ALKANE

Hydrocarbon :
Hydrocarbon Functional group Example

Alkanes none (no double or triple bonds) CH3–CH2–CH3, propane

Alkenes C=C double bond CH2=CH–CH3, propene

Alkynes –CC– triple bond H–CC–CH3, propyne

C CH2CH3
C C
Aromatics benzene ring C C ethylbenzene
C

* Introduction :
Alkanes
(i) The alkanes or the paraffins are the saturated hydrocarbons. Many occur naturally, and the chief
source of the alkanes is mineral oil or petroleum, which occurs in many parts of the world.
(ii) These are also called as ‘Paraffins’ (Parum + Affinis i.e. less reactive).
(iii) General formula is CnH2n+2 .
(iv) Hybridisation state of carbon is sp3.
(v) Geometry of carbon is tetrahedral.
(vi) Bond angle is 109º28’.
(vii) Number of bond angle in methane are six, while in ethane are twelve.
(viii) C – C bond length is 1.54 Å while C – H bond length is 1.11 Å.
Preparation of alkane
1. From aldehydes and ketones :
1.1 Clemmensen Reduction :
The Clemmensen reduction is most commonly used to convert acylbenzenes to alkylbenzenes, but it also
works with other ketones and aldehydes that are not sensitive to acid. The carbonyl compound is heated
with an excess of amalgamated zinc (zinc treated with mercury) and hydrochloric acid. The actual reduction
occurs by a complex mechanism on the surface of the zinc.
O
||
Ph  C  CH 3 Zn ( Hg
) or Na ( Hg )
  Ph–CH2–CH3
HCl

1.2 Wolf-Kishner Reduction.

Compounds that cannot survive treatment with hot acid can be deoxygenated using the Wolf-Kishner
reduction. The ketone or aldehyde is converted to its hydrazone, which is heated with a strong base such
as KOH or potassium t-butoxide.
O N–NH2 +
H
N 2H 4 t-BuO¯ K H
 + N2
O
cyclohexanone hydrazone CH3—S—CH3
(DMSO, a solvent)
cyclohexane (80%)
2. From alkene :
2.1 By catalytic hydrogenation : Addition of H2 on alkene takes place in cis manner in the presence
of Pt or Ni or Pd to give alkane.
eg.
Ni
(i) CH2 = CH2 + H2   CH3 – CH3
R D

H
R R
(ii) C=C  D R
 H2
D D
(meso)
H

D R
C=C + H2
R D

[NOTE : This reaction is exothermic and heat of hydrogenation depend upon no. of ‘’ hydrogen.]
2.2 By hydroboration followed protolysis
H BH2
| | 
H
(i) CH2 = CH2 + H — BH2  CH2 — CH2   CH3 — CH3

BH2
HBH2 |
(ii) CH3—CH = CH2    CH3 –  CH – CH – CH2
3 2
2CH3 CHCH2 H2O /H
CH3–CH2–CH2 – BH2    (CH3–CH2–CH2–)3B    3CH3–CH2–CH3

NOTE : Long- chain alkane may be possible by the coupling of alkyl boranes.
If (AgNO3 + NaOH) is used instead of H2O / H+
(R–)3B R–R

3. From alkyne - by reduction :

H H H
Pt/Pd/Ni. | Pt/Pd/Ni. + H2 | |
R – C  C – H + H2 2CH –OH R– C =C –H R C  C H
3 | 200º–300ºC | |
H H H
alkene alkane
in the above reaction if catalyst palladium or platinum is used, it is called simple reduction and if Ni is
used, it is called Sabatier - Senderen’s reaction (catalytic hydrogenation).
4. From Alkyl halides :
4.1 By reduction :
When alkyl halide is reduced with Zn-Cu couple + ROH or Na + EtOH/ Na- Hg + H2O or
LiAlH4, then we get respective alkane.
Zn  Cu
R–X + 2 H R RH + HX
 OH

CH3Cl + 2H  CH4 + HCl

CH3I + 2H  CH4 + HI

4.2 Wurtz’s reaction :

When alkyl halide reacts with Sodium in presence of dry ether then we get higher alkane. Mechanism
of the reaction is based on ionic and free radical both.
R – X + 2 Na + X–R dry ether
   R – R + 2 NaX
CH3Cl + 2 Na + ClCH3  CH3–CH3 + 2 NaCl
NOTE : * Current reaction is not suitable for tertiary alkyl halides

Mechanism : Two mechanism has been suggested for Wurtz’s reaction.

(i) Via organometallic compound
 
  C 2 H 5 Na
C2 H5 Br + Na NaBr CH3 – CH2 – CH2 – CH3

(ii) Via free radical

2NaBr
2C2 H5 – Br + 2Na    2C2H5•  C2H5–C2H5

(iii) Corey-House Synthesis

Li CuX
R—X Ether R—Li R2CuLi

R2 CuLi + 2 R—X 2 R—R + LiX + CuX

Lithium dialkyl (should be 1°)
cuprate
This method is better than Wurtz and can be used for preparing symmetrical as well as unsymmetrical
alkanes (having an odd number of carbon atoms).

4.3 Frankland’s reaction/Method :

If alkyl halide is treated with Zn dust in closed tube then higher symmetrical alkanes will be formed.
2RX + Zn  R – R + ZnX2
This is known as Frankland’s reaction, in this reaction first Frankland’s reagent (R–Zn–R, dialkyl Zinc)
is formed. Which then react with alkyl halide to give higher alkanes.
[Where R = CH3]
For Example:-
CH3 – Br + 2Zn + Br – CH3  CH3 – Zn – CH3 +ZnBr2
CH3 – Zn – CH3 + CH3 – Br  CH3 –CH3 +CH3ZnBr
ethane
4.4 Grignard reagents.
Alkyl halides in ether react with magnesium to form alkyl magnesium halides or Grignard reagents which,
on treatment with water or dilute acid, are decomposed to alkanes.
ether 
(i) RI + Mg   R — Mg — I H RH
H O  H
(ii) CH3MgBr    CH4
(iii) CH3MgBr R
O  H
 CH4
(iv) CH3MgBr CH
3CO2 H
 CH4
Preparation by Wurtz reaction, Kolbe electrolysis Ullmann, Fitting reaction, Wurtz-Fitting reaction already
discussed in free radical reaction.
5. From Red P + HI :
It is a powerful reducing agent which will convert, aldehyde, ketone, alcohol, carboxylic acid to alkanes
with same number of carbon.
Re d P  HI
(i) R–CH3–OH     R –CH3

O
|| Re d P  HI
(ii) R  C  H     R –CH3

O
|| Re d P  HI
(iii) R  C  R     R –CH2– R

(iv) R–COOH Re

dP  HI
 R –CH3

SOLVED EXAMPLE
Ex.1 Grignard reagent give alkane with :
(A) C2H5OH (B) C2H5NH2 (C) H2O (D) None of these
Sol. (A, B, C)
Grignard reagent react with compounds containing active hydrogen like, NH3, H2O, ROH, RNH2,
R2NH to form alkane [Zerewitnoff method]

R–Mg–X + C2H5OH  RH + Mg

R–Mg–X + C2H5NH2  R'–H + Mg

R''–Mg–X + HOH R''–H + Mg

Ex.2 The Kolbe synthesis of alkane using a sodium salt of butanoic acid gives -
(A) n-hexane (B) isobutane (C) n-butane (D) propane
Sol. (A)
Ex.3 Electrolysis of an aqueous solution of sodium butanoate gives main product –
(A) octane (B) heptane (C) hexane (D) butane
Sol. (C)
Koble electrolysis
2CH3CH2CH2COONa  2CH3CH2CH2COO– + 2Na+

2CH3CH2CH2COO + e–

2CH3CH2CH2 + 2CO2

CH3–CH2–CH2–CH2–CH2–CH3
Hexane.
Ex.4 Corey House reaction involves :
(A) An organometalic compound (B) A nucleophilic substitution
(C) A nucleophilic addition reaction (D) Development of a C–C bond.
Sol. (A, B, C)
Corey House reaction :
I– step : Formation of organometallic compound
R – X + 2Li  R – Li + LiX
II–step : Organometallic compounds react with cuprous iodide (CuI) to form alkyl lithium cuprate.
2R – Li + CuI  R2CuLi + LiI
III step : This compound react with another alkyl halide by nucleophilic substitution reaction to form
higher alkane containing both even and odd number of carbon atom which was otherwise not possible
with Wurtz reaction.
R2CuLi + R' – X  R – R + LiX + CuR
overall this results in development of C – C bond.
Li
CH3–CH2–Br  CH3CH2–Li CuI
 (CH3CH2)2CuLi


CH 3  CH 2  CH 2  C H  CH 3 CH 3  C H  CH 2  Br
|
|
CH 3 CH 3
isohexane
Ex.5 In Wurtz reaction if we take CH3Cl & C2H5Cl then product will be -
(A) Propane + Ethane
(B) Propane
(C) Propane + Ethane + Butane + Ethene + CH4
(D) Propane + Butane
Sol. (C)
(1) CH3 Cl + 2Na + Cl–CH3  CH3–CH3
Ethane
(2) CH3 Cl + 2Na + Cl–CH2–CH3  CH3–CH2–CH3
Propane
(3)  CH3–CH2–CH2–CH3
butane
Ex.6 Butane cannot be obtained by :
(A) Action of soda lime on sodium butanoate
(B) Clemmensen reduction (Zn amalgam in conc. HCl) of butanone
(C) Action of water on butyl magnesium iodide
(D) Sabatier Senderens hydrogenation of butene.
Sol. (A)
(A) RCOONa + NaOH  R – H + Na2CO3
CH3–CH2CH2COONa + NaOH CaO
 CH3CH2CH3 + Na2CO3
(B) Clemmensen reduction :
CH 3  C CH 2  CH 2 + 4H conc
  CH –CH –CH –CH
. HCl 3 2 2 3
||
O

(C) R–MgI + HOH  R – H + Mg

CH3CH2CH2CH2– MgI + HOH  CH3CH2CH2CH3 + Mg

(D) Sabatier – Senderens hydrogenation of alkene

NI
R–CH=CH2 + H2 
200º C R – CH2–CH3

NI
CH3–CH2–CH = CH2 + H2  CH3–CH2–CH2–CH3
200º C

Ex.7 Which of the following compounds cannot be prepared by Wurtz reaction :

(A) CH3CH3 (B) CH 3 C H  CH 3 (C) (CH3)2CHCH3 (D) CH3CH2CH2CH3
|
CH 3
Sol. (B, C)
Alkane containg even number of carbon atom can be obtained in good yield by Wurtz reaction.
2 Na
R – Cl + R – Cl ether
  R – R

Alkane containing odd number of carbon atom can't be obtained in good yield by Wurtz reaction.
2 Na
R'–Cl + R – Cl   R– R+ R' – R' + R – R'
Ethane and butane are obtained in good yield because they contain even number of carbon atom.
2CH3 – Cl + 2Na ether
 CH3–CH + 2NaCl
2CH3CH2–Cl + 2Na ether  CH3–CH2–CH2CH3 + 2NaCl
Isobutane contain odd number of carbon atom
ether
CH3–CH2–Cl + C H 2 –Cl + 2Na CH 3  C H  CH 3  CH 3  C H  C H  CH 3  CH 3  CH 3
| | | |
CH 3 CH 3 CH 3 CH 3
Ex.8 Hydrolysis of an ester gives a carboxylic acid which on Kolbe's electrolysis yields ethane. The ester is :
(A) Methyl ethanoate (B) Methyl methanoate
(C) Ethyl methanoate (D) Methyl propanoate
Sol. (A)
Hydrolysis of ester forms carboxylic acid which on kolbe's electrolysis produces ethane.

3 H O
CH 3  C OCH 3   CH 3COOH  CH 3OH
||
O
Methyl ethanoate

CH3COOH NaOH
 2CH3COONa  CH3–CH3
H O
(B) H  C O  CH 3 3  H  C OH  CH 3OH
|| ||
O O

3 H O
(C) H  C O  CH 2CH 3   H  C OH  CH 3CH 2OH
|| ||
O O
3 H O
(D) CH 3  CH 2  C OCH 3 
  CH 3  CH 2  C OH  CH 3OH
|| ||
O O

Physical Properties
1. The first four alkanes (from methane to butane) are colourless and odourless gases. The next thirteen
(from pentane to heptadecane) are colourless and odourless liquids. And, the rest of higher alkanes (having
18 carbon atoms or more) are colourless solids at ordinary temperature.

2. Alkanes being non-polar molecules, are soluble in non-polar solvents like benzene, ether, and chloroform.
However, they are insoluble in polar solvents like water. Their solubility decreases with increase in their
molecular weight.

3. Melting and boiling points are increases with molecular mass and decreases with No. of branches.
As far as melting point is concerned the alkane having even carbons has more M.P. than odd
carbons, since the intermolecular forces in a crystal depend not only upon the size of the molecules
but also upon how well they fit into a crystal lattice.
For e.g.
C C C C C C
> n- pentane n- hexane
C C C C C
(both methyl group (both methyl group
are directed on are adversely
same side) directed)
M.P. even > odd no. of atoms
4. Boiling point decreases with the increment of branches.
(n-Pentane is liquid but neo-Pentane is gas due to increase in branching, surface area decrease
therefore intermolecular forces & Vander Waals forces of attraction decreases).

5. Physical state:
Alkanes
C1– C4  Gaseous state
C5 – C17  Liquid state
(except neo-pentane)
C18 & above  Solid like wax

6. Alkanes are lighter than water, so it floats over water.

Chemical Properties :
1. Halogenation
Alkanes react with bromine or chlorine in the presence of sunlight or UV light or in dark at high temperatures
(250°C– 400°C) forming a mixture of substituted products. For example,
Cl2 Cl2 Cl2 Cl2
CH4 h CH3Cl h CH2Cl2 h CHCl3 h CCl4

The yield of monohalogenated product can be increased by using substrate (alkane) in excess.
The reactivity of halogens follows the order: F2 > Cl2 > Br2 > I2.
Direct fluorination is explosive and can be achieved by the action of inorganic fluorides on bromo or iodo
derivatives.
2C2H5Br + HgF2  2C2H5 F + HgBr2
Bromination is slower than chlorination and is carried out at higher temperatures.
Iodination is reversible and can be carried out sufficiently in the presence of strong oxidising agents like
iodic acid (HIO3) or nitric acid that destroys hydroiodic acid (HI) and shifts the equilibrium towards the
right.
CH4 + I2 l CH3 I + HI
5 HI + HIO3  3I2 + 3H2O
Alkyl iodides can be prepared conveniently by Finkelstein reaction which involves treating chloro or
bromoderivative with NaI in acetone or methanol solution.
acetone
R Cl + NaI RI + NaCl 
Alkyl chlorides or bromides cannot be prepared by this method of halide exchange because NaCl and
NaBr are insoluble in acetone.

Mechanism of Halogenation :
The mechanism involves the following steps:

(i) Chain-initiation step

250-400°C
X2 or UV light 2X•
(ii) Chain-propagation step
X• + R—H  H—X + R•
R• + X2  R—X + X•

(iii) Chain-termination step

X• + X•  X2
•
R +X •  R—X
•
R +R •  R—R
Radical inhibitors stop chain propagation by reacting with free radical intermediates. Example:
R• + •O—O •• •••  R— O—O
•• ••
•• ••
•• ••
.
Peroxy radical
2. Nitration.
Under certain conditions, alkanes react with nitric acid, a hydrogen atom being replaced by a nitro-group,
NO2. This process is known as nitration. Nitration of the alkanes may be carried out in the vapour phase
between 150° and 475 °C, whereupon a complex mixture of mononitroalkanes is obtained.
NO 2
HNO3
|
CH3CH2CH3 400
 CH CH CH NO + CH 3CHCH 3 + C H NO + CH NO
C 3 2 2 2 2 5 2 3 2

3. Sulphonation.
Sulphonation is the process of replacing a hydrogen atom by a sulphonic acid group, SO3H. Sulphonation
of a normal alkane from hexane onwards may be carried out by treating the alkane with oleum (fuming
sulphuric acid). It has been shown that in concentrated sulphuric acid, hydrocarbons containing a tertiary
hydrogen atom undergo hydrogen exchange (Ingold et. al. 1936). The mechanism is believed to occur
via a carbonium ion:
R3CH + 2H2SO4  R3C+ + HSO 4 + SO2 + 2H2O
R3C+ + R3CH  R3CH + R3C+ , etc.
This reaction is of particular interest since optically active hydrocarbons have been racemised in sulphuric
acid; e.g., Burwell et. al. (1948) have shown that optically active 3-methylheptane is racemised in sulphuric
acid.
CH 3 CH 3
| |
CH 3  C  H H 2SO 4
 CH 3  C  SO 3H
| |
CH 3 CH 3
Imp. Rate of sulphonation 3° > 2° > 1°.

4 Chlorosulphonation/ Reaction with SO2 & Cl2:

The reaction is also called as Reed’s reaction. When propane reacts with SO2 and Cl2 in presence of
ultraviolet light then propyl sulphonyl chlorides are formed.
U.V. Light
CH3-CH2-CH3 + SO2 + Cl2    CH3–CH2–CH2SO2Cl + HCl
This process is used in the commercial formation of detergents.
O O
CH 3 – (CH2)10 – CH2– CH2–S–Cl + NaOH Alkaline
    CH3 – (CH2)10 – CH2–S–O– Na+
hydrolysis
O O
(detergent i.e. sodium salt of sulphonic acid)
5. Isomerisation : Lower alkanes are not isomerised but butane or higher number of alkanes if heated with
aluminium chloride at high temperature then they convert into stable isomers by the rearrangement reaction.
Isomerisation is also held by heating alkane with- (AlX3 + HX; X = Cl, Br, I or Al2(SO4)3 + H2SO4) at
200ºC.

AlCl
3
 CH3  CH  CH3
400 º C |
CH3
n - butane iso butane
If we take n- heptane then it converts into most stable form Triptane (trivial name).

CH3-(CH2)5 –CH3 

n- heptane 2, 2, 3- trimethyl butane

If we take n - octane, iso-octane is formed.

CH3(CH2)6 CH3 

n- octane iso-octane
Reaction is used in Petroleum industries, with the help of this reaction, we can convert unbranched
alkane to branched alkane (i.e. lower octane number alkane to higher octane number alkane) or bad fuel
to good fuel.

