Semester - I C) Aliphatic Hydrocarbons by Dr Pramod R Padole

1
B.Sc. First year
Students
Semester – I
C) Aliphatic Hydrocarbons
by
Dr Pramod R Padole

Organic Chemisrtry
Unit-III
A) Electronic Displacements:
B) Reactive Intermediates:
C) Aliphatic Hydrocarbons:

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Organic compounds containing carbon
and hydrogen only are called
hydrocarbons.

Do you know?
What are Aliphatic Hydrocarbons?

What are Aliphatic Hydrocarbons?
 Aliphatic hydrocarbons are the organic molecules containing Carbon (C)
and Hydrogen (H) atoms in their structure; in straight chains, branched
chains or non-aromatic rings.
 Aliphatic hydrocarbons can be categorized into three main groups;
alkanes, alkenes and alkynes.
Two C=C bonds
ALKADIENES or
Dienes:
X

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By Dr Pramod R Padole
First Second Third Fourth

LOGO
Alkanes
organic compounds containing carbon and hydrogen only
Saturated Hydrocarbons

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i) Alkanes:
(Saturated Hydrocarbons)
Alkanes:
The open chain saturated organic compounds
containing carbon and hydrogen only are
called as alkanes.
(i.e., Characteristic of C-C & C-H linkage)
They have the general formula CnH2n+2.
Where, n = 1,2,3,…, etc.
Examples:
1) CH4 - Methane, 2) C2H6 – Ethane,
3) C3H8 – Propane, 4) C4H10 – Butane,…… etc.

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In general, alkanes are non-reactive (inert)
compared to other classes of organic compounds.
1
They don’t
have
a functional
group
2
There bonds are
quite strong,
i.e., C-C (strong
bond).
Large energy is
needed/required
to break
the bond hence
less reactive
3
These two bonds are
almost non-polar and
therefore, neither
electrophilic nor
nucleophilic substitution
reaction can take place.
Can’t react with electron-
loving species or a proton
loving species
(Nucleophilic)
Reasons Reasons Reasons

i) Alkanes:
(Saturated Hydrocarbons)
 Alkanes contain strong C-C and C-H covalent bonds.
 Therefore, this class of hydrocarbons
is relatively chemically inert.
 Hence they are also called as Paraffins
(Latin, parum = little, affinis = affinity; Chemical attraction).
 Sources: The most important source of alkanes is petroleum or
the natural gas associated with it. Fuel gases obtained from coal,
 e.g. Coal gas, etc. contain these hydrocarbons in small amounts.
 Methane (called as Marsh gas) is formed during the decay of
plants or animal tissues.
 Butane is used as a fuel in lighters.
It is also used in same
camping stoves.

Why methane is called marsh gas?
Methane (called as Marsh gas) is formed during the decay
of plants or animal tissues.

LOGO
Methods of Preparation:
Reactions that produce a particular functional group are called
preparations.

Methods of Preparation:
i) Wurtz Reaction
or Synthesis:
or Preparation of
Higher
symmetrical
alkanes:
Preparation
Of
Alkanes
ii) Corey-House
Reaction or Synthesis:
or Preparation of Higher
unsymmetrical alkanes:
or
Coupling of alkyl halides
with organometallic
compounds:

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Wurtz Reaction:
or Preparation of Higher symmetrical alkanes:
The Wurtz reaction is a very useful reaction in the fields of organic
chemistry and organometallic chemistry for the formation of alkanes.
In this reaction, two different alkyl halides are coupled to yield a
longer alkane chain with the help of sodium and dry ether solution.
Unhydrous Diethyl ether (dry ether) is an especially
good solvent for the formation of Higher
symmetrical alkanes using Wurtz Reaction
because ethers are non-acidic (aprotic).

i) Wurtz Reaction or Synthesis:
Q.1) What happens when alkyl halide is reacted with sodium metal in presence of
ether?
Q.2) How will you convert : Alkyl halide to Higher alkanes.
Q.3) Explain: Wurtz reaction with suitable example. (W-17, 4 Mark)
Q.4) What happen when, methyl bromide is reacted with sodium metal in
presence of dry ether? (W-13 & S-16, 2 Mark)
Q.5) What happen when, ethyl bromide is reacted with sodium metal in presence
of dry ether? (W-14, 2 Mark)
Q.6) How will you prepare the Butane from ethyl chloride? (Wurtz Reaction)
(S-18, 2 Mark)
Q.7) How will you convert: Methyl chloride to Propane? (W-19, 2 Mark)
Wurtz Reaction:
When alkyl halide (two molecules) is reacted or
treated with sodium metal in presence of dry ether
as a solvent; to form higher alkanes.

