LECTURE NOTES

Session - 2009-10

ORGANIC CHEMISTRY
TOPIC : GOC - I
CONTENTS :
1 IUPAC Nomenclature

Refer sheet GOC- I

JEE Syllabus [2009]
Concepts: Hybridisation of carbon; Sigma and pi-bonds; Shapes of molecules; Structural and
geometrical isomerism; Optical isomerism of compounds containing up to two asymmetric cen-
ters, (R,S and E,Z nomenclature excluded); IUPAC nomenclature of simple organic compounds
(only hydrocarbons, mono-functional and bi-functional compounds); Conformations of ethane
and butane (Newman projections);

Page No.1
 IUPAC NOMENCLATURE
1.1 INTRODUCTION TO ORGANIC COMPOUNDS:-
Organic compounds are compounds of carbon and hydrogen and the following elements may also be
present: (Halogens, N, S, P, O). There are large no. of organic compounds available and large no of organic
compounds are synthesized every year.The most important reason for large no of organic compounds is the
property of catenation (self–linkage) in carbon.

Element Bond Energy

C C–C (strongest bond)
Si Si–Si

Ge Ge–Ge  Decreasing order

Sn Sn–Sn
Pb Pb–Pb
Bond energy depends on
(i) Size of atom (Inversely proportional) Size : C–C>Si–Si
(ii) E.N difference along period (Directly) E.N Diff : C – H < N – H < O – H < H – F
(iii) Bond order (no. of covalet bonds b/w two atoms) (Directly) Bond order : C – C < C = C < C  C
Catenation in carbon:-
 The element carbon has strongest tendency to show catenation or self–linkage due to –
(a) its tetravalency so that it can form  bonds with many elements as well as carbon itself.
(b) It can form multiple bonds (C = C, C  C). Due to its small size, there is efficient colateral overlapping
b/w two P–orbitals.

Does not exist

(c) High bond energy of C – C bond so that it can form strong bonds in long chains and in cylic compounds.

1.2 CLASSIFICATION OF ORGANIC COMPOUNDS :

Page No.2
 Ex. Compounds Classification

Unsaturated

Heterocyclic , saturated

Unsaturated

Saturated

Saturated, Alicylic

1.3 IMPORTANT TERMS:

Saturated compounds:
When all the valencies of an element are satisfied by  covalent bonds.
Unsaturated compound:
When a compound contains one or more  bonds (C = C, C = N, N = N, C = O or C  C, C  N, N  N)
Molecular Formula (M.F.) :
The molecular formula of a compound indicates the actual number of atoms of each element present in one
molecule.
Structural Formula (S.F.):
It indicates the linkage due to covalent bond between different atoms in a molecule.
(i) Expanded Structural Formula (E.S.F.)
(ii) Condensed Structural Formula (C.S.F.)
(iii) Bondline Structural Formula (B.S.F.)
Ex. M.F. C3H8 Ex. M.F. C4H10
H H H H H H H
E.S.F. H – C – C – C – H E.S.F. H – C – C – C – C – H
H H H H H H H
C.S.F. CH3 – CH2 – CH3 C.S.F. CH3 – CH2 – CH2 – CH3
B.S.F. or B.S.F.

Ex. B.S.F. Ex. B.S.F.

CH 3 – CH – CH 2 – CH 3
C.S.F. M.F. C6H12
CH3
M.F. C5H12

Homologous series :

Homologous series may be defined as a series of similarly constituted compounds in which the members
possess the same functional group, have similar chemical characteristics and have a regular gradation in
their physical properties. The two consecutive members differ in their molecular formula by CH2.

Calculation of Degree of Unsaturation (DU):-

(a) It is the hydrogen deficiency index (HDI) or Double Bond Equivalence (DBE)

Page No.3
 –2H –2H
(b) H3 C – H2C – CH3     CH3 – C  CH or CH2 = C = CH2 or
(DU  O)

That means Deficieny of 2H is equivalent to 1 DU

(c) (i) 1DU = Presence of 1 Double Bond or Presence of 1 Ring closure
(ii) 2DU = Presence of 2 Double bond or 1 Triple bond or two ring closure or 1 double bond + 1 ring.
(d) G.F. D.U.
y
(i) CxHy (x + 1) –  
2

yo
(ii) CxHyOz (x + 1) –  
 2 

ys
(iii) CxHyXs (x + 1) –  
 2 

y–w
(iv) CxHyNw (x + 1) –  
 2 

ys– w
(v) CxHyOzXsNw (x + 1) –  
 2 
Ex Calculate DU of following compounds

(a) C6H6O DU = 4

(b) C6H5Cl DU = 4

(c) C6Br6 DU = 4

(d) C5H11OCl DU = 0

(e) C9H12N2 DU = 5

(f) C6N6 DU = 10

(g) C10H8SO5 N4Cl2 DU = 8

e.g. M.F. DU S.F.
H H
C–C=C=C
1. C4H6 2 H H
H H
C–C–CC

C–CC–C
C=C–C=C
C

CH2

Page No.4
 2. C2H2Cl2 =1

Total isomers = 3

3. C7H6O (Aromatic) =5

4. C7H8 (Aromatic) =4

Note : In case of the aromatic Compounds minimum DU = ‘4’. That means at least 1 Benzene ring is
present.

Degree of carbon :
It is defined as the number of carbon atoms attached to a carbon atom.

H
|
Primary (1°) carbon :
| 1°
H

Secondary (2°) carbon :

C
|
Tertiary (3°) carbon :
|
3°
H

C
|
Quaternary (4°) carbon :
| 4°
C

Superprimary : (1°)

Ex.

10 – 5 carbon
CH3 CH3 0
2 – 1 carbon
1 2 | 3| 4 5
CH3 – C – CH – CH2 – CH3
| 30 – 1 carbon
CH3
40 – 1 carbon

Page No.5
 Ex.

*Degree of hydrogen is same as the degree of carbon to which it is attached.

Type of C – C bonds & type of replaceable H-atoms in saturated hydrocarbon.
Note : Total no. of monosubstituted product does not depend upon type of C-atoms but depends upon type
of replaceable H-atoms.

1.4 IUPAC NOMENCLATURE OF ORGANIC COMPOUNDS:

General Scheme of Naming:-
Secondary Prefix + Primary Prefix _ _Word Root_ _Primary Suffix_ _Secondary suffix

The organic compound is always named according to the general scheme as given by IUPAC. In every
compound, two parts, viz, word root and primary suffix, always exist.

Word Root: It indicates the no. of carbon atoms present in the main chain. It is represented as Alk.

Prefix : It is the first part of the name.

(i) Primary Prefix : ‘Cyclo’
(ii) Secondary Prefix : Normal substitutents and junior functional groups are treated as a substituent than
their name is treated as secondary prefix.

The following substituent groups are always cited in the prefix.

(i) R – Alkyl
(ii) R – O – Alkoxy
(iii) X halo (fluoro, chloro, bromo, iodo)
(iv) –NO2 Nitro
(v) –N = O Nitroso
(vi) Junior functional group
 The prefixes are always written in alphabetical order
 The position of substituent group in the main carbon chain is mentioned by writing the number just before
the name of substituent by writing a small dash (–).

Suffix :
(i) Primary Suffix:- It indicates saturation or unsaturation existing in the main chain.
ane  single bond (saturated)
ene  >

diene  If two double bonds

Polyene  if plenty of double bond are present.

yne  one  bond

diyne  two  bond

(ii) Secondary Suffix:- It is used for the principal functional group.

1.5 NAMING OF SATURATED HYDROCARBONS
Rules :(Branched and substituted Alkanes)
(1) Selection of parent chain 
(a) Chain with maximum number of ‘C’ atoms (longest chain).

Page No.6
(b) If number of carbon atoms are same in more than one longest chain then more substituted longest chain
will be the parent chain.

(c) If number of side chain are also same then that will be parent chain having its substituent at lower number.

C C
C–C–C–C–C–C–C–C
C–C–C
C
(2) Numbering
(a) Numbering is done from that side of the parent chain having it substituent at lower number
(lowest set of locant)

(b) If position of substituent are same from both the end of the parent chain, then numbering is done
from alphbetical order.

* If alphabets are also same then numbering is done from that side of the parent chain having its
substituent of substituent at lower number.