6. Combustion :
Combustion is a rapid oxidation that takes place at high temperatures, converting alkanes to carbon
dioxide and water. Little control over the reaction is possible, except for moderating the temperature and
controlling the fuel / air ratio to achieve efficient burning.
CnH2n+2 + excess O2 heat
 nCO2 + (n+1) H2O
CH3CH2CH3 + 5 O2 heat
 3 CO2 + 4 H2O

7. Aromatization :
Cr2O3  Al2 O3
(i) n-Hexane  

CH3
Cr2O3  Al2 O3
(ii) n-Heptane  

(iii) n-Octane Cr

2O
3  Al2 O3
 O + M + P (xylene)

8. Cracking and Hydrocracking :
Cracking of large hydrocarbons at high temperatures produces smaller hydrocarbons. The cracking
process usually operates under conditions that give the maximum yields of gasoline. In hydrocracking ,
hydrogen is added to give saturated hydrocarbons; cracking without hydrogen gives mixture of alkanes
and alkenes.
Catalytic hydrocracking

H
2 , heat

catalyst
C12H26
long-chain alkane

Catalytic cracking

C12H26 H
2 , heat

catalyst
long-chain alkane

Knocking :
Knocking may be defined as the pre-ignition of fuel-air mixture in the cylinder aheadof the flame. It
reduces the efficiency of the engine and also causes damage to the cylinder and piston of the engine.
1. The extent of knocking depends on the quality of the fuel used. In other words, a fuel that produces
minimum knocking is considered as a good-quality fuel
2. The anti-knocking property or the quality of a fuel is usually expressed in terms of the octane number.
Octane Number :
Octane number is a scale that is used to determine the quality of a fuel.
It may be defined as the percentage of isooctane by volume in a mixture of isooctane and n-heptane that
has the same anti-knocking properties as the fuel under examination.
1. Straight-chain aliphatic hydrocarbons have a higher tendency to knock than branched-chain hydrocarbons.
2. Two pure hydrocarbons have been selected as standards :
(a) n-Heptane has poor anti-knocking property, therfore, arbitrarily it has been assigned an octane
number of zero.
(b) 2,24-Trimethylpentane (isooctane), a branched-chain hydrocarbon, has the highest anti-knocking
property. Therfore, it has been given the octane number of 100. The octane number of nonane

is –45 and that of 2,2,3-trimethylpentane , called as triptane, is 116.

n-Heptane, octane number = 0

2,2,4-Trimethylpentane or isooctane, octane number = 100

 Alkanes Octane number
CH4 122
C 2H 6 101
C 3H 8 96
C4H10 89

3. Straight-chain alkanes have very low octane numbers. As the length of the chain increases, the octane
number decreases.
4. Branched-chain alkanes have higher octane number. As the branching increases, the octane number
increases.
5. Cycloalkanes have higher octane number. As the extent of branching increases, octane number further
increases.
6. Unsaturated hydrocarbons, i.e., alkenes and alkynes, have higher octane numbers than the corresponding
straight-chain alkanes.
7. Aromatic hydrocarbons have very high octane numbers.
8. It has been found that gasoline obtained by creacking has a higher octane number than gasoline obtained
by direct distillation. This is because creacked gasoline contains higher precentage of alkenes, branched-
chain aliphatic hydrocarbons, and aromatic hydrocarbons.
The tendency of knocking decreases in the order :
straight-chain alkane > branched-chain alkane > alkenes > cycloalkanes > aromatic
hydrocarbon.
Anti-knocking agent or Gasoline additives :
Compounds that are added to gasoline and reduce knocking are called anti-knocking agents. The problem
of knocking in engines can be checked by using fuels of high octane number and by addition certain
compounds (anti knocking agents) that reduce knocking.
The best anti-knocking agent is tetraethyl lead (TEL) or Pb(C2H5)4. This is added to the extent of
0.001% in gasoline. Such a gasoline is called ethyl gasoline or leaded gasoline. In the cylinder of an
engine, TEL decomposes to produce ethyl radicals that combine with the radicals produced due to
irregular combustion. As a reslt, reaction chains are broken and smooth burning of the fuel occurs. This
prevents knocking.

Pb(C2H5)4 Heat
 Pb + 4C2H5 (Ethylradical)
The lead deposited in the cylinder is removed by adding 1,2-dibromoethane (ethylene bromide) that
decomposes into ehene (ethylene) and bromine. The bromine thus obtained combines with lead to form
lead bromide which, being volatile, is carried of from the engine by exhaust gases.
BrCH2CH2Br  CH2=CH2 + Br2
Pb + Br2  PbBr2 (Volatile)
Cetane Number :
1. Cetane number is a scale that is used to determine the quality of a diesel fuel.
Cetane number is the percentage of cetane by volume in a mixture of cetane and -methylnaphthalene
that has the same ignition properties as the diesel under examination in a test engien.
C6H34 (n-hexadecane), cetane number = 100 (ignites spontaneously)

-Methylnapthalene, cetane number = 0 (ignites slowly)

or
1-Methylnapthalene

2. LPG and CNG : The term LPG (liquified petroleium gas) refers to the mixture of hydrocarbons containing
three or four C atoms. This includes propane (C3H8), propene (C3H6), n-butane (C4H10), isobutane
(methylpropane), and various butenes with a little amount of ethane (C2H6). The major sources of LPG
are natural gas and refining and creacking of petroleium. Alkenes are mainly produced during creacking.
LPG is mainly used in the manufacturing of chemicals, as a fuel for households, and in the petro-chemical
industry.
CNG (compressed natural gas) is a highly compressed form of natural gas. Many vehicles are designed
to operate on CNG. Natural gas has an octane rating of 130.
 SOLVED EXAMPLE
Ex.1 The reaction conditions leading to the best yields of C2H5Cl are –
(A) C2H6 (Excess) + Cl2 u
.v.light
  (B) C2H6 + Cl2 dark
 room temp.
 
(C) C2H6 + Cl2 (Excess) u .v.light
  (D) C2H6 + Cl2 u .v.light
 
Sol. (A)
Chlorination of ethane takes place by free radical substitution reaction can be prevented by using excess
of C2H6.
C2H6 + Cl – Cl u
.v.Light
 C2H5Cl + HCl
(excess)

Ex.2 C2H6 + SO2 + Cl2 U.V.Light

  product. In this reaction product will be -
(A) C2H4 (B) CH3CH2Cl (C) CH3CH2SO2Cl (D) C2H2
Sol. (C)
Above reaction is Reed reaction and product is sulphonyl chloride.

Ex.3 Product of the Wolff-Kishner reduction is –

(A) Alkene (B) Alkyne (C) Alkane (D) Amine
Sol. (C)
In the Wolff-Kishner reduction carbonyl compound is converted into alkane by intermediate hydrazone.
Ex.4 Which of the following should be subjected to Wurtz reaction to obtain the best yield of n-hexane?
(A) Ethyl chloride and n-butyl chloride
(B) Methyl bromide and n-propyl bromide
(C) n-Propyl bromide
(D) Ethyl bromide and n-butyl bromide
Sol. (C)

Ex.5 2-Methylbutane on reacting with bromide in the presence of sunlight gives mainly -
(A) 1-Bromo-3-methylbutane (B) 2-Bromo-3-methylbutane
(C) 2-Bromo-2-methylbutane (D) 1-Bromo-2-methylbutane

Sunlight
Sol. (C) + Br
2   .

Ease of substitution of H atom is 3º > 2º > 1º.

Ex.6 Which of the following alkyl bromides may be used for the synthesis of 2, 3-dimethylbutane by Wurtz
reaction ?
(A) n-Propyl bromide (B) Isopropyl bromide
(C) Isobutyl bromide (D) s-Butyl bromide
Sol. (B)
 REACTION CHART FOR ALKANES
GMP GR
H , Ni
(1) R–C  CH 2  (1) X , h or UV light or 400 C
2         
 RX
200300C
or
R–CH=CH2 Sabatier senderens
(2)  R-N
Nitration
reaction

(2) R–X Zn

Cu  HCl
 Sulphonati on H S O
Re d P  Hi , LiAlH4 (3) 2
       2
7  Alkyl Sulphonic acid

(3) RX Na
,dry ether

Wurtz reaction Reed reaction
(4) SO2 + Cl2    RSO2Cl
h
(4) RX  Zn

Frankland 's reaction
R–H
AlCl / HCl
or (5) 3 branched alkanes
(5) RX  R 2CuLi
  Isomerisation
( Corey  House reaction
) R–R

 HOH or ROH
or
(6) R–Mg–X      CnH2n+2 (6) Pyrolysis
 
or NH or RNH
3 2 500 700 C
Alkenes + CH4 or C2H6

Cr or Mo or V oxide
       
(7) R–OH, R–CHO Re d P / HI (7)  Al2O 3 500C Aromatic compound
 
R  C  R , RCOCl, RCOOH
|| CH N
2
  2

O (8) step up reaction
Higher alkane

(8) R CR Zn

Hg / Conc. HCl
 
|| Clemension 's reduction O
2
(9) 
 CO2 + H2O
O 
Combustion
H 2 N  NH 2
(9)        

R CR Wolf / Kishner reduction
||
O
or
(RCH2CH2)3B H O
2

(10) RCOONa NaOH  CaO



(11) RCOONa Kolbe

 'selectrolytic synthesis
  
 ALKENE

1. GENERAL INTRODUCTION
 Hydrocarbons that contain a carbon–carbon double bond are called alkenes.
 Each double-bonded carbon of an alkene has three sp2 orbital of another atom to form a  bond.
 Thus, one of the carbon–carbon bonds in a double bond is a  bond, formed by the overlap of an sp2
orbital of one carbon with an sp2 orbital of the other carbon.
 The second carbon-carbon bond in the double bond (the  bonds) is formed from side-to-side overlap
of the remaining p-orbitals on the sp2 carbons. Because three points determine a plane, each sp2 hybridized
carbon and two of the three atoms bonded to it are in a plane.
 In order to achieve maximum orbital-orbital overlap, the two p-orbitals must be parallel to each other.
Therefore, all six atoms of the double bond system are in the same plane. A molecular orbital description
of a carbon-carbon double bond.
 It is important to remember that the  bond represents the cloud of electrons that is above and below the
plane defined by the two sp2 carbons and the four atoms bonded to them.

H3C CH3
C—C
H3 C CH3
p orbitals overlap
to form a  bond

 Because the two p orbitals that form the  bond must be parallel to one another to achieve maximum
overlap, rotation about a double bond does not occur. If rotation about a doubel bond did occur, the
two p-orbitals would ceases to overlap and the  bond would be destroyed.

 bond broken
H3C CH3 H3C H
H3C
C—C H C—C
C—C
H H CH3 H CH3
H
cis isomer trans isomer

2. Characteristics of Alkene :
(a) General formula : CnH2n.
(b) Functional group : in alkenes is >C=C<
(c) The double bond is made up of one sigma and one pi bond.
(d) The doubly bonded carbon atoms are sp2 hybridized.
(e) Geometry of unsaturated ‘C’ carbon is trigonal planar.
(f) C = C bond length is 1.34 Å.
(g) C–H bond length is 1.10 Å.
(h) C = C bond energy is 143.1 k cal mol–1.
(i) C – H bond energy is 98.69 k cal mol–1.
(j) Alkene shows chain, position, functional (ring chain), optical & geometrical isomerism.

ALKENE [17]
NOTE : Cumulated polyene can exhibit geometrical as well as optical isomerism. for eg. Isomers of C4H8:
(i) Chain isomers  CH3 – CH2 – CH = CH2 & CH3  C  CH2
|
CH3

(ii) Position isomers  CH2 = CH – CH2 – CH3 & CH3 – CH = CH – CH3

(iii) Functional isomers (ring chain isomerism)

CH2  CH2
CH3 – CH2 – CH = CH2 & | |
CH2  CH2
1-butene cyclobutane

(iv) Geometrical isomers  (v) Optical isomerism 

&

cis -2-butene Trans-2-butene

3. METHODS OF PREPARATION OF ALKENES

3.1 ELIMINATION REACTIONS

Elimination reactions consist in removing the two groups (generally one being a proton and other is
leaving group from one or two carbon atoms of a substrate to form an unsaturated linkage.
Elimination reactions are classified under two general categories.

TYPES OF ELIMINATION REACTIONS

1. -elimination (1, 1)
2. -elimination (1, 2)
1. -elimination : When the two atoms or groups are eliminated from the same carbon, the process is
called –elimination.
2. -elimination : When the two groups or atoms are removed from the two adjacent carbon atoms, the
process is known as -elimination.
SOME IMPORTANT TERMS
(1) Substrate : That molecule or ion which undergoes change is called substrate. Substrate may be alkyl
halide, alcohol etc.
(2) Base : It is negative ion or neutral molecule which abstract the proton from the substrate.
(3) Leaving Group : It is an ion or molecule which leaves the substrate with a pair of electron. The tendency
of leaving group is directly proportional to strength of its conjugated acid. Leaving group ability of
different halide ions follows the sequence
I> Br> Cl> F

ALKENE [18]
-ELIMINATION CAN OCCUR BY THREE DIFFERENT MECHANISM
(1) E-1 mechanism
(2) E-2 mechanism
(3) E1 cB mechanism

CHARACTERISTICS OF E-1 MECHANISM :

(a) It is called unimolecular elimination reaction.
(b) It takes place in two steps.
Ist Step : Departure of leaving group from a molecule to form carbocation.

IInd Step : Carbocation formed above loses a proton to the base and forms the alkene.

(c) Ist step of E-1 mechanism is the rate determining step (R.D.S.)
(d) Since in R.D.S. only substrate undergoes covalency change.
i.e. it follows first order kinetics with respect to substrate
Rate = K [substrate]
(e) Since in R.D.S. carbocation is formed as an intermediate so the reactivity order of different substrate
follows the stability of order of carbocation formed.
Allylic > Tertiary > Secondary > Primary
(f) Loss of leaving group is the part of R.D.S. therefore the order of reactivity of alkyl halide in elimination
reaction is
R–I > R–Br > R–Cl > R–F
Above order is parallel to order of leaving group ability
I> Br > Cl > F
(g) E-1 mechanism does not show isotopic effect as the loss of hydrogen is not the part of R.D.S.
KH
1
KD
(h) Since the ionization take place in step I. Therefore polar solvent favors the E-1 mechanism.
(i) E-1 mechanism involves carbocation as an intermediate so rearrangement of carbocation can take if
possible. More stable carbocation forms the major product in the reaction.
E-2 MECHANISM
(a) It is known as bimolecular elimination mechanism.
(b) It take place in single step.
(c) It involves the formation of transition state

H   B....H 
| |  | |  |
Slow
 C C     C  C   – CC –
B |
| |  | | 
  
L  L 

Transition State

ALKENE [19]
(d) Formation of transition state is the rate determining step.
(e) Rate of reaction depends on the concentration of substrate as well as of base i.e. it follows the second-
order kinetics.
Rate = [Substrate] [Base]
(f) E-2 mechanism shows the isotopic effect as the loss of hydrogen is a part of R.D.S.
KH 7

KD 1
(g) E-2 mechanism shows the halogen effect as loss of leaving group is a part of R.D.S.
 Reactivity of different alkyl halides follows the sequence
R – I > R – Br > R – Cl > R – F
Reactivity order of different substrate follows
3º > 2º > 1º
(h) The rate of reaction increases with increasing strength and concentration of the base.
(i) Non-polar solvent favors the E-2 mechanism.
(j) Since E-2 mechanism involves the transition state ; rearrangement is not possible.
E1 cB MECHANISM
(a) In this mechanism, the reaction take place in two steps.
(i) The first step involves rapid removal of a proton from the -carbon (carbon adjacent to the
carbon bearing halogen atom) forming a carbanion.
(ii) The carbanion formed then loses the halides ion in the second rate determining step.
B......H
| 

R  C H  C H2 R  C H  C H 2  C 2 H 5OH
| |
Br Br

Slow
 R – CH  CH 2  Br 

(b) The overall rate of the reaction is limited to slower second step & hence the rate of reaction depends
only on the concentration of carbanion.
(c) Since carbanion is the conjugate base of the alkyl halide and rate of reaction depends on concentration
of carbanion hence the mechanism is designated as E1cB mechanism.
MECHANISM OF –ELIMINATION
-elimination does not occur frequently and completes in two stages, the second of which is rate
determining.
for eq.
CHCl3 + OH– 
CCl3 + H2O

CCl3 :CCl2 + Cl–
dichlorocarbene
Another example of -elimination reaction is formation of diphenylacetylene from 2, 2-diphenyl vinyl
bromide and sodamide.

N H2
 C6H5–CC–C6H5

ALKENE [20]
3.2 From Alkynes
H
|
Lindlar sCatalyst
R – C  C – H + H2 
PdBaSO4
 R  C  C H
|
H
poison of catalyst such as BaSO4, CaCO3 are used to stop the reaction after the formation of alkene,
otherwise alkanes are formed.

Mechanism
(i) The reaction takes place at the surface of Pd, that is why it is cis addition and the product
is cis form eg.
H /Pd
2
R – C C –R  
 BaSO4
(ii) Alkyne can be reduced to trans alkene by using Na + NH3, or Li AlH4

R – C  C – R Na

NH3 liq.
Terminal alkynes are not reduced by the Na – NH3 untill presence of (NH4)2SO4.

3.3 From mono halides

When mono halide react with alcoholic KOH or NaOH then respective alkenes are formed
H
H H
| | |
(i) R C  C H+  
Alc.KOH –HX R  C  C H
| | |
H X H

Alkene

(ii) CH3 – CH2 – C H – CH3  

| NaOH alc.
Cl

NOTE : CH3–CH2CHX–CH3  CH3–CH2–CH=CH2  CH3 – CH =CH – CH3

Minor Major
CH3—CH = CH —CH3 (Trans) is the major product because it is more stable as having
six ‘  ’ Hydrogen in the comparison to CH3 –CH2 – CH=CH2 having only two ‘  ’ hydrogen.
This is in accordance with the saytzeff rule.

Mechanism : E2 and E1 are possible.

E 2 Mechanism :
Cl Cl
|
C —C
Slow
C— C – Cl– C= C
| – OH – H2O
H H
T.S. OH
As molecularity of slowest step is two i.e. E2 mechanism.

ALKENE [21]
 E1 Mechanism :-
CH2  H CH2
| ||
Cl
CH3  C  Cl  Fast CH3 — C
  
|  Slow – H |
CH3 CH3
As molecularity of slowest step is one. That is why it is termed as E1.
Remember : Possibility of E1 increases with increasing stability of carbo cation . similarly possibility
of E2 increases with decreasing stability of expected carbocation.
3.4 From Dihalides
3.4.1 From gem dihalides : When gem dihalide is heated with Na in ether then higher alkenes are formed.
X X Zn
R–CH ++ CH–R  R –CH = CH–R
– 2 Znx
X Zn X 2
Cl Zn Cl
CH3–CH + + CH–CH3     CH3–CH=CH–CH3
Cl Zn Cl  2 ZnCl2
2 – butene
Conclusion – If we take two different types of gemdihalides then we obtain three different types
of alkenes.
Note : The above reaction is used in the formation of symm. alkenes only, because if we take two
different types of halides then mixture of alkenes is obtained so the yield of an individual alkene is
reduced and it is improper to separate each alkene from the mixture because the difference of boiling
points in alkenes is very less.

3.4.2 From vicinal dihalides :

When vicinal dihalides are heated with Zn dust, alkene of same no. of carbon is obtained.
H H H H
| |  | |
R  C  C  H  Zn dust  R  C  C  H  ZnX
2
| | 300º C
X X

H H H H
| | | |
H  C  C  H  Zn dust  H  C  C  H  ZnCl2
| |
Cl Cl
Note : Alkene is not formed from 1,3 dihalides. Cyclo alkanes are formed by dehalogenation of it.
For eg.