i) Wurtz Reaction or Synthesis:
{Note that: This method is particularly suitable for the preparation of symmetrical alkanes.}

LOGO
ii) Corey-House Reaction or
Synthesis:
or
Preparation of Higher unsymmetrical
alkanes:
or
Coupling of alkyl halides with organometallic
compounds:
alkyl halide is reacted with lithium metal in presence of solvated ether

Corey-House Reaction or Synthesis:
(also called the Corey–Posner–Whitesides–House
reaction and other permutations)
 Corey-House Reaction:
This method was developed in late 1960s by E.J. Corey
and Herbert House, called as Corey-House synthesis.
 The coupling reaction is a good synthetic way to join two
alkyl groups together to produce higher alkanes.
 This versatile method is known as the Corey-House
reaction. (also called the Corey–Posner–Whitesides–
House reaction and other permutations)
Corey
Q.1) What happens when alkyl halide is reacted with lithium metal in presence
of solvated ether?
Q.2) How will you convert :- i) Alkyl halide to Higher alkanes.
Q.3) Explain: Corey-House reaction with suitable example.
Q.4) How will you prepare the Propane from ethyl bromide?
(Corey-House Reaction) (S-18, 2 Mark)
Q.5) How will you convert: Methyl chloride to Propane? (W-19, 2 Mark)

 When alkyl halide is reacted or treated with
lithium metal in presence of ether as a
solvent; to form first alkyl lithium, further
it is reacted with cuprous halide; to form lithium dialkyl
cuprate (LiR2Cu). This is then treated with the second alkyl
halide; to form a new alkane as a final product along with
organocopper compound (R-Cu) and a lithium halide (LiX).
Corey
Where, R’-X should be a primary halide (second alkyl halide);
The alkyl group R- in the organometallic may be primary, secondary, or tertiary.
{Note that: This method is particularly suitable for the preparation of unsymmetrical alkanes.}

 When alkyl halide is reacted or treated with lithium metal in presence of ether as a
solvent; to form first alkyl lithium, further it is reacted with cuprous halide; to form
lithium dialkyl cuprate (LiR2Cu). This is then treated with the second alkyl halide;
to form a new alkane as a final product along with organocopper compound (R-Cu)
and a lithium halide (LiX).
Corey
CH3
-Br + Li CH3
-Li + LiBr
Example:
ether
2
Methyl bromide Methyl lithium
Lithium
1)
CH3 -Li + CuI Li(CH3)2Cu + Li-I
CH3CH2 -Br + Li(CH3)2Cu CH3 -CH2CH3 + CH3Cu + Li-Br
2)
Methyl lithium
Cuprous iodide
Lithium dimethyl
cuperate
3)
Ethyl bromide Lithium dimethyl
cuperate
Propane
2
(second alkyl halide)

Do you know?
Why is Corey House reaction a better method for preparing
alkane than Wurtz reaction?

Q.1) Why is Corey House reaction a
better method for preparing alkane
than Wurtz reaction?
From Wurtz reaction we can prepare only even
carbon number of alkane but from
Corey House reaction
we can prepare odd and even carbon number alkane.
So Corey House reaction is preferred over Wurtz
reaction.
1) First four alkanes methane, ethane, propane and butane are gases.
Next thirteen members (C5 to C17) are colourless liquids.
Higher alkanes are wax like solids.
2) Alkanes are nonpolar compounds.