CH3
–

Ex. CH3 – CH – CH3 Methylpropane

CH3 CH3
–

CH3 – CH – CH – CH3 2, 3-Dimethyl butane

CH3
– –

CH3 – C – CH3 Dimethyl propane

CH3

2-Methylbutane

Page No.7
 C

–
C–C–C–C–C 3-Ethyl-2-methyl hexane

–
C–C–C

C–C–C

– –
C–C–C–C–C 4, 4-Diethyl heptane
C–C–C

C – – –
C–C–C
C–C–C–C–C 4, 4-Diethyl-2, 5-Dimethyl heptane
C–C–C
–

C–C–C–C–C 3-Ethyl hexane

–

C–C–C
CH3 – CH2 – CH2 – CH2Cl 1-Chlorobutane

C–C–C–C–C 3-Bromo-2-chloropentane
–
–

Cl Br

C–C–C–C–C 2-Bromo-4-chloro pentane

–

Cl Br

Cl
– –

C–C–C–C–C 4-Bromo-2, 2-dichloro pentane

–

Cl Br

C–C–C–C–C–C 5-Bromo-2, 3-dichloro hexane

–
–

Cl Cl Br
(3) Rules for writing Alkyl Radicals:-
–H
Radicals:- (a) CnH2n + 2  CnH2n + 1 Alkyl Radical

–H
CnH2n  CnH2n – 1 Alkenyl Radical

–H
CnH2n – 2  CnH2n – 3 Alkynyl Radical

R (structural formula) Name

CH3 – Methyl
CH3 – CH2 – Ethyl

Page No.8
 CH3
–H 3 2| 1
CH3 – C – CH2 – Isobutyl
CH3 |
| H
(b) CH3 – C – CH3
| CH3
H |
–H CH3 – C – CH3 Tert butyl
| (Tertiary butyl)

(4) Systematic Names of Radicals :

Complex Radical :

Ex 5 – ( 1,2 – Dimethylpropyl) nonane

5 – (1,1 – Dimethylpropyl) nonane

5 – (1 – Ethylpropyl) nonane

5 – (Dimethylethyl) nonane

Alkene / Alkyne radicals

Alkene Alkenyl

Alkyne Alkynyl

CH2 = CH2 CH2 = CH –

ethenyl

CH3 – CH = CH2 Prop – 1 – enyl

Methylethenyl

Prop – 2 – enyl

Page No.9
CH3 – C  CH CH3 – C  C –
Prop-1-ynyl

CH3 – C  CH – CH2 – C  C – H
Prop-2-ynyl
(5) Benzene Radical

1. Phenyl

2. Benzyl

3. Benzal

4. Benzo

5. Tolyl

CH3
(o, m, p)

 Methylene

7. Alkylidene

8. - naphthyl

9. -naphthyl

(6) Numeral Prefixes

(1) Following prefixes are considered for alphabetization :

(a) Iso

(b) neo

(c) Di, Tri, Tetra of complex radical are considered for alphabetization

(2) Following are not considered for alphabetization

(a) Di, Tri, Tetra for simple radicals.
(b) Sec, Tert are not considered for alphabetization
(c) Bis , Tris

Page No.10
 (7) Retained Names of Alkanes:

(1) Normal (n) : Radical or hydrocarbon which has straight chain and if it has free valency it must be present
at either of ends.
C – C – C – C (n butane)
(2) Iso : Two methyl group at the end of linear chain (unbranched chain)
C
–

(i) C– C –C Isobutane or methyl propane

C
–

(ii) C– C –C–C Isopentane or methylbutane

C
–

(iii) C– C –C–C–C Isohexane or 2-methylpentane

Conditions :-
 There should be 4 to 6 carbon atoms only
 There should not be any other alkyl group present in the chain.
Q. Iso-octane (commercial name) in petroleum industry
C
– –

Ans. C– C–C– C–C IUPAC name is 2, 2, 4-trimethylpentane

–

C C

(3) Neo :  There should be 5 to 6 carbon atoms

 2, 2-dimethyl
 No other substituent.
C
– –

Ex. C– C–C Neopentane

C
C
– –

C – C – C – C Neohexane
C

Page No.11
 (4) Secondary : It is applicable only for radical.

Ex.1

(5) Tertiary :

First member

1.6 NAMING OF UNSATURATED COMPOUND (ALKENES AND ALKYNES) :

(A) General formula:- CnH2n and CnH2n – 2 respectively

(B) Rules for selection of main chain:-

 Longest carbon chain with a multiple bond.
 Longest carbon chain with maximum number of multiple bonds.
 If first and second factor are common then chain with lowest locant (multiple bond) is selected as main
chain.
 Lowest locant Rule is followed till first point of difference. (Multiple bond prior to substituent).
 Alphabetization.
(C) Rules for Numbering:-
 Lowest locant Rule till first point of difference (irrespective of double bond or triple bond).
 Then double bond is prefer over triple bond in numbering, naming and longest chain selection, If all the
other factors are common

Ex. CH3 – CH2 – CH = CH2 But-1-ene

CH3 – CH = CH – CH3 But-2-ene

CH3
–

4-Methylpent-1-ene
CH3 – CH – CH2 – CH = CH2

CH3
|
CH3 – CH – CH = CH – CH3 4-Methylpent-2-ene

Cl
– –

6 5 4 3 2 1
CH3 – C – CH2 – CH2 – CH = CH2 5, 5-Dichlorohex-1-ene
Cl

(a) C = C – C = C Buta-1,3-diene

(b) C  C – C  C Butadiyne

(c) C = C – C  C Butenyne

CC–C–C=C–C Hex-4-en-1-yne
1 2 3 4 5 6

2 4 6
1
3 Hexa-1, 3, 5-triene
5

Page No.12
Ex.1 3-(2-Methylpropyl) hept-1-ene

Ex.2 4 -(1,1-Dimethylpropyl)-4-ethenylhepta-1,5-diene

Ex.3 3-Ethynylhexa-1, 5-diene

Ex.4 4-Ethenylhept-2-en-5-yne

Ex.5 4-(1, 2-Dimethylbutyl) hept-2-en-5-yne

Ex.6 2, 4-Dimethylpenta-1, 3-diene

Ex.6 2,3-Dimethylhex-1-en-4-yne

1.7 NAMING OF CYCLIC HYDROCARBON (ALICYCLIC COMPOUNDS) :

(A) Main chain selection:
(a) Multiple Bond > No. of carbon atoms > Maximum no. of substituents > Nearest locant > Alphabetization.
(b) If all factors are similar in cyclic and acyclic part, then Cyclic > Acyclic
(B) Numbering:
(a) Lowest Locant (b) Alphabetization
(C) Naming:
 Prefix ‘cyclo’ is used just before the word root if it constitutes the main chain.
 If cyclic part is the main chain then the prefix ‘cyclo’ is not considered for alphabetical order..
If cyclic part constitutes the side-chain (substituent) then prefix cyclo is considered for alphabetization:-

Ex:- Cyclopropane Cyclobutane

Page No.13
 Cyclopentane Cyclohexane

Cycloheptane Cyclooctane
–

Methylcyclopropane Ethylcyclopropane

Propylcyclopropane 1-Cyclopropylbutane

C – C – Cl
–

2-Chloroethylcyclopropane

Cl – C – C
–

Chloroethylcyclopropane

Cl
–

C–C–C
1-Chloro-3-cyclopropylpropane
–

3 2 1

Cl
–

1 3 2
C–C–C
2-Chloro-1-cyclopropylpropane
–

Cl
– –

C–C–C
1 2 3 1-Chloro-1-cyclopropylpropane

C–C–C
–

1-Chloro-2-propylcyclopropane
–

Cl

Cl
–

1
Br
2 –
3 – 2-Bromo-1-chloro-3-fluoro-4-iodocyclohexane
4 F
–

Methylethylcyclopropane or isopropylcyclopropane

Cl
–

2
Br
–

3 1
4 6
1-Bromo-2-chloro-4-iodocyclohexane
5
–

I
6
1 5

2 4
3 3-Chlorocyclohex-1-ene
Cl
6 Br
1 5
2
3
4 5-Bromo-3-chlorocyclohex-1-ene
Cl

Page No.14
1.8 FUNCTIONAL GROUP TABLE (Seniority order):

Class Name Suffix Prefix

1. R – COOH Alkanoic acid – oic acid (carboxylic acid) Carboxy

2. R – SO3H Alkane sulphonic acid – sulphonic acid sulpho

3. R–C–O–C–R Alkanoic anhydride – anhydride ------------

|| ||
O O

4. R – COOR Alkyl alkanoate – alkanoate (carboxylate) Alkoxy carbonyl

5. R–C–X
Alkanoyl halide – oyl halide (carbonyl halide) halo carbonyl
||
O

6. R – C – NH2
|| Alkanamide – amide (carboxamide) Carbamoyl
O

7. R–CN Alkanenitrile – nitrile (carbonitrile) cyano

8. R–C–H
Alkanal – al (carbaldehyde) formyl / Oxo
||
O

9. R–C–R Alkanone – one Oxo / Keto

||
O

10. R – OH Alkanol hydroxy

– ol

11. R – SH Alkanethiol – thiol mercapto

12. R – NH2 Alkanamine – amine amino

1.9 NAMING OF FUNCTIONAL GROUP CONTAINING COMPOUNDS :

(A) Selection of Main Chain:
Senior F.G. > Max. no. of F.G. (Similar group) > Multiple bond > Max. no. of ‘C’ atoms > Max. no. of locants
> lowest locant > alphabetization

(B) Numbering (See F.G. + Sub.):

The carbon atom bearing functional group (or C – atom of terminal functional group) is given lowest possible
number.
(i) Lowest locant (F.G. > M.B. > substituent > Alphab)

(C) Naming:- General scheme :

The senior most functional group constitutes secondary suffix. Other junior F.G.’s are written as prefix in
alphabetical order.

Page No.15
1.9.1 Carboxylic acid

F.G. Prefix Suffix IUPAC name

Oic acid - 'C' of COOH considered
Alkanoic acid
in the parent chain
– COOH Carboxy
Carboxylic acid - 'C' of COOH is not
Alkane carboxylic acid
considerd in parent chain

Rule : If first alphabet of sec. suffix is begin from a, i, o, u, y then ‘e’ of primary suffix will be dropped.