Zn dust
CH2  CH2  CH2   
| |
X X
3.5 From Alcohols
When alcohol is heated with conc. H2SO4 at about 160º C, alkenes are formed after dehydration.
H H H H H H
| |  | |  160 ºC | |
R C  C H + H – SO4 – H   R  C  C  H   R  C  C  H
| | 160ºC | |  H2SO 4
H OH H SO 4H
(Alkyl hydrogen sulphate) Alkene

ALKENE [22]
 
CH3 CH2 CH2 CH2 OH + H2 SO4    CH3 –CH2 – CH =CH2 + CH3 –CH=CH–CH3
160ºC
1–butene 2–butene
(major product)

H OH H
| | | 
CH3 – CH– CH  CH2 + H2SO4   CH3 –CH = CH – CH3 + H2SO4 + H2O
160ºC
2–butene
(chief product)

OH
|
CH3  C  CH3 
+ H2SO4  
 CH3  C  CH2 + H2SO4 + H2O
| 160ºC |
CH3 CH3
Iso butylene

Mechanism :-

H O
2
 

+
(+) or (–) –H – H+
H
|
CH3  CH  CH  CH2 CH3–CH=CH–CH3

 Trans 
(Minor)  
 major 
Note : Since the mechanism proceeds via- C+ ion therefore rearrangement is also possible eg.:
CH3
H2SO4 Conc. |
  + CH3 — C — CH  CH2
160 º
|
CH3
(Major) (Minor)
Negligible

Mechanism :

..
CH3:OH CH3 CH3
| | + | |
CH3 — C — CH — CH3 H CH3– –CH3  H
– H2O   CH3 — C — C — CH3
|
CH3

ALKENE [23]
Characteristics of dehydration of alcohol
(1) Since the carbocation is formed as an intermediate in the R.D.S. of dehydration of alcohol The relative
reactivity of alcohols decreases in the order.
Tertiary > Secondary > Primary
(2) Since the R.D.S. is reversible it is necessary to remove the H2O molecule produced in the reaction.
Therefore in experimental condition conc. H2SO4 is added in a regular interval.
(3) For dehydration, different dehydrating agent like Al2O3, ThO2 can also be employed.
(4) Rearrangement of carbocation can take place.
(5) –OH is not a good leaving group, in alcohols so turn it into a good leaving group, it is converted into –
H2O by protonation
(6) H2O is a good leaving group.

Rearrangement in Alcohol Dehydration

(1) Some alcohols undergo dehydration to form alkene having carbon skeleton different from those of the
starting alcohol.
(2) This is due to rearrangement of carbocation formed in the reaction to more stable carbocation.
(3) An example of alcohol dehydration that is accompanied by rearrangement is.

CH 3 CH 3
| |
2 H SO
CH 3  C  C H  CH 3  4  CH 3  C CH  CH 2 +

heat
+
| | |
CH 3 OH CH 3
(3%)
(4) Percent yield of alkenes formed from rearranged carbocation is greater than the percent yield of
alkene obtained from unrearranged carbocation.
(5) Rearrangement of carbocations can also lead to a change in ring size, as the following example shows

 
H,
heat
   H


(  H 2O )

Regioselectivity in alcohol dehydration :

1. In alcohols such as 2-methyl-2-butanol, dehydration can occur in two different directions to give alkenes
that are constitutional isomers.
2. More substituted alkene forms the major product and is called Saytzeff alkene or Zaitsev alkene.
3. Less substituted forms the minor product and is called Hofmann’s alkene.
OH
| H SO
CH 3  C CH 2  CH 3 2 
4  +
|
CH 3
(Minor) (Major)
3.6 Kolbe’s Synthesis
When aqueous solution of K or Na succinate is electrolysed, ethylene is released at anode.
O O
||  ||
CH2  C  O K CH2  C  O . .
|   | + 2K+
CH2  C  O K CH2  C  O . .
|| ||
O O
ALKENE [24]
 At Anode
2
CH COO
| 2  CH2 COO•
  – 2e –
 |  CH2 = CH2 + 2CO2
CH2COO CH2 COO•
unstable
At Cathode
2K + + 2e –  2K
2K + 2H2O  2KOH + H2
Note – If we use methyl succinic acid as reactant then propylene is formed.

3.7 From Esters

When esters are heated in presence of liq. N2 and glass wool, then alkyl part of ester converts
into respective alkene while alkanoate part of ester is converted into respective acid.
CH3 — CO — O H
| | Glass wool 450 º
   CH3–COOH + CH2=CH2
CH2 — CH2 liq. N2

The reaction is called as ‘Pyrolysis of Ester’.

Mechanism : Pyrolysis of Ester is a type of E1 elimination which proceed via cyclic T.S. that is
why the product becomes ‘cis’

CH2=CH–CH2–CH3+ CH3–CH2– C –O–H

||
O
Cyclic Transition state
* It is interesting to note that, in this case major product is 1- alkene in the comparision to 2-alkene.
O
||
H O — C — CH3
| |
CH3 — CH — CH — CH3  CH3 —CH2—CH =CH2 + CH3–CH=CH–CH3

(Major) (minor)

3.8 From quaternary ammonium hydroxide

When quaternary ammonium hydroxide is heated strongly it decomposes to give alkene. eg.
+
H


(i)
CH2–CH2 OH  CH2 = CH2 + CH3 – N  CH3 + H2O
|
CH3
CH3 CH3 CH3
(ii) N H  CH – CH –CH=CH + CH – CH = CH – CH
3 2 2 3 3
| |
CH3— CH2 — CH — CH2
(major) (minor)
This elimination is called Hoffman’s elimination. It is markable that in this reaction more acidic
H is preferably eliminated.

ALKENE [25]
3.9 The wittig reaction
In this reaction methylene triphenyl phosphorane is reacted with carbonyl compound to give alkene.
eg. :
(i) (Ph)3 P = CH2 + CH – R (Ph)3 P=O + R – CH
|| ||
O CH2

O
||
(ii) (Ph)3 P=CH –R + CH – R  (Ph)3 P =O + R – CH =CH – R

3.10 The Cope Elimination :

Tertiary amine oxide undergoes the elimination of a dialkylhydroxyamine when they are heated. This
reaction is called the Cope elimination.

CH 3 CH 3
| |
 
Re action : CH 3  CH 2  CH 2  N  CH 3  CH 3  CH  CH 2  N  CH 3
| |
O OH

Mechanism :
CH 3
|

 CH 3  CH  CH 2  N  CH 3
|
OH

SOLVED EXAMPLE

Ex. 1 The minimum number of C atoms required to be present in an optically active alkene are:
(A) 4 (B) 6 (C) 8 (D) 10 (Ans.B)

Sol. The minimum number of C atoms required to be present in an optically active alkene are 6.

H
|
e.g. CH3  C  CH  CH2
|
CH2  CH3

Here the central carbon atoms is an asymmetric carbon atom.

Ex.2 Chloroethane reacts with alcoholic KOH to form –
(A) Propyl chloride (B) Ethyl chloride
(C) Ethene (D) None (Ans.C)
H H H
| | |
Sol. H  C  C  H +Alc.KOH   H C  C H
HX
| | |
H Cl H
ethene

ALKENE [26]
Ex.3 The synthesis of ethene from electrolysis of an aqueous solution of potassium succinate is known
as:
(A) Faradays electrolysis (B) Kolbe - Schmidt reaction
(C) Hoffmann’s rearrangement (D) Kolbe’s electrolytic Synthesis
(Ans. D)
Sol. The synthesis of ethene from electrolysis of an aqueous solution of potassium succinate is known
as Kolbe’s Electrolysis synthesis. The reaction takes place as follows :


–
CH 2COO
| At anode
–2e   +2CO2
CH 2COO – 

2K+ + 2e  2K (at cathode )

2K + 2H2O  2 KOH + H2

H Cl
| | NaOH
Ex.4 CH2 — CH2   CH2 = CH2. Most probable mechanism for this reaction is -
alk.
(A) E 1 (B) E 2 (C) E1CB (D)  elimination

Sol. (B) 1º halide generally gives E 2 mechanism.

Ex.5 RCH = CH2 can be obtained by :

O
||
(A) R  C  H and (C6H5 )3 P = CH2 (B) By heating RCH2CH2N(CH3)2

 
(C) By heating RCH2CH2OCOCH3 (D) All of these (Ans. D)

Sol. RCH = CH2 can be obtained by all above reagents as follows :

O
||
(A) R  C  H and (C6H5 )3P=CH2  PO(C6H5)3 + RCH=CH2

(B) RCH2CH2 N (CH3)2 Cope    RCH = CH + (CH ) NOH
Re acn. 2 3 2

O
(C) RCH2CH2OCOCH3   RCH=CH2 + CH3COOH


Ex.6 If we take ethylidene chloride and isopropylidene chloride with zinc dust then product will be –
(A) 2-butene
(B) 2-butene + 2,3-dimethyl -2-butene
(C) 2-methyl -2-butene
(D) 2-butene, 2-methyl -2-butene, 2,3-dimethyl -2-butene (Ans. D)

Cl CH3
Sol. C     CH3  CH  C  CH3
2ZnCl2
Cl CH3 |
CH3
2-methyl-2-butene

ALKENE [27]
    CH – CH = CH – CH
2ZnCl2 3 3

2-butene
CH 3 Cl Zn
C + +    CH3  C  C  CH3
2 ZnCl 2
CH 3 Cl Zn | |
CH3 CH3
2,3-dimethyl-2-butene

Ex. 7 Identify ‘Z’ in the following reaction series,

Al2 O 3
NaOH( aq) Cl /H O
CH3 CH2 CH2Br   2
 ( X )  ( Y ) 2
 (Z)
heat

(A) Mixture of CH3  CH  CH2 and CH3  CH  CH2

| | | |
Cl Cl OH Cl

(B) CH3  CH  CH2

| |
OH Cl

(C) CH3  CH  CH2 (D) CH3  CH  CH2 (Ans.B)

| | | |
Cl OH Cl Cl

Al3 O 3 HOCl
Sol. CH3CH2CH2Br NaOH( aq.)
 CH3CH2CH2OH    CH CH = CH  CH3 CHCH2 Cl
3 2  |
OH
Cl
| Zn/ dust
Ex.8 CH3 – CH – CH – CH3   CH3 –CH =CH –CH3.
|
Cl
This 2-Butene is -
(A) Cis 2-Butene (B) Trans -2-Butene
(C) Depend upon reactant (D) Recemic mixture

Sol. (C)
Cl
|
CH3 – CH – CH – CH3 may be of two types. It may be asymmetrical or meso and they can give
|
Cl
different compounds.
Cl Cl
| | CH3 CH3
CH3 — CH — CH — CH3  Zn

dust
 C=C
H H
Cl Cl
| | CH3 H
Zn
CH3 — CH — CH — CH3  dust C =C

H CH3
meso

ALKENE [28]
4. PHYSICAL PROPERTIES
(i) From C2–C4 they are colourless, odourless gases, from C5–C17 they are colourless liquids, C18
onwards alkenes are solids.
(ii) Alkenes are practically insoluble in water because they cannot form hydrogen bonds with H2O
molecule. They dissolve freely in organic solvents like benzene, chloroform, CCl4 petroleum
ether, etc. ( Like dissolves like )
(iii) B.P and M.P. decreases with increasing branches in alkene.
(iv) The melting points of cis isomers are lower than trans isomers because cis isomer is less
symmetrical than trans. Thus trans packs more tightly in the crystal lattice and hence has a
higher melting point.
(v) The boiling points of cis isomers are higher than trans isomers because cis–alkenes has
greater polarity (Dipole moment) than trans one.
(vi) These are lighter than water.
(vii) The increase in branching in carbon chain decreases the boiling point among isomeric alkenes.

5. CHEMICAL PROPERTIES
The main reactions of alkene are electrophilic addition reaction and the mechanism is based on ionic.
It was proved by Francis.

Thus the electrophilic addition on alkene is trans addition.

6. FRANCIS EXPERIMENT
According to Francis first attacks electrophile on the olefinic bond
CCl4
CH2 =CH2 + Br –Br    CH2  CH2
| |
Br Br
NaCl

CH2  CH2 CH2  CH2

| | | |
Br Br Br Cl

7. CHEMICAL REACTIONS
7.1 Reaction with Hydrogen –

Mechanism : The reaction takes place at the surface of Ni, therefore the addition is cis addition.
eg :-

ALKENE [29]
 CH3 CH3 H2 / Ni
C =C   
D D 

CH3 D H /Ni
2
C=C  
D 
CH3

enontriomeric pair
Note : Hydrogenation (catalytic hydrogenation) of alkene is a cis addition and is an exothermic
reaction. Thus the heat evolved decreases with increasing stability of alkene.
Remember, Stability of alkene depends upon hyperconjugation and type of geometrical isomerism
7.2 Reduction of alkene via hydroboration
(i) Alkene can be converted in to alkane by hydroboration followed by protolysis

BH2
| H / H2O
R—CH — CH2 H–BH
 2  R—CH2— CH2  
  R—CH2—CH3

This reaction is also represented as

H—BH2 H /H O
2
RCH = CH2   (R – CH2 –CH2)3B   3R – CH2 – CH3 + H3BO3
(ii) Alkene can be converted into alkane by hydroboration followed by treatment with AgNO3 + NaOH.
This method gives coupling.

6CH3 –CH2 –CH2 – CH2 B

2H
6  2[CH – (CH ) –
3 2 2
CH2]3B

CH3–CH2–CH2– CH2–CH2– CH2– CH2–CH3

7.3 Birch Reduction

A terminal double bond may be reduced by sodium in liq. NH3 in the presence of alcohol. This
method is known as the Birch Reduction and is believed to proceed Via an anionic free radical
Na
R – CH =CH2  
 R –
e
– CH2Et
— O —H

Na Et – O–H
R– – CH3   R – CH – CH3 
e
  R –CH2 –CH3
7.4 Halogenation –
In presence of polar medium alkene form vicinal dihalide with halogen.
H H H H
| | | |
CCl4
R  C  C  H  X  X   R  C  C  H
| |
X X
Vicinal dihalide
Order of reactivity of halogens is : F2 > Cl2 > Br2 > I2

ALKENE [30]
 Mechanism : It is an electrophilic addition by molecular attack in which the addition takes place
in trans manner.
–Br
| Br
Br |
C — C Br Br
  C—C
C——C |
Br
that is why-

CH3 CH3 Br
2
C=C 

CCl4
H H

cis

Br
2

CCl4

Note : For suitable condition

(i) Cis addition on cis form results meso (ii) Cis addition on trans form results (±)
(iii) Trans addition on cis form results (±) (iv) Trans addition on trans form results meso.

7.5 Reaction with HX (Hydrohalogenation) –

H
|
C =C + H X   C– C
|
X
alkene alkyl halide

7.5.1 Markovnikoff’s Rule

When an unsaturated unsymmetrical hydrocarbon reacts with HX then halogen goes on that unsaturated
carbon which has minimum number of hydrogen atom. Mechanism of the reaction is based on — C — .
|

CH3 – CH = CH2 H–X



Mechanism :

H X–
CH3 –CH=CH2  
 (Minor)
less stable

H X–
CH3 –CH=CH2  


more stable major

(i) Formation of Carbocation is Rate determining step.

(ii) Rearrangement of carbocation to more stable carbocation can occur to form a more stable product.

ALKENE [31]
(iii) Regiochemistry - Markownikoff’s Addition of HX.
Addition of HX on unsymmetrical alkenes (R–CH=CH2) takes place according to Markownikoff’s rule
which states that, “the negative part of abbendum is added on the carbon atom carrying lesser number of
hydrogen atoms”.
(iv) Reactivity order of different hydrogen halide towards addition
HI > HBr > HCl > HF

7.5.2 Anti Markownikoff’s Principle / Kharasch Effect / Peroxide Effect –

To understand antimarkownikoff’s principle let us consider the following reactions
H H
| |
CH3 – CH = CH2 + HBr  CH3  C  C  H
| |
Br H
(Markownikof f s rule)
H H H H
| | | |
Peroxide
CH3  CH  CH2  HBr     CH3  C  C  H  CH3  C  C  H
| | | |
Br H H Br
(minor) (major)
(Anti markownikoff’s rule )
It is based on free radical mechanism.

(a) Mechanism Concept –

(i) R – O – O – R  RO• + •OR
Peroxide
(ii) RO• + H – Br  ROH + xBr
H H H H
| | | |
(iii)   CH3  C  C  H + CH3  C  C  H
| • • |
Br Br
minor 20% major 80%
H H H H
| | | |
(iv) CH3  C•  C  H  H Br   CH3  C  C  H  xBr
| | |
Br H Br
×
(v) Br + ×Br  Br2
Note : It is interesting to note that anti markovnikov addition in the presence of peroxide is not applicable for
HCl and HI
(i) In the case of H – Cl, the step
Cl Cl H
|  | |
CH2 – CH2 H— Cl
 CH2 — CH2 + Cl


is endothermic (as  H = + 12.6 KJ mol–1)

ALKENE [32]
 (ii) In the case of H – I, the step
I
 |
CH2 = CH2 + I  CH2 – CH2 is
endothermic (as H = + 46 KJ mol–1)
But in the case of HBr both of the steps are exothermic, which results spontaneous reaction.
* CCl4, CBrCl3 etc. can also be added to alkene in anti markovnikov manner.
7.6 Reaction with Hypohalous Acid –
 
R – CH=CH2 + HO – Cl 

(i) When chlorine water or bromine water is used.

Cl Cl
Cl Cl
| |
R—CH — CH2    R — CH — CH2   R — CH — CH2
 Cl ¯ | H |
 O —H OH
|
H

(ii) When aq. solution of HOCl is added in the presence of strong acid.

Cl
|  H2O – H
CH2 — CH — R    

Cl
| Zn dust
CH2 — CH — R     C—C
| CH3COOH
OH
halohydrine
7.7 Reaction with H2SO4 –
(a) With conc. H2SO4
H H H H
| |   | |
R ——
 C — C — H  H — HSO 4  R — C — C — H
| |
SO 4H H
alkyl hydrogen sulphate
(b) With dil. H2SO4
4 H– SO H
R–CH=CH2 
H2O


Mechanism:

–H
R — CH — CH3   R — CH — CH3
| |
O — H OH
|
H

ALKENE [33]
* The hydration of alkene is not stereoselective, as in the case of HCl addition. This fact can be
explained on the basis of classical carbon cation formation.

* Since hydration proceeds via carbocation intermediate therefore rearrangement is always probable.


H
 

CH3 CH3 CH3 CH3

| | | |
– H
CH3 — C — C — CH3   CH3 — C — C — CH3
| | | |
H — O H OH H
|
H

* Rearrangement, however, may be avoided by treatment of alkene by oxymercuration-demercuration

method

CH3 CH3 HgOOC — CH3 CH3 H

| Hg(OOC—CH3 )2 | | | |
 Hg
CH3 — C — CH — CH2 
THF
 CH3 — C ——CH —— CH2     CH — C — CH — CH
3 2
| | | NaBH 4 / NaOH
CH3 O — C — CH3 | |
CH3 CH3 OH
||
O

7.8 Addition of Nitrosyl halide –

Alkene with nitrosyl bromide or nitrosyl chloride (Tillden reagent) react according to Markovinikoff’s
rule to give alkene nitrosobromide and alkene nitrosochloride respectively.
 
R – CH + O= N – Cl  

 
R–CH + O= N – Br  

 
CH3 – CH + O= N – Cl  CH3 – CH  CH2
| |
Cl NO
propelene nitrosochloride
7.9 Oxidation
Oxidation is completed by the following ways.

ALKENE [34]
7.9.1 With Acidic KMnO4 / Hot KMnO4
H H O
| | ||
R C  C H  O acidic
 R  C  O  H  CO 2  H2O
KMnO4
H H
| | O
H  C  C  H 
 2CO 2  2H2 O
ethene
H
[ O]
|
  H  C  C  O  CO 2  H2 O
| |
H OH
propenes acetic acid
H H
| |
[ O]
CH3  C  C  CH3   2CH3COOH
2-butene
[ O]



isobutylene

[ O]



7.9.2 With alkaline KMnO4 / Bayer’s reagent

H H
| |
Alk.KMnO
  4
R  C  C  H  O  H  OH  OH

glycol
Mechanism: Hydroxylation by alkaline KMnO4 (Bayer’s reagent) is cis addition and the mechanism
is cyclic.