Chemical Reactions of Alkanes:
Halogenation
(Addition of Halogen): 2) Aromatistion:
Formation of
Aromatic compound:
Chemical Reactions:
a) Chlorination
b) Reaction with
Iodine
in presence of
HIO3:
i) Reaction with
Chlorine:
ii) Reaction with
Excess of Chlorine:

Replacement of hydrogen atoms from alkane by halogen atom is
known as Halogenation
i) Reaction with
Chlorine
in presence of U.V.
light or heat at
623-673 K:
Chlorination
Of
Alkanes
ii) Reaction with
Excess of
Chlorine
in presence of U.V. light or
heat at
623-673 K:

a) Chlorination:
1) Reaction with Chlorine:
 When alkane is treated or heated or reacted with chlorine
in presence of UV light or at high temperature (623-673 K);
to form haloalkane.
Q.1) What happens when methane is treated with Cl2 in presence of U.V. light or heat at 623-673 K?
(W-06, W-11 & S-15, 2 Mark)
Q.2) Compete the following reaction: (W-17, 2 Mark)
Q.3) How will you prepare Ethyl chloride from Ethane? (S-19, 2 Mark)

a) Chlorination:
1) Reaction with Chlorine:
 When alkane is treated or heated or reacted with chlorine in presence of UV
light or at high temperature (623-673 K); to form haloalkane.
Q.1) What happens when methane is treated with Cl2 in presence of U.V. light or heat at 623-673 K?
(W-06, W-11 & S-15, 2 Mark)
Q.2) Compete the following reaction: (W-17, 2 Mark)
Q.3) How will you prepare Ethyl chloride from Ethane? (S-19, 2 Mark)

LOGO
Reaction with excess of
Chlorine:
excess of chlorine in presence of UV light or
at high temperature (623-673 K)
Reaction with excess of Chlorine:

Reaction with excess of Chlorine:
 When methane is treated or heated or reacted with excess of chlorine in
presence of UV light or at high temperature (623-673 K); hydrogen atoms of
methane can be replaced by chlorine atoms one by one; to form different
product of chloromethane.
CH3-H + Cl-Cl CH3
-Cl + H-Cl
UV light
or High Temp.
Examples:
1)
623-673 K
Methane Methyl
chloride
excess
CH3-Cl + Cl-Cl CH2
Cl2
+ H-Cl
UV light
or High Temp.
2)
623-673 K
excess
Methyl
chloride
Methylene dichloride
CH2
Cl2
+ Cl-Cl CHCl3
+ H-Cl
CHCl3
+ Cl-Cl CCl4 + H-Cl
UV light
or High Temp.
3)
623-673 K
excess
Methylene dichloride Chloroform
4)
623-673 K
excess
Chloroform
UV light
or High Temp.
Carbon
tetrachloride

b) Iodination (in presence of HIO3):
Q.1) What happens when- i) Methane is treated with iodine in presence of HIO3 (Oxidizing Agent)
(S-04, W-04, S-05 & W-06, 1-2 Mark)
Q.2) Complete the reaction: CH4 + I2 HIO3 ? (W-09, 2 Mark)
Q.3) How will you convert: Methane to iodomethane? (S-10, 1 Mark)

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b) Iodination (in presence of HIO3):
When alkane is treated or reacted with iodine in
presence of strong oxidizing agent, such as HIO3,
HNO3 or HgO, etc.; to form alkyl iodide.
R -H + I -I + HIO3
R -I + H2O
CH3 -H + I -I + HIO3 CH3 -I + H2O
5 2 5 3
Alkane Alkyl iodide
General Reaction:
Examples: 1) 5 2 5 3
Methane Methyl iodide
5 C2H5 -H + I -I + HIO3
C2H5 - I + H2O
2 5 3
Ethyl iodide
Ethane
2)

LOGO
Mechanism of
Chlorination of
Methane

LOGO
Mechanism of Chlorination of Methane:
 When methane is treated or heated or reacted
with chlorine in presence of UV light or at high
temperature (623-673 K); to form chloro-methane
or methyl chloride.
CH3-H + Cl-Cl CH3
-Cl + H-Cl
Examples:
623-673 K
UV light
or High Temp.
Methane Methyl
chloride
Q.1) Discuss / Explain the mechanism of chlorination of Methane.
(S-04, S-06, W-07, W-13, W-15 & S-17, 4 Mark)
Q.2) Discuss the mechanism of the following reaction: (S-05, 4 Mark)
CH4 + Cl2 U.V light CH3Cl + HCl
Q.3) Discuss / Explain free radical mechanism of chlorination of Methane.(S-07, S-08, W-09 & W-10, 3 Mark)