Ex. (1) HCOOH Methanoic Acid

(2) CH3 – COOH Ethanoic Acid

(3) C – C – COOH Propanoic Acid

(4) C – C – C – COOH Butanoic Acid

2 3 4 5 6
(5) C – C – C – C – C – C – C – C 2-propylhexanoic acid
1
COOH

COOH
–

(6) Cyclopropanecarboxylic acid

2 1
CH2 – COOH
–

(7) Cyclopropylethanoic acid

COOH
–

1' 2'
3'
C–C–C

(8) 4-(2-carboxypropyl) cyclohexane-1-carboxylic acid

–

COOH

(9) HOOC – COOH Ethanedioic acid

(10) HOOC – CH2 – COOH Propanedioic acid

(11) HOOC – CH2 – CH2 – CH2 – CH2 – COOH Hexanedioic acid

(12) Cycloprop-1-ene-1, 2, 3-tricarboxylic acid

1.9.2 Sulphonic acid

F.G. Prefix Suffix IUPAC name

– SO3H Sulpho Sulphonic acid Alkanes sulphonic acid

Ex. (1) CH3 – SO3H Methanesulphonic acid

(2) C – C – C – C – C Pentane-2-sulphonic acid

–

SO3H

Page No.16
 SO H
– 3
(3) Cyclobutanesulphonic acid

1 2 3 4 5
(4) HOOC – C – C – C – C – SO 3H 5-Sulphopentanoic acid

1.9.3 Anhydride

Heat
(a) R – C – O – H + H – O – C – R  
 R– C –O– C –R
–H2O
O O O O
Alkanoic anhydride
(b) Mixed Anhydride (according to alphabet)
CH3 – COOH + CH3 – CH2 – COOH  CH3 – C – O – C – C2H5
Ethanoic Pr opanoic
Acid O O
acid
Ethanoic propanoic anhydride

Ex. (1) H– C –O– C –H Methanoic Anhydride

(unstable )
O O
C C
C–C–C–O–C–C–C
(2) 2-Methylpropanoic anhydride
O O

(3) Butanedioic anhydride

(4) C – C – C – C – O – C – C – C – C Butanoic anhydride

|| ||
O O

O
||
C
(5) O Cyclohexane-1, 2-dicarboxylic anhydride
C
||
O

(6) –C–O–C– Cyclopropane carboxylic anhydride

O O

(7) Benzene dicarboxylic anhydride (Phthalic anhydride)

Page No.17
1.9.4 Ester :
F.group Prefix Suffix Name

– C – OR Alkoxy Carbonyl Oate Alkyl....alkanoate

O
or or or
(Ester) Alkanoyloxy carboxylate Alkyl.....Alkanecarboxylate

–H2 O R – C – O – R'
R – C – O – H + H – O – R'   
O
O
Example:- (1) H – C – O – CH3 Methyl methanoate
O
(2) C – C – C – O – C – C – C Propyl propanoate
O
(3) C – C – O – C – C – C – C Ethyl butanoate
O
C
–

(4) C – C – C – O – C – H Isobutyl methanoate

O
3 2 1
(5) C – O – C – C – C – COOH 3-(Methoxycarbonyl)propanoic acid
O
(6) C – C – O – C – C – COOH 3-(Ethanoyloxy)propanoic acid
O

(7)

2-(Methoxycarbonyl)-4-(Propanoyloxy) cyclohexan-1-carboxylic acid

O
C
–

–
O–
(8) Cyclopropyl benzene carboxylate

O
(9) – O – C – CH2 – CH3 Methyl-3-propanoyloxy cyclohexane carboxylate
–
O=C
–

OCH3

1.9.5 Acid halide:-

F.Group Prefix Suffix Name

–C–X Halo Carbonyl Oyl-halide Alkanoyl halide
O or or
Carbonylhalide Alkane carbonyl halide

O = C – Cl
–

Example:- (1) Benzene carbonyl chloride

Page No.18
 (2) CH3 – C – Cl Ethanoyl chloride
O
(3) Prop-2-enoyl bromide

(4) 6-Methyl hept-4-en-2-ynoyl chloride

Cl O

–
C

–
(5) CH2 – CH2 – CH2 – CH – C – Cl 5, 5-Dichlorocarbonylpentanoic acid
–

COOH O

COOH
–

(6) 3-Chlorocarbonyl cyclopropane-1, 2-dicarboxlic acid

–

COOH C – Cl
O

O
(7) C – CH2 – CH – CH = CH – CH 2 – COCl 5-Methylhept-3-ene-1, 7-dioyl Chloride
–

Cl CH 3

(8)

4-Bromocarbonyl-2-chlorocarbonyl-5-propanoyloxycycloexane-1-carboxylic acid
1.9.6 Amide:-

F.Group Prefix Suffix Name

– C – NH2 Carbamoyl amide Alkanamide

O or or
Carboxamide Alkane Carboxamide

Example:- (1) H – C – NH2 Methanamide (Formamide)

O
(2) CH3 – C – NH2 Ethanamide (Acetamide)
O
(3) CH3 – C – NH – CH3 N-methylethanamide
O

(4) CH3 – C – N – C – C – C N-butyl-N-propylethanamide

–

O C
C
C
C

Page No.19
 O = C – NH2

–
(5) Benzene carboxamide

O = C – NH2

(6) – Benzene-1, 3-dicarboxamide

–
C – NH2
O

COOH
–

(7) – 3-Carbamoyl cyclohexane-1- carboxylic acid

C – NH2
O

O
C – NH – CH3
(8) CH2 N, N’-Dimethyl propane-1, 3-diamide
C – NH – CH3

O
1.9.7 Nitrile:-

Example:- (1) H – CN Methanenitrile

(2) CH3 – CN Ethanenitrile

(3) CH3 – CH – CH2 – CN 3-Methyl butanenitrile

–

CH3

CN
–

(4) Cyclopropane-1, 2, 3-tricarbonitrile

–
–

CN CN

CN
–

(5) Benzene carbonitrile

CH2 – CH2 – COOH

(6) 3-Cyano propanoic acid
–

CN

CH2 – CH – CH2
–

(7) Propane-1,2,3-tricarbonitrile
–

CN CN CN

Page No.20
1.9.8 Isocyanide :-

F.Group Prefix Suffix Name

–NC Isocyano Isocyanide Alkyl Isocyanide

Example:- (1) CH3 – N  C Methyl isocyanide

(2) CH3 – CH2 – N  C Ethyl isocyanide

NC
–

(3) Phenyl isocyanide

1.9.9 Aldehyde:-

F.Group Prefix Suffix Name

– CH = O Formyl or oxo al or carbaldehyde Alkanal or Alkane

Carbaldehyde

Example:- (1) HCHO Methanal

(2) CH3 – CH2 – CH = O Propanal

(3) CCl3 – CH2 – CH2 – CH2 – CH = O 5, 5, 5-Trichloro pentanal

CH = O
–

(4) Cyclohexane carbaldehyde

CH = O
–

(5) Benzaldehyde/Benzene carbaldehyde

Ph – –CH = O
(6) CH – CH – 2-Chloromethyl-3, 3-diphenyl propanal
Ph – CH2Cl

CH = O
–

(7) Benzene-1, 3-dicarbaldehyde

–
CH = O
(8) HOOC – CH2 – CH2 – CH2 – CH = O 5-Oxopentanoic acid

(9) HOOC – CH2 – CH – CH2 – COOH 3-(2-Oxoethyl) pentane-1, 5-dioic acid

–

CH2 – CH = O

COOC2H5
–

(10) – ` Ethyl 3-(3-oxopropyl) cyclohexane-1- carboxylate

CH2– CH2 – CH = O

Page No.21
1.9.10 Ketone:-

F.Group Prefix Suffix Name

–
C=O Keto or oxo one Alkanone
–

Example:- (1) CH3 – C – CH3 Propanone

O
O

(2) Cyclohexanone

(3) Cyclohexane-1, 3, 5-trione

O O
O
CH2 – C – CH3
–

(4) 2-(2-Oxopropyl) cyclohexanone

O

COOH
–

(5) 3-(3-Oxobutyl) cyclohexane -1-carboxylic acid

–

CH2 – CH2 – C – CH3

O

(6) OHC – CH2 – C – CH3 3-Oxobutanal

O
1.9.11 Alcohol:-

F.Group Prefix Suffix Name

–OH Hydroxy ol Alkanol

Example:- (1) CH3 – CH – CH2 – OH 2-Methyl Propanol

–

CH3

(2) CH 3 – CH – CH 2 – CH 2 – CH – CH 3 2, 5-Dimethylhexane-1, 6-diol

–

CH 2 – OH CH 2 – OH

OH
–

HO OH
–

–
(3) Cyclohexane-1, 2, 3, 4, 5, 6-hexaol
–

–
HO OH
–

OH
(4) CHO (Glyceraldehyde) 2, 3-Dihydroxy propanal
– –

CH – OH
CH2 – OH

OH
–

Cl – CH2 – CH – CH2 –
(5) CH – CH2 – CH2 – CH2 – CH2 – OH
–

CH3 – CH2 – CH2

–
–

Cl OH
7-Chloro-5-(3-Chloro-2-hydroxy propyl) octane-1, 6-diol

Page No.22
 O
HO OH

–
–
(6) 2, 4, 6-trihydroxy cyclohexane-1, 3, 5-trione
O O

–
OH

(7) CH2 – CH – CH2 (Glycerol) Propane-1, 2, 3-triol

–

–
–
OH OH OH
1.9.12 Amines:-

F.Group Prefix Suffix Name

–NH2 amino amine Alkan amine

(a) R – NH2 1º = amine Alkanamine

(b) R – N – R' 2º = amine N  Alkylalkan a min e

H (R' ) (R )

(c) R – N – R' 3º = amine N, N-Dialkylalkanamine

R'