H2O /H
+     

Similarly

KMnO4 / OH
  

ALKENE [35]
7.9.3 Hydroxylation by OsO4 :

+    H / H2O
  

¯
OH
+ OsO4  

7.9.4 Ozonolysis :-
This is the two step reaction
(i) Ozonide formation
(ii) decomposition of ozonide (reductive hydrolysis)
eg:-

O3 H2O / H / Zn
C=C  
I
  
II
 ZnO +

Remember : Ozonolysis gives oxidative cleavage of alkene to form two carbonyl group for each
. Whether it is in acyclic or cyclic or in aromatic compound. eg :-
O O O O
|| || || ||
(a) CH2 =CH – CH = CH2 Ozonolysis
 H — C — H + H — C — C — H  H — C — H

(b) Ozonolysis
 

Trans

7.9.5 Epoxidation by O2 /Ag

(i) CH2=CH2 + ½O2  Ag



(ii) CH3 – CH = CH2 + ½ O2  Ag

  CH3—
7.9.6 Epoxidation by performic acid or
perbenzoic acid

(i) CH2 = CH2

ALKENE [36]
 O
||
H— C — O — O —H
(ii) CH3 – CH = CH2   CH3—
O
||
It is observed that CF3 C – O – OH is an excellent reagent for epoxidation

R–CH = CH – R CF3 COOOH

  R– –R

7.10 Hydroboration
Alkene with borane hydride form an important compound called trialkyl borane.
3R–CH=CH2 + BH3  ( R – CH2 – CH2)3B

(a) Trialkyl borane is an important compound because it gives respective alkane on acidic hydrolysis.

HCl


(b) It gives respective alcohol on alkaline hydrolysis [Hydroboration oxidation (HBO)]

OH
 


(c) It gives 1º amine on alkaline hydrolysis with chloramine

 3R – CH 2 CH2NH2 + 3NaCl + H 3BO3

AgNO3
(d)    3R – CH2 – CH2 – CH2 – CH2 – R
NaOH

(e) In the overall hydroboration-oxidation reaction, three moles of alkene react with one mole of BH3 to
form three moles of alcohol.
(f) Because carbocation intermediates are not formed in the reaction carbocation rearrangement do not
occur.
(g) Regiochemistry Anti-markownikoff’s Addition of water.
(h) Alkyl boranes undergo coupling by means of silver nitrate in the presence of NaOH at 25°C. Higher
alkanes are the products of this reaction.

ALKENE [37]
7.11 Hydroformylation / Reaction with CO and H2
H H
CoH( CO)4 | |
R – CH = CH2 + CO + H2    R C  C H
| |
H CO
|
H

Note : If CO + H2O is taken then respective acid is formed

R – CH = CH2 + CO + H2O CoH(CO)4
 

The above reaction is also called as ‘Oxo reaction’ or ‘Carbonylation’.

7.12 Addition of formaldehyde :– In the presence of dil. mineral acids alkene react with formaldehyde
to give 1, 3-diol or cyclic acetal. Mechanism, can be given as follows :
  
RCHCH2 
H2C=O + H [H2C = O H  H2 C –OH]    R– C H–CH2–CH2–OH


HOH HCHO / H
R–CH–CH2–CH2 
 

   
H | |
OH OH

1, 3-diol cyclic acetal

 
HCHO / H O HCHO / H
CH2=CH2    3 CH2  CH2  CH2    

| |
OH OH

1, 3 Propane diol 1, 3-dioxane

7.13 Polymerisation of Alkene
(a) Mechanism is based on ionic or free radicals both.
(b) In alkene polymer name of polymer is given on the basis of name of monomer, only ‘Poly’ prefix
is used.
(c) If in polymerisation Zeigler-Natta catalyst (trialkyl aluminium + Titanium tetrachloride-
(C2H5)3 Al + TiCl4.) is used than polymerisation is named as Ziegler-Natta Polymerisation. e.g.
H H H H
| | | |
C  C  C  C 
+  Trace O2  catalyst
  | | | |
1500ºC /high pr.
H H H H n

Polymer or
(– CH2 – )n polythene
Note : (i) If propelene is reacted then polypropelene is formed its commercial name is ‘Koylene’.
(ii) By the use of vinyl chloride polyvinyl chloride is formed.
(iii) By the use of Vinyl cyanide (Acrylonitrile), polyacrylonitrile or orlon is formed.
(iv) Zeigler-Natta catalyst is used in addition polymerisation.

ALKENE [38]
7.14 Substitution reaction –
(a) Except ethene other higher alkene having allyl hydrogen when treated with chlorine or bromine ‘’
H is substituted.

500  600 C
CH3 – CH = CH2 + Cl2      Cl–CH2– CH=CH2 + HCl
allyl chloride (3-chloropropene)
(b) Allylic bromination can be easily done by NBS (N–bromo succinamide)

+ H–CH2–CH=CH2   + Br– CH2 – CH = CH2

NBS propene allyl bromide

7.15 Isomerisation :-
Alkenes isomerises when heated at high temperature or at lower temperature in the presence of
various catalysts as AlCl3
eg :-

(i) CH3 – CH2 – CH2 – CH = CH2 CH3 – CH2 – CH = CH – CH3

CH3 CH3
| |
(ii) CH2 – CH = CH2 CH2 = C — CH3
the mechanism proceeds through carbocation

ALKENE [39]
 Solved Example
Na/NH3 (l)
Ex.1 R — CH = CH2 
C H OH
RCH2CH3 is called
2 5

(A) Clemmensen’s reduction (B) Fisher–Spier reduction

(C) Birch reduction (D) Arndt-Eistert reduction (Ans.C)
Sol. It is the name of reaction
Ex.2 What would be the product when 2-pentene reacts with HBr –
(A) 2-bromo pentane (B) 3-bromo pentane
(C) Both A and B (D) 1-bromo pentane (Ans.C)
 
HBr
Sol. CH3 – CH2 – CH = CH – CH3 

3-bromo pentane
 
HBr
CH3 – CH2 – CH = CH – CH3 

2-bromo pentane
Ex.3 A hydrocarbon reacts with HI to give (X) which on reacting with aqueous KOH forms (Y). Oxidation of
(Y) gives 3-methyl-2-butanone. The hydrocarbon is :
(A) CH3 CH  C  CH3 (B) CH2  CH  CH  CH3
| |
CH3 CH3
(C) (D) CH  C  CH  CH3 (Ans. B)
|
CH3
H  KOH( aq.) [O ]
Sol. 
 CH –CH–CH–CH   CH 3–CH–CH–CH 3

3 3
| | |
CH3 OH CH3

Ex.4 What would be the product when ethene is oxidised with ozone and forms ozonide which is hydrolysed
in the presence of Zn and acetic acid?
(A) H  C  H (B) CH3OH (C) H  C  OH (D) H  C  C  H
|| || || ||
O O O O
(Ans. A)
Zn  CH3 COOH
Sol.  
 H C H  H C H
H OH || ||
O O

Ex.5 Reaction of CH3CH = CH2 with Br.CCl3 in the presence of a peroxide yields the following product.
(A) CH3  CH  CH2  CCl3 (B) CH3  CH  CH2Br
| |
Br CCl3
(C) BrCH2 – CH = CH2 and CHCl3 (D) No reaction takes place (Ans. A)
Sol. Compounds like CCl4, CHCl3, Br.CCl3 etc also show kharash effect, hence they will show Anti
Markownikoff’s addition in the presence of peroxides. The reaction with Br. CCl3 takes place as
CH3 CH = CH2 + BrCCl3 Peroxide
  CH3  CH  CH2  CCl3
|
Br

ALKENE [40]
Ex.6 What would be the product when propene reacts with chlorine in presence of CCl4 –
H H
| |
(A) CH3  C  C  H (B) Cl– CH2 – CH = CH2
| |
Cl Cl

H H H H
| | | |
(C) CH3  C  C  Cl (D) CH3  C  C  H (Ans. A)
| |
Cl H
H H
CCl4
| |
Sol. CH3–CH = CH2 + Cl
2
 CH3  C  C  H
| |
Cl Cl
Ex.7 Propene on reaction with methylene iodide in presence of Zn–Cu couple gives :
(A) Cyclopropane (B) Cyclopropen
(C) Methyl Cyclopropane (D) Cyclobutene (Ans.C)
Sol. CH3CH = CH2+ CH2 I2  
Zn


–
Ex.8 RCH = CH2 + BH3  intermediate OH
/ NH 2 Cl
 product,
product will be –
(A) RCH2–CH2–NH2 (B) RCH  CH2
| |
NH2 NH2
(C) R–CH2–NH2 (D) RNH2+CH2=CH–NH2 (Ans. A)
Sol. Intermediate trialkyl, borane react with NH2Cl in NaOH and gives same Carbon no alkyl amine.

Ex.9 The addition of OsO4 on an alkene followed by hydrolysis produces the following product
(A) H4OsO4 (B) H3OsO4 (C) Os2O3 (D) H2OsO4 (Ans. D)
Sol. The addition of OsO4 on an alkene followed by hydrolysis produces Osmic acid (H2OsO4). The reaction
takes place as follows :
CH3  CH CH3  CH  OH
|| +OsO4  H2O |  + H2OsO4
CH2  
 CH2  OH

Cyclic Osmium Ester Propan 1,2 diol Osmic acid

Ex.10 Propene + HOCl  A  Final product. In the above reaction A will be

H H H
| | |
(A) CH3  C  C  H (B) CH3  C  C  H
| – – |
OH Cl
H H
| |
(C) CH3  C  C  H (D) CH3— (Ans. D)
– |
OH


Sol. CH3 — CH = CH2 + HO — Cl  CH3 —

ALKENE [41]
Ex.11 In hydroboration it is evident that in the overall reaction a molecule of a water has been added to
propene and the addition is :
(A) According to markownikoff’s rule (B) Contrary to Markownikoff’s rule
(C) Not concerned with markownikoff’s rule (D) None of above (Ans. B)
Sol. In hydroboration it is evident that in the overall reaction a molecule of water has been added to propene
and the addition is contrary to Markownikoff;s rule
3CH3 CH = CH2 + BH3 0C (CH3CH2CH2)3B
Tripropyl borane
(CH3CH2CH2)3B + 3H2O2 + 3NaOH  3CH3CH2CH2OH
n–propyl alcohol
Co / ThO 2 R  CH2
Ex.12 R–CH=CH2 + CO + H2O 200
 
300 C
| ,
COOH
Above reaction is known as –
(A) Oxo reaction (B) Carbonylation
(C) Both of the above (D) None of these (Ans.C)
Sol. Reaction is named as Oxo and Carbonylation. If CO + H2 is taken than the reaction is named as a
hydroformylation.
Ex.13 Ethylene adds on a molecule of sulphur monochloride to give a poisonous gas used in war, known as:
(A) Calor gas (B) Isocyanide gas
(C) Liquified petroleum gas (D) Mustard gas (Ans. D)
Sol. Ethylene adds on a molecule of sulphur monochloride to give a poisonous gas, known mustard gas and
was used for the first time in first world war (1914-18)
CH2 Cl Cl CH2
|| | | ||  Mustard gas
CH2 + S S + CH2
(,’dichloro diethyl sulphide)
Ex.14 NBS react with 1-butene to give –
(A) 3-bromobutene-1 (B) 1,2-dibromobutane
(C) 1-bromobutene (D) 1,2-dibromobutene-1 (Ans. A)
Sol. NBS is used for the bromo substitution of allylic hydrogen.

Ex.15 Isobutane reacts with the following to form isooctane :

(A) Isobutylene (B) n-Butene (C) n-Pentene (D) Isopentene (Ans. A)
Sol. Isobutane reacts with isobutylene in the presence of conc. H2SO4 to form isooctane

CH3 CH3 CH3 CH3

| | | |
H2 SO 4
CH3  C  H  CH2  C  CH3  CH  C  CH  CH  CH3
or HF 3 2
| |
CH3 CH3
Isobutane Isobutylene 2,2,4 Trimethyl pentane(Isooctane)

Ex.16 Reaction of alkene and peracid gives oxyrane. This reaction is named as –
(A) Peroxidation (B) Oxidation
(C) Priles chaiev (D) None (Ans. C)
Sol. Reaction is known as Prileschaiev reaction.

ALKENE [42]
Ex.17 The compound which reacts with HBr obeying Markownikov's rule is -
H3C CH3
(A) CH2 = CH2 (B) C=C
H H

H3C H H3C H
(C) C=C (D) C=C (Ans. D)
H CH3 CH3 H

Ex.18 Alkene and alkyne gives following types of polymerisation –

(A) Addition (B) Condensation
(C) Substitution (D) Replacement (Ans.A)
Sol. Due to unsaturation alkene and alkyne gives addition polymerisation.

Ex.19 Cyclic osmium ester of alkene after reacting with aqueous solution of sodium bisulphite gives –
(A) Diol (B) Cis-diol
(C) Trans-diol (D) Salt (Ans. B)
Sol. Cis-diol is the end product.

ALKENE [43]
 REACTION CHART FOR ALKENES
GMP GR
H Ni
(1) 2, R–CH2–CH3
(1) R–H Pyrolysis
 200300 C

(2) X
2 R–CHX–CH2X
Ni, H
2
(2) R–C  CH 
R–CH=CH2 (3) 
HX
 R–CHX–CH3
200300C
 or 
(4) HBr
 , Peroxide
 R–CH2–CH2Br
C n H 2n
(3) R–CH2–CH2–X alc
. KOH
 (5) HOCl
 R–CH(OH)–CH2Cl
 HX
dil. H SO
2 4
Zn dust (6)  HO R–CH2(OH)–CH3
 
 2
(4) R–CH2–CH for higher alkene 1 / 2 O2
X 2 (7) Ag 
300C
R  CH  CH 2
(5) | | Zn
dust

 (8) 
CH 2 N 2

X X
(9) BH
3  (RCH2CH2)3B
(6) CH2=CHCl CuR 2
 R  CH  CH 3 R  CH 2  CH 2
CO  H 2
   
(10) HCo (CO)  | + |
conc. H SO 4 CHO CHO
(7) R–CH2–CH2–OH  24  O2
 H 2O (11)  CO2 + H2O


(8) R  C  O  CH 2  CH 2  R Pyrolysis
 R  CH  CH 2
|| (12) O
S O4
 | |
O
OH OH
RCH  COOK
(9) | Kolbe's electrolytic synthesis

RCH  COOK Bayer reagent R  CH  CH 2
(13)    
1% alkalineKMnO | |
4
OH OH
(10) (C2H5)4N+OH 


(14) strong R  C  OH + CO + H O
 oxidant
 || 2 2

O
Per acid
(15) Pr iles  
 chalev 's reaction

3 O H O
2
  
(16) Ozonolysis +
2 O

(17) 200 Polyalkene
C high P
Cl
(18)  2 Substitution product

500 C

Al (SO )
2  
(19) 200
4 3
 Isomerisation
300 C
(20) acetic
 anhydride
 R–CH2=CH–COCH3
Methyl alkenyl ketone
Alkane
(21)   Higher alkane

ALKENE [44]
 ALKYNE
1. GENERAL INTRODUCTION :
(1) The chemistry of the carbon-carbon triple bond is similar to that of the double bond. In this chapter,
we see that alkynes undergo most of the reactions of alkenes, especially the additions and the
oxidations. We also consider reactions that are specific to alkynes : some that depend on the unique
characteristics of the C  C triple bond, and others that depend on the unusual acidity of the
acetylenic C– bond.
(2) Alkynes are hydrocarbons that contain carbon–carbon triple bonds. Alkynes are also called acetylenes
because they are derivatives of acetylene, the simplest alkyne.
(3) Alkynes are not as common in nature as alkenes, but some plants do use alkynes to protect themselves
against disease or predators. Cicutoxin is a toxic compound found in water hemlock, and capillin protects
a plant against fungal diseases. The alkyne functional group is not common in drugs, but parsalmide is
used as an analgesic, and ethynyl estradiol (a synthetic female hormone) is a common ingredient in birth
control pills.
(4) Acetylene is by far the most important commercial alkyne. Acetylene is an important industrial feedstock,
but its largest use is as the fuel for the oxyacetylene welding torch. Acetylene is a colourless, foul-smelling
gas that burns in air with a yellow, sooty flame. When the flame is supplied with pure oxygen, however,
the color turns to light blue, and flame temperature increases dramatically.
(5) Bond angle in alkyne is 180º.
(6) Their general formula is CnH2n–2
(7) C – C triple bond length is 1.20 Å.
(8) C – H bond length is 1.08 Å.
2. METHODS OF PREPARATION
2.1 From Gem Dihalides (Dehydrohalogenation) :
H X X
| | | NaNH
 R  CH  C  H   2
R C  C H + alc. KOH.   R– C C– H
| | HX HX
H X

Note: Alc.KOH is not used for elimination in second step because in this case elimination takes
place from doubly bonded carbon atom which is stable due to resonance so strong base NaNH2
is used for elimination of HX.

2.2 From vicinal dihalides :

H H H
alc. KOH NaNH2
—C—C— –HX
—C = C— —C  C—
–HX
X X X
vic-dihalide vinyl-halide alkyne
(very unreactive)

The elimination of one molecule of hydrogen halide yields vinyl halide which is very unreactive. Under
mild conditions, the dehydrohalogenation stops at vinylic halide stage but more vigorous conditions—
like the use of a stronger base like amide ion (NH2–)—are required for alkyne formation.

ALKYNE [45]
2.3 From Tetrahaloalkanes (Dehalogenation) :

 
+ Zn(dust)  

300ºC
 R – C  C–H
Zn dust

In the above reaction it is necessary that the four halogen atoms must be attached at vicinal carbons.
If they are attached at the two ends then the product cyclo alkene is obtained.
2.4 From Kolbe's Synthesis :

O O
||  || 

HC C  O K H C  C  O
||  ||  + 2K+

HC C  O K H C  C  O
|| ||
O O

Potassium Malaete
At Anode :

– 2e –  

H–C
|||
H–C
+ 2CO2 A

At Cathode :
2K+ + 2e–  2K•
2K• + 2H2O  2KOH + H2 A
2.5 Laboratory method of preparation of Acetylene :
(a) In laboratory acetylene is prepared by hydrolysis of calcium carbide.

   +

(b) It can also be prepared from CHCl3 with Ag dust.