LOGO
Mechanism of Chlorination of Methane:
1) Homolytic fission can be done by using UV light or at high
temperature or peroxide.
2) Generally, a covalent bond between two same atoms or two atoms
having nearly same eletronegativity, undergo homolysis.
3) Generally, homolysis is favoured by following factors:
a) Gaseous state of reactants.
b) At high temperature or in presence of sunlight, UV light, etc.
c) Non-polar solvents.
d) In presence of peroxide.
Fission
Homolytic fission or Homolysis
(Homo = same, lysis = breaking)
Heterolytic fission or Heterolysis
(Hetero = different, lysis = breaking)

Mechanism: (Free Radical Mechanism)
 Chlorination of methane follows Free radical mechanism and involved
three steps.
Step-I) Chain Initiation: where a radical species is generated, generally by
heat, light, or other catalytic process.
When chlorine molecule is heated in presence of UV light or at high
temp. (623-673 K), undergoes homolytic fission; to form chlorine free
radicals.
Cl-Cl Cl + Cl
* *
623-673 K
UV light
or High Temp.
Chlorine
molecule Chlorine free radical
Homolysis

Step-II) Chain Propogation: here one radical species interacts with
another molecule to create another radical species.
a) When chlorine free radical attacks on methane; to form
methyl free radical.
Cl*
CH3
-H CH3
+ H-Cl
*
Chlorine free radical
+
Methane Methyl
F.R.

Step-II) Chain Propogation: here one radical species interacts with
another molecule to create another radical species.
b) When methyl free radical is reacted with chlorine; to form
methyl chloride & Cl free radical.
CH3 + Cl-Cl
*
CH3-Cl + Cl
*
Methyl chloride Chlorine free radical
Methyl
F.R.
In above reaction, regeneration of chlorine free radical is take place.
This two reactions (a & b) repeated over & over again.
So, this step is known as chain propogation step.

Step-III) Chain Termination: where two radical species interact and
quench the radical reaction an form a stable product.
Chain reaction can be stop or end by the combination of two F.R.
as shown below,
Cl + Cl Cl2
CH3
+ Cl CH3
-Cl
CH3
+ CH3
CH3
-CH3
i)
* *
ii)
* *
iii)
* *
The final stage of a radical reaction is the termination reaction which
quenches the radical species present. For the methane – chlorine
reaction this is the formation of the chloromethane (CH3Cl).

LOGO
Aromatization:
Formation of Aromatic Compound:
Cr2O3 catalyst
supported over Al2O3
at 873 K (600oC)

Aromatization:
Formation of Aromatic compound:
Q.1) Explain the term:- Aromatisation of an alkane with a suitable example.
(S-04, W-04, W-05, S-06, W-06, W-08, S-09, S-10, W-10, S-11 & S-15, 2 Mark)
Q.2) How can you bring the following conversions? (S-07 & S-09, 2 Mark)
i) Benzene from n-Hexane.
Q.3) How will you convert: n-hexane to benzene? (W-11 & S-18, 2 Mark)
Q.4) Complete the following reaction:
Q.5) Write short note on: Aromatisation of an alkane. (W-16, 2 Mark)
Q.6) What happen when n-hexane heated with Cr2O3 supported over alumina?(S-17, 2 Mark)

Aromatization:
Formation of Aromatic compound:
 e.g. 1) When vapours of n-hexane is passed over Cr2O3
catalyst supported over Al2O3 at 873 K (600oC); to form
Benzene (& H2 is set free).
e.g. 2) When vapours of n-heptane is passed over Cr2O3
catalyst supported over Al2O3 at 873 K (600oC); to form
Toluene (& H2 is set free).
CH3
CH2
CH2
CH2
CH2
CH3 Cr2
O3
/ Al2
O3
n-Hexane
, at 873 K
+ 4 H2
Benzene
CH2
CH2
CH2
CH2
CH2
CH3 Cr2
O3
/ Al2
O3
CH3
CH3
n-Heptane
, at 873 K
+ 4 H2
Toluene

 Aromatization in alkane can be brought by using Cr2O3
catalyst supported over Al2O3 at 873 K (600oC).
 This process is known as reforming.
 In this reaction, one hydrogen atom is removed from each
terminal carbon atom of n-hexane to form cyclohexane.
This is cyclization reaction.
 This is followed by the removal of three hydrogen
molecules to form benzene. This is aromatization
reaction.