(d) R – N – CH3 N-Ethyl-N-methylalkanamine

CH2 – CH3

Ex:- (1) CH3 – NH2 Methanamine

(2) C – C – C – C – C Pentan-3-amine
–

NH2

(3) C – C – C – NH – C – C N-Ethylpropan-1-amine

(4) C – C – C – N – C N-Ethyl-N-methylpropan-1-amine
–

C
C

(5) C – C – C N-Ethylpropan-2-amine
–

NH – C – C

OH
–

(6) CH2 – CH = CH – CH – CH3 4-(N-Ethyl amino) pent-2-en-1-ol

–

NH – CH2 – CH3

(7) OHC – CH2 – CH – CH2 – CH3 3-(N-Phenyl amino) pentanal

–

NH – Ph
If hetero atom’s are count as a ‘C’ atom in the parent chain then they are written as
(1) –NH–  Aza
(2) –O–  Oxa
(3) –S–  Thia
(4) –Se  Selena

Example:- (1) HOOC – CH2 – NH – CH2 – CH3 3-Aza pentanoic acid

H
–

N
(2) Aza Cyclo pentane

Page No.23
 H

–
N
(3) 3-Methyl aza cyclo hexane
–
CH3

1.10 ETHERS (R – O – R’):-

F.Group Prefix Suffix Name

R – O – R' alkoxy – Alkoxy alkane

Alkoxy + Alkane
 
less no. more no.
of ' C' of ' C'

(i) Acyclic Ethers:-

(1) C – O – C Methoxymethane

(2) C – O – C – C Methoxyethane

(3) C – O – C – C – C 1-Methoxypropane

(4) C – C – C 2-Methoxypropane
–

O
–

C
C
–

(5) C – C – C – O – C – C 1-Isopropoxypropane
1-(Methylethoxy) propane

O–C–C
–

(6) Ethoxycyclohexane

O–
–

(7) Cyclopropoxycyclohexane

Br
–

3 O–C–C–C
–

4 2
(8) 5 1
3-Bromo-6-chloro-2-propoxycyclohexan-1-ol
–

6
OH
–

Cl
 In cyclic system, numbering always starts from senior most functional group.
OH
–

Br
–

2 1
6
(9) 6-Bromo-5-methoxycyclohex-2-en-1-ol
3
–

5
4 O – CH3

OH
–

Br
–

2 1
6
(10) 6-Bromo-5-ethoxycyclohex-3-en-1-ol
3 5
–

4 O–C–C

Page No.24
 (ii) Cyclic Ether: (3-membered Ring)
Hetero cyclic compounds
(1) C – C Oxirane or Epoxyethane
O

(2) C – C – C 1, 3-Epoxypropane
O

(3) C – C – C 1, 2-Epoxypropane
O

Cl
–

(4) C – C – C 1-Chloro-2, 3-Epoxypropane

O
 Polyethers:-
(1) C – C – O – C – C – O – C – C
1 2
1, 2-Diethoxyethane

1 2 3 4 5 6 7 8 9 10 11
(2) C – C – O – C – C – O – C – C – O – C – C
3, 6, 9-Trioxaundecane

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
(3) C – O – C – C – C – O – C – C – O – C – C – C – O – C – C
2, 6, 9, 13-Tetraoxapentadecane
2 3
1 4
O O 5
12
(4)
11 6
10 O O7
9 8

1, 4, 7, 10-Tetraoxacyclododecane
1.11 AROMATIC COMPOUNDS:
Classification:

Non-Benzenoid:
H H
–

 + Cation 2 e– Cyclopropenium ion

–

H
H H
–

–
+
 Cation 2 e– Cyclobutenium dication
+
–
–

H H

Page No.25
 H H

–
–
 Anion 6 e– Cyclopentadienyl anion
– ..

–
H - H

–
H
H H
H – –
–H
–
 6 e– Cycloheptatrienyl cation (Tropylium cation)
–

H + – H
–

H
Benzenoid Aromatic Compounds:
All organic compounds which contain atleast one benzene ring are known as benzenoid aromatic com-
pounds.
Naming of Aromatic Hydrocarbons (Arenes):
CH3
–

(1) Methylbenzene (Toluene)

CH3
–

CH3
–

(2) (Ortho) 1, 2-Dimethylbenzene (Ortho-xylene)

CH3
–

(3) (Meta) 1, 3-Dimethylbenzene (Meta-xylene)

–
CH3

CH3
–

(4) (Para) 1, 4-Dimethylbenzene (Para-xylene)

–

CH3
Q. Write structures of
C–C
–

(i) Metadiethylbenzene
–

C–C
C–C
–

(ii) 1-ethyl-4-isopropylbenzene
–

C–C–C

(iii) Orthodiethenylbenzene
–

C=C
–

C=C
Trialkyl Substituted Benzene:
If all the three substituents are similar, then only 3 trisubstituted benzene derivative are possible.
C C
C
–

C
–
–

C –
–
– –
–

C C
–

C C
1, 2, 3-Trimethyl benzene 1, 2, 4-Trimethyl benzene 1, 3, 5-Trimethyl benzene
(Vicinal Trimethyl benzene) (Unsymmetrical Trimethyl benzene) (Symmetrical Trimethyl benzene)

Page No.26
 Examples :

1. Isopropylbenzene (Cumene)

– –
C–C
C
C
–

2. 1-Chloro-4-methyl-2-nitrobenzene
–

NO2
–

Cl
C=C
3. Phenylethene (Double bond > Phenyl) (Common name : Styrene)

C–C=C
–

4. 3-Phenylprop-1-ene (Allylbenzene)

CH = CH – CH3
–

5. 1-Phenylprop-1-ene

COOH
–

6. Benzene carboxylic acid (Benzoic acid)

O O
C–O–C
–

7. Benzene carboxylic anhydride (Benzoic anhydride)

SO3H
–

8. Benzene sulphonic acid

O
C – O – C2H5
–

9. Ethyl benzene carboxylate (Ethyl Benzoate)

O – C – CH3
–

10. O Phenyl ethanoate

O
C – Cl
–

11. Benzene carbonyl chloride ( Benzoyl chloride)

O
NH – C – CH3
–

12. N-Phenylethanamide

O
C – NH – CH3
–

13. N-Methylbenzene carboxamide

Page No.27
 CN

–
14. Benzene carbonitrile (Benzonitrile – popular)

CH = O
15. – Benzene carbaldehyde (Benzaldehyde – popular)

O
C
–
–

16. Diphenyl ketone (Benzophenone)

O
C
–
–

17. CH3 Methyl phenyl ketone (Acetophenone)

OH
–

18. Phenol

OH
–

19. ––– CH3 Methylphenol [o-Cresol, m-cresol, p-cresol]

OH
–

OH
–
20. o-Hydroxyphenol (Catechol)

OH
–

21. m-Hydroxyphenol (Resorcinol)

–

OH
OH
–

22. p-Hydroxyphenol (Quinol)(Hydroquinone)

–

OH
NH2
–

23. Benzenamine (Aniline)

SH
–

24. Benzenethiol

O
O – C – CH3
O
25. –C–O–H 2-Ethanoyloxy (Acetoxy) benzene -1-carboxylic acid (Aspirin)

Page No.28
 LECTURE NOTES
Session - 2009-10

ORGANIC CHEMISTRY
TOPIC : GOC - I
CONTENTS :
1 Isomerism
* Structure Isomerism
* Stereo Isomerism
Geomertical
Optical
Conformational

Refer sheet GOC- I

JEE Syllabus [2009]
Concepts: Hybridisation of carbon; Sigma and pi-bonds; Shapes of molecules; Structural and
geometrical isomerism; Optical isomerism of compounds containing up to two asymmetric cen-
ters, (R,S and E,Z nomenclature excluded); IUPAC nomenclature of simple organic compounds
(only hydrocarbons, mono-functional and bi-functional compounds); Conformations of ethane
and butane (Newman projections);

Page-29
 Isomerism

Isomers:
Compounds with same general formula or molecular formula but different physical and chemical property.
Ex:- CH3 – CH2 – OH and CH3 – O – CH3

Ex:- CH3–COOH and HCOOCH3

Homologs:
Compounds with same general formula differing by same structural unit – CH2 – or molecular weight by 14
unit.

Ex:- CH3 – OH and CH3 – CH2 – OH

Difference between isomers and homologs:-

Page-30
Structural isomers:
When two or more number of organic compounds have same molecular formula but different structural
formula these are called structural isomers.

Stereo isomers:
When two or more compounds have same Molecular Formual (M.F.) and same Structural Formula (S.F.) but
have different stereochemical formula (S.C.F.), these are called stereoisomers.
Stereo Chemical Formula (S.C.F) :
It indicates different arrangements of atoms or groups in space around a stereo centre or it indicates different
spatial orientations of atoms or groups around a stereo centre.
M.F. C4H8
CH3

–
S.F. (i) CH3 – CH2 – CH = CH2 (ii) CH3 – CH = CH – CH3 (iii) CH 3 – C = CH 2
These are structural isomers.
For (ii), S.C.F. are
H3C CH3 H3C H
C=C and C=C
H H H CH3
Stereocentre
These are stereoisomers.
Structural isomers are also known as:-
(i) Skeletal isomers (ii) Linkage isomers (iii) Constitutional isomers

Chain isomers :
They have different size of main carbon chain and / or side alkyl chain
(i) The two chain isomers should have same nature of F.G./multiple bonds/substituents (except –R group)
(ii) The position of F.G./M.B./substituent (locants) is not. considered here.

