
this is because 1-phenylcyclohexene, being a conjugated alkene, is thermodynamically
more stable than the isomeric 3-phenylcyclohexene.
In the case of acyclic systems, the Cope elimination have a carbanion-like transition state.
That is, the major product of the Cope elimination, like that of Hofmann elimination, is
the one obtained by removing a hydrogen from the b-carbon bonded to the most hydrogens.
For example:
CH3
|!
D
CH 3CH 2 NCH 2CH 2CH 3 ææ Æ CH 2 == CH 2 + CH 3N(OH)CH 2CH 2CH 3
|
:O @
However, in the case of cyclic compounds, the transition state of the Cope elimination
possesses considerable double bond character, i.e., the Saytzeff rule is followed. For this
reason, although the formation of both the alkenes I and II from the following tertiary
amine oxide involves planar five-membered transition state, the transition state leading
to the formation of the more substituted alkene I is relatively more stable than the
transition state leading to the formation of the less substituted alkene II. Because of this,
the thermodynamically more stable alkene I is obtained nearly exclusively.
4.1.5 Evidence in Favour of the E2 Mechanism

(a) The 1,2- or b-elimination reactions which are designated as E2 reactions have been
found to follow second-order kinetics.
(b) If the C — H bond cleavage take place in the rate-determining step of a reaction,
then the same reaction of the deuterated compound will take place at a slower rate
(C — D bond is stronger than the C — H bond). The variation in the reaction rate is
called primary kinetic isotope effect. For second-order E2 eliminations, an isotope
effect of 3 to 8 has been observed. This provides a very strong support for the rate-
determining cleavage of C — H (or C — D) bond in E2 reactions.
(c) In a second-order elimination, two-step mechanism involving reversible formation
of a carbanion may also operate.
If a reaction which proceeds by this mechanistic pathway is carried out in a solvent

which could act as a deuterium source (e.g., EtOD) and the reaction is interrupted
before completion, the recovered substrate is expected to contain deuterium.
The recovered substrates in typical second-order elimination reactions are actually

found to be free of any deuterium. It is inconsistent with the mechanism in which
carbanions are formed reversibly but consistent with the concerted E2 mechanism
in which there is no scope of deuterium incorporation.
However, the mechanism in which the substrate slowly loses a proton to form a
carbanion and the carbanion then rapidly loses the halide ion (X①) cannot be ruled
out in the absence of hydrogen exchange. The first step is essentially irreversible.
The rate of C — X bond cleavage must have no effect on the overall reaction rate if
this mechanism operates. Actually, the rates of typical second-order elimination
reactions were found to depend on the strength of the C — X bond, i.e., the rate
varies with X. It is inconsistent with the (irreversible) carbanion mechanism but
consistent with the E2 mechanism in which the C — X bond breaking is involved in
the rate-determining step.
(d) Rearranged products are never obtained in typical second-order E2 eliminations.
This is consistent with the concerted (one-step) mechanism because it provides no
opportunity for rearrangement.
(e) E2 reactions are stereospecific. The threo isomers give trans-olefins and the erythro-
isomers give cis-olefins usually by the E2 pathway. This indicates that not only a
trans-elimination occurs but also the process of elimination is a concerted one.
4.1.6 Factors Influencing E2 Reaction Rate or E2 Reactivity

(a) The effect of substrate structure on rate: The major structural features that
can increase the rate of E2 reaction are those that serve to stabilize the resultant
alkene or more particularly, the transition state that precedes it. Therefore,
increasing alkyl substituent at both a- and b-carbon atoms of the substrate molecule
facilitates the E2 reaction, because the corresponding alkene is thermodynamically
more stable and the competing SN2 reactions become progressively slower due to
steric reasons.
(b) The effect of base on rate: Since the base appears in the rate equation of E2
reaction, therefore, the reaction rate increases as the strength and concentration
of the base increases. Stronger the base faster the rate of the reaction. E2 reactions
①
are generally run with strong negatively charged bases like OH①, EtO①, NH2 ,
①
etc. Since their basicity decreases in the order: NH2 > OEt① > OH①, therefore,
the E2 reaction rate decreases in the same order. Strong sterically hindered bases
like Me3CO①, DBN, DBU, etc. also favour E2 reaction. The bulky t-butyl group of
the tert-butoxide ion inhibits the reaction by substitution, allowing elimination
reaction to take precedence.
(c) The effect of leaving group on rate: Because the bond to the leaving group is
partially broken in the transition state, the better the leaving group the higher the
rate of the E2 reaction. The increasing order of reactivity of R — X is R — F < R — Cl
< R — Br < R — I. When PhCH2CH2X (X = halogen), for example, is treated with
EtO①/EtOH, the following relative rates have been observed.
PhCH 2CH 2F Ph CH 2CH 2 Cl PhCH 2CH 2Br PhCH 2CH 2l

1 70 4.2 ¥ 103 2.7 ¥ 104
(d) The effect of solvent on rate: The E2 reactions are favoured by a decrease in the
polarity of the solvent because the charge in the E2 transition state is considerably
dispersed. In fact, polar aprotic solvents increase the rate of E2 reactions. Because
polar aprotic solvents like HCONMe2 (DMF), Me2SO(DMSO) and Me2CO (acetone)
do not solvate anions well, a negatively charged base is not trapped by the strong
interaction with the solvent molecules and the base is stronger. A stronger base
increases the reaction rate.
(e) The effect of temperature on rate: The rate of E2 reaction increases with
rise in temperature. The number of particles increases in E2 reactions and so,
E2 reactions have the more favourable entropy term, and because this (DS=| ) is
multiplied by T in the relation of free energy of activation, DG=| (DG=| = DH=| – DS=| ),
it increasingly outweighs a less favourable DH=| as the temperature rises.
4.1.7 Factors that Govern the Proportions of E2 and SN2 Reactions

Since the reactive part of both nucleophiles and bases is an unshared pair of electrons,
therefore, all nucleophiles are potential bases and all bases are potential nucleophiles.
Then, it should not be surprising that nucleophilic substitution reactions and elimination
reactions often compete with each other.
The following factors govern the proportion of E2 and SN2 reactions:

(a) Substrate structure: When the substrate is a primary (1°) halide and the base is
strong and unhindered (e.g., EtO①), substitution is highly favoured because the base can
easily approach the carbon bearing the halogen atom (the leaving group). For example:
C H OH
CH 3CH 2O* Na≈ + CH 3CH 2Br æææææ
2 5
55∞ C
Æ CH 3CH 2OCH 2 CH 3 + CH 2 == CH 2
Primary ( - NaBr) SN 2 - product E2 - product
substrate (major; 90%) (minor; 10%)
When the substrate is a secondary (2°) halide, a strong base favours elimination and this
is because steric hindrance in the substrate makes substitution more difficult.
Br
| C2H5OH
CH 3CH 2O* Na≈ + CH 3 CHCH 3 æææææ55∞ C
Æ (CH 3 )2 C — O — CH 2 CH 3 + CH 2 == CH CH 2
Secondary ( - NaBr) SN 2 - product E2 - product
substrate (major; 20%) (minor; 79%)
When the substrate is a tertiary (3°) halide, an SN2 reaction cannot take place due to severe
steric hindrance in the substrate. Elimination reaction is highly favoured, especially when
the reaction is carried out at higher temperatures. If any substitution occurs, it must take
place through an SN1 mechanism.
At room temperature:
C H OH
CH 3CH 2O* Na≈ + (CH 3 )3 C — Br æææææ
2 5
25∞ C
Æ CH 2 == C(CH 3 )2 + (CH 3 )3 C — O — CH 2 CH 3
Tertiary E2 + E1- product SN1 - product
substrate (major; 91%) (major; 9%)
At higher temperature:
C H OH
CH 3CH 2O* Na≈ + (CH 3 )3 C — Br æææææ
2 5
55∞ C
Æ CH 2 == C(CH 3 )2
Tertiary E2 + E1- product
substrate (100%)
(b) Base: One way of favourably influencing an elimination reaction of an alkyl halide
is to use a strong sterically hindered base such as the tert-butoxide ion (Me3CO①). The
bulky methyl groups of the Me3CO① ion inhibit its reaction by SN2 and as a consequence,
elimination reaction takes place. This can be well demonstrated by the following two
reactions. The relatively unhindered methoxide ion (MeO①) reacts with 1-bromooctadecane
mainly by substitution (SN2), whereas the bulky tert-butoxide ion reacts with the same
substrate mainly by elimination (E2).
CH OH
CH 3O* + CH 3 (CH 2 )15 CH 2CH 2Br ææææ 3
65∞ C
Æ
an unhindered 1 - Bromooctadecane
base/nucleophile
CH 3 (CH 2 )15 CH 2CH 2OCH 3 + CH 3 (CH 2 )15 CH == CH 2
SN 2 - product E2 - product
(major; 99%) (minor, 1%)
CH OH
(CH 3 )3 CO* + CH 3 (CH 2 )15 CH 2CH 2Br ææææ 3
65∞ C
Æ
a hindered 1 - Bromooctadecane
base/nucleophile
CH 3 (CH 2 )15 CH == CH 2 + CH 3 (CH 2 )15 CH 2CH 2OC(CH 3 )3
E2 - product SN 2 - product
(major; 85%) (minor, 15%)
The relative rates of E2 and SN2 reaction also depend on the relative basicity and
polarizability of the base/nucleophile. The likelihood of elimination (E2) increases when a
strong, weakly polarizable base (the attacking atom is more electronegative and smaller in
size) such as OH①, NH2① or RO① (especially a hindered one) is used. On the other hand, the
likelyhood of substitution (SN2) increases when weakly basic ions such as Cl①, CHCOO①
or weakly basic and highly polarizable ions (the attacking atom is less electronegative and
larger in size) such as Br①, I① or RS① are used. For example, the acetate ion reacts with
2-bromopropane nearly exclusively by the SN2 mechanism.
Br
|
CH 3COO* + CH 3 CHCH 3 Æ CH 3COOCH(CH 3 )2 + : Br:*
ææ
weakly 2-Bromopropane SN 2 - product
basic (∼ 100%)
On the other hand, the more strongly basic ethoxide ion (EtO①) reacts with 2-bromopropane
mainly by the E2 mechanism.
(c) Solvent: The bimolecular reactions are favoured by decrease in the polarity of the
solvent and since the charge in the E2 transition state is more dispersed than in the
SN2 transition state due to involvement of five atoms, decreasing solvent polarity favours
elimination (E2) more than substitution (SN2).
The influence of solvent on the E2 – SN2 competition is illustrated by the following reaction.
The E2 dehydrobromination of 2-bromopropane (CH3CHBrCH3) increases as the medium
is changed from polar aqueous ethanol to less polar ethanol.
Br
|
60% EtOH + 40% H2O
CH 3 CHCH 3 + NaOH æææææææææ
55∞ C
Æ CH2 == CHCH 3 + (CH 3 )2 CHOC2H5 + (CH 3 )2 CHOH
E2 – product
SN 2 - product
(54%)
(46%)
Br
|
C2H5OH
CH 3 CHCH 3 + NaOEt æææææ
55∞ C
Æ CH2 == CHCH 3 + (CH 3 )2 CHOC2H5
E2 - product SN 2 - product
(71%) (29%)
(d) Temperature: Elimination – whether E1 or E2 – is favoured with respect to
substitution by rise in temperature. Because more bonding changes occur during
elimination, these reactions have greater free energies of activation than substitution
reactions. When the reaction is carried out at a higher temperature, the proportion of
molecules able to surmount the barrier of activation energy for elimination increases more
than the proportion of molecule able to undergo substitution, even though the rate of both
reactions will be increased. Furthermore, elimination reactions are favoured entropically
over substitution because the products of an elimination reaction are greater in number
than the reactants, i.e., elimination leads to an increase in the number of particles,
whereas substitution does not. Since DS=| is multiplied by T in the relation for the free
energy of activation, DG=| (DG=| = DH=| – TDS=| ), therefore, it will increasingly outweigh a
less favourable DH=| term as the temperature rises.
(e) Leaving group: In second-order reactions, the elimination/substitution ratio is not
greatly dependent on a halide leaving group, although a slight increase in elimination
in the order I > Br > Cl has been observed. When OTs① is the leaving group, there is
usually much more substitution. For example, when n–C18H37 Br is treated with t-BuOK,
85% elimination occurs. However, n–C18H37OTs gave, under the same reaction conditions,
99% substitution. The positively charged leaving groups, on the other hand, increase the
amount of substitution.
4.1.8 Regioselectivity in b-elimination Reactions (Orientation of

p Bond in the Product Alkene)
Symmetrical substrate undergoes b-elimination reaction to yield a single olefinic product.
But, unsymmetrical substrates, i.e., substrates containing two types of b-hydrogens,
undergo E2 elimination reaction to yield more than one alkene. Two empirical rules, the
Hofmann and the Saytzeff rules, govern the orientation of elimination of these reactions.
The Saytzeff rule: This rule states that neutral substrates (alkyl halides except fluorides
and sulphonates) possessing two different types of b-hydrogens yield predominantly the
more substituted alkene. For example, when 2-bromobutane is treated with NaOEt in
EtOH, it undergoes E2 dehydrobromination following the Saytzeff rule because it produces
75% of the more substituted alkene but-2-ene and 25% of the less substituted alkene, but-
1-ene.
The predominant formation of Saytzeff product can be explained as follows:

Neutral substrates (except fluorides) undergo ideal E2 elimination reaction in which both
the C — H and C — LG bonds are being broken and the carbon–carbon double bond is
being formed simultaneously. For this reason, the transition state possesses considerable
double bond character. Therefore, any effect that stabilizes the product alkene also
stabilizes the transition state. Because of hyperconjugation effect, the stability of a double
bond increases progressively with increase in the number of alkyl groups bonded to it,
i.e., with increase in the number of a-H atoms. The disbustituted alkene, but-2-ene (with
six hyperconjugable a-H-atoms) is, therefore, thermodynamically more stable than the
monosubstituted alkene, but-1-ene (with only two hyperconjugable a-H atoms) and so,
but-2-ene is formed more rapidly through the transition state of lower energy than but-1-
ene, i.e., the reaction produces the more substituted alkene but-2-ene predominantly.
Thus, an E2 reaction is regioselective because more of one constitutional isomer is formed

than the other.
The Hofmann rule: This rule states that E2 elimination produce predominantly the
less-substituted alkene if the eliminations are carried out with charged substrates. For
≈
example, when sec-butyltrimethylammonium hydroxide, (CH 3 )3 NC(CH 3 )CH 2CH 3 :OH * ,
is heated, 95% of the less substituted alkene, but-1-ene and 5% of the more substituted
alkene, but-2-ene are obtained.
The predominant formation Hofmann product can be explained as follows:

≈ ≈
Substrates containing charged leaving group (e.g ., — N(CH3 )3 ), — SMe2 , etc.) undergo
E2 elimination reaction in which breaking of the b C — H bond starts well before the
breaking of the C— LG bond. For this reason, the transition state of such a reaction
possesses little alkene character but considerable carbanionic character. Therefore, any
factor that stabilizes a carbanion also stabilizes the transition state. The stabilities of
carbanions decrease progressively with increase in the number of electron-releasing (+I)
alkyl groups attached with the negative carbon, i.e., the order of carbanion stability is
methyl > primary (1°) > secondary (2°) > tertiary (3°). The abstraction of b-H by base thus
takes place preferentially form the methyl group because this leads to the formation of a
more stable (lower energy) carbanion-like transition state. The reaction thus produces a
predominance of the less substituted alkene but-1-ene.
The formation of the less substituted terminal alkene but-1-ene may also be explained in
terms of steric effect. The charged groups are generally bulkier than the neutral ones. From
the Newman projections for the conformation required for anti-elimination it becomes
≈
clear that the bulky — N(CH 3 )3 group comes into steric interaction with the adjacent
methyl group (gauche to each other) in the conformations B and C leading to the formation
of the more substituted Saytzeff product. B also suffers from an additional butane–gauche
interaction, consequently, a large number of molecule react through the less crowded and
more stable conformation A to give the Hofmann product but-1-ene predominantly.
Steric effect due to branching in the alkyl group, in the group which is going to leave
*
with its bonding electrons and in the base, i.e., R, LG and B :, plays an important role
in increasing the proportion of Hofmann product over Saytzeff. It has been found that
an increase in the size of LG, or more particularly branching in it, leads to an increasing
proportion of Hofmann elimination with the same alkyl group. For example:
The amount of Hofmann product is also found to increase with increasing branching in the
alkyl group of the substrate (with the same LG and base), and with increasing branching
in the base used. For example, the proportion of Hofmann elimination increases with
increasing branching in the base from a bromide substrate like 2-bromo-2,3-dimethyl
butane from which preferential Saytzeff elimination would normally be expected.
These several steric effects are explainable on the basis of the fact that any crowding,
irrespective of its origin, will make the T.S. I that involves the removal of proton (2) form
the substrate R2CHCMe2LG (Saytzeff elimination) relatively more crowded than the T.S.
II that involves removal of proton (I) (Hofmann elimination). The differential increases
as the crowding increases in R, LG or B① and Hofmann elimination will be progressively
favoured over Saytzeff.
Saytzeff elimination:
Regioselectivity in E2 dehydrohalogenation
The distribution of products obtained from the E2 reaction of MeONa/MeOH and
2-halohexanes are given in the following table:
CH 3 (CH 2 ) CH 2 CHCH 3 More substituted product Less substituted product
CH 3ONa/CH 3OH
2 | ææææææææ æÆ CH3(CH2)2CH==CHCH3 + CH3CH2CH2CH== CH2
X E2 Hex-2-ene
2- Halohexane (dehydrohalogenation) Hex-1-ene
(mixture of cis and trans)
X Conjugate acid pKa
I HI –10 81% 19%
Br HBr –9 72% 28%
Cl HCl –7 67% 33%
F HF 2.2 30% 70%
When 2-halohexanes undergo dehydrohalogenation in the presence of MeO①, the major
product of the chloride, bromide and iodide is the more substituted alkene, but that of
fluoride is the less substituted alkene. Also, there is a steady increase in the fraction
of the less substituted alkene hex-1-ene along the series I, Br, Cl and F. In the case of
2-iodohexane, breaking of both the C—H and C—I bonds occurs simultaneously in the
transition state, i.e., the transition state possesses considerable double bond character.
So, the formation of the more thermodynamically stable, i.e., more substituted alkene
hex-2-ene is preferred and the orientation is Saytzeff. As the series is traversed, the C—X
bond becomes progressively stronger (as evidenced by their pKa values) and its tendency
to be broken in the transition state decreases. At the same time, the electron-attacking
–I effect of X increases. This progressively favours the development of negative charge
on b-carbon. Therefore, the carbanionic transition state is gradually favoured and so, the
percentage of the less substituted alkene hex-1-ene is increased. Due to much stronger
–I effect of fluorine and much stronger C—F bond, the C—H bond-breaking occurs
considerably before C—F bond breaking. Therefore, the transition state possesses little
alkene character but considerable carbanionic character. Since a methyl hydrogen is more
acidic, it is preferentially abstracted by the base to give the less substituted alkene hex-1-
ene predominantly. Hence, the orientation is Hofmann.
Violation of Saytzeff Rule

In cyclic systems, the usual simple requirements of Saytzeff or Hofmann rules may be
overridden by other special requirements of the system. For example, there is a preference
for elimination from the trans-diaxial conformation in cyclohexane derivative. Another
such limitation is that it is not normally possible to carry out on elimination reaction so as
to introduce a double bond on a bridgehead carbon atom in a bridged bicylic compound with
small ring (Bredt’s rule). For example, when the bromide I is treated with NaOEt in EtOH
it produces the Hofmann product II exclusively instead of III (the Saytzeff product).
This is because the developing p orbitals in this E2 elimination, far from being coplanar,
would be virtually at right angles to each other, and so, significant orbital overlap does
not take place to allow development of a double bond. The bicyclic ring system is rigid
enough to make the distortion required for effective p orbital overlap, i.e., such distortion
is energetically unattainable. With bigger rings, for example, the bicyclononene (IV), or
more flexible system (V), sufficient distortion is possible to allow the introduction of a
double bond at the bridgehead position by an E2 reaction.
In the following reaction, Saytzeff rule does not lead to the more substituted alkene because
the rule does not take into account the fact that conjugated double bonds are more stable
than isolated double bonds. Because the conjugated alkene is the thermodynamically more
stable alkene, it is the one that is more easily and readily formed and is, therefore, the
major product of the reaction. Hence, if there is a double bond (or a benzene ring) in the
alkyl halide, Saytzeff rule is not to be used to predict the major product of the elimination
reaction.
*
OH
CH 3CH == CHCH 2CH ClCH(CH 3 )2 æææ Æ CH 3CH == CH — CH == CHC(CH 3 )2
5-Chloro-6-methyl-2-heptene 6-Methyl-2,4-heptandiene
(a conjugated diene)
major product
+ CH 3CH == CHCH 2CH == C(CH 3 )2
6-Methyl-2,5-heptadiene
(an isolated diene) minor product
+ H2O + Cl*
4.1.9 Hofmann Exhaustive Methylation or Hofmann Degradation

Hofmann exhaustive methylation or Hofmann degradation is a widely used process for
structure elucidation of alkaloids. It is actually a three-step process. In the first step, a 1°,
2° or 3° amine is converted to a quaternary ammonium iodide salt by treatment with an
excess of methyl iodide. For example:
In the second step, the iodide salt is treated with moist silver oxide (Ag2O) to yield the
corresponding hydroxide.
+ – + –
2 —CH2CH2N(CH3)3I + Ag2O + H2O Æ 2 —CH2CH2N(CH3)3OH + AgIØ
In the third step, an aqueous or alcoholic solution of the hydroxide is distilled. As a result,
the hydroxide undergoes thermal decomposition to form an alkene, a tertiary amine and
water. A specific E2 reaction in which a quaternary ammonium hydroxide undergoes
thermal decomposition to yield an alkene, a tertiary amine and water is called Hofmann
elimination.
Complete Hofmann degradation of some secondary amines containing two types of b-H
atoms are shown below:
4.1.10 Fragmentations
When carbon is the positive leaving group (the electrofuge) in an elimination reaction,
the reaction is called fragmentation. These processes take place on substrates of the type
≈ ≈
W-C-C-X, where X is a normal nucleofuge (e.g., halogen, OH 2 , OTs, NR 3 , etc.) and W is a
positive carbon electrofuge. In general, W is HO— C or R 2N— C, so that the positive charge
on carbon is stabilized by the unshared pair on the oxygen or nitrogen atom. For example:
The mechanism are mostly E2 or E1.

Some examples of fragmentation reaction are given below:
(1) When the following 1,4-dibromide is treated with zinc, it undergoes debromination
accompanied by cleavage of a C — C bond to give two molecules of 2-methyl-1-
m-tolylpropene. This takes place through the conformation in which each of the
departing Br atoms is anti to the C-2 — C-3 bond which breaks.
(2) When trans-1,4-dibromo cyclohexane is treated with zinc, it undergoes

fragmentation reaction to give hexa-2,5-dideuterio-1,5-diene. Since in the stable
diequatorial conformation of this dibromide, each C — Br bond is anti to the
C-2 — C-3 bond, therefore, it undergoes debromination by zinc accompained by
fission of the C-2 — C-3 bond.
[Debromination does not take place through the trans-diaxial conformation because
in that conformation anti relationship between the C-2 — C-3 bond and the axial
C — Br bonds is not maintained.]
(3) When the following bromo derivative of quinuclidine (a bicyclic compound) is
heated, it undergoes fragmentation reaction to give an iminium salt and this is
because the C — C bond is anti to both the sp3 orbital of nitrogen containing an
unshared electron pair and the departing Br atom.
4.1.11 Summary of the E2 Reaction

The E2 reaction is a b-elimination reaction. The key points about this reaction are as
follows:
1. The rates of E2 reaction are second order overall: first order in base and first order
in the alkyl halide.
2. E2 reactions normally occur with anti stereochemistry.
3. The E2 reaction occurs at a faster rate with better leaving groups, i.e., those that
give the weakest bases as products.
4. The rate of E2 reactions shows substantial primary kinetic isotope effects
(deuterium isotope effects at the b-H atoms).
5. When an alkyl halide has more than one type of b-H atoms, more than one alkene
products can be formed. In that case, the major product is the more substituted
alkene (the Saytzeff product) unless the base is large, the alkyl halide is an alkyl
fluoride and the alkyl halide contains one or more double bonds.
6. E2 reactions compete with SN2 reactions. Elimination is favoured by alkyl branches
in the alkyl halide at the a- or -b-carbon atoms, by alkyl branches in the base, by
stronger base and by rise in temperature.
1. When (R)-1,2-di-p-tolyl-1-chloroethane is treated with NaOEt in EtOH,

it undergoes base-induced dehydrochlorination to give trans-1,2-di-p-
tolylethene as the major product and cis-1,2-di-p-tolylethene as the minor
product. Explain this observation.
Solution (R)-1,2-Di-p-tolyl-1-chloroethane contains two b-H atoms. So, anti-elimination
of HCl may take place from two different conformations. The gauche-conformation (the two
bulky p-CH3C6H4-groups are gauche to each other) gives rise to cis-1,2-di-p-tolylethene
via the transition state I, while the anti-conformation (the two bulky p-CH3C6H4-groups
are anti to each other) gives rise to trans-1,2-di-p-tolylethene via the transition state II.
The stereochemical course of the reaction may be shown as follows:
The population of the anti-conformation (B) is greater than the gauche-conformation (A)
because the former conformation is relatively more stable. On that basis, it is normally
expected that the trans-product would predominate. However, this reasoning is incorrect.
The activation energy of the reaction is very large compared to the barrier to internal
rotation. In such a case according to Curtin–Hammett principle, the relative amounts of
the two disastereoisomeric products do not depend on the relative populations of the two
conformations, but depends only on the relative stabilities of the transition states (i.e., on
the values of activation energies) leading to the products. Because of steric strain between
the two bulky p-tolyl groups on the same side of the molecule, the transition state I is
less stable than the transition state II in which they are placed in opposite sides. The
activation energy for the formation of the trans-alkene is, therefore, lower than for the
formation of the cis-alkene and consequently, the former alkene is formed readily as the
major product.
2. Predict the two elimination products expected to be formed in the
following reaction:
*
OEt /EtOH
CH 3CH 2CHDCH 2Br ææææææ
E2
Æ
Which is formed as greater yield and why?

Solution The substrate CH3CH2CHDCH2Br undergoes base-induced dehydrobromination
(loss of HBr) and dedeuteriobromination (loss of DBr) to yield 2-deuteriobut-1-ene and
but-1-ene, respectively:
Since the C — D bond is stronger than C — H bond, therefore, the activation energy for
the elimination of DBr is greater (less stable T.S.) than that for the elimination of HBr.
Because of this, elimination of HBr takes place at a faster rate to produce 2-deuteriobut-
1-ene in greater yield.
3. (1R,2R)-1-2-Dibromo-1,2-dicyclohexylethane undergoes dehydrobromina-
tion in the presence of pyridine ( )
N: to yield trans-1-bromo-1,2-dicy-
clohexylethene while (1R, 2S)-1,2-dibromo-1,2-dicyclohexylethene under-
goes debromination to yield trans-1,2-dicylohexylethene. Explain these
observations.
Solution In the presence of the base pyridine, the (1R, 2R)-isomer of 1,2-dibromo-1,2-
dicyclohexylethene undergoes dehydrobromination instead of debromination because
the reaction proceeds through a more stable transition state in which the two bulky
cyclohexyl groups are anti to each other (too far from each other to become involved in
steric interaction).
The (1R, 2S)-isomer, on the other hand, undergoes debromination instead of dehydrobro-
mination because this proceeds through a more stable transition state where the cyclo-
hexyl groups are anti.
4. Dehydrohalogenation of vic-dibromides leads to the formation of alkynes.