 H–C  C – H
 AgCl

2.6 From Alkynes : (To form higher Alkynes)

2.6.1 With Na : When acetylene or 1- alkyne react with Na in presence of liq. NH3 then an intermediate
compound sodium acetylide or sodium alkynide is formed which gives higher alkyne with alkyl halide.

liq.NH X –R
3 2 H – C  C Na  

2H – C  C – H + 2 Na  
H
 H – C  C – R
2

liqNH
. 
X –R
3 2R – C C Na  
2R – C C – H + 2Na  
H2
 R – C C –R

ALKYNE [46]
2.6.2 With GR : When acetyline or 1- alkyne react with GR then alkane and unsaturated GR is formed
which further react with alkyl halide and form higher alkyne.
XR
 H  C – MgX   H – C  C – R
H –C CH + R – Mg-X  
 RH  MgX 2
acetylene
R– C  C–H + R – Mg – X    R –C  C – MgX R X
  R–C  C–R'
 RH

SOLVED EXAMPLE
Ex.1 What is the chief product of reaction between -butylene chloride and alc. KOH/NaNH2
(A) 1,2-butadiene (B) 1,3-butadiene (C) 2-butyne (D) 1-butyne (Ans.C)
Sol. 2-butyne is the chief product according to Saytzeff's rule -

Ex.2 Chloroform is heated with Ag powder in laboratory what will be the product -
(A) Acetylene (B) Ag2O (C) CH2Cl2 (D) CH4 (Ans.A)

Sol.   HC  CH
6 AgCl

acetylene

Ex.3 Acetylene can be prepared from –

(A) Potassium fumarate (B) Calcium carbide
(C) Ethylene bromide (D) All Ans. (D)
Sol. These are methods of preparation of C2H2. Potassium fumerate on electrolysis, CaC2 on hydrolysis
and ethylene bromide on elimination gives acetylene.
Ex.4 Ethylene dibromide on treating with alcoholic KOH gives –
(A) C2H6 (B) CH4 (C) C2H4 (D) C2H2 Ans. (D)
Sol. Alkynes can be prepared by dehydrohalogenation of alkyl dihalides
CH2Br KOH alc.
 
 CHBr KOH alc.
 
 CH
| KBr
|| KBr |||
CH2Br H2O CH2 H2O CH
Ethylene dibromide Vinylbromide Acetylene
Vinyl bromide being less reactive and thus to get better yield a stronger base NaNH2 is used in
second step.
Q.5 Which of the following on ozonolysis gives three different compound
(A) hex-1-en-4-yne (B) Penta-1,3-diene (C) oct-2, 5-diyne (D) Toluene
Sol. (A,B,C)
CHO
(A) CH2=CH–CH2–C  C–CH3 O 3 / H 2O
 H  C H + CH2 + CH3CHO
Zn || CHO
O
hex-1-ene-4-yne
3 2 O /H O
(B) CH2=CH–CH=CH–CH3  
Zn
 H  C H + C HO + CH3CHO
|| |
O CHO
penta 1, 3-diene

ALKYNE [47]
 3 2 O /H O CHO
(C) CH3–CH=CH–CH2–CH=CH–CH2–CH
3  CH3CHO+CH2 + CH3CH2CHO
Zn CHO
3O /H O
2
(D)    CH3  C CHO  C HO  C HO
Zn || || |
O CHO CHO
Q.6 Treatment of an alkyne with H2 in presence of palladised coke results in the formation of-
(A) Alkane (B) Alkane vigourously
(C) Trans-alkene (D) cis-alkene
Sol. (D)
Alkyne on treatment with H2 in presence of palladised coke (Lindlar catalyst) Pd/C result in the formation
of cis alkene
Pd / C
CH3 – C  C–CH3 + H2 
BaSO4

2-butyne cis-2-butene

3. PHYSICAL PROPERTIES :
(a) Alkynes are colourless, odourless and tasteless.
(b) Lower alkynes are partially soluble in H2O. (It is due to its polarisibility).
(c) Higher alkynes are insoluble in water due to more % of covalent character.
(d) Completely soluble in organic solvents.
(e) Melting point and boiling point increases with molecular mass and decreases with number
of branches.
(f) Upto C4 alkynes are gaseous.C5–C11 are liquid, C12 & above are solids.
(g) Acetylene & 1- alkyne are acidic in nature. It is due to greater electronegativity of sp
hybridised 'C'.

4. CHEMICAL PROPERTIES
The chemical behaviour of alkynes is similar to that of alkenes. Alkynes form addition products with
two or four univalent atoms or groups. They are generally less reactive than alkenes towards
electrophilic addition reactions even though  electron density is higher in alkynes. This is because
on moving from alkene to alkyne, the C—H bond has more of s character (33% in alkenes as
compared to 50% in alkynes). Therefore, the -electrons are more firmly held by carbon nuclei
in alkynes and are thus less reactive to an electrophile.
4.1 Hydrogenation
H H H
Ni, Pd, Pt | | |
R–CC –H + H2   R  C  CH 
  RC  C H
High Temp. | | |
H H H

The above reaction is called as Sabatier Senderson's reaction.

4.1.1 Hydrogenation in the presence of Lindlar Catalyst :
Addition up to alkene takes place in cis manner.

(i) H–CC–H + H2 Pd BaSO 4





ALKYNE [48]
 (ii) CH3 –C C – CH3 + H2 PdBaSO4
 

4.1.2 Hydrogenation by Na + NH3 (liq.) :

Addition upto alkene takes place in trans manner

Na
(i) CHCH  

NH3 (liq.)

(ii) R–CC –R Na /NH3liq.

 

Mechanism :

NaNH Na/NH3 2 NH2 H

3      
R – C  C – R  


 e  2 NH2

e /  Na

4.1.3 Hydrogenation by LiAlH4 :-

Addition upto alkene takes place in trans manner by LiAlH4 also.

LiAlH4
R – C  C–R   

4.1.4 Reduction with the help of B2H6 :

Alkyne is first reacted with B2H6 and is follwed by acidic hydrolysis, cis alkene is obtained.
1. B2H6
R–CC–R   
2.H2O / H

4.2 Halogenation :
In presence of Lewis acid as a catalyst alkyne form tetrahaloderivative with halogen.
X X X
FeX 3 | x2 | |
   R  C  C  H  R  C  C  H

R–C  C–H + X–X X  F,ClBr
,
| | |
X X X
Example :
Br Br Br
Br2 Br2
—C  C— —C = C— —C—C—
CCl4 CCl4
Br Br Br
Alkynes add two molecules of Br2 in CCl4 and decoloration of bromine water is used to detect the
presence of a double or triple bond.

4.3 Reaction with HX (Hydrohalogenation) :

Alkyne form gem dihalide with HX because reaction follows markownikoff's Ist and IInd rule both.
X X
| H X |
R – C  C – H + H – X    R  C  CH2   R  C  CH3
2
Hg |
X
gem dihalide

ALKYNE [49]
 Mechanism : The reaction takes place in the presence of Hg2+ ion.
Hg2+

2 – R Hg H R H
R –C C – R Hg
  R– C C –R
HX C=C  2
 Hg  C=C
X R X R

HX
 

Note : (i) First step is faster than the second step among addition of two HX on alkyne.

X H X H
| |
/ Hg2 | | HX / Hg2
R – C  CH HX
  R — C — CH    R — C — C — H
Faster slower | |
X H
Slower rate of addition of 2nd molecule is caused by lower electron cloud density on
C = C, due to – I effect of 'Cl'.
(ii) peroxide give same effect as in alkene
H Br H Br
HBr | | HBr
| |
   
R–CCH Peroxide R — C — CH     R — C — C —H
Peroxide
| |
H Br

4.4 Reaction with Hypochlorous acid or Chlorine water :

Hypochlorous acid is broken into OHCl& ions & give addb. according to markonicoff's rule.
OH OH
| RC  O
  | |
RC   RC R  C  OH  
||| + H O  Cl H  C  Cl
HO  Cl  ||   |  H2O
H C H C |
H  C  Cl Cl
| |
Cl Cl
unstable
4.5 Hydration : Addition of water to alkynes is carried out in the presence of acid and mercuric sulphate.

H2O H H
HC  CH HgSO4, H2SO4
H—C = C—H Tautomerizes
H—C—C = O
H O—H
vinyl alcohol H
(enolic form-less stable) Acetaldehyde

H 2O
CH3—CH2—C  CH CH3—CH2—C = CH2 CH3CH2—C—CH3
HgSO4, H2SO4
O—H O
(enol) (ketone)

C  CH HO—C = CH2 O = C—CH3

H 2O
HgSO4, H2SO4
Acetophenone

ALKYNE [50]
4.6 Addition of alcohol
O  R' O  R'
RC  
| |
||| + H O R NaO
 R  C 
   R  C  O  R'

—R || HO R |
H C
C H HC H
| |
H H
olefinic ether acetal / ketal
4.7 Addition of carboxylic acid :
In the presence of Hg2+ unioxylation of carboxylic acid takes place.
O O
|| 2 ||
H–C  C – H + H – O – C – CH3 Hg
  CH2 = CH – O – C – CH3
4.8 Adition of Boron Hydride (Hydroboration) :
(i) With mono-alkyl acetylene, R2BH gives an intermediate which on hydrolysis gives alkene but on
alkaline oxidation yields aldehyde.

R H CH3COOH
R—C  C—H + R2BH  C=C RCH = CH2
hydrolysis
Dialkyl H BR2
borane oxidation H2O2, NaOH

RCH2CHO

(ii) With dialkyl acetylenes, the product of hydrolysis is cis-alkene and that of oxidation is a ketone.

R R
R—C  C—R + BH3  C=C B
H
3
Dialkyl acetylene vinyl borane
H2O2, CH3COOH
NaOH oxidation hydrolysis
R R R
C=O C=C
R H H
ketone cis-alkene

4.9 Addition of HCN :

Addition takes place in the presence of CuCl

H – C  C – H + H – CN CuCl
 CH2 = CH – CN
770º
4.10 Addition of AsCl3 :
Lewisite is obtained
R– CC –H R— C  C —H
| |
Cl AsCl2

Lewisite

ALKYNE [51]
4.11 Oxidation :
4.11.1 With acidic or alkaline KMnO4 alkyne break into two parts from triply bonded carbon and every
part forms respective acid.
R – CC –H + [O]  Acidic / Alk.KMnO4
  R  C  OH + H  C  OH
|| ||
O O
H–CC–H + [O]  Acidic

KMnO4
 2H  C  OH
||
O
CH3 – C C – H + [O]  Acidic / Alk.KMnO4
  CH3  C  OH + H  C  OH
|| ||
O O
CH3 –CC–CH3 + [O]  Acidic / Alk.KMnO4
  2CH3  C  OH
||
O
Exception : Acetylene forms oxalic acid with alkaline KMnO4 exceptionally.
Alk.KMnO4 COOH
H –C C–H + [O]    |
COOH
4.11 Ozonolysis
The ozonolysis of alkynes yields a mixture of carboxylic acids.
O
H2O H2O2
R—C  C—R + O3  R—C—C—R R—C—C—R RCOOH + R COOH
Hydrolysis oxidation
O—O O O

4.12 Acidic nature of 1- Alkyne or Acetylene

In 1- alkyne or acetylene, the H which is linked with sp hybridised carbon is called as acidic or
active H. It can easily be substituted by metal or alkaline species. Hence 1- alkyne or acetylene
are acidic in nature. eg.
4.12.1 Reaction with Na :
LiqNH
. 3
2R – CC–H + Na  
 H2
 2R – C  CNa

sodium alkynide
OR
2R – C C – H + NaNH2    2R–C  CNa
NH3

H– C C – H + Na  NaC CNa

disodium acetylide
Note : Where this alkynide is treat with alkyl halide higher alkyne is obtained.
R–CCNa + X–R  R–CC–R' + NaX
Na – C  CNa + 2R – X  R'C  C – R' + 2NaX
4.12.2 Reaction with Ammonical Silver Nitrate
solution : (Tollen's Reagent)
2R – C C – H + AgNO3 + NH4OH    2R – CCAg
H O 2
silver alkynide (white ppt)
H–C C – H + AgNO3 + NH4OH   AgC  CAg
 H2O
silver acetylide (white ppt)

ALKYNE [52]
4.12.3 Reaction with Ammoniacal Cuprous
Chloride solution :
2R – C C – H + Cu2Cl2 + NH4OH   2R – CCCu
 2HCl
cuprous alkynide (red ppt.)
 CuC  CCu
H –CC – H + Cu2Cl2 + NH4OH  
 2 HCl
dicuprous acetylide (red ppt.)
4.12.4 Reaction with NaOCl :
Acidic hydrogen is substituted by 'Cl'
(i) H – CC–H NaOCl
 Cl – CC – Cl

(ii) R – CC–H NaOCl

 R – C C–Cl
Mechanism :

R–C  R – CC–Cl

Note : These (5.12.2) and (5.12.3) both reactions are used in the test of terminal alkynyl hydrogen.
These ppt or metal acetylide or alkynide are insoluble in solution and in dry condication explode,
therefore they are destroyed with HNO3 before dryness (they form same compound again)
4.13 Polymerisation Reactions : Alkyne mainly shows addition polymerisation reactions.
4.13.1 Dimerisation and Cyclysation
(a) Dimerisation :
Two mole acetylene reacts with Cu2Cl2 & NH4Cl and forms vinyl acetylene.
Note : If acetylene would be in excess then product would be divinyl acetylene and the reaction
is called trimerisation.
H H H
| | |
+ H–C  C–H Cu2Cl2  NH4Cl H  C  C  C  C  H
HC  C  
vinyl acetylene or butenyne
H H H H H H H
| | | | | | |
H C  C  C  C H + C  C  H    H  C  C  C  C  C  C  H
divinyl acetylene
(b) Trimerisation :
If three mole of acetylene is passed into red hot iron or Cu or quartz tube, then a cyclic trimer is formed
which is called benzene.
CH
CH
CH Re d hot Fe / Cu / Quartz
 
CH Co(CO)8  Br2
CH

(Octa Carbonyl cobalt

CH

with bromine)
benzene

ALKYNE [53]
 CH3
| CH3
|
CH C
||| \\\ C
CH3  C CH CH CH

C  CH3 
 CH3 –C C– CH3
/// CH
HC
mesitylene (having 3-1º, 3-2º & 3-3º carbons)
Important : Mesitylene can also be obtained from acetone by condensation polymerisation.
(c) Tetramerisation : According to the name four moles of acetylene are heated with nickel tetra
cyanide, then acetylene forms a cyclic tetramer cyclo octa tetraene.

CH  CH CH=CH
CH CH CH CH
||| ||| Ni(CN)4
  
CH CH CH CH
CH  CH CH=CH
(Nonaromatic) cyclooctatetraene
4.13.2 Coupling:-
(a) Alkyne form respective cuprous alkynide with ammonical cuprous chloride solution. When cuprous
alkynide is reacted with pottasium ferri cyanide [K3Fe(CN)6] they converted into conjugated diyne.
R – C  CH + Cu (NH3)2Cl  R–C C.Cu
cuprous alkynide
2R – C  C.Cu O
2  R – C C – C C – R

diyne
(b) Coupling is also done easily by 1-alkyne in the presence of Cu2Cl2 (cuprous compound) and amine
(ie. pyridine + air) cuprous alkynide is formed (this coupling is known as oxidative coupling or glaser
coupling)
1
2R – C  CH + O2 Cu2Cl2 /NH
 3 R – C C – C  C – R + H2O
2

2 CH3 – CCH + ½ O2 Cu2Cl2 /NH

 3  CH – C C– C  C – CH + H O
3 3 2

2,4–Hexadiyne.
4.14 Formation of heterocyclic compounds :

CH CH Al2O3
(a) ||| + NH3 + |||    (Pyrrole)
CH CH 

CH CH Al2O3
(b) ||| + + |||   (Thiophene)
CH CH 

ALKYNE [54]
4.15 Reaction with Formaldehyde
1– Alkyne in the presence of copper react with methanal to form alkynol.

H
|
H  C  C + CH2 = O 
 CH  C – CH2OH
prop-2-yne-1-ol
Note : Acetylene react with two mole of HCHO (methanal) to give 2–butyne–1,4–diol. This reaction
is used in the formation of 1,3 butadine. Which is used in the formation of buna rubber.
4.16 Isomerisation :–
(a) When 1-alkyne is treated with alcoholic KOH 2-alkyne is formed.

R–CH2–CCH  Alco

.KOH
 R – C  C – CH3

1–alkyne 2–alkyne
(b) When 2-alkyne is treated with sodamide then it is converted into 1-alkyne.
H2O
CH3 – C  C – CH3 NaNH
 2  CH – CH – CCNa  
 3 2   CH3 – CH2 – CCH
NH3 NaOH

SOLVED EXAMPLE
Ex.1 What happens when 2- butyne reacts with H2 in presence of Nickle Boride or lindlar's catalyst
(Pd/CaCO3–PbO) -
(A) CH3  C  H (B) CH3  C  H
|| ||
CH3  C  H H  C  CH3

(C) CH3 – CH2– CH2– CH3 (D) CH2  CH Ans.[A]

|
CH  CH2

Sol. 2-butyne forms cis-2-butene with lindlar's catalyst.

Ex.2 What will be the product of chlorine water and acetylene -

(A) Dichloro propanol (B) Dichloroethanal
(C) Propanol (D) 2-Chloro ethanol Ans.[B]
Sol. Chlorine water (HO O  Cl ) when react with acetylene dichloro ethanal is formed


OH OH
| | H C  O
H C    
H  C  OH 
||| + H C HO  Cl |
HO  Cl  ||   | H2O H  C  Cl
H C H  C  Cl
H C |
| | Cl
Cl Cl

Ex.3 Product formed by the oxidation of acetylene in the presence of alkaline KMnO4
(A) Glyoxal (B) Oxyrane (C) CO2 + H2O (D) Oxalic acid Ans.[D]
Sol. This is the exceptional case of oxidation in which triple bonded carbon is not separated. The product
is oxalic acid.