Ex:- Alkanes:-

(a) C1 – C3 chain isomers not possible

(b) C4H10 H3C – CH2 – CH2 – CH3 and CH3 – – CH3

(c) C5H12 CH3 – CH2 – CH2 – CH2 – CH3 , CH3 – CH2 – – CH3 , CH3 – – CH3

(d) C6H14 (i) CH3 – CH2 – CH2 – CH2 – CH2 – CH3, (ii) CH3 – CH – CH2 – – CH3,

(iii) CH3 – CH2 – – CH2 – CH3, (iv) CH3 – – – CH3,

(v) CH3 – CH2 – – CH3

(i) and (ii) – Chain isomers

(i) and (iii) – Chain isomers
(i), (iv) – Chain isomers
(i), (v) – Chain isomers
(ii), (v) – Chain isomers
(ii), (iii) – Position isomers
(iv), (v) – Position isomers

Ex. and shows which types of isomerism ?

Ans. Chain Isomers

Page-31
Q. Write chain isomers of N-alkanamine (1º) containing 4 carbon atoms ?
C

–
Ans. C – C – C – C – NH2, C – C – C – NH2

Positional isomers:
They have different position of locants Functional group (F.G.) or Multiple Bond (M.B.) or substituents in
the same skeleton of C-atoms.

Nature of F.G. or M.B. should not change.

The skeleton of C-atom should not change.

Ex. (1) C – C – C – C – C , C–C– C –C–C

–

–
C C

(2) C = C – C – C , C–C=C–C

(3) C  C – C – C – C , C–CC–C–C

(4) C – C – C – C , C – C – C – C – OH
–

OH
Q. Write all Positional isomers of Dichlorobenzene ?
Cl
Cl
–

Cl
–
–

Cl
–

Ans. , ,
–
–

Cl Cl
Q. Write all Positional isomers of Dichlorocyclopropane ?
Cl Cl– Cl
–

Ans. – ,
Cl

Q. Write all Positional isomers of Dichlorocyclobutane ?

Cl– Cl Cl– Cl– Cl
–

–
Ans. , ,
–

Cl
Q. Write all Positional isomers of Dichlorocyclopentane ?
Cl Cl Cl Cl
–
–

–
Cl
–

Ans. , ,
–
Cl
Ex. (i) C – C – C – C – C – C (ii) C – C – C – C – C – C, (iii) C – C – C – C – C – C
–

C–C C C C C
(i), (ii) – Chain
(ii), (iii) – Position
(i), (iii) – Chain

Ex. C6H12 (Cycloalkanes):-

–

– – –
(1) , (2) , (3) , (4) ,
–

Page-32
 C

– –
C–C–C C–C

–
–

–
–
(5) , (6) , (7) , (8) ,

–
–
– C–C C–C

–
–

–
(9) – , (10) , (11)
–
Ans. (3) and (4) – Positional isomer
(3) and (6) – Positional isomer
(4) and (6) – Positional isomer
(5) and (10) – Positional isomer

(All are chain isomers of 1)

Remaining are chain isomers

Functional isomers:
They have different nature of functional group (F.G.). The chain and positional isomerism is ignored (not
considered).
Compound Functional isomer
C–C–C–C Nil

C–C–C=C (Ring-chain isomers are Functional isomers)

Ex. Compound Isomer Functional Remarks

(1) C–C–CC (a) C = C – C = C Alkadiene (a), (b) are not functional isomers
Alkyne (b) Cycloalkene among themselve Why.....??

(c) Bicyclo

(2) C – C – C – OH C–C–O–C
Alcohol Ethers

(3) C – C – CH = O C– C –C
O
Aldehydes Ketones

(4) C – C – COOH C– C –O–C

O
Carboxylic acids Esters

.. ..
(5) C – C – C – NH2 (a) C – C – NH – C (b) C – N – C 1º, 2º, 3º amines are functional isomers
–

C
1º amine 2º amine 3º amine

(6) C–C–CN C – C – NC
Cyanide Isocyanide
(Propanenitrile) (Ethane isocyanide)

Important point :- Following compounds don’t exist at room temperature therefore not consider as a structure
isomer
–

(i) – C = C – OH (ii) – C  C – OH (iii) – C – OH

–

OH
–

(iv) – C – OH (v) – C – O – C = C (vi) Any peroxy compound

–

OR OH

Page-33
Q. Write acyclic isomers of C5H12O
C

–
Ans. (i) C – C – C – C – C – OH, C – C – C – C – C, C – C – C – C – C, C – C – C – C – OH,

–
OH OH
C
– C C

– –

– –
C – C – C – C – OH, C – C – C – OH, C – C – C – OH (7 alcohols)
C C
C

– –
C – C – C – C – O – C, C – C – C – O – C, C – C – C – O – C, C – C – O – C,

–
C C C
C

–
C – C – C – O – C – C, C– C –O–C–C (6 ethers).

Q. G.F. n M.F.
CnH2n + 3N (1) 3 C3H9N

1º Amine (R – NH2) 2º Amine (R – NH – R) 3º Amine R – N – R

–
R
Q G.F. n M.F.
CnH2n + 1N (1) 2 C2H5N
2

1º Amine 2º Amine 3º Amine

(1) C = C – NH2 C–C=N C–N=C
Ethenamine Ethanimine N-Methylene methanamine
(Unstable at room temperature)
H2C– – CH2 Azine
–

N
–

H
Q. G.F. n M.F.
CnH2n – 1N 2 C2H3N
Ans. Cyanides and isocyanides
H3 C – C  N CH2 = C = NH HC  C – NH2


H3 C – N C

Metamers:
When two isomers have same functional group (containing a hetero atom –O, N, S) but have different nature
of alkyl or aryl (aromatic radical) group attached to hetero atom, then these are called metamers.

Conditions:
(a) Same functional group
(b) Chain or position isomerism is not considered.
(c) >C = O (keto) group does not show metamerism

Following functional groups show metamerism

(a) R – O – R’ Ethers (All isomeric ethers are metamers)

(b) R’ – NH – R 2º Amines

(c) R’ – N – R 3º Amines
–

R''
(d) R – S – R’ Thioethers
Page-34
 (e) R – C – O – R’ Esters
O

(f) R – C – O – C – R’ Anhydrides
O O

O
(g) R – S – O – R’ Sulphonate esters
O

Q. How many esters are possible for C3H6O2 ?

Ans. C  C  O  C  H and C  C  O  C are metamers
|| ||
O O
Q. 3º Amines of M.F. C5H13N (All metamers)
C C C C
–

–
–
Ans. C–C– N–C–C C–C–C– N–C C– C– N–C

O–C–C O–C
–
–

Q. and Show which type of isomerism ?

–

C
Ans. Metamerism

Q. Acyclic compound (A) contains 18 1º H atoms, two types of ‘C’ atoms. All H are identical. Identify (A).
C C
– –

– –

Ans. C – C – C – C  C8H18
C C

Q. Cyclic compound (P) contains 18 1º H atoms, two types of ‘C’ atoms. All H are identical. Identify (P).

C
–

C – C
–
Ans. or
– –
C C
–

C
Q. Compound X is an ether. It has 12 1º H atoms. It has 2 types of H-atoms and 3 types of C-atoms. Write it
Structural formula ?
CH3
– –

Ans. H3C – O – C – CH3

CH3

Page-35
 Stereoisomers (Classification)

Stereoisomers :
The stereoisomers has different orientation of groups along a stereo centre. A stereocentre can be C = C
(any double bond), a ring structure, asymmetric carbon atom (*Cabcd). These isomers has same general
formula, structural formula and molecular formula but different stereochemical formula.

E.g. CH3 CH3 (I)

C=C
H H
CH3 H
C=C (II)
H CH3
CH3–CH2–CH=CH2 (III)

I, III  Positional
II, III  Positional
I, II  Stereoisomer
Configurational isomers :
Configurational isomerism arises due to different orientations along a stereocentre and these isomers can
be seperated and these isomers do not convert into one-another at room temperature. Therefore, they are
true isomers. They can separated by physical and chemical method.
E.g. Cis–2–Butene
Trans-2-Butene
Conformational isomers :
When different orientations arise due to the free-rotation along a sigma covalent bond. Such isomers are
called conformational isomers.
E.g. eclipsed ethane and
staggered ethane
These isomers change into each other at room temperature and can never be isolated. So these are not
considered as true isomers.
Geometrical Isomerism
1. Cause of Geometrical isomerism :
Geometrical isomerism arises due to the presence of a double bond or a ring structure
C = C, C = N, N = N, or Ring structure (Stereo centres)

Due to the rigidity of double bond or the ring structure to rotate at the room temperature the molecule exist
in two or more orientations.
This rigidity to rotation is described as restricted rotation/hindered rotation/no rotation.
E.g. a a
C=C
b (I) b
The root form of geometrical isomers lie in restricted rotation.

a a

b (I) b

Page-36
 Condition
(i) Restricted rotation
(ii) The two groups at each end of restricted bond must be different.
1 3
C=C
2 4
Caa Caa X
Caa Cbd X
Cae Cbb X
Cab Cab 
Cab . Cbd 
Cab Cde 
(iii) In two geometrical isomer the distance between two particular groups at the ends of the restricted bond
must be changed.

Q. Which of the following compounds show G.I.

(1) Ethene (2) Propene

(3) 2-Methylbut-2-ene CH3  C  CH  CH3 (4) But-2-ene CH3 – CH = CH – CH3

|
CH3

(5) Penta-1, 3-diene C = C – C = C – C (6) 1, 2-Dideuteroethene

H2C = CH
(7) Phenylethene (8) Buta-1, 3-diene C = C – C = C

Ans. 4, 5, 6

1. Geometrical isomerism across

– Nil

and By E / Z

2. Geometrical isomerism across

(a) Imine ( )
Imine compounds are produced from carbonyl compounds on reaction with ammonia.