Suggest a mechanism of the following reaction:
Solution The reaction proceeds through the steps as follows:

Step 1: The strongly basic amide ion brings about an E2 reaction to form a bromalkene.
Step 2: A second E2 reaction produces the alkyne.
In both the steps, the reaction takes place smoothly because the stereoelectronic
requirement of E2 process is satisfied.
5. Show how you might synthesize cyclohexylacetylene from cyclohexyl
methyl ketone.
Solution Cyclohexyl methyl ketone is first treated with phosphorus pentachloride to
yield a gem-dichloride. The gem-dichloride is then heated with 3 equivalents of NaNH2
in the presence of mineral oil. The terminal alkyne thus obtained is deprotonated by the
third equivalent of base. Aqueous NH4Cl is finally added to the reaction mixture to convert
the sodium alkynide to the desired product, cyclohexylacetylene.
6. Suggest a mechanism for each of the following reactions along with the
stereochemical course involved when it is appropriate:
+ ≈
H H
(a) HO — — OH (b) Ph 2C(OH)CMe2 (OH) Me2 æææÆ
H Me
N D
(c) (d) Cl
Solution
(a) The cyclic diol fragments in the presence of acid (acid-catalyzed dehydration
involving fission of a C — C bond) to give a b, g-unsaturated aldehyde.
(b) This 1,3-diol undergoes fragmentation reaction in the presence of acid to yield
the alkene Ph2C == CMe2 and acetone. Due to stabilizing effect to the two phenyl
groups, this particular — OH is expelled as water.
(c) The tosylate undergoes fragmentation reaction in the presence of the base t-BuO①
and this is because the C-2—C-3 bond is anti to both the O — H and C — OTs
bonds.
(d)
This heterocyclic ring system undergoes concerted fragmentation reaction because

the stereoelectronic effect is helpful in this case also.
7. How would you carry out the following transformations?
(a) (b)
(c) CH3CH2CH2CH2CH2Br Æ CH3CH2CH2COCH3

(d) CH3CH2CH == CH2 Æ CH2 == CH—CH == CH2
Solution
(a)
(b)
(c)
(d)
8. Predict the major product in each of the following reactions and account
for its formation:
(a) (b)
OH H
H – OH –
OH OH
Ph H Ph H
H Br H Br
(c) (d)
OH H
H OH
– OH OH
–
Ph Ph or Ag2O/H2O
Br Br
H H H H
Solution
(a) The –OH group is in proper position for backside attack and therefore, SN2 type
displacement leading to the formation of an epoxide is the main reaction.
(b) Since H and Br are anti-periplanar, therefore, E2 elimination is stereoelctronically

favourable. An enol is formed as an intermediate, which readily tautomerizes to
the more stable keto form (a cyclohexanone derivative).
(c) This more stable conformation of the compound has no proper stereoelctronic
situation for either epoxidation or E2 elimination. Elimination of HBr is possible
only from the other conformation in which the stereoelectronic requirement is
satisfied. However, the rate is slow because this conformation is thermodynamically
less stable as it suffers from syn-axial interactions. The intermediate enol readily
tautomerizes to the more stable ketone.
(d) In the presence of OH①, the compound undergoes ring inversion followed by an
internal SN2 attack to give an epoxide. However, because of syn-diaxial interactions,
the conformation is less stable and as a consequence, the reaction is slow.
In the presence of Ag2O (Ag≈), a cyclopentanecarboxaldehyde derivative is obtained

by bond migration and this is because Ag≈ induces the departure of equatorial
bromine by acting as a Lewis acid.
9. Draw the structure of an isomer of benzene hexachloride which does not

undergo E2 dehydrochlorination and explain why it is unreactive.
Solution The (e,e,e,e,e,e) isomer of benzene hexachloride or its conformational
diastereoisomer (a,a,a,a,a,a) does not undergo E2 dehydrochlorination. This is because
in both forms, no Cl and H atoms are diaxially placed, i.e., in none of two forms, the
stereoelectronic requirement for the E2 elimination is satisfied.
10. The compound A readily forms an alkene with I①, but the isomeric
compound B does not. Explain these observations.
Solution The tert-butyl group ( ) is particularly bulky and due to sever 1,3-diaxial
interactions it cannot take up the axial position. A tert-butyl group-substituted cyclohexane,
therefore, remains fixed in the chair conformation where the Me3C-group remains in the
equatorial position. Hence, the compound A and its isomer B containing equatorial tert-
butyl group exist in these particular conformations and they are expected to react through
these conformations. Now, in A, the two Br atoms are in axial positions, i.e., they are anti-
periplanar. Therefore, the stereoelectronic requirement for E2 elimination is satisfied.
Hence, the compound A readily undergoes debromination by I① to produce an alkene. On
the other hand, in the compound B, the two Br atoms are in equatorial positions. Therefore,
the stereoelectronic requirement of E2 debromination is not satisfied in this case. For this
reason, the compound B does not undergo debromination to yield an alkene.
11. Predict the products A and B and suggest the mechanism for the conversion
of A to B:
Solution
The thiocarcbonate A is converted into B (cyclohexene) through the steps as follows:
12. Predict the product and suggest a mechanism for each of the following
reactions:
(a)
*
OH /H2O Mg
(b) ClCH 2C ∫∫ CCH 2Cl æææææ
D
Æ (c) BrCMe2CMe2OCH 3 æææÆ
ether
*
OH
(d) HOCMe2CH 2CMe2Br æææÆ (e) Me2C CMe2
Mg
C ether
(f)
Br Br
Solution
(a)
Since the methylene hydrogen adjacent to the electron-withdrawing p-nitrophenyl

group is more acidic than the other methylene hydrogen, therefore, the transition
state leading to this alkene is relatively more stable than the transition state
leading to the formation of the isomeric alkene trans-p-CH3OC6H4CH == CHCH2–
C6H4NO2-p and so, this alkene is obtained predominantly by an E2 process.
(b) This dichloride undergoes repeated E2 dehydrochlorination in the presence of

alkali to form butadiyne.
(c) When 2-bromo-3-methoxy-2,3-dimethylbutane is treated with Mg in ether, a

Grignard reagent is obtained. Since the carboanionic centre of this Grignard reagent
has a good leaving group (— OMe) adjacent to it, therefore, it is unstable and so,
it undergoes ready elimination (E2) of MgBr(OMe) to give 2,3-dimethyl-2-butene.
(d)
The compound undergoes dehydrobromination along with fragmentation in the

presence of alkali to yield 2-methylpropene (Me2C ==CH2) and acetone (Me2C ==0).
(e)
The initially formed Grignard reagent is unstable. It readily dissociates to yield an

allene.
(f) When treated with acetic anhydride (Ac2O) in the presence of acid, the aldoxime
undergoes acetylation. The — OCOCH3 group then undergoes protonation and
becomes a very good neutral leaving group. Subsequent E2 elimination of AcOH
and H% yields acetonitrile (CH 3C ∫∫ N).

(a) 3-Bromobutanal undergoes base-promoted dehydrobromination at a very much
faster rate than 2-bromobutanal under similar conditions.
NaOEt/EtOH NaOEt/EtOH
CH3CHBr CH2 CHO CH3 CH2 CHBr CHO
fast slow
3-Bromobutanal 2-Bromobutanal
CH3CH == CH — CHO
2-Butenal
(b)
D NaOEt D D NaOEt
;
EtOH EtOH
Cl Cl
D D D D
I II
(c) The following compound gives no elimination product when treated with NaOMe
in MeOH:
(d)
Solution
(a) The electron-attracting group is placed a to the bromine atom in 2-bromobutanal,
but b to the bromine atom in 3-bromobutanal. The — CHO group could stabilize
the incipient double bond by conjugation from either position. However, it can
make the b-H atom in 3-bromobutanal somewhat acidic due to its – I effect. As
a result, the transition state for E2 dehydrobromination of 3-bromobutanal is
relatively more stable than that for dehydrobromination of 2-bromobutanal and so,
3-bromobutanal undergoes base-promoted dehydrobromination much faster than
2-bromobutanal under similar reaction conditions.
(b) The compound I undergoes base-induced E2 elimination of HCl through that

conformation in which Cl atom and an adjacent H(or D) atom are anti-periplanar,
i.e., through that conformation in which the stereolectronic requirement is satisfied,
to form 1,3-dideuteriocyclohexene.
Similarly, the diasteremeric compound II undergoes base-induced elimination of

DCl to yield 3-deuteriocyclohexene.
(c)
There is no axial b-H atom in the conformation B in which Br is axial. For this
reason, this compound III does not undergo E2 dehydrobromination when treated
with NaOMe in MeOH.
(d)
The compound IV undergoes E2 dehydrobromination because in this case the

stereoelectronic requirement of E2 reaction is satisfied. On the other hand, because
of its rigidity, the isomeric compound V cannot flip and so, Br cannot go to the
axial position. Since the stereoelectronic requirement is not satisfied, therefore,
this compound does not undergo base-induced dehydrobromination.
(a) 2-Bromo-3-methyl-1-phenylbutane reacts with NaOEt in EtOH
to give 3-methyl-1-phenylbut-1-ene as the major product and
3-methyl-1-phenylbut-2-ene as the minor product, i.e., the less and
more substituted products become the major and minor products,
respectively.
(b) 2-Bromo-2-methylbutane undergoes dehydrobromination in the

presence of t-BuO① to give an excess of the less substituted alkene,
2-methylbut-1-ene, even though the substrate is a bromide.
Solution
(a) The transition state of this E2 elimination involving an alkyl bromide possesses
considerable double bond character. Since a phenyl group stabilizes a double bond
by conjugation, the transition state leading to 3-methyl-1-phenylbut-1-ene (the
Hofmann product) is much more stable than that leading to 3-methyl-1-phenylbut-
2-ene (the Saytzeff product). It is for this reason, the less substituted product
3-methyl-1-phenylbut-1-ene is obtained predominantly.
(b) Because of the steric bulk of the base t-BuO①, the transition state of the expulsion
of one of the less exposed methylene hydrogens leading to the formation of the
more substituted alkene (the Saytzeff product) becomes highly crowded. Since the
steric effect then outweighs the hyperconjugation effect, therefore, the transition
state energy become higher than that of the expulsion of the more exposed methyl
hydrogen leading to the formation of less substituted alkene (the Hofmann product).
As a consequence, the less substituted alkene (the Hofmann product) is obtained
as the major product, even though this bromide substrate is expected to give the
more substituted alkene predominantly.
15. Out of the three diasteroisomers of 2-chloro-1,3-dimethycyclohexane

only one cannot undergo E2 dehydrochlorination and the only product
obtained from the other two is 1,3-dimethylcyclohexene. Identify these
diastereoisomers and account for these observations.
Solution The three diastereoisomers of 2-chloro-1,3-dimethylcyclohexane are as follows:
Among the three diastereoisomeric substrates only II does not undergo dehydrochlorination
because the vicinal hydrogens are equatorial and anti-periplanarity with Cl cannot be
achieved. Also, an axial Cl atom cannot be syn-coplanar with an adjacent-equatorial H
atom. In I the axial Cl atom is anti-periplanar with two axial H atoms and in III, it is
anti-periplanar with one axial H atom. So, they undergo E2 dehydrochlorination to yield
1,3-dimethylcyclohexene.
16. Predict the major product and suggest a mechanism for each of the
following reactions:
(a)
(b)
Solution
(a) Charged substrates violated Hofmann rule when an electron-withdrawing group
remains attached to one of the b-carbons. In this quaternary ammonium hydroxide,
the elelctron-attracting C == O group makes methylene b-hydrogens more acidic
than methyl b-hydrogens. Furthermore, due to conjugation, the carbanion-like
transition state leading to the elimination product, CH2== CHCOCH3 is relatively
more stable than the carbanionic transition state leading to CH2== CH2, the
other elimination product. So, this quaternary ammonium hydroxide on thermal
decomposition produces CH2== CHCOCH3 predominantly.
(b) Due to — I effect of the — C6H4NO2-p group, the hydrogens on the carbon containing
the p-nitrophenyl group are somewhat more acidic than methyl hydrogens of the
ethyl group. Also, due to conjugation, the transition state leading to the formation
of the elimination product p-O2N—C6H4—CH== CH2 is thermodynamically more
stable than the transition state leading to the other product CH2== CH2. Therefore,
in this case also, the quaternary ammonium hydroxide on thermal decomposition
produces the more substituted alkene p-O2NC6H4CH== CH2 predominantly.
17. 3-Bromopentane undergoes E2 dehydrobromination at a rate much faster

in DMSO-tert-BuOK than in EtOH-KOH. Explain.
Solution A less polar solvent favours E2 over SN2. Since ethanol (e = 24) is less polar
than DMSO (e = 45), elimination is expected to be favoured more in ethanol. But
dehydrobromination of 3-bromopentane (CH3CH2CHBrCH2CH3) occurs much more readily
in DMSO-t-BuOK than in EtOH-KOH. This can be explained in terms of the strength of
the base which is a more important factor that promotes elimination reaction. A strong
base favours E2 more than SN2. t-BuOK is a stronger base than OH① or OEt① and in the
aprotic solvent DMSO, its effectiveness as a base increases because solvation through
H-bonding does not take place. Elimination is, therefore, favoured very much by using
t-BuOK in DMSO. Furthermore, E2 is also favoured because backside attack by the bulky
base (nucleophile) Me3CO: * to bring about substitution (SN2) is disfavoured.
18. The following reaction is not an useful method for the synthesis of t-butyl
isopropyl ether (Me3COCHMe2) — Why?
* ≈ EtOH
(CH 3 )2 CH — O Na + (CH 3 )3 C — Br ææææ Æ (CH 3 )3 C — O — CH(CH 3 )2
Solution When a tertiary halide is treated with a strong base, it give almost exclusively
the elimination product mainly by E2 mechanism. This is due to the fact that tertiary
substrates are very much reluctant to undergo an SN2 reaction due to steric hindrance. A
very small amount of substitution product is obtained by an SN1 reaction. t-Butyl bromide,
therefore, reacts with sodium isopropoxide (a strong base) to produce 2-methylpropene as
the major product by the E2 mechanism. Hence, the given reaction is not an useful one for
the synthesis of t-butyl isopropyl ether.
19. How would you carry out each of the following transformations:
Me Me
C== C
Et Et Et Et
(a) Me cis
C—C Me
Me Et
H NMe2
C== C
Et Me
trans
(b)
Solution
(a)
(b) The alkene I can be prepared by syn-elimination of –NMe2 and the b-D as follows:
The alkene II can be prepared by anti-elimination of –NMe2 and the b-H as

follows:
Solution When dehydrobromination of 1-bromo-1-methylcyclopentane is carried out

with the bulky base potassium tert-butoxide (t-BuOK) in tert-butyl alcohol (t-BuOH), the
formation of the less substituted alkene methylencyclopentane is favoured. This is due to
the steric bulk of the base and to the fact that in tert-butyl alcohol the base is associated
with solvent molecules and thus made even larger. The large tert-butoxide ion faces much
difficulty in removing one of the ring b-H atoms because of greater crowding at that site in
the transition state. It removes one of the more exposed methyl hydrogens instead to give
methylenecyclopentane as the major product.
21. Selenoxides have recently been introduced as excellent alkene precursors.