ALKYNE [55]
Ex.4 Acetylene on treating with HI gives –
(A) 1,1-Diiodoethane (B) Ethylene (C) Iodoethane (D) Ethane Ans. (A)
Sol. Acetylene on treating with HI gives 1,1-diiodoethane
CH2 CH3
CH ||
dark
||| + HI   CHI dark |
  CH  I2
CH vinyl iodide ethylidene diiodide (11
,  diiodoethane)
The reactivity of halogen acids is
HI > HBr >HCl
Ex.5 Acetylene and ethylene react with alk KMnO4 to give –
(A) Oxalic acid and formic acid (B) Acetic acid and ethylene glycol
(C) Ethyl alcohol and ethylene glycol (D) None Ans. (A)
Sol. Acetylene and ethylene react with alk. KMnO4 to give oxalic acid and formic acid respectively.
CH COOH
||| + 4 [O] |
alk. KMnO 4


CH COOH
acetylene oxalic acid
CH2
|| + 4 [O] alk . KMnO 4
CH2   2HCOOH

ethylene formic acid

Ex.6 When 2-butyne reacts with sodamide in an inert solvent in the presence of dilute HCl, the product formed
is:
(A) n-Butane (B) 2-Butene (C) 1-Butyne (D) 1-Propyne Ans. (C)
Sol. When 2-butyne is heated with NaNH2 in an inert solvent, the sodium derivative of but-1-yne which
is converted into but-1-yne by the action of dil HCl.
dil. HCl
CH3 – C  C – CH3 + NaNH2 Paraffin
 NH3+ CH3CH2C  CNa  CH3CH2CCH
1-butyne
Ex.7 1-Pentyne reacts with :
(A) Sodium (B) Sodamide
(C) Ammonical silver nitrate (D) Ammonical cuprous chloride
Sol. (A, B, C, D)
Terminal alkyne having acidic hydrogen atom react with certain reagent like Na, NaNH2 ammonical
silver nitrate, ammonical cuprous chloride
CH3–CH2–CH2–CCH + Na CH3–CH2–CH2–CCNa + 1/2H2
CH3–CH2–CH2–CCH + NaNH2 CH3–CH2–CH2–CCNa + NH3
CH3–CH2–CH2–CCH + Ag(NH3)2+ CH3–CH2–CH2–CCAg
Ex.8 Which of the following can not react with ammonical silver nitrate-
(A) Acetylene (B) Hexyne-1 (C) Phenyl acetaldehyde (D) p-tolualdehyde
(E) hept-4-yne-2-ene
Sol. (E) Alkyne having acidic hydrogen atom react with ammonical silver nitrate i.e. Tollen’s reagent.
(A) CHCH
(B) CHC–CH2–CH2–CH2–CH3
Aldehyde reduces Tollen reagent to metallic silver.
(C) C6H5CH2CHO + Ag2O C6H5CH2COOH + 2Ag
Phenyl acetaldehyde

ALKYNE [56]
 (D) + Ag2O + 2AgCl

p-tolualdehyde
Note : Hept 4-yne-2-ene (CH3–CH=CH–CC–CH2–CH3) does not contain acidic hydrogenso does
not reacts with Tollen reagent.
Ex.9 The hydrocarbon that reacts with ammonical cuprous chloride is-
(A) Essentially aromatic (B) Ethane (C) Ethyne (D) Ethene
Sol. Ans. (C)
Terminal alkyne react with ammonical cuprous chloride CH CH

CH3–CH3, CH2=CH3 and does not give test

Ex.10 The product of reaction CH3CH2CH2 MgBr + HC  CCH3 

is
(a) CH3 CH2 CH3 (b) CH3 CH2 CH2 C  C CH3
(c) CH3 CH2 CH2 OH (d) CH3 CH2 CHO
Sol. Grignard reagent reacts with any compound having active hydrogen producing an alkane.
n—C3H7 MgBr + HC  CCH3  n—C3H8 + CH3C  CMgBr
The answer is (a)

Ex.11 1-butyne can be distinguished from 2-butyne by using

(a) bromine water (Br2 in CCl4) (b) cold alk. KMnO4 (Baeyer’s Reagent)
(c) ammoniacal solution of silver (Tollen’s Reagent) (d) diethyl ether
Sol. Terminal alkynes form a precipitate with Tollen’s reagent.
1
RC  CH + Ag+  RC  C–: Ag+ + H2
2
ppt. of Silver acetylide
1-butyne, a terminal alkyne gives this reaction but
2-butyne, a non-terminal alkyne does not give this reaction.
The answer is (c)
Ex.12 The ozonolysis of a triple bond produces
(a) a mixture of aldehydes and ketones (b) a mixture of ketones and carboxylic acids
(c) a mixture of carboxylic acids (d) none of the above
O
O3
Sol. R—C  C—R R—C — C—R
O O
H2O, H+
RCOOH + RCOOH
The answer is (c)

ALKYNE [57]
 REACTION CHART FOR ALKYNES
GMP GR
(1) CH2Br–CH2Br alc . KOH or NaNH 2 H2
  (1)  C2H4, C2H6
Ni
(2) CH3–CHBr2 alc. KOH , NaNH 2 X2
     (2)  C2H2X4
HBr
(3) CHCl3 Ag
powder
  (3) 
 CH BrCH Br
3 2
 Peroxide
 C 2H 2

(4) CHBr2–CHBr2 Zn
dust

 HBr
(4) No   CH –CHBr
 Peroxide 3 2

CHBr (5) HOCl

 Cl2CHCHO
(5) || Zn
(6) HCN , Ba ( CN ) 2
CHBr
  2  CH2=CHCN
CH COOH , Hg
(7) 3    CH3CH(OCOCH3)2
(6) CH2=CH–Cl alc. KOH , NaNH
    2  Hg 2 , 80C , dil. H SO
(8)     2 4  CH CHO
HC  COONa Kolbe's electrolytic synthesis ( Kucherov 's reaction ) 3

(7) ||       
HC  COONa (9) Conc

. H 2SO 4
 CH3CH(HSO4)2
H O AsCl

2
 (10) Ca det 3  CHCl=CHAsCl2
(8) CaC2 & Bunsen reaction

(9) 2C + H2 electric
  arc ,1200
 C
 C2 H5OH / H 2O
Berthelot 's process (11)   
HgSO
 CH CHO
3
4
(10) CH3–CCH ( i ) Na (ii ) R X
   
 CO  HOH
(10) CH3–CCH (i ) CH 3MgI (ii ) R X (12)    CH =CH–COOH
Ni ( CO ) 2
     4

CO  EtOH
(13)    CH =CH–COOEt
Ni ,160C 2

(14) NaNH
 2  Na–CC–Na
AgNO3  NH 4OH
(15) (     Ag–CC–Ag
Tollen 's Re agent )
(16) Cu Cl  NH OH
22 4   Cu–CC–Cu
Combustion
(17)     O2
 CO2 + H2O
CHO
(18) Bayer
 Reagent
 |
  HCOOH
CHO
3O

(19) Ozonolysis H O
2
  HCOOH
Trimerisat ion
(20)    benzene
(Re d hot iron tube )

Trimerisation
(21)    C8H8or 1,3,5,7-cyclo octa tetraene
[ Ni ( CN )2 ]
Dimerisati on
(22)    butenyne
[ Cu ( NH 3 ) 2 ]
s
(23) 



CH OH
(24)  3  methylal
( BF  HgO )
3

ALKYNE [58]
 EXERCISE # I
ALKANE
Q.1 The simplest alkane which exhibits chain isomerism has how many carbon atoms :
(1) 4 (2) 5 (3) 6 (4) 3

Q.2 The reaction of chlorine with methane is carried out in sunlight. The role of sunlight is
(1) To break Cl2 molecules into free radicals (2) To break Cl2 molecules into ions
(3) To heat up the mixture (4) To break C – H bonds in CH4

Q.3 Which has least carbon bond length -

(1) Ethane (2) Ethyne (3) Ethene (4) Ethanol

Q.4 Alkyl halides on reduction with Zn-Cu couple and alcohol give :
(1) Alkanes (2) Alkenes (3) Alkynes (4) Cyclic compounds

Q.5 Isopropyl bromide undergoes Wurtz reaction to form -

(1) Hexane (2) 2, 3 - Dimethyl butane
(3) Propane (4) Neohexane

Q.6 Alkanes can be prepared from Grignard reagents by reacting with -

(1) Alcohols (2) Primary amines (3) Alkynes (4) All of them

Q.7 Which of the following will have least hindered rotation about carbon-carbon bond ?
(1) Ethane (2) Ethylene (3) Acetylene (4) Hexachloroethane

Q.8 Which reducing agent is used in Clemmensen reduction –

(1) Zn / HCl (2) LiAlH4 (3) Zn-Hg / HCl (4) Na / C2H5OH

Q.9 The order of reactivity of halogens towards halogenation of alkanes is :

(1) F2 > Br2 > Cl2 (2) F2 > Cl2 > Br2 (3) Cl2 > F2 > Br2 (4) Cl2 > Br2 > F2

Q.10 CH4 & C2H6 both are prepared from following compound is -
(1) CH3COONa (2) CH3Cl (3) CH3OH (4) Both A and B

Q.11 The chlorination of an alkane involves :

(1) Cl free radicals (2) Cl+ species (3) Cl– species (4) CH4 free radicals

Q.12 Kolbe's reaction is convenient for the preparation of :

(1) Methane
(2) Alkanes containing even number of carbon atoms
(3) Alkanes containing even as well as odd number of carbon atoms
(4) Alkanes containing odd number of carbon atoms

Q.13 Isomerisation in alkane may be brought about by using

(1) Al2O3 (2) Fe2O3 (3) Al Cl3 and HCl (4) concentrated H2SO4

Q.14 The percent of methane in coal gas is about

(1) 25 (2) 35 (3) 45 (4) 55

ALKANE, ALKENE, ALKYNE [59]

Q.15 Alkanes are readily attacked by -
(1) Electrophiles (2) Nucleophiles (3) Free radicals (4) bases

Q.16 The most important method of preparation of hydrocarbons of lower carbon number is :
(1) Pyrolysis of higher carbon number hydrocarbons
(2) Electrolysis of salts of fatty acids
(3) Sabatier Senderen’s reaction
(4) Direct synthesis

Q.17 In Reed’s reaction alkane reacts with Cl2 and SO2 in the presence of -
(1) UV light (2) IR light (3) Visible light (4) Dark

Q.18 Formation of alkane by the action of Zn on alkyl halide is called-

(1) Frankland reaction (2) Wurtz reaction
(3) Cannizzaro’s reaction (4) Kolbe’s reaction

Q.19 The thermal decomposition of alkanes in the absence of air is known as :

(1) oxidation (2) Combustion (3) Hydrogenation (4) pyrolysis

Q.20 An alkane is most likely to react with :

(1) A free radical (2) An alkali (3) An electrophile (4) A nucleophile

900 K
Q.21 Name the reaction : C10H22  C4H8 + C6H14
(1) Alkylation (2) Cracking (3) Dehydrogenation (4) Fractionation

Q.22 The hydrocarbon which is a liquid at room temperature is :

(1) butane (2) propane (3) decane (4) neopentane

Q.23 Petroleum is a mixture of :

(1) Aromatic hydrocarbons with small amounts of aliphatic compounds.
(2) Aliphatic hydrocarbons with small amounts of aromatic compounds.
(3) Mixture of equal amounts of aliphatic and aromatic hydrocarbons.
(4) Alcohols and fatty acids.

Q.24 The most volatile alkane is :

(1) n-pentane (2) isopentane (3) neopentane (4) n-hexane

Q.25 A sample of petrol has an octane number of 60. It means it has the same knocking as a mixture
of :
(1) 40% n-heptane and 60% isooctane (2) 40% isooctane and 60% petrol
(3) 40% isooctane and 60% n-heptane (4) 60% isooctane and 60% petrol

Q.26 Alkanes have almost no solubility in water because of the following except -
(1) Nonpolarity (2) Inability to form H-bonds
(3) Zig-zag structure (4) No H-bonding and nonpolarity

Q.27 To enable easy detection of leakage of gas from cylinders the substance added to LPG is-
(1) Glycols (2) Phenols (3) Glycerols (4) Thioalcohols

ALKANE, ALKENE, ALKYNE [60]

Q.28 Which of the following compounds has an octane number of 100 ?
(1) 2, 2-Dimethyl pentane (2) 2, 2, 4-Trimethyl hexane
(3) 2, 2, 4-Trimethyl pentane (4) None

Q.29 Pyrolysis of alkanes is a -

(1) Nucleophilic addition reaction (2) Free radical substitution reaction
(3) Electrophilic addition reaction (4) Free radical elimination reaction

Q.30 Methane can be prepared by :

(1) Wurtz reactions (2) hydrogenation
(3) decarboxylation (4) dehydrohalogenation

Q.31 Which of the following alkyl halides is not suitable for Corey-House synthesis of alkanes-
(1) CH3I (2) C2H5Br
(3) CH3CH2CH2CH2I (4) (CH3)3CBr

Q.32 Which of the following reactions does not involve a C – C bond formation ?
(1) Hydrolysis of a Grignard reagent (2) Combination of two alkyl free radicals
(3) Corey-House synthesis of alkanes (4) RNa + R – Br R – R + NaBr

Q.33 Which reducing agent is used in Clemmensen reduction ?

(1) Zn / HCl (2) LiAlH4 (3) Zn-Hg / HCl (4) Na / C2H5OH

Q.34 Which of the following alkanes cannot be formed by Wurtz reaction as well as Sabatier and Senderens
reaction?
(1) Methane (2) Ethane (3) Propane (4) Butane

Q.35 The reaction/method that does not give an alkane is-

(1) Catalytic hydrogenation of alkenes (2) Wurtz reaction
(3) Hydrolysis of alkylmagnesium bromide (4) Dehydrohalogenation of an alkyl halide.

Q.36 Two different compounds are obtained by mono bromination of isobutane. The names of the compounds
are-
(1) Butyl bromide & isobutyl bromide
(2) sec. butyl bromide & butyl bromide
(3) Isobutyl bromide and tertiary butyl bromide
(4) sec. butyl bromide and tert. butyl bromide

Q.37 Three monosubstituted halogen derivatives would be obtained from :

(1) n-pentane (2) Neopentane (3) 2-methylpentane (4) Isobutane

Q.38 Select the group of alkanes which on chlorination gives two isomeric monochloro products :
(1) Methane, ethane, propane (2) n-Butane, isobutane, propane
(3) Pentane, neopentane, isopentane (4) Isobutane, isopentane, neopentane

Q.39 Alkane reacts with which of the following halogens in dark ?

(1) F2 (2) Cl2 (3) I2 (4) Br2

ALKANE, ALKENE, ALKYNE [61]

 ALKENE
Q.40 Which of the following must be an alkene ?
(1) C2H4 (2) C3H6 (3) C4H8 (4) All of these

Q.41 The number of s–sp3, s–sp2 and sp2–sp2 bonds in propylene are given by the set-
(1) 1, 3, 2 (2) 1, 2, 3 (3) 3, 2, 2 (4) 3, 3, 1

Q.42 The compound 1, 2,-butadiene has

(1) only sp hybridised carbon atoms (2) only sp2 hybridised carbon atoms
(3) both sp and sp2 hybridised carbon atoms (4) sp, sp2 and sp3 hybridised carbon atoms

Q.43 The General formula of alkenes & cycloalkane will be -

(1) CnH2n + 2 (2) CnH2n + 3 (3) CnH2n (4) CnH2n – 2

Q.44 Baeyer’s reagent is

(1) alkaline potassium permanganate solution (2) acidified potassium permangnate solution
(3) neutral potassium permangnate solution (4) aqueous bromine solution

Q.45 Which one of the following has the smallest heat of hydrogenation per mole ?
(1) 1- Butene (2) trans-2- Butene (3) cis-2- Butene (4) 1,3- Butadiene

Q.46 Alkene can be formed from carbonium ion by

(1) Combination of proton (2) Elimination of hydride ion
(3) Elimination of proton (4) First combination of H then removal of H

Q.47 The compound 1,3-butadiene has-

(1) Only sp hybridized carbon atoms (2) Only sp2 hybridized carbon atoms
(3) Both sp and sp2 hybridized carbon atoms (4) sp, sp2 and sp3 hybridized carbon atoms.

Q.48 In ethylene dichloride there are -

(1) One – bond and five – bond (2) Zero – bond and seven – bond
(3) One – bond and six – bond (4) One – bond and six –bond

Q.49 The presence of unsaturation in an organic compound can be tested with-

(1) Schiff’s reagent (2) Tollen’s reagent (3) Fehling solution (4) Baeyer’s reagent

Q.50 A carbocation undergoes following reactions except -

(1) Combination with a nucleophile
(2) Rearrangement to form a less stable carbocation
(3) Rearrangement to form a high stable carbocation
(4) Elimination of a proton to form C = C

Q.51 Baeyer’s reagent is used in the laboratory for-

(1) Detection of double bonds (2) Detection of glucose
(3) Reduction (4) Oxidation

Q.52 The disappearance of the characteristic purple colour of KMnO4 in its reaction with an alkene is the test
for unsaturation. It is known as :
(1) Markownikoffs test (2) Baeyer’s test (3) Wurtz test (4) Grignard test

ALKANE, ALKENE, ALKYNE [62]

Q.53 The relative stability of the compounds
CH3 CH3 CH3
| | |
CH3 — C — C — CH3 CH3 — C — CH — CH3 CH 3 — CH — CH 2 CH 2 — CH 2
I II III IV
is in the order -
(1) I > II > III > IV (2) IV > III > II > I
(3) I > III > II > IV (4) II > I > IV > III

Q.54 Propyl bromide on reaction with alcoholic KOH gives -

(1) Propane (2) Propene (3) Butane (4) Acetylene

Q.55 The formation of 2-butene from 2-butanol in the presence of conc. H2SO4 is in accordance with :
(1) Hoffmann’s rule (2) Saytzeff’s rule
(3) Markownikoff’s rule (4) Kharasch rule

Q.56 In dehydrohalogenations, the base (alcoholic KOH) abstracts -

(1) The halide ion.
(2) The proton present on the carbon next to the carbon to which the halogen is attached.
(3) The proton present on the carbon to which the halogen is attached.
(4) The proton on the -carbon.

Q.57 Which of the following compound undergoes dehydrochlorination most easily when treated with
alcoholic KOH -
(1) CH3  CH  CH2Cl (2) CH3  CH  C 2H5
| |
CH3 Cl
(3) CH3 – CH2 – CH2Cl (4) (CH3)3C – Cl

Q.58 The ease of dehydrohalogenation of alkyl halide with alcoholic KOH is

(1) 3º > 2º > 1º (2) 3º < 2º < 1º (3) 3º > 2º < 1º (4) 3º < 2º > 1º

Q.59 Ethyl chloride on heating with alcoholic potash gives :

(1) C2H4 (2) C2H2 (3) C2H6 (4) CH4

Q.60 The ease of dehydration of an alcohol with concentrated H2SO4 is :

(1) 3º > 2º > 1º (2) 3º < 2º < 1º (3) 3º > 2º < 1º (4) 3º < 2º > 1º
base
Q.61 Consider the following reaction : (CH3)3CCH2 – C(CH3)2  The major alkene product is -
Br

(1) (CH3)3CCH = C(CH3)2 (2) (CH3)3

CH3

(3) (CH3)2C = C – CH(CH3)2 (4)

Q.62 The dehydrohalogenation of 2- bromobutane with alcoholic KOH gives

(1) only 2- butene (2) only 1- butene
(3) 2- butene as the major product (4) 1- butene as the major product

ALKANE, ALKENE, ALKYNE [63]

Q.63 BrCH2—CH2—CH2Br reacts with Na in the presence of ethanol at 100ºC to produce -
(1) BrCH2—CH=CH2 (2) CH2 = C = CH2
(3) H2C — CH 2 (4) all of these
CH2

Q.64 The addition of halogen to an alkene involves the formation of

(1) carbocation as the intermediate (2) carbanion as the intermediate
(3) free redical as the intermediate (4) halonium ion as the intermediate

Q.65 Which alcohols can be turned most easily to alkenes -

(1) Ethyl alcohol (2) n-Butyl alcohol (3) Sec. butyl alcohol (4) Tert. butyl alcohol

Q.66 Ethylene is formed by the dehydration of -

(1) CH3CHO (2) C2H5OH (3) Propyl alcohol (4) Ethyl acetate

Q.67 The most common reactions of alkenes are-

(1) Nucleophilic substitution (2) Electrophilic substitution
(3) Electrophilic addition (4) Nucleophilic addition

Q.68 The addition of HCl in the presence of peroxide does not follow anti-Markownikoffs rule because
(1) HCl bond is too strong to be broken homolytically
(2) Cl atom is not reactive enough to add on to a double bond
(3) Cl combines with H to give back HCl
(4) HCl is a reducing agent

ALKYNE
Q.69 Which order is correct for bond length -
(1)  C – H > – C – H > = C – H (2) – C – H <  C – H < = C – H
(3)  C – H < = C – H < – C – H (4) None of these

Q.70 Which can yield acetylene in single step

(1) Propyne (2) Ethene
(3) Ethylene dichloride (4) Sodium acetate

Q.71 Acetylene is prepared industrially by passing electric discharge through graphite electrods in the at-
mosphere of -
(1) Air (2) N2 (3) H2 (4) CO2

Q.72 Which one of these will react with sodium metal -

(1) Ethyne (2) Ethene (3) Ethane (4) Ether

Q.73 Ethene can be separated from ethyne by passing the mixture through
(1) concentrated H2SO4 (2) ammonical Cu2Cl2
(3) pyrogallol (4) charcoal powder

Q.74 Lindlar’s catalyst is :