(Syn and anti)

Imines prepared from unsymmetrical addehydes and ketones, always show geometrical isomerism.
Page-37
Q. Which of the following compounds show geometrical isomerism after reaction with NH3.
O O O
|| || ||
(a) CH3  C  CH3 (b) H  C  H (c) H  C  D

O
||
(d) CH3  C  H (e) Ph  C  CH3 (f) Ph  C  Ph
|| ||
O O

(g) (h) (i)

Ans. c, d, e, g, i

(b) Oximes C = N – OH :

These are prepared by reacting carbonyl compuond with hydroxyl amine (NH2 – OH)
R R
H2O C = N–OH
C=O + H2 N–OH  
H H
(Aldoxime)

R R OH
C=N C=N
and
H OH H
(I) (II)
(syn) (anti)
Syn and anti in aldehyde only not for ketones.

* Except formaldehyde (CH2O) All other aldehyde form two oximes.

* Unsymmetrical ketoes form two oximes.

Eg. Which of the following ketones will form two oximes.

O
|| O
||
(1) Propanone CH3  C  CH3 (2) Butanone
CH3  CH2 – C  CH3
(3) 3-Pentanone C–C–C–C–C (4) Acetophenone C 6H 5–C–CH 3
O O
O
(5) Benzophenone C 6H 5–C–C 6H 5 (6) Cyclohexanone

O
O
Me
(7)
Methyl Cyclohexanone
Ans. 2, 4, 7

Q. The lowest molecular weight of acyclic ketone and its next homologue are mixed with excess of NH2 – OH
to react. How many oximes are formed after the reaction ?
Ans. 3

Page-38
 (c) Hydrazones C = N – NH2 :

R R
H2O
C = O + H2N.NH2  C = N–NH2

H hydrazine H

R R NH2
C=N + C=N
H NH2 H
(I) (II)
(Geo. diastereomers)

Q. Two chain isomer of a cycloalkanone which are next higher homologue of lowest molecular weight,
cycloalkanone reacted with hydrazine. Identify the structure and number of isomer of hydrazones prepared ?
..
N .. NH2
O +
NH2 N N ..
Ans. + H2N – NH2 + +
NH2
CH O CH3 CH 3
3

Number of isomer = 3

(3) Geometrical isomerism across azo compounds (– N = N – )

(i) H – N = N – H (H2N2)

(ii) Ph2 N2 (Azobenzene)

.. Ph ..
..

N=N N=N
Ph Ph .. Ph
syn anti

(4) Geometrical isomerism across ring structure

(i)

Restricted rotation

(ii)

(iii)

Page-39
(5) Geometrical isomerism in cycloalkenes across double bonds :
In cycloalkenes, G.I. exists across double bonds with ring size equal to or greater then 8 carbon atoms (due
to ring strain)

Stereocentre:- It may a double bond , a ring or a asymmetric carbon

An atom or bond across which stereoisomerism exists (either G.I. or optical isomerism)

Those stereoisomers which are not mirror images of each other are called diastereomers. Those com-
pounds which are non-superimposable mirror-images of each other are called enantiomers

are enantiomers

E/Z Nomenclature :

Z (Zussamen = together) a > b and e > d

E (Entegegen = opposite) a > b and e > d

Rules :(i) The group with the first atom having higher atomic number is senior.
Thus – F > – OH > – NH2 > – CH3
(ii) If the first atom is identical, then second atom is observed for deciding the seniority of the group.

(a)

(b) <

(c) <

(d) >

Page-40
Ex. Z

Ex. E

(iii) If the first atom has same atomic number but different atomic mass, that is isotopes. Then heavier
isotope has higher seniority.

(iv) If the group has unsaturation, then a hypothetical hypothetical equivalent in drawn for it and it is com-
pared with other group for seniority.

(1)

(2) – CH = CH2 < – C  CH

(Hypothetical)

(v) Bond pair is always senior to lone pair. lp: lowest priority

CH 2 = CH C(CH3)3
C == C
Ex. (1) HC C E
CH2 – CH = CH2

(2) Z

N C CH2 – C(CH3)3
C == C
(3) H2C = C = CH E
C CH

(4) Z

(5) E

(6) E

Page-41
Number of geometrical isomers
Case I : Compounds having dissimilar ends
No. of G.I. = 2n
where n = number of stereocentre

Ex.
Ans. Stereocentre = 2
Geometrical isomerism = 4

(a) (b)

(c) (d)

Case II : Compounds with similar ends & even no. of stereocentre.

n
1
No. of Geometrical isomerism = 2n–1 + 2 2
Ex.1 CH3 – CH = CH – CH = CH – CH3
Ans. Stereocentre = 2
Geometrical isomerism = 21 + 20 = 3

Case III : Compounds with similar ends but odd number of stereocentre.
n 1
No. of Geometrical isomerism = 2n–1 + 2 2

Ex.1 CH3 – CH = CH – CH = CH – CH = CH – CH3

Ans. Stereocentre = 3
Geometrical isomerism = 6

Physical properties of Geometrical isomers :

The physical properties of organic compounds can be compared through the knowledge of their molecular
formula, structural formula and stereochemical formula (S.C.F.)

More Polar geometrical isomer is more soluble in water.

(i) > Cis > Trans

(ii) < Trans > Cis

COOH COOH H COOH

(iii) C=C > C=C Cis > Trans
H H COOH H

Page-42
Page-43
 Optical Isomerism
Some organic compounds can rotate the plane of plane-polarised light. Such compounds are called
optically active compounds.
The optically active compound can show optical isomerism.
Measurement of optical activity
It is measured by an instrument called polarimeter.

(1) (5)
(3) (6)

Recorder

(8)

(1)  Source of light

(2)  Polychromatic Non-polarised light
(3)  Slit (Monochromator)
(4)  Monochromatic Non-polarised light
(5)  Polariser (Prism-setting)
(6)  Monochromatic plane-polarised light
(7)  Sample tube (  = path length)
(8)  Recorder for measurement of optical rotation
Observations:- When plane polarised light is passed through sample tube, then following changes can be observed
in the recorder:-
Angle of rotation () Inference
(1)  = 0º Compound is optically inactive.
(2)  = +xº (clockwise rotation) Optically active, dextrorotatory or d or (+).
(3)  = –xº (anticlockwise rotation) Optically active, laevorotatory or l or (–).
Specific Rotation:- The specific rotation of a compound indicates the optical rotation of unit concentration (1 g/mL)
present in a sample tube of 1 dm of path-length at given temperature and given wavelength of light. It is
represented as follows:-

t  25 º C 
[ ]  580 nm 
c
where  = observed angle of rotation
c = concentration in g/mL (or density)
 = Path length of Sample tube in dm
Q. Compound ‘X’ has  = +70º for 2 g/mL solution in sample tube of  = 1 dm. Calculate  ?
70
Ans. = = +35º
2
(1) Its concentration is made twice, the observed rotation() will be 140º but  = 35º
(2) If  = +70º, it will be a
(i) d compound or (ii)  compound of  = –(360º – 70º) = –290º
It can be decided by changing the concentration or by changing the length of the tube (c or  ).
If concentration is reduced to half, d will have +35º and  will have –145º (not distinguish) still halfed, it will
give +17.5º and –72.5º to distinguish.

(I) (II)
With symmetrical With asymmetric
molecule molecule
[] = 0 [] 0
Optically inactive compounds:-
For symmetrical molecule optical rotation observed after interaction of light is zero.
Page-44
 But in case of optically active compounds. The molecules are asymmetric in nature and show non-zero
optical rotation.
Symmetry of elements :-
(1) Centre of symmetry (C.S.)
(2) Plane of symmetry (P.S.)
(3) Axis of symmetry
(4) Alternating axis of symmetry

Plane of symmetry : - The imaginary plane which divides a molecule into two equal halves which are
related as mirror image is known as plane of symmetry

Centre of symmetry : - A centre of symmetry in a molecule is said to exist if a line is drawn from any atom
or group to this points and then extended to an equal distance beyond this point, meets the identical atom
or group.
Cl
R H
H H
C=C
H R
Cl
Centre of symmetry Centre of symmetry
360 º
Axis of symmetry : It is defined as Cn axis of symmetry that means of the molecule is rotated by
n
angle, then its original or identical or superimposable. The molecule is said to have n-fold axis of symmetry

Ex. 
2C

(180 ºrot )

Ex. (C4 axis) Ex. (axis)

There is no relation whatsoever with chirality and axis of symmetry. Meso molecule does not have axis of
symmetry

 no axis of symmetry .

Page-45
 Alternating Axis of Symmetry :
360º
When a molecule is rotated by angle and its mirror image is taken in the perpendicular plane of
n
rotation, then its original molecule is obtained. It is represented by Sn

H Me
Me

Me H
H 180º
Ex. Me Reflextion H
H rotation

H (II) Me (III)
Me (I)

I and III are equivatent.

Dissymmetry:- The molecules or objects in which minimum two elements of symmetry (Centre of symmetry and
Plane of symmetry) are absent are called dissymetric molecules/objects.

Asymmetry:- When all the elements of symmetry (23 including Centre of symmetry and Plane of symmetry are
absent) the molecule is said to be asymmetry.
So, all asymmetric molecules are always dissymmetric, but the reverse is not true.

Dissymmetry and chirality:- All dissymmetric molecules are chiral and are optically active.

Dissymmetry, Chirality and Optical Activity:- Dissymmetry/Chirality is the minimum and sufficient condition for a
molecule to be optically active. So, if in a molecule, a C.S. and P.S. are absent, the molecule will be optically
active. All asymmetric molecules are also optically active.