They readily undergo syn-elimination at or near room temperature
to give good yields of alkenes. How would you carry out the following
transformation through the formation of a selenoxide?
Give the mechanism of syn-elimination process involving the selenoxide.
Solution This transformation may be carried out as follows:

22. Suggest a mechanism for each of the following pyrolytic eliminations:

Br
600°C
Me H
(a) (b) H Me
H
An alkyl bromide
(c)
Solution
(a)
(b)
(c)
23. When heated, the salt I undergoes decarboxylative debromination at a

rate faster than its isomeric salt II. Explain.
Solution In the salt I, the Br atom and the — COONa group are anti to each other. Since
the stereoelectronic requirement for normal E2 elimination in anti-fashion is satisfied,
therefore, it undergoes decarboxylative debromination through a more stable transition
state. On the other hand, in the salt II, the Br atom and the — COONa group syn to each
other. So, it undergoes decarboxylative debromination through a less stable syn-coplanar
transition state. It is for this reason, the salt I undergoes decarboxylative debromination
as a rate faster than does the salt II.
1. Chlorofumaric acid undergoes dehydrochlorination at about 50 times faster than

does chloromalic acid to give the common product acetylenedicarboxylic acid.
[Hint: Chlorofumaric acid is (Z)-2-chlorobutenedioic acid in which H and Cl are
anti to each other and chloromalic acid is (E)-2-chlorobutenedioic acid in which H
and Cl are syn to each other.]
3. Predict the product and suggest a mechanism for each of the following fragmentation
reactions:
@
OH
(a) (b) HOCR 2CH 2CR 2¢ Br æææÆ
(c) D (d)
R 2NCR 2CH 2 CR 2¢ Br ææ Æ
4. Dehydrohologenation of vic-dihalides (with the elimination of 2 moles of HX)

normally leads to the formation of an alkyne rather than to the formation of a
conjugate diene. However, when 1,2-dibromocyclohexane is dehydrolalogenated,
1,3-cyclohexadiene is obtained in good yield. Account for this observation.
[Hint: A six-membered ring cannot accommodate a triple bond because of severe
strain that would be introduced.
5. Predict the product and give the mechanism of each of the following reactions:
CH3 CH3
Me Me | |
Zn
(a) Zn (b) Br CH 2C == CCH 2 Br æææ Æ
Br Br Br O
| ||
Zn
(c) Zn
Br CH 2C ∫∫ CCH 2Br æææ Æ (d) Me2C — C — Br æææ Æ
6. When the compound I (and its enantiomer) is treated with C2H5OH in C2H5OH,
cis-2-butene is obtained without loss of deuterium and trans-2-butene is obtained
with loss of deuterium. However, its diastereoisomer II (and its enantiomer) gives
cis-2-butene with loss of deuterium and trans-2-butene without loss of deuterium.
How do you account for these findings.
CH3 CH3
H D D H
H Br H Br
CH3 CH3
I II
7. When the following deuterium-labelled compound is subjected to dehydrobromina-
tion using NaOEt in EtOH, the only alkene product is 3-methylcyclohexene (which
contains no deuterium). Provide an explanation for this result.
Br H
H H
D
H3C
8. Predict the product of the following reaction? What is the purpose of sulfuric acid
in the second step?
Br
H CO2H 1. KOH / EtOH, D
HO2C H 2. dil. H2SO4
Br
9. Compare the substitution/elimination ratio for the reactions of ethyl bromide
(CH3CH2Br) and 2,2,2-trideuterioethyl bromide (CD3CH2Br) with potassium
tert-butoxide in tert-butyl alcohol.
10. On E2 elimination of HCl with n-C5H11ONa/n-C5H11OH to yield 2-chloronorbornene
(I), II reacts about 100 times as fast as its diastereoisomer III. Account for this
observation.
11. Dehydrochlorination of 1-chloro-1-methylcyclopropane produces two alkenes (I and

II). Explain why I is obtained as the major product despite the fact that it contains
less substituted double bond.
– +
CH3 Me3COK
+ CH3
Cl Me3COH
1-Chloro-1-methyl- I II
cyclopropane (major) (minor)
[Hint: The alkene II containing two sp2 hybridized carbon in a three-membered

ring is thermodynamically very unstable due to severe strain that has been
introduced.]
O Et
12. Threo- PhCHDCHPhO—C— , when treated with NaOEt in EtOH, retained
Et
practically all the D. However, the erythro-isomer, when similarly treated, lost
practically all its D. Write a mechanism consistent with this observation.
[Hint:
13. Which isomers of C4H9Br yield only a single alkene on dehydrohalogenation? Give
the structures of the alkenes.
14. Convert: (a) Ethylbenzene Æ Styrene; (b) 2-Butyne Æ 1-Butyne.

15. Write the product when b-phenylethyl bromide is treated with sodium ethoxide in
the presence of C2H5OD. Give the mechanism of the reaction.
16. Describe how you can convert (Z)-2-butene to (E)-2-butene.
[Hint: Epoxidation of (Z)-2-butene by PhCOOOH produces an oxirane which on
treatment with triphenyl phosphine (Ph3P) gives a charge-separated intermediate
(a betaine). This undergoes cyclization followed by elimination of Ph3PO to give
(E)-2-butene.
(a)
(b)
[Hint:
(a)
(b)
18. E2 elimination of HX involving enantiotropic hydrogens must produce homomers,

while that involving diastereotopic hydrogens must produce diastereoisomers.
Illustrate with suitable examples.
19. Write a mechanism of the following transformation:
[Hint:
The thermodynamic stability of the acetylide anion will drag the overall reaction
far to the right.]
20. Which of the following compounds (A or B) eliminates HBr more readily when
heated with a base and why?
21. 1-Bromobicyclo[2.2.1]heptane does not undergo dehydrobromination when heated

with a base. Explain.
22. Explain the following reaction and write the product:
Br
|
Et3CO:@
Me2CCH 2CH 3 æææææ
Et COH
Æ (only the major product)
3
23. Which alkyl halide and what conditions should be used to prepare the following
alkene in good yield by an E2 elimination? Explain your choice.
==
[Hint: 1-Bromo-1-methylcyclohexane is to be treated with a sterically hindered

base like Me3CO@ or Et3CO@ in the corresponding alcohol.]
24. Which alkyl halide would you expect to be more reactive in an E2 reaction?
(a) —CH2CHBrCH2CH3 or —CH2CH2CHBrCH3
or
(b) Br
Br
(c) Me2CCHBrCH2CH3 or Me2CHCH2CHBrCH3
[Hint: Alkyl halides leading to the formation of conjugated double bonds and more
substituted double bond are relatively more reactive.]
25. Write down the major elimination product expected to be obtained from an E2
elimination of each of the following alkyl halides with OH@ ion:
(a) CH3CHClCH2CH == CH2 (b) (CH 3 )2CHCHCH 2CH 3 (c)
|
Br Br
Br
(d) (e) CH3CHFCH2CH3
CH3
26. Predict the product of each of the following reactions:
(a) (b)
D (d) Me2CH Me
1. RLi (2 equiv.)
(c) 2. H2O
SO2 N—NHTs
27. Explain the results of the following reactions:

H
Ph D Ph Ph
(a) H +
OC—SCH3
3-Phenylcyclohexene 1-Phenylcyclohexene
(>96%) (<4%)
erythro S
(b)
28. Give the steps involved in the complete Hofmann degradation of each of the
following compounds:
(a) NH
(b)
Me
H3C N
CH3
(c)
NHMe (d)
H
MeHN
H
29. Write down the structures of the three alkenes I, II and III when the following
quaternary ammonium hydroxide is heated. Arrange the products in the increasing
order of their relative amounts and explain this observation.
30. Explain the distribution of products in the following dehydrobromination using

various bases:
31. Arrange the following compounds in the order of decreasing E2 reaction rate and
explain the order:
CH3 CH3 CH3

, ,
Br Br Br
CH3
I II III
I, II, III
[Hint: ææææææææææ
E2 reaction rate decreases
Æ; this is because the conformation in which
the stereo-electronic requirement of E2 is satisfied becomes gradually more
unstable.]
32. Dehydrohalogenation of isopropyl bromide, which requires several hours of refluxing
in alcoholic KOH, is brought about in less than a minute at room temperature by
t-BhOK in DMSO. Suggest a possible explanation for this.
33. Explain why (CH3)2CHCHBrCH2CH3 reacts faster than (CH3)2 CHCH2CHBrCH3
in an E2 reaction, even though both alkyl halides are secondary.
34. DBN and DBU are two strong sterically hindered nitrogen bases useful for E2
reactions. Write their structures and explain why they are strong bases.
35. Give the products that would be obtained by treating the following tertiary amines
with hydrogen peroxide followed by heat:
(a) (CH3)2NCH2CH2CH3 (b) H3C— —N(CH3)CH2CH2CH3
CH3
| (d)
(c) N CH3
CH 3 CH 2NCH 2 CH (CH 3 )2
CH3
36. How would you expect the ratio of SN2 to E2 product formed from the reaction of
CH3CH2CH2Br with CH3O@ in CH3OH to change when the nucleophile is changed
to CH3S@?
[Hint: CH3O@ is a strong base, but a weak nucleophile. On the other hand, CH3S@
is a strong nucleophile, but a weak base. Therefore, the SN2/E2 ratio is higher in
the presence of CH3S@, but lower in the presence of CH3O@.]
37. Which member of each of the following pairs of compounds results in higher ratio
of elimination (E2) to substitution (SN2) when treated with NaOEt/EtOH?
(a) (CH3)2CHBr or (CH3)3 CBr
Br
Br
(b) or
Me Me
(c) CH CH CHBrCH or CH CH CHCH
3 2 3 3 2 3
|
!SMe I@
2
[Hint:
(a) (CH3)3CBr gives higher E2/SN2 ratio because SN2 is inhibited due to steric
hindrance.
Br
(b) gives higher E2/SN2 ratio because the resulting double bond is
Me
stabilized by conjugation.
(c) Being a charged substrate CH3CH2CH(CH3)S≈Me2I@ gives higher E2/SN2
ratio.]
38. The following two isomeric compounds yield the same products when treated with
Me3CO@ in MeCOH. Predict the products and sketch the mechanisms involved.
H
OTs
(a) (b)
H
OH
[Hint:
(a)
(b)
39. Which one of the following hexachlorocyclohexanes is the least reactive in an E2

reaction and why?
Cl Cl Cl Cl
Cl Cl Cl Cl Cl Cl Cl Cl
Cl Cl Cl Cl Cl Cl Cl Cl
Cl Cl Cl Cl
I II III IV
[Hint: I; because in both of its conformations there is no diaxially placed Cl and H

atoms.]
40. Give the dehydrochlorination products obtained when the following isomeric
substrates are treated with high concentration of MeO@:
(a) (2S, 3R)-2-Chloro-3-methylpentane
(b) (2R, 3R)-2-Chloro-3-methylpentane
41. Indicate which of the species in each pair would give a higher substitution/
elimination ratio when it reacts with isopropyl bromide (MeCHBrMe):
@ @
(a) OCN or SCN (b) Cl@ or Br@ (c) C2H5O@ or Me3CO@
42. Explain why 2-bromopropane gives different amounts of substitution and
elimination products when the nucleophile/base is changed from acetate (CH3COO@)
to ethoxide (C2H5O@).
Me2CHBr + CH 3COO@ ææ
Æ Me2CH OCOCH 3
(100%)
Me2CHBr + C2H5O@ ææ Æ Me2CHOC2H5 + MeCH == CH 2
(20%) (80%)
43. Predict the products of the following reaction:
[Hint: One of the products is Me3SiOCH3.]

44. Would you expect the following two structures to form the identical or different
elimination product?
D Ph
OTs H
Ph D
H OTs
I II
45. E2 elimination of the compound I is ten times slower than that of the compound
II. Explain.
CH 3CH 2CH C (CH 3 )3 CH 3 — CHC(CH 3 )3
| @ |
!N(CH ) OH
3 3
!
N(CH3 )OH@
I II
46. The rate of pyrolytic elimination of 1-phenylethyl acetate is about eleven times
faster than for 2-phenylethyl acetate. Explain.
47. Which one of the following bromides will undergo dehydrobromination at a faster
rate in sodium hydroxide-isopropanol and why?
H Br H H
C==C C==C
H Br
O2N NO2
A B
[Hint:
49. Suggest a mechanism for the preparation of (E)-1,2-diphenypropene form erythro-

1,2-diphenyl-1-propanol by the Chugaev reaction.
50. Predict the major and minor products obtained in the following reaction and explain
such distribution of products:
ª 150∞C
CH 3CH 2CH CH 3 ææææ Æ
|
! @
(CH3 )2 N — O:
51. When optically active coniine is subjected to Hofmann degradation, the major
amine product (C10H21N) is found to be optically active. Predict the structure of
the product and give the mechanism of the reaction involved.
N CH2CH2CH3
H
Coniine
(an alkaloid)
(a)
D H
(b) + – H
N(CH3)3OH
(E)-Cyclooctene
(principal product)
53. Explain why weakly ionizing solvent promotes syn-elimination when the leaving
group is uncharged but anti-elimination when the leaving group is charged.
[Hint: Ion pairing occurs extensively in non polar solvents. This can cause syn-
elimination with an uncharged leaving group through the following fashion:
]
The effect is reversed when the leaving group is positively charged. Ion pairing
favours anti-elimination in this case. A relatively free base can be attracted to
the charged-leaving group and as a result, it places itself in a favourable position
for attack on the syn b-H atom. Ion pairing reduces this attraction and thereby
disfavours the arrangement for syn-elimination as shown below:
54. The ratio of syn-to-anti elimination when 1,1,4,4-tetramethyl-7-cyclodecyl tosylate

is treated with t-BuOK in the nonpolar solvent benzene is much higher. But when
the crown ether dicyclohexano-18-crown-6 is used, the syn/anti ratio decreases to a
much lower value. Explain these observations.
[Hint: Ion pairing causes syn-elimination somewhat greater than anti-elimination
because the leaving group is uncharged and the reaction is carried out in nonpolar
@ !
solvent benzene. The crown ether selectively removes K≈ from t-BuO K by forming
a stable complex with K≈. The free base t-BuO@ then causes predominant anti-
elimination.]
55. Predict the products in each of the following reaction sequences:
(a)
(b)
Give the mechanism of the last step in each sequence.