D. Con. H2SO4
B. Fe2O3
A. Al2O3
Aromatization in alkane can be
brought by using:
C. Cr2O3 /Al2O3 at 873 K (600oC)

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ii) Alkenes:
Alkenes are also called olefins
because the lower members forms an oily products
on treatment with chlorine or bromine.
(Latin word : Olefiant - oil forming) or
( Oleum = oil ; fiacre = to make).

ii) ALKENES:
Q.1) Define Alkenes.
Q.2) Aliphatic unsaturated hydrocarbon containing one carbon-carbon double bond (C = C bond) is
called _____. (W-11 & S-15, ½ Mark)
Defination:
Aliphatic unsaturated hydrocarbons containing one carbon-carbon
double bond (C = C bond) are called alkenes. (W-11, ½ Mark)
Or
Aliphatic unsaturated hydrocarbon containing one carbon-carbon
double bond (C = C bond) are called alkenes. (S-15, ½ Mark)
They are represented by the general formula CnH2n
Where, n is the number of C-atoms & n = 2, 3, 4, etc.
e.g. 1) CH2= CH2 2) CH3CH= CH2
Ethylene Propylene
Alkenes are also called olefins because the lower members forms
an oily products on treatment with chlorine or bromine. (Latin word :
Olefiant - oil forming) or ( Oleum = oil ; fiacre = to make).

Q.1) Write the IUPAC / Common names.
According to the common name system olefins are named as
'alkylenes'.
In these suffix “ane” of the corresponding alkane is replaced by “ylene”
In the IUPAC system their names are obtained by replacing the suffix
'ane' of an alkane by 'ene'.

Sources of Alkenes or Preparation of Alkenes:
Dehydration of
alcohols:
a) By using conc. H2SO4
(as a dehydrating agent):
(Liquid Phase Dehydration)
&
b) By using Alumina,
(Al2O3):
(Vapour Phase Dehydration)
Alkenes
Preparation:
Dehydrohalogenation
of
Alkyl Halides:

Dehydration of Alcohols:
Preparation of Alkenes form Alcohols:
Dehydration of
alcohols:
a) By using
conc. H2SO4 :
(as a dehydrating
agent):
(Liquid Phase
Dehydration)
Dehydration
of
Alcohol:
Dehydration of
alcohols:
b) By using
Alumina,
(Al2O3):
(Vapour Phase
Dehydration)

Preparation of Alkenes form Alcohols
or Dehydration of Alcohols:
1) Dehydration of alcohols:
Dehydration means removal of water molecule from adjacent
C-atoms (α , β carbon atoms) and produces alkenes.
Dehydration of alcohol is a β-elimination reaction.
 Alcohols are dehydrated by conc. H2SO4 or Alumina (Al2O3) or
H3PO4 or P2O5.
 The decreasing order of reactivity of alcohols in dehydration is,
Tertiary (3o) > Secondary (2o) > Primary (1o)
Q.1) How are alkenes prepared from alcohols? Give mechanism.
Q.2) Discuss the mechanism of dehydration of alcohols.
Q.3) Explain the mechanism of dehydration of ethyl alcohol (ethanol). (S-12 & S-13, 4 Mark)
Q.4) Give the preparation of ethylene form ethyl alcohol and discuss its mechanism. (S-14, 4 M)
Q.5) How will you prepare 1-Propene from 1-Propanol? (S-18, 2 Mark)

a) By using conc. H2SO4 (as a dehydrating agent):
(Liquid Phase Dehydration)
Primary Alcohol Secondary Alcohol
3o Alcohols
When alcohols on heating with conc. H2SO4 (at different temp.)
undergo dehydration; to form alkenes
conc. H2SO4
(95% H2SO4)
423-453 K
(150oC-180oC)
60% H2SO4
373 K
(100oC)
20% H2SO4
at 363 K (90oC)

pramodpadole@gmail.com Dr Pramod R Padole
a) By using conc. H2SO4 (as a dehydrating
agent): (Liquid Phase Dehydration)
When alcohols on heating with conc. H2SO4
(at different temp.) undergo dehydration;
to form alkenes.

pramodpadole@gmail.com Dr Pramod R Padole
a) By using conc. H2SO4 (as a dehydrating
agent): (Liquid Phase Dehydration)
CH3CH2CH2OH
95% H2SO4
CH3CH CH2
1-propanol
1-propene
423-453 K
+ H2O

 From 2o Alcohols:
{Note that:- Certain alcohols on dehydration produce two isomeric alkenes.}

When tert-butyl alcohol is heated with 20% H2SO4 at 363 K
(90oC); to form iso-butylene.