Chirality (Hand-like property or Handedness):- The human hand does not have Centre of symmetry and Plane
of symmetry , so it is dissymmetry. It does not have any element of symmetry, so it can be called
asymmetry. Any molecule which has this property is called chiral.

Note : All dissymmetric compounds are chiral.

Optical Isomers:-
Enantiomers:- The mirror-image stereoisomers are called enantiomers.
Two types:-
(a) Dextrorotatory (b) Laevorotatory
 Enantiomers are always mirror image isomers.
 They have same molecular formula, structural formula but have different orientation in space.
 They have dissymmetry/chirality.
 Every enantiomer is optically active.
 The two enantiomers can rotate the plane-polarised light with equal magnitude and opposite signs.
 They have similar physical properties except the sign of optical rotation.
 These are the isomers which have maximum resemblance with each other.
 These can be distinguished only by polarimeter.
Chirality (Dissymmetry) and Optical Isomers:-
(a) The dissymmetric molecules have two orientations in space. These two orientations are called
stereoisomers, optical isomers, enantiomers.
(b) These are mirror images (enantiomers).
(c) Non-superimposability of enantiomers:- The dissymmetric molecules are always non-superimposable
on their mirror-image orientations.
(d) The non-superimposable orientations are non-identical orientations, so are called isomers.
(e) Because of dissymmetry, these two isomers are capable of rotating plane-polarised light. So, these are
called optical isomers.

Page-46
Optically Active Carbon Compounds:-
If a carbon-atom is attached with four different group, then if does not have any element of symmetry. It is
known as asymmetric carbon atom, which is represented as *Cabde.
a

C*
e b
d
If a molecule contains only one asymmetric carbon atom, then the molecule as a whole becomes chiral and
optically active and show optical isomers.
Centre of symmetry Plane of symmetry Optical active
H
(1) C Absent Yes No
H H
H
H
(2) C Absent Yes No
H Cl
H
H
(3) C Absent Yes No
Br Cl
H
H
(4) C Absent No Yes
Br Cl
F
Ex.1:- Mark the chiral objects:-
(i) Cup (ii) Plate (iii) Letter A (iv) Letter G (v) Fan (vi) Door
(vii) Shoe (viii) Glove
Ans. IV, VII VIII
Projection Formula of Chiral Molecules:-
(i) Wedge-Dash Projection formulae
down

up

Ex:- (i) Butan-2-ol

CH3

C
C2H5
H OH
(ii) Fisher Projection formula
Ex:- (i) Butan-2-ol (CH3 – CH2 – CH – CH3)
–

OH
Rules of writing Fisher Projection formula :-
(i) It is represented by a cross (+).
(ii) Groups at Vertical line are away from observer.
(iii) Groups at Horizontal line are towards the observer.
(iv) Central ‘C’ atom of the cross is chiral.
(v) High priority group lies at the top of vertical line (IUPAC Numbering starts from top).
///////////////////////

CH3

H OH

CH2CH3

Page-47
Ex:- Draw Fisher Projection formula of following molecules:-

(1) 2-chlorobutane

(2) Pentan-2-ol

*
(3) CH2 – CH – CH
–

OH OH O

Glyceraldehyde

1 2 3 4 5
* *
(4) CH3 – CH – CH – CH2 – CH3 , ,
–
–

OH OH

(I, II), (III, IV)  Enantiomers.

All other diastereomers.
They are true isomers (not superimposable)

(5) Glucose CHO

– – – – –

CHOH
CHOH
CHOH
CHOH
CH2OH

Mark:- (i) No. of C*  4

(ii) Draw one Fisher Projection Formula
CH = O

H OH

H OH

H OH

H OH

Page-48
Configuration nomenclature in Optical Isomers :-
Relative configuration : The experimentally determined relationship between the configurations of two
molecules, even though we may not know the absolute configuration of either. Relative configuration is
expressed by D-L system.
Absolute configuration : The detailed stereochemical picture of a molecule, including how the atoms are
arranged in space. Alternatively the (R) or (S) configuration at each chirality centre.

(I) D - L System (Relative configuration) : Application on correct Fisher Projection Formula

This method is used to relate the configuration of sugars and amino acids to the enantiomers of glyceralde-
hyde. The configuration of (+)-glyceraldenyde has been assigned as D and the compounds with the same
relative configuration are also assigned as D, & those with (-) glyceraldehyde are assigned as L.

Left=L

Examples :

Sugars have several asymmetric carbons. A sugar whose highest numbered chiral centre (the penultimate
carbon) has the same configuration as D-(+)-glyceraldehyde (– OH group on right side) is designated as a D-
sugar, one whose highest numbered chiral centre has the same configuration as L-glyceraldehyde is desig-
nated as an L-sugar.

e.g.

(I) R/S Configuration :

R  Rectus S  Sinister

Examples :

d (1) Seniority order a > b > c > d

b c
(2) Put junior most group at dotted line
b

 c
a d
(3) Now go from a to b, b to c.
(i) If it follows clockwise route, it is (R).
(ii) If it follows anticlockwise route, its configuration is (S).
Br Br

Q. Cl Cl
I I
F F
(R) (S)

Page-49
R/S Configuration in Fisher Projection Formula :-

Me(c) (c)

Ex:- H(d) Et(b)  (d) abc R

(b)     
Clock wise
Pr (a)
(a)

 If the junior most group is at horizontal line clockwise  S & anticlockwise  R

 If juniormost at vertical line, then clockwise  R & anticlockwise  S

Q. (1) Ans. R

(2) Ans. R

(3) Ans. S

Properties of enantiomers :-
One chiral carbon :
(i) Number of optical isomer = 2 (d or  )
(ii) Number of Racemic Mixture  Equimolar mixture of d and  . [] = 0 due to external compensation of
optical rotation.

Properties:-

(i) Dipole moment same

(ii) Boiling Point same
(iii) Melting Point same
(iv) Solubility same
(v) Specific rotation [] different

Molecules with more than one chiral carbons:-

(I) Calculation of No. of optical isomer :-
Case I:- When all Chiral carbon atoms are differently substituted (all are dissimilar C*)
optical isomer = (2)n
Case II:- When all the Chiral carbon atoms are similar at ends.
n
–1
(i) n = even  x = 2n – 1 + 2 2 (ii) n = odd  x = 2n – 1

Page-50
Molecules with two Asymmetric Carbon Atoms of Dissimilar Nature:-
(i) Structural formula CH3 – CH – CH – CH2 – CH3

–
Cl Cl
(ii) Optical isomer = 2n = 22 = 4
(iii) Stereochemical Formula:-

CH3 CH3 CH3 CH3

H Cl Cl H H Cl Cl H

H Cl Cl H; Cl H H Cl
C2H5 C2H5 C2H5 C2H5
I II III IV
[] = +xº –xº +yº –yº

Analysis:-
(a) I, II – Enantiomers III, IV – Enantiomers
I, III – Diasteromers I, IV – Diasteromers
II, III – Diasteromers II, IV – Diasteromers

(b) No. of Racemic Mixture – Two (I + II, III + IV).

Q. CH3 – CH – CH – CH – CH2 – CH3

–

Cl Cl Cl
Calculate total number of optical isomer.

Ans. n = 3,
x = 23 = 8 (4 d-  pairs)

(i) (ii) (iii) (iv)

and their mirror images.

(ii) 4 Racemic Mixtures (4 different fractions on fractional distillation)

Compounds with 2 Asymmetric Carbons of Similar Nature:-

Ex:- CH3 – CH – CH – CH3
–

OH OH
(i) n = 2
n
–1
(ii) x = 2n – 1 + 2 2 = 3 (Optical stereoisomers).
(iii) Stereo chemical formula :-

Me
H OH

OH H
Me
(I) (II) (III) (IV)
In (I and II)  Plane of symmetry present.
 Superimposable on its mirror image.
 Thus, I and II are identical.
 [] = 0, optically inactive.
 Meso isomer
In (III) Specific rotation [] = +xº

In (IV)  Specific rotation [] = –xº
Page-51
 No. of d –  pairs = 1 (III + IV) = Racemic mixtures
In (I) and (II) : [] = 0 due to internal compensation and Non-Resolvable
In (III) and (IV) : Resolvable [can be separated into two isomers (enantiomers)]
Meso isomers:-
 The optical stereoisomers which have more than one asymmetric carbon atoms but have a plane of
symmetry are called meso compound.
 They are achiral (optical rotation = 0).
 They have [] = 0 due to internal compensation of optical rotation.
 They are diastereomer of d –  pair. So, it has different physical properties than d –  -pair..
 Presence of more than one asymmetric ‘C’ atoms.
 They are non resolvable.
Q. Mark meso compound among following

COOH H
H OH Cl Cl OH
(1) (2) (3) (4) H
(5)
H OH Me Me *
COOH OH
Ans. (1) (2) (4) and (5)
Molecules with three similar chiral carbon:-

Ex:- CH3 – CH – CH – CH – CH3  M.F..
–

Cl Cl Cl
n=3
x = 2n – 1 = 4 stereoisomers.
Stereo Chemical Formula :-
\\\\\\\\\\\\\\\\\\\\

(I) (II) (III) (IV)

2S 2S 2S 2R
3S 3R 3  achiral 3  achiral
4R 4R 4S 4R
Why they r optical isomer Optically active
(i) & (ii) Achiral due to Plane of symmetry. (Optically inactive).
Properties of Optical Diastereomers:-
The optical isomer which are neither mirror image nor superimpossible to each other are called optical
diastereomers.