[Hint:
Ph D Ph H
C fi ; F fi
H Ph H Ph
[Hint: The b-D seems to be sandwiched between the bulky t-butyl groups and
the methyl groups of trimethylamine so that an external base (i.e., OH@) cannot
approach it but an intramolecular attack by the anion (obtained by abstraction of
a proton from the — CH3 group attached to positive nitrogen) is favoured.]
4.2 THE E1 REACTION

Example kinetics, mechanism, stereochemistry, evidence in favour of the
mechanism, factors influencing E1 reactivity, i.e., the E1 reaction rate and other
relevant topics.
4.2.1 Example of E1 Reaction

When tert-butyl chloride is treated with 80 percent aqueous ethanol at 65°C, 36 percent
2-methylpropene is obtained by unimolecular elimination (E1) and 64 percent tert-butyl
alcohol and tert-butyl ethyl ether by unimolecular substitution (SN1).
E1 reactions almost always accompany SN1 reactions.
4.2.2 Kinetics of E1 Reaction

The E1 reaction exhibits first-order kinetics and obeys the following rate law:
Rate = k[(CH3)3 Cl]

Like the SN1 mechanism, the kinetics suggests that the mechanism of this elimination
reaction involves more than one step, and that the slow step is unimolecular, involving
only the alkyl halide.
4.2.3 Mechanism of E1 Reaction

The most straightforward explanation for the observed first-order kinetics is a two-step
reaction in which the bond to the leaving group breaks followed by the p bond is formed.
The mechanism of this reaction is designated as E1 (E stands for Elimination and 1 stands
for unimoleculare) because only one reactant (the substrate) is involved in the transition
state of the rate-determining step.
The E1 mechanism of the formation of 2-methylpropene form tert-butyl chloride involves
the following two steps:
Step 1:
Step 2:
In the first step, tert-butyl chloride slowly dissociates under the influence of solvent forces
to form tert-butyl cation (a stable 3° carbocation) and chloride ion. It is the rate-determining
step of the reaction and it is the same first step as the SN1 mechanism. In the second step,
the solvent H2O (or Cl@) acts as a base and removes a proton from a carbon adjacent to
the positive carbon of the carbocation (a b-carbon) to form the alkene. During proton loss
the vacant p orbital of the positive carbon and the C — H bond become coplanar because
in such a geometry maximum interaction between the C — H bonding electrons and the
vacant p orbital occurs.
The energy profile diagram for this E1 reaction is as follows:

The carbocation formed in the first step of an E1 reaction is planar. This means that the
electrons from a departing proton can move towards the positively charged carbon from
either side. Therefore, both syn- and anti-elimination can take place. Because both syn-
and anti-elimination can occur, an E1 reaction forms both the E and Z products regard-
less whether the b-carbon atom from which the proton is removed is bonded to one or two
hydrogens. The major product is the one with the bulkiest groups on opposite sides of the
double bond because this alkene is thermodynamically more stable. For example:
(i)
(ii)
It is to be noted that an E2 reaction forms the E and Z products only if the b-carbon
from which the proton is expelled is bonded to two hydrogens. If it contains only one
hydrogen, only one product is obtained because anti-elimination takes place. The
configuration of the product then depends on the configuration of the substrate.
4.2.5 Evidence in Favour of the E1 Mechanism

(a) The E1 reactions have been found to follow first-order kinetics and obeys the rate
law: rate = k[RX]. This is consistent with the E1 mechanism in which only one
species, the substrate, is involved in the rate-determining step (the first step).
(b) Primary isotope effect is not observed in first-order eliminations. This is consistent
with the E1 mechanism where the proton is lost in the second, fast step.
(c) The first-order eliminations show the same effect of structure on reactivity as SN1
reactions do. E1 involves exactly the same first step as SN1. Since the first step is
rate-determining, it follows that the order of reactivity of alkyl halides (R–X) in E1
reaction must be same as in SN1 reaction (i.e., 3°> 2°> 1°).
(d) The substitution/elimination ratio remains the same when two substrates which
differ only in the leaving group undergo first-order elimination reaction. This
supports the E1 mechanism in which the competition between substitution and
elimination cannot be influenced by the nature of the leaving group.
(e) Where the structure permits, first-order eliminations often result in formation of
rearranged products. A rearrangement product can be obtained if the intermediate
is a carbocation which can rearrange to give a more stable carbocation. This is also
in agreement with the E1 mechanism which involves rate-determining formation
of a carbocation.
4.2.6 Factors Influencing E1 Reaction Rate or E1 Reactivity

(a) The effect of substrate structure on rate: The order of reactivity is primary
< secondary < tertiary and this is because the stability of the carbocations formed
in the rate-determining step increases on going from primary to tertiary due to +I
and hyperconjugative effects.
(b) The effect of the base on rate: Since the base is not involved in the rate-
determining step of the E1 reaction, therefore, the E1 reaction rate does not depend
on base concentrations. Weak bases like H2O and EtOH favour E1 reactions.
(c) The effect of the leaving group on rate: Since a better leaving group causes
facile ionization in the rate-determining step, the more the leaving ability of a
group increases the more the E1 reaction is favoured.
(d) The effect of solvent on rate: E1 reactions are favoured by increasing the polarity
and ion-solvating ability of the solvent. This is because the transition states (step
1) of E1 reactions involving ionic intermediates are considerably polar.
(e) The effect of temperature: Increasing the temperature increases the percentage
of elimination product. This is because elimination leads to an increase in the
number of particles and thus has the more favourable entropy term.
4.2.7 Regioselectivity of E1 Reactions

In an E1 reaction, the leaving group departs in the first step and a proton is lost from
an adjacent carbon in the second step. If the substrate contains more than one type of
b-hydrogen, then which alkene is produced predominantly is clearly determined by which

b-proton is lost faster from the carbocation in the second step. The transition step of
the second step possesses considerable alkene character. Therefore, the transition state
leading to a more substituted alkene is more stable than the transition state leading to
a less substituted alkene and because of this, the more substituted alkene is produced
predominantly. Thus, for E1 reactions, the Saytzeff rule governs the orientation of
elimination. For example, the E1 elimination of HBr from 1-bromo-1-methylcyclopentane
yields 1-methylcyclopentene as the major product and methylenecyclopentane as the
minor product.
Similarly, 2-bromo-2-methylbutane produces 2-methylbut-2-ene as the major product and

2-methylbut-1-ene as the minor product.
If in an E1 reaction the transition state leading to the more substituted alkene suffers
from severe steric strain, it becomes the minor product. For example, 2-chloro-2,4,4-
trimethylpentane undergoes E1 reaction to form an excess of the less substituted alkene
2,4,4-trimethylpent-1-ene (the Hofmann product).
When a substituted cyclohexane undergoes an E1 reaction, the two groups that are
eliminated do not have to both be in axial positions because the elimination reaction is not
concerted. The carbocation formed in the first step of the following reaction loses a proton
to yield the Saytzeff product predominantly.
4.2.8 Rearrangement of the Carbocation Intermediate Involved in an

E1 Reaction
Because the E1 reaction involves the formation of a carbocation intermediate,
rearrangement of the carbon skeleton may take place before the loss of a proton. For
example, the secondary carbocation obtained from 3-chloro-2-methyl-2-phenylbutane
undergoes a 1,2-methyl shift to form a more stable tertiary benzylic cation
In the following example, the secondary carbocation undergoes a 1,2-hydride shift to yield
a more stable secondary allylic cation.
Similarly, in the following reaction, the secondary carbocation undergoes a 1,2-hydride

shift, forming a more stable tertiary carbocation.
Some examples of carbocations which have the potential to undergo rearrangement to

form a more stable carbocation are given below:
(a)
(b)
CH3 –
+ CH3- shift +
(CH3)2 C——CHCH3 (CH3)2 C——CH(CH3)2
A 2° carbocation A 3° carbocation
(c)
(d)
(e)
4.2.9 Acid-Catalyzed Dehydration of Alcohols

Most alcohols undergo dehydration (loss of a molecule of H2O) to form an alkene when
heated with a strong acid.
This is an elimination reaction which is favoured at higher temperature. The most

commonly used acids are sulphuric acid (H2SO4) and phosphoric acid (H3PO4).
The reaction temperature and the concentration of acid required to dehydrate an alcohol
depends on the structure of the alcohol used.
(a) Primary alcohols are very much difficult to dehydrate. For example, dehydration of
ethyl alcohol requires concentrated H2SO4 and a temperature of 180°C.
conc.H2SO4
CH 3 — CH 2 — OH ææææææ
180∞C
Æ CH 2 == CH 2 + H 2O
Ethyl alcohol Ethylene
(a 1° alcohol)
(b) Secondary alcohols usually require milder conditions for dehydration. Dehydration of
cyclohexanol, for example, requires 85 percent H3PO4 and a temperature of 165–170°C.
(c) Dehydration of tertiary alcohols is usually so easy that extremely mild conditions
can be used. tert-Butyl alcohol, for example, undergoes dehydration in only 20%
aqueous sulphuric acid at a temperature of 85°C.
20% H2SO4
(CH 3 )3 C — OH æææææ
85∞C
æÆ CH 2 == C(CH 3 )2 + H 2O
tert-Butyl alcohol Methylpropene
(a 3° alcohol)
Thus, the relative ease with which alcohols undergo dehydration is in the following
order:
R 3C — OH > R 2CH — OH > RCH 2OH
3∞ alcohol 2∞ alcohol 1∞ alcohol
This behaviour is, in fact, related to the relative stabilities of the intermediate
carbocations.
Mechanism of dehydration The dehydration of alcohols is generally an E1 reaction of

their conjugate acids and involves the steps as follows:
Step 1:
H H
+
—C——C— + H—OH2 —C——C— + H2O
+
OH OH2
In this step, protonation converts the very poor leaving group (OH@) into a good neutral
leaving group (H2O).
Step 2:
In this step (the rate-determining step), the protonated alcohol dissociates to form a
carbocation and water.
Step 3:
In this step, a base in the reaction mixture (either H2O or HSO4@) removes a proton
from a carbon adjacent to the positively charged carbon, forming the alkene product and
regenerating the acid catalyst.
Because the rate-determining step is the formation of a carbocation, the ease of dehydration
follow the same order as the ease of fromation of carbocations: 3° > 2° > 1°. Hence, the
case of dehydration of alcohols follows the order: 3° > 2° > 1°. In fact, the primary (1°)
carbocations are too unstable to be fromed and dehydration of primary alcohols actually
takes place by the E2 mechanism (as in the case of CH3CH2OH) or by the E1 mechanism
in which the incipient 1° carbocation undergoes rearrangement.
Acid-catalyzed dehydration of primary alcohols is rarely a good method for converting them
to 1-alkenes because rearrangement and isomerization of the products in the presence of
acid are very common with primary alcohols. 1-Butanol, for example, undergoes acid-
catalyzed dehydration with rearrangement to give a mixture of 1-butene and 2-butene.
The more substituted alkene 2-butene is thermodynamically more stable and so it is
obtained as the major product following the Saytzeff rule.
The formation of 2-butene as the major product can also be explained by an E2 elimination
to form 1-butene followed by addition of a proton from the acidic solution to the double
bond of 1-butene following Markownikoff rule, thereby forming a stable carbocation and
finally loss of a proton form the carbocation following the Saytzeff rule.
Rearrangement during dehydration Dehydration of secondary and tertiary alcohols in-

volves the formation of a carbocation intermediate. If that carbocation is less stable it may
rearrange to a more stable one. For example:
An example of a ring-expansion rearrangement may be given as follows. Both the initially

formed carbocation and the carbocation it rearranges to give are 2° carbocations. However,
because of strain in the four-membered ring, the initially formed carbocation is less stable.
Rearrangement relieves this strain. The rearranged 2° carbocation can further rearrange
by a 1,2-hydride shift to form a relatively more stable 3° carbocation.
4.2.10 Dehydration using POCl3 and Pyridine

The relatively harsh conditions (acid and heat) required for dehydration of alcohols can
cause low yield of a desired alkene. The sever conditions required for dehydration of an
alcohol can be avoided, however, if the reaction is carried out in the presence of phosphorus
oxychloride (POCl3) and pyridine (C H N) . For example:
6 5
OH
|
POCl3
CH 3CH 2CHCH 3 ææææææ
pyridine,0∞C
Æ CH 3 CH == CH CH 3
2-Butanol 2-Butene
(major product)
Mechanism: The mechanism of the reaction is as follows:
@
Phosphorus oxychloride (POCl3) converts the — OH group of the alcohol into O POCl2 (a
good leaving group) in the presence of pyridine. The resulting compound undergoes E2
reaction under the mildly basic reaction conditions to give 2-butene as the major product
and 1-butene (CH3CH2CH == CH2) as the minor product.
4.2.11 Factors Influencing the Extent of E1 and E2 Reactions

The following factors affect the extent of E1 and E2 eliminations:
(a) Substrate structure: The increasing order of reactivity by both the mechanisms is
primary < secondary < tertiary. (In fact, primary alkyl halides undergo only E2 reactions
or E1 reactions with rearrangement. They cannot undergo normal E1 reactions because of
the difficulty encountered in forming primary carbocation.) The reasons for the reactivity
order are, however, different. The rate of elimination by E2 increases mainly because of
the alkene being formed increases as the number of alkyl group attached to the double-
bonded carbon increases. The rate of E1 reaction, on the other hand, increases because the
stability of the carbocation to be formed in the rate-determining step increases.
(b) Base: The base is involved in the rate-determining step of an E2 reaction but not
in the rate-determining step of an E1 reaction. Because of this, it is only the rate of E2
reaction which depends on the concentration of the base used. In fact, an E2 reaction
is favoured by a high concentration of a strong base. In E1 reaction, it is generally the
solvent which acts as the base. Hence, when external bases are added, the mechanism is
shifted towards E2.
(c) Leaving group: Since a better leaving group causes facile ionization, the more the
leaving ability of a group increases, the more the E1 reaction is favoured over the E2
reaction.
(d) Solvent: E1 reactions are favoured by increasing the polarity and ion-solvating ability
of the solvent, i.e., these are generally favoured by protic polar solvents like H2O, AcOH
ROH etc. E2 reactions, on the other hand, are favoured by a decreasse in the polarity of
the solvent. This is because the transition states of E1 reactions (step 1) involving ionic
intermediates are more polar than the transition states of E2 reactions. In fact, aprotic
polar solvents like DMSO, DMF, etc. have a very pronounced effect to increase the rate of
the E2 reaction.
4.2.12 Factors that Govern the Proportions of E1 and SN1 Reactions

The following factors govern the proportion of E1 and SN1 reactions:
(a) Substrate structure: The first step of these two unimolecular reactions is the same
(formation of a carbocation) and it is the second step which determines the proportion
of elimination to substitution. The extent of E1 elimination depends upon the fact that
how fast the carbocation intermediate loses a proton, and the extent of SN1 substitution
depends upon the fact that how fast the carbocations undergoes attack by a nucleophile.
The tendency of a carbocation to lose a proton increases in the order: primary < secondary
< tertiary. This is because the stability of the resulting alkanes increases progressively
on going from primary to secondary to tertiary. The tendency of the carbocation to be
attacked by the nucleophile (usually the solvent), on the other hand, increases in the order:
tertiary < secondary < primary. This is because the stability of the carbocation decreases
progressively, i.e., the carbocation becomes progressively more reactive on going from
tertiary to secondary to primary. The other reason for such order is purely steric. The bond
angles of the carbocation remain the same on elimination (120°–120°), but decreases on
substitution (120°–109.5°). Therefore, steric crowding increases when a carbocation with
a trigonal planar arrangement of three groups is converted to the substitution product,
with a tetrahedral arrangement of four groups. The extent of substitution thus decreases
as the size of the groups attached to the positive carbon increases, i.e., on going from
primary to secondary to tertiary and as a consequence, the extent of elimination increases
in this order.
(b) Base: At low base concentrations or in the absence of base altogether, in ionizing
solvents, unimolecular mechanisms are favoured, and it is SN1 which is favoured with
respect to E1.
(c) Leaving group: For first-order reactions, the leaving group has nothing to do with the
competition between elimination and substitution, since it is gone before the decision is
made as to which path to take. However, where ion pairs are involved, this is not true, and
it has been observed that the nature of the leaving group does affect the product ratio.
(d) Solvent: The unimolecular reactions are favoured by an increase in the polarity
and ion-solvating ability of the solvent and in most solvents SN1 reactions are favoured
over E1. However, the extent of elimination somewhat increases by a decrease in solvent
polarity and this is because the charge is more dispersed in the transition state leading to
elimination than in the transition state leading to substitution.
(e) Temperature: In most unimolecular reactions, the SN1 reaction is favoured over the
E1 reaction, especially at lower temperatures. However, increasing the temperature of
the reaction favours the E1 reaction at the expense of the SN1 reaction. This is because
elimination leads to an increase in the number of particles and thus has the more favourable
entropy term.
1. Some primary and secondary alcohols undergo rearrangement of their

carbon skeleton during dehydration. Write structures of the alkenes
expected to be formed in each of the following dehydration reactions and
provide mechanism for their formations.
CH3
| H2SO4
(a) CH 3CH 2CHCH 2OH ææææH2SO4
Æ (b) (CH 3 )2CH CHOH CH 3 ææææ
D
Æ
D
(c) H2SO4 (d)