(1) Na in liquid NH3 (2) Pt in ethanol (3) Ni in ether (4) Pd with BaSO4

Q.75 The compounds 1- butyne and 2- butyne can be distinguished by using

(1) bromine water (2) KMnO4 solution
(3) Tollens reagent (4) chlorine gas

ALKANE, ALKENE, ALKYNE [64]

 EXERCISE # II
ALKANE
Q.1 Both ionic and free radical mechanism involve in the reaction :
(1) Chlorination of alkane (2) Williamson's synthesis
(3) Electrolysis of potassium acetate (4) Friedel-Crafts reaction

Q.2 Which of the following alkanes may be synthesized from a single alkyl halide by a process involving
coupling reaction -
(1) 2-Methylbutane (2) 2-Methylpropane (3) 2, 3-Dimethylbutane (4) Propane

Q.3 An aqueous solution containing sodium acetate and sodium propionate is electrolyzed. Which of the
following alkanes is expected as product -
(1) Ethane (2) Propane (3) Butane (4) All of these

Q.4 Electrolysis of concentrated solution of sodium propanoate produces the hydrocarbon-

(1) Methane (2) Ethane (3) Propane (4) Butane

Q.5 A mixture of ethyl iodide and n-propyl iodide is subjected to Wurtz reaction. The hydrocarbon that
will not be formed is-
(1) n–Butane (2) Propyne (3) n–Pentane (4) n–Hexane

Q.6 Decarboxylation of isobutyric acid leads to -

(1) Isobutane (2) Propane (3) Butane (4) None

Q.7 A chain reaction is that in which -

(1) A carbocation is formed (2) A carbanion is formed
(3) There is a series of steps and in each step a reactive intermediate is formed
(4) A free radical is formed

Q.8 Which of the following reactions can be employed for getting unsymmetrical alkanes in good yield?
(1) Wurtz reaction (2) Corey-House reaction
(3) Both (4) None

Q.9 Which of the following on reaction with RMgX does not give RH ?
(1) Ethylamine (2) Dimethylamine (3) Trimethylamine (4) Phenol

Q.10 False statement is :

(1) Ethane can be prepared by Sabatier and Senderen's reaction
(2) Carbon in CH4 and CH3 possess the same hybridisation state
(3) Methane can be prepared by Wurtz reaction
(4) Methane is one of the component of natural gas

Q.11 During the preparation of ethane by Kolbe’s electrolytic method using inert electrodes the pH of the
electrolyte –
(1) Increases progressively as the reaction proceeds
(2) Decreases progressively as the reaction proceeds
(3) Remains constant throughout the reaction
(4) May decrease of the concentration of the electrolyte is not very high

ALKANE, ALKENE, ALKYNE [65]

 ALKENE
Q.12 Addition of HI on double bond of propene yields isopropyl iodide and not n-propyl iodide as the major
product, because addition proceeds through :
(1) A more stable carbonium ion (2) A more stable carbanion
(3) A more stable free radical (4) None

Lindlar 's Na /NH

Q.13 B   R  C  C  R  
3
 A . A and B are geometrical isomers (R–CH = CH – R) -
Catalyst

(1) A is trans, B is cis (2) A and B both are cis

(3) A and B both are trans (4) A is cis, B is trans

Q.14 CH2 = CHCl reacts with HCl to form -

(1) CH2Cl – CH2Cl (2) CH3 –CHCl2 (3) CH2 = CHCl.HCl (4) None

Q.15 What is the slow, rate-determining step, in the acid-catalyzed dehydration of 2-methyl-2- propanol?
H SO
(CH3)3COH 2  4 
(1) Protonation of the alcohol to form an oxonium ion
(2) Loss of water from the oxonium ion to form a carbocation.
(3) Loss of a -hydrogen from the carbocation to form an alkene.
(4) The simultaneous loss of a -hydrogen and water from the oxonium ion.

Q.16 Propene when heated with chlorine at about 500ºC forms :

(1) CH2Cl.CH = CH2 (2) CH3CHCl.CH2Cl
(3) CH2Cl.CHClCH2Cl (4) All the three

Q.17 When ethyl alcohol is heated with conc. H2SO4 at 433 K, ethylene is formed by :
(1) Intramolecular hydration (2) Intermolecular hydration
(3) Intermolecular dehydration (4) Intramolecular dehydration.

Q.18 Addition of halogen acid is least in :

(1) CH2 = CHCl (2) CH2 = CH2 (3) CH3– CH = CH2 (4) (CH3)2C = CH2

Q.19 When two alkenes may be formed by dehydrohalogenation of an alkyl halide, the alkene which is more
substituted is the major or preferred product. This generalization is known as :
(1) Markownikoff’s rule (2) Anti-Markownikoff’s rule
(3) Saytzeff rule (4) None of these is correct

Q.20 The compound formed when 2-butene is treated with hot alk KMnO4 is -
(1) Acetaldehyde (2) Acetic acid (3) CH2OH.CH2OH (4) CH3.CH2.CO.CH3

Q.21 For hydroformylation, the suitable reagent is :

(1) CO + H2O (2) H–CHO (3) CO + H2 (4) HCOOH

Q.22 The hydroboration-oxidation reaction can be characterized as the .................... to an alkene :

(1) Anti-Markovnikov syn addition of water (2) Anti-Markovnikov anti addition of water
(3) Markovnikov syn addition of water (4) Markovnikov anti addition of water

Q.23 The simplest branched chain alkene cannot be prepared by heating which of the following with
ethanolic potassium hydroxide -
(1) t-Butyl halide (2) s-Butyl halide
(3) Isobutyl halide (4) 2-Halo-2-methylpropane

ALKANE, ALKENE, ALKYNE [66]

Q.24 Arrange the following alkanols 1, 2 and 3 in order of their reactivity towards acid catalysed dehydration:
OH OH
| |
(1) CH3  CH  CH2  CH2 (2) CH3  C  CH2  CH3 (3) (CH3 )2 CH  CH  CH3
| | |
CH3 OH CH3

(1) 1 > 2 > 3 (2) 2 > 1 > 3 (3) 2 > 3 > 1 (4) 3 > 2 > 1
Q.25 Which of the following alkanols is dehydrated most readily -
(1) 2-Methyl-2-propanol (2) 2-Methyl-1-propanol
(3) 1-pentanol (4) 1-Butanol

Q.26 + Br2  A. A will have configuration :

(1) (2) (3) both true (4) none is true

Q.27
I II III
which is most easily dehydrohalogenated -
(1) I (2) II (3) III (4) all with same case

Q.28 alc. KOH Product. The product can be–

   

(1) (2) (3) (4)

Q.29 HCl does not form Anti-Markownikoff’s product with propene, because -
(1) HCl is a polar covalent molecule
(2) Dipole-Dipole attraction exists between HCl molecules
(3) Formation of Cl• is energetically unfavourable
(4) Chlorine has the highest electron affinity
Q.30 Choose the correct statement -
(1) Addition of HCl to vinyl chloride yield vicinal dichloride
(2) There is difference between peroxide effect and Kharash effect
(3) Peroxide is a ready source of free radical in the Markownikoff’s reaction
(4) Propene with HCl in presence of peroxide gives isopropyl chloride
2 Conc. H SO
Q.31 CH 3  C H  CH 2  CH 3    4  Product(I) + Product (II)
|
OH
What is not true regarding the products :
(1) Product–I and II are position isomers
(2) Product–I and II contains the same number of sp3 and sp2 carbon atoms
(3) The yield of the product I and II is same
(4) Reaction obeys Saytzeff rule

ALKANE, ALKENE, ALKYNE [67]

Q.32 Suitable reagent to ascertain double bond in C4H8 will be -
(1) Bromination (2) Hydroxylation (3) Ozonolysis (4) Hydroboration

Q.33 The modern explanation of Markownikoff’s rule is in greater stability of the -

(1) Primary carbocation (2) Secondary carbocation
(3) Tertiary carbocation (4) Secondary and tertiary carbocations

Q.34 2-methyl propene is isomeric with butene-1. They can be distinguished by -

(1) Baeyer’s reagent (2) Ammonical AgNO3 (3) Br2 solution (4) O3, Zn/H2O

Q.35 Which of the following alkenes gives acetaldehyde only on ozonolysis -

(1) 1-Butene (2) -Butylene (3) Isobutylene (4) -Butylene

Q.36 In which of the following Kharasch effect operate -

(1) CH3CH2CH = CH2 + HCl (2) CH3CH2 – CH = CH2 + HBr
(3) CH3CH = CH – CH3 + HBr (4) CH3CH2CH = CH2 + HI

Q.37 1-Butene on heating with aluminium chloride at high temperature ( 200º-300º) isomerises to give-
(1) 2-Butene only (2) Isobutylene only (3) Both A and B (4) Cyclobutane.

alc.KOH 3 (i) O
Q.38 A (C4H9Br) B   CH3CHO
 (ii) Zn,H2O

A and B in the above reaction sequence are-

(1) s-Butyl bromide, -Butylene (2) t-Butyl bromide, Isobutylene
(3) s-Butyl bromide , -Butylene (4) n-Butyl bromide, -Butylene

C 2H5 O ¯Na
Q.39 Identify Z in the sequence, CH3  CH2  CH  CH2 HBr

/H2 O 2
 Y    Z

(1) CH3  CH  CH2  O  CH2  CH3 (2) CH3  CH2  CH  O  CH2  CH3
| |
CH3 CH3

(3) CH3  (CH2 ) 3  O  CH2  CH3 (4) CH3  (CH2 ) 4  O  CH3

CCl Br
3 product. The major product is :
Q.40 CH2 = CH – CH =CH2  
(1) Br –CH2 – CH=CH–CH2 – CCl3 (2) CH2 = CH– – CH2 – CCl3

(3) CH2 =CH– – CH2 – Br (4) None is correct

(1) Hg(OAc) / H O/ THF

2 2
Q.41    A. A is-
i
(2) NaBH4 / NaOH/H2O

(1) (2) (3) (4)

ALKANE, ALKENE, ALKYNE [68]

Q.42 Which of the following reactions will lead to the creation of two chiral centres in the product ?
(1) CH3CH = CHCH3 + Br2 CCl 4  (2) CH3CH2CH = CH2 + Br2 CCl 4 
(3) CH3CH = CHCH3 + HBr  (4) CH3CH2CH = CH2 + HBr 
Q.43 The reaction of ethylene with Br2 in water in the presence of NaCl gives -
(1) 1,2-dibromoethane (2) 2-bromoethanol
(3) 1-bromo-2-chloroethane (4) all of these
ALKYNE
Q.44 2-Butyne and 1-Butene show resemblance in all except –
(1) Both decolourise alkaline KMnO4 (2) Both turn bromine water colourless
(3) Both undergo addition reaction (4) Both from white precipitate with Tollen’s reagent
Q.45 Degradation of propyne is affected by all except –
(1) Ozonolysis (2) Oxidation by KMnO4
(3) Combustion (4) Red-hot iron tube
Q.46 Consider the reaction : CH  CH(g) + 2 HCl(g)  product. The product of this reaction is :
(1) CH3CHCl2(g) (2) CH2 = CHCl(g) (3) CHCl = CHCl(g) (4) CH2ClCH2Cl(g)
Q.47 Ethylidene diacetate on strong heating produces -
(1) ethyl acetate and ethylene (2) acetylene and acetic acid
(3) acetic anhydride and acetaldehyde (4) ethylene and acetic acid
Q.48 The end product of the following sequence is :
H2SO 4
CaO + C Heat
 A H
2O
  B HgSO
  C
4

(1) ethanol (2) ethyl hydrogen sulphate

(3) acetaldehyde (4) ethylene glycol

Q.49 The reduction of an alkyne to alkene using Lindlar's catalyst results into -
(1) Cis addition of hydrogen atoms
(2) Trans addition of hydrogen atoms
(3) A mixture obtained by cis and trans additions of hydrogen which are in equilibrium with each
other
(4) A mixture obtained by cis and trans additions of hydrogen atoms which are not in equilibrium
with each other

Q.50 The products obtained via oxymercuration (HgSO4 + H2SO4) of 1-butyne would be -
(1) CH2CH2COCH3 (2) CH3CH2CH2CHO
(3) CH3CH2CHO + HCHO (4) CH3CH2COOH + HCOOH
Catalyst
Q.51 The catalyst required for the reaction : HC  CH + dil. H2SO4  CH3CHO is -
(1) HgSO4 (2) Pd (3) Pt (4) AlCl3

Q.52 When acetylene is passed into a solution of cuprous chloride and ammonium chloride, it forms finally:
(1) Vinyl acetylene (2) Divinyl acetylene (3) Chloroprene (4) Poly vinyl acetylene

Q.53 Which of the following reagents cannot be used to locate the position of triple bond in
CH3 – C  C – CH3 -
(1) Br2 (2) O3 (3) Cu2+ (4) KMnO4

ALKANE, ALKENE, ALKYNE [69]

Q.54 The reduction of an alkyne to alkene using lithium metal in liquid ammonia as solvent results into-
(1) cis addition of hydrogen atoms
(2) trans addition of hydrogen atoms
(3) Both cis and trans additions of hydrogen atoms. The relative amounts of the two depends on
temperature.
(4) Both cis and trans additions of hydrogen atoms. The relative amounts depend on the nature of
alkyne.

Q.55 Which of the following statements is correct ?

(1) Alkynes are more reactive than alkenes towards halogen additions
(2) Alkynes are less reactive than alkenes towards halogen additions.
(3) Both alkynes and alkenes are equally reactive towards halogen additions.
(4) Primary vinylic cations (RCH = CH+) is more reactive than secondary vinylic cation ( RC+ = CH2)

Q.56 A gas which reacts with aqueous KMnO4 solution but does not give precipitate with ammonical Cu2Cl2
solution is :
(1) ethylene (2) methane (3) ethane (4) acetylene

Q.57 Select the structure of the major product formed in the following reaction.

(1) (2) (3) (4)

Q.58 Addition of 2 mol of HCl to 1-butyne would yield :

(1) CH3CH2CH2CHCl2 (2) CH3CH2CCl2CH3
(3) CH3CH2CHClCH2Cl (4) CH3CH2CH=CHCl

Q.59 Which of the following reductions of an alkyne is not correct :

2H
2  CH3CH2CH2CH2CH3
(1) CH3CH2CCCH3 
Pt

2 H
(2) CH3CH2CCCH3 NiB 2

Li
(3) CH3CH2CCCH3  
liq. NH3

(4) All of the above are correct

Q.60 C3H8 + Cl2   C3H7Cl + HCl is an example of which of the following types of reactions -
Light

(1) Substitution (2) Elimination (3) Addition (4) Rearrangement

Q.61 Alkane is prepared by -

(1) Wurtz (2) Reduction to alkyl halide
(3) By grignard reagent (4) All the above

Q.62 Butene-1-may be converted to butane by reaction with -

(1) Pd/H2 (2) Zn - HCl (3) Sn - HCl (4) Zn - Hg.

ALKANE, ALKENE, ALKYNE [70]

Q.63 On mixing a certain alkane with chlorine and irradiating it with ultraviolet light, it forms only one
monochloroalkane.
This alkane could be -
(1) neopentane (2) propane (3) pentane (4) isopentane

Q.64 W hich one of the following is reduced with Zn & HCl to give the corresponding hydrocarbon :
(1) Butane-2-one (2) A cetic acid (3) A cetamide (4) Ethyl acetate

Q.65 A lkyl halides react with dialkyl copper reagent to give

(1) alkyl copper halides(2) alkenes (3) alkeny halides (4) alkanes

Q.66 W hich set of products is expected on reductive ozonolysis of the following olefins -

(1) CH3CHO ; CH3COCH = CH2 (2) CH3CH = C(CH3)CHO ; CH2O

(3) CH3CHO ; CH3 COCHO ; CH2O (4) CH3CHO ; CH3COCH3 ; CH2O

Q.67 When propene is treated with HBr in the dark and in absence of peroxide the main product is -
(1) 1-Bromopropane (2) 2-Bromopropane
(3) 1, 2-Dibromopropane (4) 1, 3-Dibromopropane

Q.68 Ethene is given by the following compound on dehydration -

(1) Ethyl acetate (2) C2H5OH (3) HCHO (4) 1 and 2

Q.69 Favourable conditions of the polymerisation of ethene is -

(1) Only high temperature (2) Only catalyst
(3) Only high pressure (4) High temperature and High pressure

Q.70 Acid catalyzed hydration of alkenes except ethene leads to the formation of
(1) secondary or tertiary alcohol (2) primary alcohol
(3) mixture of secondary and tertiary alcohols (4) mixture of primary and secondary alcohols

Q.71 When an alkyne, RC  CH, is treated with cuprous ion in an ammoniacal medium, one of the products
is -
(1) RC  CCu (2) CuC  CH (3) CuC  CCu (4) RC=CR

(i) X
Q.72 In the reaction CH3—C  C—CH3 (ii)

HO 2
, X is -

(1) HNO3 (2) O2 (3) O3 (4) KMnO4

Q.73 Benzene is a polymer of -

(1) Ethyne (2) Ethylene (3) Methane (4) Ethane

ALKANE, ALKENE, ALKYNE [71]

Q.74 Compound ‘C’ can be distinguished from the other three compounds by the reagent -
(A) CH3C  C—CH3 (B) CH3—CH2—CH2—CH3
(C) CH3CH2C  CH (D) CH3CH=CH2
(1) Bromine in CCl4 (2) Bromine in acetic acid
(3) Alkaline KMnO4 (4) Ammoniacal silver nitrate

Q.75 But-1-ene and propyne are distinguished by -

(1) Baeyers reagent (2) Hinsbergs reagent (3) Tollen’s reagent (4) None

Q.76 Arrange the following compounds in the order of increasing rate of reaction with HCl -

(I) (II) CH3CH = CH2 (III) (IV) CH2 = CH2

(1) IV < II < I < III (2) III < IV < II < I (3) III < I < II < IV (4) I < II < III < IV

ALKANE, ALKENE, ALKYNE [72]

 EXERCISE # III
SECTION–A
Q.1 The compound
H3C  C = CH– CH2 – CH3Vigorous
 oxidation
 product, here product is [AIEEE-2002]
|
H 3C

(1) CH3COOH & CH3 – C –CH3 (2) CH3 – CH2 – COOH & CH3– C  O
|| |
O CH 3
(3) CH3 – CH2 – COOH only (4) HCOOH & CH3 – C – CH3
||
O

Q.2 Reaction
H – C  C – H + HOCl  product, here product will be - [AIEEE-2002]
(1) CHCl2 – CHO (2) CHO – CHO (3) CH – Cl = CHCl (4) CHCl2 – CHCl2

Q.3 Acetylene does not react with - [AIEEE-2002]

(1) NaNH2 (2) NaOH (3) Na metal (4) Ammonical AgNO3

Q.4 Butene-1 may be converted to butane by reaction with – [AIEEE-2003]

(1) Zn – Hg (2) Pd / H2 (3) Zn – HCl (4) Sn – HCl

Q.5 During dehydration of alcohols to alkenes by heating with conc. H2SO4 the initiation step is -
(1) Elimination of water (2) Formation of an ester [AIEEE-2003]
(3) Protonation of alcohol molecule (4) Formation of carbocation

Q.6 On mixing a certain alkane with chlorine and irradiating it with ultraviolet light, it forms only one
monochloroalkane. This alkane could be – [AIEEE-2003]
(1) Isopentane (2) Neopentane (3) Propane (4) Pentane

Q.7 Which one of the following has the minimum boiling point ? [AIEEE-2004]
(1) n – Butane (2) 1– Butyne (3) 1– Butene (4) isobutane