Conclusion:- Optical Diastereomers have different physical properties, so these can be separated by normal
physical methods of separation.
(i) By fractional distillation (different B.P.)
(ii) By fractional crystallization (different solubility)
(iii) Chromatography (different solubility)
(iv) Differential Melting (M.P. diff.)
Page-52
 Assertion:- All diastereomers have different polarities (Both Geometrical and Optical) – True
 Assertion:- Geometrical isomer are always diastereomers – True
Chemical method of separation (resolution) by using optically active reagent:-
Separation of Racemic Mixture:- The enantiomers in a d –  pair have identical physical properties, so these
cannot be separated by normal physical methods of separation. The special method is used for separation
of d –  pair known as optical resolution.
 General scheme of resolution:-

d+ + d'
Racemic Optically
mixture active
Reagent

Hydrolysis Hydrolysis
d + d'

 
Example:- (I) Resolution (by using Esterification) of RCOOH and R'OH

  Esterifica tion 
General Reaction:- R – C – O – H + H – O – R '      R* – C – O – R'
(H2 SO 4 )
O O
– H2 O

Me

H COOH
(d)
Me Me Me Et Me
Et
+ OH H H C–O H + H C–O H
Et
O (dd') O (d' )
(d')
H COOH D Et D Me D
( ) (d') (dd') ( d ')
Me

Pair of Diastereomers
Ester upon still hydrolysis will give back the carboxylic acid and alcohol.

(II) Separation of (d ) pair of alcohol by using optically active acid:-

Ph D

+ H O–C H
O (dd')
Me Me

Me Ph

+ H C–O H
O
D (d' ) Me

Page-53
(III) By salt formation (RCOOH + R’NH2)

Diastereomers (Separable)

Optical Purity or Enantiomeric Excess (O.P. or E.E.):-

Ex:- Compound X has [] =  70º
(obs)
We have mixture of enantiomers in different percentage. %E.E.   100%
[]specific

Page-54
Optically active compounds without a chiral carbon:-
Rapid
R R
– Inversion –
 Asymmetric N:- R' – N   N – R'

–
R'' R''
 A nitrogen atom attached with 3 different groups (sp3) is chiral, asymmetric.
 In case of acyclic compounds, the Nitrogen, having trigonal pyramidal shape undergoes rapid inversion of shape
at room temperature. So at room temperature, everytime a racemic mixture of ‘d’ and ‘  ’ forms exists.
This racemic mixture can never be resolved at room temperature.
 Nitrogen converts into its enantiomer..

So, acyclic molecules with chiral nitrogen are chiral but optically inactive ([] = 0) and are non-resolvable.

If asymmetric Nitrogen atom is present in ring structure, then inversion does not take place and such molecules are
optically active [ ]  0  . (otherwise ring will break)

 Me
N
Ex:- [ ]  0

Optical activity without asymmetric carbon :

CCC–C–C CC–CC–C C C–C–C C

(Cumulated double bond) (Conjugated double bond) (Isolated double bond)
  
One after another Alternate Separated

(I) Case of allene :

(a) Allenes with even  bonds : e.g.

the orbital diagram of this structure will be

Since the groups at the end of allene are in perpendicular plane, it will not show geometrical isomerism. The
molecule lacks centre of symmetry as well as plane of symmetry. Overall the structure has molecular
dissymmetry which is the sufficient condition for optical activity. The molecule will exist in two enantiomeric
forms.

(b) Allenes with odd  bonds : e.g.

the orbital diagram of this structure will be -

The groups at the end of allene structure lie in same plane (ZX plane). Therefore it will have a plane of
symmetry (ZX plane). The molecules lacks molecular dissymmetry & it will not show optical activity hence
optical isomerism. But the compound will exist in two geometrical diastereomeric forms.

Page-55
(II) Case of spiranes : A similar case like allenes is observed in spiranes. The spiranes with even rings and
different groups at terminal carbons show optical activity & optical isomerism, while the spiranes with odd
rings shows geometrical isomerism.
(a) spiranes with even rings :

shows optical isomerism.

(b) spiranes with odd rings :

shows geometrical isomerism.

(III) Case of cycloalkylidene :

(IV) Case of ortho-ortho-tetrasubstituted biphenyls :

becomes non-planar at room temperature in order to have minimum electronic

repulsion among the substituent. In this orientation (phenyl planes perpendicular to each other) the free
rotation of C – C single bond is restricted and molecule shows optical activity due to molecular disymmetry.
e.g.

Page-56
 CONFORMATIONAL ISOMERISM :
1. Free Rotation : A sigma covalent bond undergoes free rotation at room temperature.
(C–C, C –O, C–N, N–N, O–O)
2. Conformers / Rotamers or conformations : The infinite number of spatial oreintation of molecule arises
due to free rotation along a sigma covalent bond
3. Conformational Isomers : Those conformations which are most stable and have minimum P.E. are defined
as conformational isomers. These are not true isomers and can never be isolated.
4. Conformational Energy : The rotational energy barrier is known as conformational energy. It is the P.E.
difference between conformational at potential energy minima and maxima.
5. Angle of rotation/Dihedral Angle (D.H.A.)/Torsion Angle : The interfacial angle between the groups
attach at two -bonded atoms is defined as dihedral angle.
Projection formula
(a) Saw horse Projection formula

Ex. CH3 – CH3

H(a) H(c) H(b)

rotation
   H H H(a)
(b) (c)
Eclipsed Staggered
(b) Newmann Projection formula

    
1

12.5 kJ / mol

Energy - Level Diagram of ethane

Ex. CH3–CH2–CH3

Page-57
 Saw horse Projection formula

Eclipsed Staggered

Newman Projection Formula

CH3 CH3 CH3
H
H H H H


 120 º H 
60 º H H   HH H H H
H H H
II III IV
Eclipsed Staggered Eclipsed Staggered

Propane :

Butane :
Ex.

Ethyl-hydrogen repulsion is less than methyl-methyl repulsion. So draw the newman between C2 - C3

CH3 CH3 CH3 CH3

CH3 H
H CH3
60 º H H

 120 º 180 º 240 º
H
     
H H H H
H H H H H CH3 H H
(I) (II) CH3
(IV) (V)

300º 360º

 


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I/VII = Fully eclipsed
II/IV = Gauche form
III/V = Partially eclipsed
IV = Anti form Staggered

Stability Order : IV > II > III > I Anti > Gauche > Partially eclipsed > Fully eclipsed

P.E. Order : IV < II < III < I

Potential Energy Diagram of butane :

Conformational Isomers : Due to free rotation from 0º to 360º those conformations which are most stable
and have minimum P.E. are defined as conformational isomers.If these conformers have same energy then
there isomers are called degenerate isomers or equienergic isomers.
The conformational isomers with different energies are called non-degenerate isomers.
The conformational isomes lie at the P.E. minima in the P.E. diagram with respect to rotational angle.

Strains :
(i) Torsional Stran (eclipsing strain) :
It is defined as the electronic repulsion between the bond-pairs electrons of two adjacent eclipsed bonds. It
is active at torsional angles 0º, 120º, 240º in which the molecule has eclipsed conformation. It is considered
almost zero in the staggered conformation. (D.H.A. = 60º, 180º, 300º)

(ii) Vander Waal strain :

It is the repulsion between the group attached at adjacent bonds. The vander waal strain is maximum in
eclipsed conformation, and minimum in anti conformation, while intermediate in gauche conformations.
It is almost zero for H-atom since the sum of vander waal radii is less than internuclear distance between two
H atom.

(iii) Angle Strain :

It arises due to distortion in normal bond-angles. In acyclic compounds, there is no distortion of bond-angle
due to free rotation. So, all conformations have zero angle strain in acyclic compound

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Ques. Draw conformation isomers of following compound (a) CH3 – CH – CH3 (b) CH3 – (CH2)3 – CH3
CH3
Ques. Draw conformation isomers of following compound CH3 – CH –CH – CH3 with respect to C2 and C3 carbon

CH3 CH3
atoms.

H H
H H H
H
H3C CH3
CH3 CH3
60 º 120 º 180 º

  
H3C CH
CH3 CH3 3 CH3 H H3C CH3
CH3 H3CCH CH3 H
CH3 3
Eclipsed Staggered

Page-60
 Case of intramolecular hydrogen bonding :
In case of G – CH2 – CH2 – OH, where G = – OH, – NH2 , – F, – NR2, – NO2, – COOH, – CHO the Gauche form
is more stable than the anti form due to intramolecular hydrogen bonding i.e.

stability : Gauche form > anti form.

Example 1. CH2–OH (Glycol).

CH2OH

OH
H OH

  

H H
H
(II) Gauche

Hydrogen Bonding : (II)

Gauche form most stable due to hydrogen bonding.

Stability Order : II > IV > III > I

Example 2. O2N – CH2 – CH2 – OH

Interconversion of projection formulae :

Example Tartaric Acid :

COOH – CHOH – CHOH – COOH

no. of Sterioisomers : 3

COOH

1. Meso H OH 
H OH

COOH
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 2. d/ 

COOH
COOH COOH
H OH H OH

Meso  COOH   H

H OH H OH H
OH COOH COOH OH

COOH COOH
COOH COOH OH

OH


H H OH H OH
H

d/
COOH COOH
COOH COOH
COOH OH H OH
OH COOH

OH   COOH  H

H OH OH
H H OH H HH OH
COOH

COOH COOH
H
COOH H

 OH
H OH
OH H COOH OH
Conformation isomers of cyclohexane : Chair form and Boat form

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