(CH 3 )2CCHOH CH 3 ææææ
D
Æ
Solution
(a) 2-Methyl-1-butanol undergoes dehydration in the presence of concentrated H2SO4
to give 2-methyl-2-butene as the major product and 2-methyl-1-butene as the minor
product.
CH3 CH3 CH3
| | |
H2SO4
CH 3CH 2CHCH 2OH ææææ D
Æ CH 3CH == C — CH 3 + CH 3CH 2 == CH 2
C
2-Methyl-1-butanol 2-Methyl-2-butene 2-Methyl-1-butene
(major product) (minor product)
Mechanism:
CH3 H
–
H—OSO3H + H shift +
CH3CH2CHCH2OH CH3CH2C—CH2—OH2 CH3CH2CCH3
D —H2O
CH3 CH3
A 3° carbocation
(very stable)
It is to be noted that carbocations rearrange only if they become more stable as

a result of rearrangement. In this reaction, the protonated alcohol undergoes
ionization assisted by a 1,2-hydride shift to give a stable 3° carbocation.
This carbocation undergoes loss of a proton forms a b-carbon to yield either
2-methyl-2-butene or 2-methyl-1-butene. The protonated alcohol may also undergo
an E2 reaction to give 2-methyl-1-butene. The more substituted alkene 2-methyl-2-
butene is thermodynamically more stable and so, it is obtained as the major product.
(b) 3-Methyl-2-butanol undergoes acid-catalyzed dehydration to give 2-methyl-2-
butene as the major product and two 1-butenes as the minor products.
H2SO4
(CH 3 )2CH CH OH CH 3 ææææ
D
Æ (CH 3 )2C == CH CH 3 + (CH 3 )2CH CH == CH 2
3-Methyl-2-butanol 2-Methyl-2-butene 3-Methyl-1-butene
(major product)
CH3
|
+ CH 3 CH 2C == CH 2
2-Methyl-1-butene
Mechanism:
The initially formed 2° carbocation rearranges to give a more stable 3° carbocation

by a 1,2-hydride shift. The resulting 3° carbocation loses different b-hydrogens
to yield 2-methyl-2-butene and 2-methyl-1-butene. The 2° carbocation also loses
different b-hydrogens to yield 2-methyl-2-butene and 3-methyl-1-butene. The more
substituted alkene 2-methyl-2-butene is thermodynamically more stable and so, it
is obtained as the major product.
(c) 3,3-dimethyl-2-butanol undergoes acid-catalyzed dehydration to yield 2,3-dimethyl-
2-butene as the major product and two 1-butenes as minor products.
CH3
|
H2SO4
(CH 3 )3CCHOH CH 3 ææææ D
Æ (CH 3 )C == C(CH )
3 2 + (CH )
3 2 CH C == CH 2
3,3-Dimethyl-2-butanol 2,3-Dimethyl-2-butene 2,3-Dimethyl-1-butene
(major product)
+ CH 2 == CH — C(CH 3 )3
3,3-Dimethyl-1-butene
Mechanism:
The initially formed 2° carbocation rearranges to give a more stable 3° carbocation

by 1,2-methanide shift. This 3° carbocation loses different b-hydrogens to yield
2,3-dimethyl-2-butene and 2,3-dimethyl-1-butene. The 2° carbocation may also
lose a proton to produce 3,3-dimethyl-1-butene. The more substituted alkene 2,3-
dimethyl-2-butene is thermodynamically more stable and so, it is obtained as the
major product.
(d) 1-Cyclohexylethanol undergoes dehydration in the presence of concentrated H2SO4
to yield 1-ethylcyclohexene as the major product.
Mechanism:
The initially formed 2° carbocation rearranges to give a more stable 3° carbocation by a 1,2-
hydride shift. The resulting 3° carbocation loses a b-hydrogen to yield 1-ethylcyclohexene.
The 2° carbocation may also lose different b-hydrogens to yield the two minor products.
C2H5 + C2H5
(a) H
CHC2H5 heat
C2H5
OH
(b)
(c)
OH
+
(d) H
heat
CH3
HBr Br
(e)
CH3
OH CH3
CH3
+
H
(f) heat
O CH2OH O CH3
OH CH3
CH3 +
(g) H
CH3 heat
H3C H3C CH3
CH3 + CH3
H
(h) CH3 heat
CH3
OH
CH2OH
(i) H
+
heat
Solution
(a)
(b)
(c)
(d)
This process is especially favourable because relief in ring strain occurs.
(e)
(f)
(g)
(h)
(i)
3. Give mechanistic explanation for each of the following observations:
(a)
(b)
Solution
(a) trans-1-Bromo-2-ethylcyclohexane undergoes E2 dehydrobromination when it is
heated with sodium ethoxide in ethanol. For a facile E2 elimination involving a low
energy T.S., the ring must flip to the diaxial conformation in which the departing
groups are anti-periplanar. Since in this conformation the only axial hydrogen b
to the axial bromine atom is situated at C-6, therefore, the reaction leads to the
formation of 3-ethylcyclohexene exclusively.
When trans-2-ethylcyclohexanol is heated in the presence of an acid, it undergoes

acid-catalyzed dehydration by the E1 mechanism. The alcohol becomes protonated
and then loses water to form a carbocation. The carbocation may lose two b-protons
(from C-2 or C-6) to form two alkenes. Since the transition state leading to the
more substituted alkene (the Saytzeff product) 1-ethylcyclohexene is relatively
more stable than the transition state leading to the less substituted alkene
3-ethylcyclohexene, therefore, the former alkene is obtained as the major product.
4. Explain why alkenes are prepared by heating alkyl halides with

concentrated alcoholic, rather than aqueous solution of KOH.
Solution Substitution is more favoured under E1/SN1 conditions than under E2/SN2
conditions. Hence, for a given alkyl halide, a higher yield of alkene is expected to be obtained
by the E2 mechanism than by the E1 mechanism. Strong bases at high concentrations and
solvents with low ionizing power favour bimolecular mechanism and it is E2 which is
favoured over SN2. Thus, an ethanolic KOH solution (solvent of low ionizing power and a
very strong base :OEt @ ) gives better yield of alkene than does an aqueous KOH solution
(solvent of high ionizing power and a relatively weak base, :OH @ ). The reactions are
carried out at higher temperature because higher temperature always favours elimination
instead of substitution. This is due to the fact that elimination leads to an increase in the
number of particles and thus has the more favourable entropy term.
Elimination:
KOH
(CH 3 )3C —Br ææææææÆ
C2H5OH,55°C
(CH 3 )2C == CH 2 + NaBr + C2H5OH
tert-Butyl bromide 2-Methylpropene
Substitution:
KOH
(CH 3 )3 C — Cl ææææææ
H2O/acetone
Æ (CH 3 )3 C — OH + Cl@
tert-Butyl chloride tert-Butyl alcohol
5. Predict the products expected to be formed when 2-bromo-3-methylbutane

is heated in ethanol.
Solution
2-Bromo-3-methylbutane is a secondary alkyl halide. The conditions involve no nucleophile
or base other than the solvent, and a polar protic solvent. Because the solvent ethanol is
a poor nucleophile and a weak base, neither SN2 nor E2 reactions can take place. Because
polar protic solvents promote the SN1 and E1 reactions, these will be the only reactions
taking place to give the products:
Br OC2H5
| |
C2H5OH
CH 3 — CH — CH(CH 3 )2 ææææD
Æ CH 3 — CH — CH(CH 3 )2
2-Bromo-3-methylbutane SN1 Product
+ CH 2 == CH — C(CH 3 )2 + CH 3CH == C(CH 3 )2
E1 Products
OC2H5 CH3
| |
+ CH 3CH 2 — C — CH 3 + CH 3CH 2 — C == CH 2
|
CH3
rearrangement products
Mechanism:
6. Write a reasonable mechanism for the following transformation:
Solution
The protonated alcohol undergoes loss of water to form a 3° carbocation. Intramolecular

cyclization produces another 3° carbocation. This finally loses a proton to yield the diene.
!
(a) (b) H
Me3CCH == CH 2 æææÆ
2. Give the major product obtained when each of the following alcohols is heated in
the presence of H2SO4:
(a) CHOHCH3 (b) (CH3)3C CHOHCH3 (c) OH
(d) CH3 (e) (f) CH3

| CH3
O CH2OH
CH 3CH 2CH 2CCH (CH 3 )2 CH3
| OH
OH

+
H
H CH2OH
O
+
(a) (b) OH H
CH==CH2
CH3
5. The following alcohol undergoes dehydration to produce three alkenes when
heated with sulphuric acid. Give mechanisms to account for the formation of these
alkenes.
[Hint: A ring-contraction rearrangement of the initially formed 3° carbocation

takes place to form another 3° carbocation. This undergoes proton loss to form II
and CH3@ shift followed by proton loss to give III.]
6. Arrange the following alcohols in the order of decreasing their reactivity towards
acid-catalyzed dehydration:
(a) 2-Methylbut-2-ol (b) Pent-1-ol (c) 3-Methylbut-2-ol
7. Provide a mechanistic explanation for each of the following reactions:
(a)
(b)
(c) H2SO4
heat
OH
CH3OH
(d) CH 3CH == CHCH 2CHBrCH 2CH 3 ææææ Æ CH 3CH == CH — CH == CHCH 2CH 3
H H
CH3
(e) CH3
CH3OH
H
Br CH3
CH3
(f)
CH3 CH3
| ! |
EtOH/H2O
CH3CHC — S (CH3 )2 Br @ æææææÆ
D
CH 3 CH == C(CH 3 ) 2 + CH 3 CH 2 C == CH2 + S(CH3 )2
8. Predict the major product that would be formed when each of the following alkyl
halides is allowed to undergo an E1 reaction:
Cl CH3 Br
(a) CH3 (b) | |
CH 3CH 2 CH CH CH 2 CH 3 (c) CH CH CH CH CHCH
3 2 2 2 3
|
I
O CMe3 Br
(d) (e) Br (f)
H3C Cl
(g) Cl CH3 CH3

(h) Cl (i) CH2CH2CH3
CH3
Cl
9. When the following alkyl bromide is treated with methanol, three substitution
products and three elimination products are obtained. Account for the formation of
these products.
10. Give detailed mechanism to account for the formation of four products in the
following reaction:
CH3 CH3 CH3
H3C Cl H3C OCH3
CH3OH
+ + +
(—HCl)
OCH3
11. When the following alkyl halide is allowed to react with methoxide ion in a solvent
that favours SN2/E2 reactions, only one product is obtained. However, when this
compound is treated with MeO@ ion in a solvent that favours SN1/E1 reactions,
eight products are obtained. Identify the products obtained under two different
sets of conditions and account for their formations.
CH3
Cl
CH3
12. Why are 1° alcohols poor starting materials for the synthesis of 1-alkenes?
13. When the following compound is heated in methanol, four products are obtained.
Propose mechanisms to account for these products.
CH3
CH2Br
CH3OH
OCH3
+ OCH3 + +
D
14. 1-Bromomethylcyclohexene undergoes solvolysis in ethanol to yield three major

products. Give mechanisms to account for these products.
CH2Br CH2OC2H5
CH3OH
+ +
D
OC2H5
15. Solvolysis of bromomethylcyclopentane in methanol produces a mixture of five

products. Give mechanisms to account for these products.
CH2Br CH3 CH2 OCH3
H3C OCH3
CH3OH
+ + + +
D
16. When neopentyl alcohol, Me3CCH2OH, is heated in the presence of acid, it is

converted into a 85:15 mixture of two alkenes. Propose mechanism to account for
the formation of these alkenes (C5H10). Which one is expected to be the major
product and why?
17. Which alcohol of each pair would you expect to be more easily dehydrated when
heated in the presence of an acid?
(a) CH3CH2CH2CH2CH2CH2OH or CH3CH2CH2CH2CH(OH)CH3
(b) (CH3)2C(OH)CH2CH2CH3 or (CH3)2CH CH (OH) CH2CH3
(c) (CH3)2CH C(OH) (CH3)2 or (CH3)2CH CH (CH3) CH2OH
18. How would you carry out the following conversion?
n-C4 H 9OH ææ
Æ CH 3C ∫∫ CCH 3
19. Give the major product obtained in each of the following reactions:
H O!
H2SO4
(b) (CH)3 CH — CH — CH 2CHO ææææ Æ
3
(a) CH 3CH 2CH 2CH 2OH ææææ
D
Æ D
|
OH
OH OH +
O
(c) H
H H
20. Branching at the b-carbon atom favours E1 elimination. Explain.
[Hint: For example, MeCH2CMe2Cl yields only 34% alkene, while Me2CHCMe2Cl
yields 62%. The reason may be party steric: the greater the branching, the
more crowding strain is released on going from sp3 hybridized halide to the sp2
hybridized carbocation intermediate. Strain is reintroduced on substitution
(SN1), but not on proton loss (E1) to yield an alkene. It is due also to the fact
that the more b-substituted alkyl halide will yield a more substituted and hence
thermodynamically more stable alkene.]
(a)
(b)
4.3 THE E1cB REACTION

Example, kinetics, mechanism and the nature of the substrate, i.e., the substrate
structure.
4.3.1 Example of E1cB Reaction

Base-promoted dehydrofluorination of 2,2-dichloro-1,1,1-trifluoroethane proceeds by the
E1cB mechanism.
:OEt@ / EtOH
F3C — CHCl2 ææææææÆ F2C == CCl2 + EtOH + :F:@
1,1-Dichloro-2,2-
difluoroethene
4.3.2 Kinetics of E1cB Reaction

The E1cB reaction exhibits second-order kinetics (first order in the substrate and first
order in the base) and obeys the following rate law (same as E2):
Rate = k[F3CCHCl2] [OEt@]
4.3.3 Mechanism of E1cB Reaction

Although the E1cB mechanism is the least common of the elimination pathways, it
provides a reasonable extension of the elimination mechanism that is not related to
substitution. This is, in fact, a base-promoted unimolecular reaction taking place through
the formation of the conjugate base (cB) of the substrate. The mechanism is designated
as E1cB (Elimination Unimolecular conjugate Base) because it is the conjugate base of
the substrate that gives up the leaving group and the conjugate base is the only species
involved in the rate-determining step of the reaction (i.e., unimolecular).
An E1cB reaction proceeds through the following two steps:
Step 1:
The base abstracts the most acidic b-hydrogen atom from the substrate to form an
intermediate carbanion. This step is fast and reversible.
Step 2:
The carbanion loses the leaving group to form the alkene product. This step is slow (rate-
determining).
– –
—C——C— C====C + LG
LG
4.3.4 The Nature of the Substrate