Q.8 Amongst the following compounds, the optically active alkane having lowest molecular mass is-
CH3
|
(1) CH3–CH2–CH2–CH3 (2) CH3  CH2  CH  CH3 [AIEEE-2004]

(3) (4) CH3–CH2 – C  CH

ALKANE, ALKENE, ALKYNE [73]

Q.9 Reaction of one molecule of HBr with one molecule of 1,3–butadiene at 40ºC given predominantly
(1)1–bromo–2–butene under thermodynamically controlled conditions [AIEEE-2005]
(2) 3–bromobutene under kinetically controlled conditions
(3) 1–bromo–2–butene under kinetically controlled conditions
(4) 3–bromobutene under thermodynamically controlled conditions

Q.10 Acid catalyzed hydration of alkenes except ethene leads to the formation of – [AIEEE-2005]
(1) secondary or tertiary alcohol (2) primary alcohol
(3) mixture of secondary and tertiary alcohols (4) mixture of primary and secondary alcohols

Q.11 Reaction of trans 2-phenyl-1 bromocyclopentane on reaction with alcoholic KOH produces
(1) 2-phenylcyclopentene (2) 1-phenylcyclopentene [AIEEE 2006]
(3) 3-phenylcyclopentene (4) 4-phenylcyclopentene

Me

Me
Q.12 N
OH
n-Bu Et
The alkene formed as a major product in the above elimination reaction is [AIEEE 2006]
Me Me
(1) CH2 = CH2 (2) (3) (4) Me

Q.13 Which of the following reactions will yield 2, 2-dibromopropane? [AIEEE 2007]
(1) CH3 – C  CH + 2HBr  (2) CH3CH = CHBr + HBr 
(3) CH  CH + 2HBr  (4) CH3 – CH = CH2 + HBr 

Q.14 In the following sequence of reactions, the alkene affords the compound ‘B’

CH3CH = CHCH3 O

3 A H2
O
 B, The compound B is [AIEEE 2008]
Zn
(1) CH3COCH3 (2) CH3CH2COCH3 (3) CH3CHO (4) CH3CH2CHO

Q.15 The treatment of CH3MgX with CH3C  C – H produces [AIEEE 2008]

H H
(1) CH3C  C – CH3 (2) CH3 C C CH3
(3) CH4 (4) CH3–CH = CH2

Q.16 The hydrocarbon which can react with sodium in liquid ammonia is [AIEEE 2008]
(1) CH3CH2 C CH (2) CH3CH = CHCH3
(3) CH3CH2C  CCH2CH3 (4) CH3CH2CH2C  CCH2CH2CH3

Q.17 Out of the following, the alkene that exhibits optical isomerism is : [AIEEE 2010]
(1) 3-methyl-1-pentene (2) 2-methyl-2-pentene
(3) 3-methyl-2-pentene (4) 4-methyl-1-penten

ALKANE, ALKENE, ALKYNE [74]

Q.18 Which branched chain isomer of the hydrocarbon with molecular mass 72u gives only one isomer of
mono substituted alkyl halide? [AIEEE 2012]
(1) Isohexane (2) Neohexane
(3) Tertiary butyl chloride (4) Neopentane.
Q.19 2–Hexyne gives trans-2-Hexene on treatment with : [AIEEE 2012]
(1) Pd/BaSO4 (2) LiAlH4 (3) Pt/H2 (4) Li/NH3
Q.20 A gaseous hydrocarbon gives upon combustion 0.72 g of water and 3.08 g of CO2. The empirical
formula of the hydrocarbon is: [JEE Main 2013]
(1) C3H4 (2) C6H5 (3) C7H8 (4) C2H4
Q.21 Which compound would give 5 – keto – 2 – methyl hexanal upon ozonolysis ? [JEE Main 2015]
CH3 CH3
CH3 CH3
(1) (2) H3C (3) CH3 (4)
CH3 CH3
Q.22 The product of the reaction given below is: [JEE Main 2016]
1. NBS/hv
X
2. H2O/K2CO3

CO2H OH O

(1) (2) (3) (4)

Q.23 The reaction of propene with HOCl (Cl2 + H2O) proceeds through the intermediate:
[JEE Main 2016]
(1) CH3–CHCl–CH2+ (2) CH3 –CH+–CH2 –OH
(3) CH3– CH+ –CH2 –Cl (4) CH3– CH(OH) – CH2+

Q.24 The major product obtained in the following reaction is : [JEE Main 2017]
Br
H t
BuOK
C6H5

C6H5
(+)
(1) C6H5CH = CHC6H5 (2) (+) C6H5CH (OtBu) CH2C6H5
(3) (–) C6H5CH(OtBu) CH2C6H5 (4) (±) C6H5CH(OtBu) CH2C6H5

Q.25 3-Methyl-pent-2-ene on reaction with HBr in presence of peroxide forms an addition product. The
number of possible stereoisomers for the product is : [JEE Main 2017]
(1) Zero (2) Two (3) Four (4) Six

Q.26 The trans-alkenes are formed by the reduction of alkynes with: [JEE Main 2018]
(1) Na/liq.NH3 (2) Sn-HCl (3) H2-Pd/C, BaSO4 (4) NaBH4

ALKANE, ALKENE, ALKYNE [75]

 (SECTION–B)
Q.1 Alcoholic solution of KOH is a specific reagent for – [IIT ‘90]
(1) Dehydration (2) Dehydrogenation
(3) Dehydro halogenation (4) Dehalogenation

Q.2 Of the following, unsaturated hydrocarbons are – [IIT ‘90]

(1) ethyne (2) cyclohexane (3) n–propane (4) ethene

Q.3 1-chlorobutane on reaction with alcoholic potash gives – [IIT ‘91]

(1) 1–butene (2) 1–butanol (3) 2–butene (4) 2–butanol

Q.4 The hybridisation of carbon atoms in C–C single bond of HCC–CH=CH2 is – [IIT ‘91]
(1) sp3–sp3 (2) sp2–sp3 (3) sp–sp2 (4) sp2–sp2

Q.5 Which of the following compounds will show geometrical isomerism? [IIT ‘98]
(1) 2–butene (2) Propene (3) 1–phenylpropene (4) 2–methyl–2–butene

Q.6 The reaction of CH3– CH=CH OH with HBr gives: [JEE 1998]

(1) CH3CHBrCH2 OH (2) CH3CH2CHBr OH

(3) CH3CHBrCH2 Br (4) CH3CH2CHBr Br

Q.7 In the compound CH2=CH–CH2–CH2–CCH, the C2–C3 bond is of the type – [IIT ‘99]
(1) sp–sp2 (2) sp3–sp3 (3) sp–sp3 (4) sp2–sp3

Q.8 Propyne and propene can be distinguished by – [IIT ‘2000]

(1) conc. H2SO4 (2) Br2 in CCl4 (3) dil. KMnO4 (4) AgNO3 in ammonia

Q.9 Which one of the following alkenes will react fastest with H2 under catalytic hydrogenation condition –
[IIT ‘2000]

(1) (2) (3) (4)

Q.10 In the presence of peroxide, hydrogen chloride and hydrogen iodide do not give anti–Markovnikov
addition to alkene because – [IIT ‘2001]
(1) both are highly ionic
(2) one is oxidising and the other is reducing
(3) one of the step is endothermic in both the cases
(4) All the steps are exothermic in both cases

ALKANE, ALKENE, ALKYNE [76]

Q.11 The reaction of propene with HOCl proceeds via the addition of – [IIT ‘2001]
(1) H+ in first step (2) Cl+ in first step
(3) OH– in first step (4) Cl+ and OH– in single step

Q.12

Hydrogenation of the above compound in the presence of poisoned paladium catalyst gives –
(1) An optically active compound (2) An optically inactive compound [IIT ‘2001]
(3) A racemic mixture (4) A diastereomeric mixture

Q.13 Consider the following reactions – [IIT ‘2002]

H 3C  CH  CH  CH 3 + ‘X’ + HBr
| |
D CH 3
Identify the structure of the major product ‘X’
• •
H C  CH  CH  C H 2 H C  CH  C CH 3
(1) 3 | | (2) 3 | |
D CH 3 D CH 3
• •
(3) H 3C  C CH  CH 3 (4) H 3C  C H  CH  CH 3
| | |
D CH 3 CH 2

Q.14 The nodal plane in the –bond of ethene is located in – [IIT ‘2002]
(1) the molecular plane
(2) a plane parallel to the molecular plane
(3) a plane perpendicular to the molecular plane which contains the carbon–carbon –bond at right angle
(4) a plane perpendicular to the molecular plane which contains the carbon–carbon –bond

Q.15 Identify a reagent from the following list which can easily distinguish between 1–butyne and 2-butyne-
(1) bromine, CCl4 (2) H2, Lindlar catalyst [IIT ‘2002]
(3) dilute H2SO4, HgSO4 (4) ammonical Cu2Cl2 solution

HgSO
4  A
Q.16 C6H5–CC–CH3   [IIT ‘2003]
H 2SO 4

(1) (2) (3) C 6 H 5  C  CHCH 3 (4) C 6 H 5  CH  C  CH 3

| |
OH OH

ALKANE, ALKENE, ALKYNE [77]

 
Q.17 H x Br2
 H 2O ( mixture)  5 compounds of molecular formula C4H8Br2

Number of compounds in X will be: [IIT ‘2003]
(1) 2 (2) 3 (3) 4 (4) 5

Q.18 2–hexyne can be converted into trans–2–hexene by the action of : [IIT ‘2004]
(1) H2–Pd-BaSO4 (2) Li in liq. NH3 (3) H2–PtO2 (4) NaBH4

Q.19 Cyclohexene is best prepared from cyclohexanol by which of the following:

(1) conc. H3PO4 (2) conc. HCl/ZnCl2 (3) conc. HCl (4) conc. HBr [IIT ‘2005]

Q.20 When Phenyl Magnesium Bromide reacts with tert. butanol, which of the following is formed?
(1) Tert. butyl methyl ether (2) Benzene
(3) Tert. butyl benzene (4) Phenol [IIT ‘2005]

Q.21 1–bromo–3–chlorocyclobutane when treated with two equivalents of Na, in the presence of ether which
of the following will be formed? [IIT ‘2005]

(1) (2) (3) (4)

Q.22 CH3–CH=CH2 + NOCl  P [IIT 2006]

Identify the adduct.

(1) CH 3  CH  CH 2 (2) CH 3  CH  CH 2
| | | |
Cl NO NO Cl

NO
| CH  CH 2  CH 2
(3) CH 3  CH 2  CH (4) | 2 |
| NO Cl
Cl

Q.23 The number of stereoisomers obtained by bromination of trans-2-butene is [IIT 2007]

(1) 1 (2) 2 (3) 3 (4) 4

Q.24 The number of structural isomers for C6H14 is [IIT 2007]

(1) 3 (2) 4 (3) 5 (4) 6

ALKANE, ALKENE, ALKYNE [78]

Q.25 The reagent(s) for the following conversion, [IIT 2007]

?

 H H
is / are
(1) alcoholic KOH (2) alcoholic KOH followed by NaNH2
(3) aqueous KOH followed by NaNH2 (4) Zn / CH3OH

Q.26 In the following carbocation, H/ CH3 that is most likely to migrate to the positively charged carbon is

H H
1 2 + 4 5
H3C—C—C—C—CH3
3 [JEE 2009]
HO H CH3

(1) CH3 at C-4 (2) H at C-4 (3) CH3 at C-2 (4) H at C-2

Q.27 The synthesis of 3-octyne is achieved by adding a bromoalkane into a mixture of sodium amide and an
alkyne. The bromoalkane and alkyne respectively are [IIT 2010]
(1) BrCH2CH2CH2CH2CH3 and CH3CH2CCH
(2) BrCH2CH2CH3 and CH3CH2CH2CCH
(3) BrCH2CH2CH2CH2CH3 and CH3CCH
(4) BrCH2CH2CH2CH3 and CH3CH2CCH

Q.28 The number of optically active products obtained from the complete ozonolysis of the given compound
is CH H [JEE 2012]
3

CH3–CH=CH–C–CH=CH–C–CH=CH–CH3

H CH3

(1) 0 (2) 1 (3) 2 (4) 4

Q.29 In allene (33H4), the type(s) of hybridisation of the carbon atoms is (are) [JEE 2012]
(1) sp and sp 3 (2) sp and sp 2 (3) only sp2 2
(4) sp and sp 3

Q.30 Isomers of hexane, based on their brancing, can be divided into three distinct classes as shown in the
figure. [JEE Advance 2014]

I and II and III

The correct order of their boiling point is

(1) I > II > III (2) III > II > I (3) II > III > I (4) III > I > II

ALKANE, ALKENE, ALKYNE [79]

Q.31 In the following reaction, the major product is : [JEE Advance 2015]
CH3
1 equivalent HBr
CH2
H2C
CH3 CH3
CH3
(1) H C (2)
2 H3C
Br Br
CH3 CH3
(3) (4)
H2C Br H3C Br
Q.32 Compound(s) that on hydrogenation produce(s) optically inactive compound(s) is(are)
[JEE Advance 2015]
H H
Br Br
(1) H C (2)
3
CH3 H2C CH3

H Br
H2C Br H
(3) CH3 (4)
H2C CH3
CH3

Q.33 The number of hydroxyl group(s) in Q is [JEE Advance 2015]

+
H aqueous dilute KMnO4(excess)
H P Q
heat 0°C
HO
H3C CH3

Paragraph for question nos. 34 & 35

In the following reactions : [JEE Advance 2015]

C8 H 6 Pd
BaSO 4
 C8 H8 i
. B2H 6
  X
H2 ii. H 2O 2 , NaOH , H 2O
 H 2O
 HgSO 4 , H 2SO 4
C8 H 8O i
. EtMgBr , H 2O
      Y
ii. H  , Heat

Q.34 Compound X is
O OH
(1) CH3 (2) CH3

OH
CHO
(3) (4)

ALKANE, ALKENE, ALKYNE [80]

Q.35 The major compound Y is

CH2 CH3
CH3
(1) CH3 (2) CH3 (3) (4) CH3

Q.36 In the following monobromination reaction, the number of possible chiral products is
[JEE Advance 2016]
CH2CH2CH3
H Br
CH3 Br (1.0 mole)
2 300
ºC 
(1.0 mole)
(enantiomerically pure)
Q.37 The correct statement(s) for the following addition reactions is(are) [JEE Advance 2017]
H3C H H 3C CH3
Br / CHCl
(i) 2  
3  M and N (ii) Br / CHCl3
2    O and P
H CH3 H H
(1) O and P are identical molecules
(2) Bromination proceeds through trans-addition in both the reactions
(3) (M and O) and (N and P) are two pairs of enantiomers
(4) (M and O) and (N and P) are two pairs of diastereomers
Q.38 Compounds P and R upon ozonolysis produce Q and S, respectively. The molecular formula of Q and
S is C8H8O. Q undergoes Cannizzaro reaction but not haloform reaction, whereas S undergoes haloform
reaction but not Cannizzaro reaction. [JEE Advance 2017]
( i ) O 3 / CH 2 Cl 2
(i) P    Q
(ii ) Zn / H O
2 ( C8H8O )
( i ) O 3 / CH 2 Cl 2
(ii) R    S
(ii ) Zn / H O
2 ( C8 H 8 O )

The option(s) with suitable combination of P and R, respectively, is(are)

(1) H3C and

CH3

H 3C
CH3 CH3
and
(2)
CH3
H3 C

H3 C CH3
CH3 CH3
and
(3)
CH3
CH3

(4) H3C and H3C

CH3
ALKANE, ALKENE, ALKYNE [81]
 ANSWER KEY
EXERCISE # I
Q.1 1 Q.2 1 Q.3 2 Q.4 1 Q.5 2 Q.6 4 Q.7 1
Q.8 3 Q.9 2 Q.10 4 Q.11 1 Q.12 2 Q.13 3 Q.14 2
Q.15 3 Q.16 2 Q.17 1 Q.18 1 Q.19 4 Q.20 1 Q.21 2
Q.22 3 Q.23 2 Q.24 3 Q.25 1 Q.26 3 Q.27 4 Q.28 3
Q.29 4 Q.30 3 Q.31 4 Q.32 1 Q.33 3 Q.34 1 Q.35 4
Q.36 3 Q.37 1 Q.38 2 Q.39 1 Q.40 4 Q.41 4 Q.42 4
Q.43 3 Q.44 1 Q.45 4 Q.46 3 Q.47 2 Q.48 2 Q.49 4
Q.50 2 Q.51 1 Q.52 2 Q.53 1 Q.54 2 Q.55 2 Q.56 2
Q.57 4 Q.58 1 Q.59 1 Q.60 1 Q.61 1 Q.62 3 Q.63 3
Q.64 4 Q.65 4 Q.66 2 Q.67 3 Q.68 1 Q.69 3 Q.70 3
Q.71 3 Q.72 1 Q.73 2 Q.74 4 Q.75 3

EXERCISE # II
Q.1 3 Q.2 3 Q.3 4 Q.4 4 Q.5 2 Q.6 2 Q.7 3
Q.8 2 Q.9 3 Q.10 3 Q.11 1 Q.12 1 Q.13 1 Q.14 2
Q.15 2 Q.16 1 Q.17 4 Q.18 1 Q.19 3 Q.20 2 Q.21 3
Q.22 1 Q.23 2 Q.24 3 Q.25 1 Q.26 2 Q.27 1 Q.28 1
Q.29 3 Q.30 4 Q.31 3 Q.32 1 Q.33 4 Q.34 4 Q.35 2
Q.36 2 Q.37 3 Q.38 3 Q.39 3 Q.40 1 Q.41 3 Q.42 1
Q.43 4 Q.44 4 Q.45 4 Q.46 1 Q.47 3 Q.48 3 Q.49 1
Q.50 1 Q.51 1 Q.52 2 Q.53 3 Q.54 2 Q.55 2 Q.56 1
Q.57 4 Q.58 2 Q.59 3 Q.60 1 Q.61 4 Q.62 1 Q.63 1
Q.64 1 Q.65 4 Q.66 3 Q.67 2 Q.68 2 Q.69 4 Q.70 1
Q.71 1 Q.72 3 Q.73 1 Q.74 4 Q.75 3 Q.76 1

EXERCISE # III
SECTION – A
Q.1 2 Q.2 1 Q.3 2 Q.4 2 Q.5 3 Q.6 2 Q.7 4
Q.8 3 Q.9 1 Q.10 3 Q.11 3 Q.12 1 Q.13 1 Q.14 3
Q.15 3 Q.16 4 Q.17 3 Q.18 4 Q.19 4 Q.20 3 Q.21 4
Q.22 3 Q.23 3 Q.24 1 Q.25 3 Q.26 1
SECTION – B
Q.1 3 Q.2 1 Q.3 1 Q.4 3 Q.5 1,3 Q.6 2 Q.7 4
Q.8 4 Q.9 1 Q.10 3 Q.11 2 Q.12 2 Q.13 2 Q.14 1
Q.15 4 Q.16 1 Q.17 2 Q.18 2 Q.19 1 Q.20 2 Q.21 4
Q.22 1 Q.23 1 Q.24 3 Q.25 2 Q.26 4 Q.27 4 Q.28 1
Q.29 2 Q.30 2 Q.31 4 Q.32 2,4 Q.33 4 Q.34 3 Q.35 4
Q.36 5 Q.37 2,4 Q.38 1,2

ALKANE, ALKENE, ALKYNE [82]