Substrates containing acidic b-hydrogens (groups like –NO2, –CHO, –COR, –CN, –COOR,
@ @ @
–SO2Ar etc. should present at Cb) and poor leaving groups like F@ , Cl@ , SePh,O Ph,OH,
etc. are very prone to react by the E1cB mechanism. In the given example, the halogens
enhance the acidity of the b-hydrogen atom and fluoride ion (F@) is a poor leaving group.
The reaction occurs as follows:
Step 1:
Step 2:
4.3.5 To Distinguish between E1cB and E2 Mechanisms

In many cases, it is not possible to distinguish between E1cB and E2 mechanism. This can
be established by the following example. Dehydrobromination of 2-phenylethyl bromide
using NaOEt as a base in ethanol can be shown to take place by E1cB or E2 mechanism
using kinetic studies.
E1cB mechanism:
Step 1:
H
–
– fast
EtO + Ph—CH—CH2Br EtOH + Ph CH CH2Br
Step 2:
– slow –
Ph—CH—CH2 Ph CH==CH2 + Br
Br
Since the reaction is carried out in ethanol medium, the concentration of the ethanol
formed in the first step of the reaction can be neglected. Hence, the equilibrium constant.
@
[Ph C H CH 2Br]
K¢ =
[Ph CH 2CH 2Br][EtO@ ]
The rate-determining step of the reaction is the loss of Br@ from the carbanion (conjugate
base). Therefore, the rate of the reaction is given by
@
rate = k[Ph CH CH2 Br] = k K ¢ [Ph CH2CH2 Br][EtO@ ]

Since k and K ¢ are both constants, their product is also a constant (say K).
Therefore, rate = k[Ph CH2CH2Br] [EtO@]
Thus, kinetically the E1cB reaction can be interpreted as a second-order reaction.
E2 mechanism: It is a one-step process and occurs through the participation of both the
substrate, PhCH2CH2Br and base, EtO@.
H
– –
EtO + Ph—CH—CH2 PhCH==CH2 + EtOH + Br
Br
Therefore, E1cB and E2 mechanisms cannot be kinetically distinguished.
Deuterium labelling can be used to distinguish the E1cB from the E2 pathway. The
reaction is carried out in a solvent which could act as a deuterium source (e.g., EtOD)
and the substrate recovered after half completion of the reaction is analysed for its
deuterium content. There is no scope of incorporation of deuterium into the substrate
if the reaction proceeds by the concerted E2 mechanism. Thus, if the substrate is found
to contain no deuterium, the reaction must take place by the E2 mechanism. On the
other hand, deuterium incorporation is expected to occur in an E1cB reaction because
it involves a carbanion intermediate. Therefore, if the recovered substrate is found to
contain deuterium, the reaction must proceed through the E1cB pathway.
E2 reaction:
E1cB reaction:
In actual experiment, it was observed that there is no deuterium incorporation when

PhCH2CH2Br is allowed to react with NaOEt/EtOH. That is, the reaction occurs by the
E2 mechanism.

F NaNH2 –
(a) + NH3 + F
liq.NH3
(b)
Ph H – Ph NO2
MeO
(c) MeO NO2 –
MeOH + MeOH + MeO
O O
|| @ ||
EtO
(d) CH 3 — S — CH 2CH 2OPh ææææ
EtOH
Æ CH 3 — S — CH == CH 2 + EtOH + PhO@
Solution
(a) Since F@ is a poor leaving group, :NH 2@ is a very strong base and ortho-H atom is
considerably acidic (attached to an sp2 hybridized carbon and adjacent to the electro
negative fluorine atom), therefore, fluorobenzene undergoes dehydrofluorination
by the E1cB mechanism.
(b) When a cyanohydrin is treated with a base, it undergoes elimination reaction by

the E1cB mechanism to give the corresponding carbonyl compound and this is
because the hydrogen attached to electronegative oxygen is considerably acidic
and the CN@ ion is not a good leaving group.
(c) Since the hydrogen atom attached to carbon containing the electron-withdrawing
–NO2 group is considerably acidic and MeO@ ( a strong base) is not a good leaving
group, therefore, this compound undergoes elimination of MeOH by the E1cB
mechanism.
(d) Since the hydrogen atom attached to the carbon a to the electron-withdrawing
S ==O group is considerably acidic and PhO@ is not a good leaving group, therefore,
this compound undergoes elimination of PhOH by the E1cB mechanism when
treated with EtO@/EtOH.
2. Reaction of Cl2C == CHCl with NaOD in D2O affords ClC ∫∫ CCl. When the
reaction is stopped short of completion, the recovered olefin contains
deuterium. Suggest a mechanism for the reaction consistent with this
observation.
Solution The observation suggests that the reaction follows the E1cB pathway and this
is because an E1cB reaction involves reversible formation of a carbanion which can take
up D≈ form D2O to form deuterium-labelled alkene.
The substrate is suitable for an E1cB reaction because the b-hydrogen is considerably
acidic (it is attached to an sp2 carbon and halogens enhance its acidity) and Cl@ ion is not
a good leaving group. The reaction occurs as follows:
3. Explain the mechanism of the following base-catalyzed dehydration

reactions:
–
OH
(a)
(—H2O)
HO Ph Ph
:OH @
(b) CH 3CH(OH)CH 2CHO ææææ
(–H O)
Æ CH 3 CH == CHCHO
2
Solution
(a) The b-H of the substrate is considerably acidic because the conjugate base is a very
stable (aromatic) cyclopentadienyl anion system. Also, the leaving group OH@ (a
strong base) is a very poor one. Therefore, this alcohol undergoes base-catalyzed
dehydration by the E1cB mechanism.
(b) Since the b-H of this substrate (a b-hydroxyaldehyde or aldol) is considerably acidic
(because the conjugate base is stabilized by resonance involving the —CHO group)
and OH@ (a strong base) is a very poor leaving group, therefore, it undergoes base-
catalyzed dehydration by the E1cB mechanism.
4. The threo as well as the erythro-isomer of 1,2-diphenylpropyltrimethyl

ammonium salt yields the same alkene, (E)-1,2-diphenylpropene when
treated with t-BuOK / t-BuOH. Explain.
Solution Since the two diasteroisomeric salts undergo E1cB reaction when treated with
t-BuOK / t-BuOH, therefore, they yield the same stable alkene, (E)-1,2-diphenylpropene.
This is a case of stereoconvertent reaction because substrates of different stereochemistry,

i.e., diastereoisomers, yield the same product.
5. E1, E2 and E1cB make a spectrum of elimination reactions. Discuss this
statement.
Solution The loss of HLG in an E2 reaction is a concerted process in which two bonds are
formed and two bonds are cleaved at the same time. However, the C — H bond cleavage may
precede C — LG bond cleavage an 'E1cB-like' E2 process) or the C — LG bond cleavage may
precede C — H bond cleavage (an 'E1-like' E2 process). In fact, the extent of concertedness
varies from one extreme, at which the C — H bond is cleaved but the C — LG bond is intact,
to another, at which C — LG bond is cleaved but the C — H bond is intact. These extremes
are clearly nothing more than a pure E1cB and pure E1 mechanistic course. Hence, all
the three mechanistic possibilities are related, differing mainly in the timing of various
bond-breaking and bond-making processes and, therefore, provide a continuum of reaction

pathway for elimination.
1. When dehydrofluorination of 1,1-dichloro-2,2,2-trifluoroethane is carried out by

treating with NaOMe in MeOD, 1,1-dichloro-1-deuterio-2,2,2-trifluoroethane is
recovered from the reaction mixture. How does this observation fit into the E1cB
mechanism proposed for this elimination reaction?
(a)
(b)
(c)
(d) O O
|| ||
t -BuOK
p-ClC6 H 4CCHClCH ClC6H 4Cl-p ææææÆ
t -BuOH
p-ClC6 H 4 C — C == CH — C6 H 4Cl-p
|
Cl
+ t - BuOH + Cl@
3. Explain why the following b-hydroxyketone does not undergo base-catalyzed
dehydration to give the corresponding a, b-unsaturated ketone.
Me
O
OH
4. Predict the mechanism of the reaction from the following observation regarding
primary kinetic isotope effect:
@
EtO
CH3CH Br CH3 ææææ
EtOH
Æ CH3CH == CH2 kH /kD = 6.7
@
EtO
CD3CH Br CD3 ææææ
EtOH
Æ CD3CH == CD2

H3C OEt OH
–
C
H3C OH
[Hint: The E1cB elimination is the usual mechanism by which a hemiacetal is
converted to a carbonyl compound under basic condition.]
4.4 a- or 1,1-ELIMINATION
Example, kinetics, mechanism and substrate structure.
4.4.1 Example of a- or 1,1-Elimination Reaction

When heated with an aqueous solution of KOH, chloroform undergoes a-elimination
via a two-step 1,1-E1cB pathway to yield dichlorocarbene, a highly reactive neutral
intermediate.
KOH/H2O
CHCl3 ææææÆ
D
:CCl2 + H 2O + KCl
4.4.2 Kinetics of a- or 1,1-Elimination Reaction

The reaction follows second-order kinetics, i.e., rate = k[CHCl3] [OH@]
4.4.3 Mechanism of a- or 1,1-Elimination Reaction

The reaction proceeds through the steps as follows:
Step 1: The base OH@ takes up the acidic proton of chloroform to form a carbanion (the
conjugate base).
Step 2: The carbanion is converted to dichlorocarbene (:CCl2) by the loss of Cl@.
Therefore, this is a case of elimination reaction where both the departing groups are lost
from the same carbon.
E2 and a-elimination reactions are kinetically indistinguishable. These two can be
differentiated by isotope labelling. The reaction shows the incorporation of D in the
unreacted substrate when D2O was used as a solvent.
4.4.4 Structure of the Substrate Involved in a-Elimination Reactions

Substrates in which a-H is strongly acidic and contains no b-H undergo a- or 1,1-elimination
reaction in the presence of strong base.
1. Explain why 2,2-dichlorobutane does not form a Grignard reagent when

treated with Mg in dry ether.
Solution Since the compound is a gem-dihalide, therefore, the initially formed Grignard
reagent readily lose a chloride ion to form a carbene, i.e., an a- or 1,1-elimination reaction
takes place.
2. What happens when sodium trichloroacetate (Cl3CCOONa) is heated?

Solution When sodium trichloroacetate is heated, it undergoes an a-elimination reaction
to form dichlorocarbene (:CCl2), CO2 and NaCl. The reaction, which is initiated due to
@
stability of Cl3 C : ion, occurs as follows:
3. Predict the products when CHClBr2 and CHF2Br are treated separately
with the base t-BuOK.
Solution When CHClBr2 and CHF2Br are treated with the base t-BuOK, bromchlorocar-
bene (:CClBr) and difluorocarbene (:CF2) are obtained respectively. This occurs because
Br@ is a better leaving group than Cl@ or F@.

(a) PhCH 2Br + PhLi ææ
Æ PhCH + PhH + LiBr
(b) Me2C == CH Br + Na NH 2 ææ
Æ Me2C == C: + NH 3 + NaBr
2. Identify the products obtained in each of the following reactions:
D or hn hn
(a) CH 2N2 ææææ Æ A+B (b) CH 2 == C == O æææ Æ C+D
(c) hn (d) D or hn
Ph2CN2 æææ Æ E+F N2CHCO2Et ææææ Æ G+H
CH2
! @
D hn
(e) Me2 S— C H 2 ææ Æ I+J (f) Ph—CH—CH2 K + L
O
(g) hn
Ph CH—CHPh M + N
INDEX
A 1,3-Butadiene 1.104
Acetylacetone 1.224 2-Bromohydrins 3.154
Acid strength and pKa 1.47 (Z)-2-Butene to (E)-2-butene 4.77
Acid-catalyzed dehydration of alcohols 4.95 p Bond as neighbouring group 3.157
Acid-catalyzed tautomerism 1.221 Base 4.35, 4.100, 4.101
Acidic Base-catalyzed racemization 2.190
character of alcohols 1.63 Base-catalyzed tautomerism 1.221
character of aromatic acids 1.126 Basic
character of imides 1.92 character of arylamines 1.93
character of phenols (Ar-OH) 1.86 character of guanidine 1.97
Acidity of carboxylic acids 1.84 strength of amines 1.65
Acids and bases 1.46 Basicity 3.25
Acraldehyde 1.16 Bicyclic systems 3.10
Acrylonitrile 1.16 Boiling point 1.150
AgSCN 3.85 Bond
Allyl bromide 3.11 angle 1.10
Allylic rearrangement 3.18 dissociation enthalpy or bond dissociation
Alternating axis of symmetry 2.37 energy 1.9
Ambident nucleophiles 3.27 energy 1.10
Ammonia 2.39 length 1.9, 1.120
Anchimeric assistance 3.147 polarity 1.22
Angle strain 1.126, 3.94 polarizability 1.36
Anthracene 2.41 Boron tifluoride 1.12
Antiaromatic carbocation 3.93 Branching at the b-carbon 3.8
Antiaromatic Compounds 1.242 Bredt’s rule 1.135
Aromatic carbocation 3.93 Bridgehead carbon 3.93
Aromaticity 1.240 Brönsted-Lowry theory 1.46
Asymmetric and dissymmetric molecules 2.38 Butane-gauche interaction 2.217
Atropisomerim 2.87
Axial approach of a nucleophile 3.72 C
Axial chirality 2.86 C—C bond as a neighbouring group 3.161
Azulene 1.26 Captodative ethylene 1.109
Captodative radical 1.217
B Carbanionic character 4.42
1-Bromotryptecene 3.135 Carbanions 1.178
I.2 Index
Carbenium ion 1.178 2,6-Di-tert-butylpyridine 3.74

Carbocation rearrangements 3.105 p-Diacetylbenzene 1.40
Carbocations 1.172 p-Dinitrobenzene 1.40
Carbonium ion 1.178 D L-system 2.109
Carbontetrachloride 2.30 DBN 4.33
Catalysis 1.283 DBU 4.33
Centre of symmetry 2.36 Debromination 4.49
Charge-transfer complex 1.26 Debye unit 1.23
Chiral and achiral molecules 2.84 Decarboxylative debromination 4.73
Chiral axis 2.86 Dehydration using POCl3 and pyridine 4.99
Chirotopic and achirotopic atom 2.131 Dehydrobromination 4.49
Chloroacetone 3.12 Deuterium oxide 2.29
Chloroform 1.92 Dextrorotatory 2.182
Chlorofumaric acid 4.74 Diastereoisomers 2.60
Chugaev reaction 4.28 Diastereotopic faces 2.138
CIP chirality rules 2.115 Diastereotopic ligands 2.135
Cis-1,2-dibromocyclohexane 4.17 Diaxial conformation 4.13
Cis-1,2-dichlorocyclohexane 1.29 Diazomethane 3.61
Cis-1,2-dideuterioethane 2.30 Dielectric constants 3.21
Cis-trans or geometric isomers 2.2 Difluoromethane 2.30
Competition experiments 3.96 Dihedral angle 2.212
Concerted reaction 1.172 Dimethyl sulphoxide (DMSO) 3.18
Configuration 2.108 Dimethylacetamide (DMA) 3.18
Configurational nomenclature 2.108 Dioxane 3.53
Conformation of Dipole moment 1.23, 1.121
acyclic organic molecules 2.211 of chlorobenzene 1.25
chloroethane 2.218 of disubstitute benzenes 1.32
ethane 2.213 Dipole–Dipole interactions 1.147
propane 2.214 Dodecadeuterocyclohexane 2.41
Constitutional isomers 2.2, 2.58 Double bond character 4.41
Contribution of resonance structures towards Double Bond Equivalent (DBE) 1.44
resonance hybrid 1.78
Coordination with an electrophilic species 3.35 E
Copper acetylide 2.31 a-Effect 3.78
Corey–House method 3.66 a-Elimination 4.127
Criteria for aromaticity 1.240 a-or 1,1-Elimination 4.126
Crown ethers 3.31 1,1- or a-Elimination reactions 4.2
Curtin–Hammett principle 4.48 1,2- or b-Elimination reactions 4.2
Cyclohexyl methyl ketone 4.52 1,2-Ethanediol 1.27
Cyclohexylacetylene 4.52 1,3- or g-Elimination reactions 4.3
Cyclopropenyl cation 3.136 E1 reaction 4.86
E1cB and E2 mechanisms 4.119
D E1CB reaction 4.118
2,2-Dimesityl ethanol 1.226 E2 reaction 4.4, 4.5
1,2-Dichloroethane 1.27, 2.218 Eclipsed conformation 2.10
1,2-Dideuterioethene 2.30 Effect of
1,2-Dimethylcyclobutane 2.40 base 4.33
1,3-Dibromoallene 2.40 hyperconjugation 1.119
1,3-Dibromocyclobutane 2.31 resonance on the properties of molecules 1.81
1,3-Di-sec-butylyclobutane 2.36 solvent 3.18, 4.34
Index I.3
on rate 3.99 Furan 1.26

substrate structure 3.7, 4.33
temperature 4.34 G
the leaving group 3.32, 4.34 Gabriel synthesis 3.36
on rate 3.102 Gas phase nucleophilicity 3.25
the nucleophile 3.22 Gas-phase acidity of alcohols 1.64
on rate 3.101 Gem-dichloride 4.52
Ei eliminations 4.24 Geometric enantiomerism 2.127
Electromeric effect 1.67 Grignard reagent 4.127
Electronegativity 1.22
Electrophiles and nucleophiles 1.193 H
Electrophilic and nucleophilic radicals 1.192 Halocyclohexanes 3.12
Electrophilic catalysis 3.102 Halocyclopropanes 3.12
Element effect 1.49 Hammond postulate 1.284
Enantiomeric excess 2.187 Hard and soft nucleophiles and
Enantiomers 2.60 electrophiles 3.26
Enantiotopic faces 2.137 Hard centres 3.28
Enantiotopic ligands 2.134 Heat of combustion 1.119
Energy diagram 1.278 Heteroatoms as neighbouring groups 3.148
Energy profile diagram 3.4 Heterolytic fission 1.171
Entropy 4.5 Hexadeuteriobenzene 2.32
Erythro and threo nomenclature 2.124 Hexamethylphosphoramide (HMPA) 3.18
Ether cleavage 3.39 Hofmann exhaustive methylation or Hofmann
Ethyl acetoacetate 1.219 degradation 4.43
Evidence in favour of Hofmann rule 4.38
E1 mechanism 4.90 HOMO 3.6
E2 mechanism 4.32 Homoallylic carbocation 1.178
SN1 mechanism 3.90 Homoaromatic compounds 1.251
Example of NGP 3.146 Homolytic fission 1.171
E-Z system of designating alkene Homomers 2.2
diastereoisomers 2.125 Homotopic faces 2.137
Homotopic ligands 2.134
F Hybrid orbital number 1.3
Face strain or F-strain 1.133 Hybridization effect 1.2, 1.51
Factors affecting acidity 1.49 Hydrogen bonding 1.148
Factors influencing Hyperconjugation 1.117
E1 reaction rate 4.90 Hyperconjugative effect 1.118
E2 reaction rate 4.33
SN1 reaction 3.91 I
SN2 reaction rate 3.7 Imidazole 1.104, 3.58
the extent of E1 and E2 reactions 4.100 Index of Hydrogen Deficiency (IHD) 1.44
Field effect 1.67 Induced moment 1.31
Fischer projection 2.4 Inductive and electrometric effects 1.60
Fluoroform 1.92 Inscribed polygon method 1.245
Flying-Wedge projection 2.3 Intermolecular
Formal charge 1.33 dehydration 3.117
Formation of forces 1.147
ethylene 1.6 hydrogen bonding 1.148
methane 1.4 Intramolecular
Fragmentations 4.45 elimination 4.25
I.4 Index
hydrogen bonding 1.149 rotation 2.185

SN2 reaction 3.15 Monochromatic light 2.182
Inversion of configuration 2.140
Invertomerism 2.220 N
Ion pairing in determining nucleophilicity 3.25 N-bromosuccinimide (NBS) 1.217
Isomers 2.2 N,N-dimethylformamide 3.18
Isotope effects and salt effects 3.113 Neighbouring group participation 3.146
Isovalent and heterovalent resonance 1.81 Neomenthyl chloride 4.15
Neopentyl bromide 3.9
K Newman projection 2.11
Ketene 1.16 Nitroform 1.92
Keto-Enol tautomerism 1.219 Nitrotoluenes 1.32
Kinetic control versus thermodynamic control No bond resonance 1.118
1.286 Nonaromatic
Kinetics of carbocation 3.93
a- or 1,1-elimination 4.126 compounds 1.242
E1 reaction 4.87 Nonclassical carbocation 1.177
E1cB reaction 4.118 Nonpolar covalent bonds 1.22
E2 reaction 4.4 Nucleophilic substitution 3.2, 4.3
SN1 reaction 3.86 Nucleophilicity 3.25
SN2 reaction 3.3 in protic solvents 3.57
KSCN 3.85
O
L Optical
Leaving group 4.37, 4.100, 4.101 activity 2.182
Levorotatory 2.182 purity 2.187
Lewis Acid-Base theory 1.52 Order of basicity of halides 3.25
Lucas test 3.130
LUMO 3.6 P
Percentage of ionic character 1.36
M Phase-transfer catalysts 3.30
Mechanism of Phloroglucinol 3.79
a- or 1,1-elimination 4.126 Plane of polarization 2.182
dehydration 4.96 Plane of symmetry 2.33
E1 reaction 4.87 Plane-polarized light 2.182
E1cB reaction 4.118 Polar and apolar solvents 3.20
E2 reaction 4.4 Polar covalent bond 1.22
Keto-enol tautomerism 1.221 Polarimeter 2.183
NGP 3.146 Principal axis 2.32
SN1 reaction 3.86 Prochiral and prostereogenic centres 2.135
SN2 reaction 3.3 Pro-R and pro-S designation 2.137
Melting point 1.150 Prostereoisomerism and topicity 2.133
Menthyl chloride 4.15 Protic and aprotic solvents 3.20
Meso Compounds 2.89 Proton sponges 1.132
Meso-Tartaric acid 2.36 Pyramidal inversion (umbrella effect) 2.186
Methods of determining mechanisms of reactions Pyrolysis of
1.298 acetates 4.24
Modern definition of aromaticity 1.243 tertiary amine oxides 4.29
Molecular xanthates 4.28
chirality 2.84 Pyrrole 1.26
orbital energy diagram 1.244
Index I.5
Q Stabilities of carbanions 1.66

Quinclidinone 1.146 Stabilities of carbocations 1.66
Staggered conformation 2.10
R Stepwise reaction 1.172
R,S-system 2.111 Stereoaxis 2.86
Ra,Sa-nomenclature of allenes 2.121 Stereocentre or stereogenic centre 2.88
Ra,Sa-nomenclature of biphenyls 2.122 Stereochemistry of
Racemic modification 2.187 E1 reaction 4.88
Racemization 2.188 E2 reaction 4.5
Radical inhibitors or ‘scavengers’ 1.192 SN1 reaction 3.87
Re and Si designation 2.138 SN2 Reaction 3.5
Reactive intermediates 1.170, 1.172 Stereoelectronic requirement 4.6
Rearrangement during dehydration 4.98 Stereoisomers 2.59
Rearrangement of the carbocation intermediates Stereospecific reaction 4.7
4.93 Steric
Regiochemistry of epoxide ring opening 3.118 acceleration 1.134
Regioselectivity in b-elimination reactions 4.37 effect 1.125
Regioselectivity of E1 reactions 4.90 retardation 1.134
Resolution of racemic modification 2.192 strain 1.125
Resonance Sulphonic acids 1.104
and resonance effect or mesomeric effect 1.75 Super acid 1.194
energy 1.77 Super leaving groups 3.33
structures 1.76 Symmetric, asymmetric and dissymmetric 2.38
theory 1.76 Symmetry elements 2.29
Retention of configuration 2.139 Syn-anti momenclature for aldols 2.128
Ring-chain tautomerism 1.228 Syn-axial interactions 4.55
Rotational axis of symmetry 2.29 Syn-elimination 4.20
Rules for writing meaningful resonance
structures 1.78 T
a-Truxillic acid 2.44
S 1,2,3-Triphenylcyclopropane 1.106
Sacrificial and isovalent hyperconjugation 1.119 1,3,5-Trihydroxybenzene 1.227
Sawhorse projection 2.10 Tautomerism 1.218
Saytzeff rule 4.37 Tetramethylallene 2.43
Scale of acidities and basicities 1.48 Thermal racemization 2.189
Selenoxides 4.71 Thermodynamically stable conformation 4.16
Sequence rules 2.115 Thiosulfate 3.79
Simple axis of symmetry 2.29 Torsional strain 1.126
SN1 reaction 3.86 Trans-1,2-dibromocyclohexane 4.17
SN1¢ mechanism 3.112 Trans-2-bromocyclopentanol 3.16
SN2 reaction 3.3 Trans-4-tert-butylcyclohexyl tosylate 4.14
SN2¢ mechanism 3.18 Trichloroacetic acid 1.63
SNi mechanism 3.109 Triphenyl phosphine 4.77
SNi¢ mechanism 3.112 Triphenylmethyl bromide 3.135
SODAR 1.44 Triphenylmethyl chloride 1.205
Sodium D line 2.185 Triphenylpyrylium salt 3.70
Sodium trichloroacetate 4.127 Tri-p-nitrophenylamine 1.104, 1.107
Soft centres 3.28 Triptycenyl cation 1.205
Source of chirality 2.85 Trityl cation 1.205
Specific rotation 2.184 Trivial axis 2.33
I.6 Index
U W
Unreactive nature of phenyl halides 3.15 Weakly polarizable base 4.36
Williamson synthesis of ethers 3.17, 3.43
V
Valence tautomerism 1.228 X
van der Waals Forces 1.149 Xenon tetrafluoride 2.35
VESPR theory 1.11
Vinylic and aryl halides 3.14, 3.95
Violation of Saytzeff rule 4.42

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ORGANIC CHEMISTRY

Nimai Tewari is a retired associate Professor, Department of Chemistry,

McGraw Hill Education (India) Private Limited

Organic Chemistry—A Modern Approach (Volume-I)

Copyright © 2017, by McGraw Hill Education (India) Private Limited.

This edition can be exported from India only by the publishers,

Visit us at: www.mheducation.co.in

1. Structure, Bonding and Properties of Organic Molecules 1.1–1.313

1.6 Resonance and Resonance Effect or Mesomeric Effect 1.75

1.12 Aromaticity 1.240

2. Principles of Stereochemistry 2.1–2.230

2.2.3 Centre of Symmetry (i) 2.36

2.7 Conformation of Acyclic Organic Molecules 2.211

3. Nucleophilic Substitution Reactions at Saturated Carbon Atom 3.1–3.180

3.3.5 Evidence for Participation by a Neighbouring Group 3.148

4. Elimination Reactions 4.1–4.128

4.3.5 To Distinguish between E1cB and E2 Mechanisms 4.119

In the course of teaching Organic Chemistry to undergraduate students, I have been

1.11.3 Position of the Tautomeric Compounds and to Classify Them

1.1 HYBRIDIZATION, BOND LENGTHS, BOND STRENGTHS OR BOND

Hybrid orbitals Hybridization Geometry Bond angle

Formation of methane (CH4) molecule: During the formation of methane molecule,

Formation of ethylene (C2H4) molecule: During the formation of ethylene molecule,

1.1.2 Bond Length

1.1.3 Bond Dissociation Enthalpy or Bond Dissociation Energy

Bond Dissociation Energies (kcal mol–1) of Some Chemical Bonds

CH2 Æ .CH + H. DH ∞ = 106 Kcal mol -1

CH4 Æ .C. + 4H. Total: 397 Kcal mol -1

Therefore, EC—H = 397/4 = 99.25 kcal mol–1.

1.1.4 Bond Angle

1.1.5 VESPR Theory and Molecular Geometry

(a) H2O (b) BF3 (c) C H 3 (d) C H 3 (e) :N H 2

(d) (e) (f)

(g) (h) (i)

(j) (k) (l)

2. Draw the structure of a hydrocarbon which contains:

(a) (b) (c)

3. Which orbitals are used to form each bond in methylamine, CH 3 NH 2?

(f) But-1, 2, 3-triene:

9. Which atoms in each of the following molecules always remain in the

(c) CH3—C—H and H—CH2OH

15. “Bromination of methane is less exothermic than that of chlorination” —

DH° (chlorination) = [(–83.5) + (103)] – [(–104) + (–58)]

4. Predict the geometry around each of the indicated atoms:

(g) H C N (h) (CH 3 )2 N H (i) Et 2 N

5. Predict the mentioned bond angles:

6. Which of the indicated bonds is shorter and why?

7. Give the state of hybridization of the indicated atoms:

(g) CH3CH == CH — NH2 (h) CH2 == C == CH2

1.2 ELECTRONEGATIVITY AND BOND POLARITY

The extent of polarity in a covalent bond is expressed quantitatively by the physical

But, it possesses dipole moment (1.85 D) because of its angular structure .

Resonance often plays an important role in determining the polarities of

(ii) The dipole moment of chlorobenzene is larger than that of ﬂuorobenzene. In

Aromaticity often plays a role in determining the polarity of a molecule. For

Formation of charge-transfer complex is often responsible for exhibiting dipole

Conformation plays an important role in determining the polarities of some

(ii) The optically inactive (meso) form of 1,2-dichloro-1,2-di-p-tolylethane has a dipole