Q.1) Explain Saytzeff rule. (S-18, 2 Mark)
Saytzeff's rule state that, in elimination reaction; hydrogen atom
from β- carbon atoms is preferentially eliminated having less no. of H-atoms &
to form major product is a more substituted alkene.
(at the University of Kazan, Russian chemist Alexander Zaitsev studied a variety of different
elimination reactions )
This is because the more substituted alkene is
more stable and it is formed more easily.
For example:
When sec-butyl alcohol is heated with 60% H2SO4 at 373 K (100oC);
to form 2-butene as a major product by Saytzeff rule.
C C
H3C
H
H OH
H
C
H
H
H
H3C CH CH CH3
H3C CH2 CH CH2
1-butene
(minor product)
2-butene
(major product)
2-Butanol
60% H2SO4
at 373K
(- H2O)
 

Zaitsev's rule
(or Saytzeff's rule, Saytzev's rule)

LOGO
Mechanism of Dehydration of Alcohol:
Q.1) Explain the mechanism of dehydration of ethyl alcohol (ethanol). (S-12 & S-13, 4 Mark)
Q. 2) Give the preparation of ethylene form ethyl alcohol and discuss its mechanism. (S-14, 4 Mark)

Mechanism of Dehydration of ethyl alcohol:
The mechanism of acid catalysed dehydration of
alcohols is illustrated here by taking example of
ethyl alcohol.
Consider the reaction:
The mechanism of dehydration involves
following three steps.
Step-1) Protonation of alcohol; to form protonated
alcohol.

alcohol.
Step-2) Dissociation of protonated alcohol; to form
carbocation.
H2
SO4
H + HSO4
+ -

Step-3) Loss of proton from carbocation; to
form alkene.
Ethyl carbocation
-----*****-----

LOGO
Dehydration of
1-Propanol
(n-propyl alcohol):
Q.1) How will you prepare 1-Propene from 1-propanol?

LOGO Mechanism of Dehydration of n-propyl alcohol:
Mechanism:
The mechanism of dehydration involves following
three steps.
alcohol.
CH3CH2CH2OH
95% H2SO4
CH3CH CH2
1-propanol
1-propene
423-453 K
+ H2O
Reaction:
H2
SO4
H + HSO4
+ -
C CH2
H3C
H
H
OH
C CH2
H3C
H
H
OH2
H
from Acid
protonated alcohol
1-Propanol

LOGO Mechanism of Dehydration of n-propyl alcohol:
Mechanism:
Step-2) Dissociation of protonated alcohol; to form
carbocation.
Step-3) Loss of proton from carbocation; to form
alkene.
C CH2
H3C
H
H
OH2
protonated alcohol
C CH2
H3C
H
H
+ H2O
n-propyl cation
C CH2
H3C
H
H
n-propyl cation
HSO4 CH3 CH CH2 H2SO4
propene

LOGO
b) Dehydration of alcohols:
By using Alumina,
(Al2O3):

b) Dehydration of alcohols: By using Alumina, (Al2O3):
Primary Alcohol Secondary Alcohol
3o Alcohols
When alcohols can also be dehydrated by passing their vapours over
heated alumina (Al2O3) tube at 623-423 K; to form alkenes.
Al2O3
623 K
(________oC)
Al2O3
523 K
(_____oC)
Al2O3
at 423 K (_____oC)
350
250
150

LOGO Dehydrohalogenation reaction
or
elimination reaction
or
α-β elimination reaction.
Dehydrohalogenation
of Alkyl Halides
Preparation of Alkenes from
Alkyl Halides:

LOGO
Dehydrohalogenation of Alkyl Halides:
Defination:
 The reaction in which hydrogen atom and halogen atom are
removed from adjacent carbon atoms (α , β carbon atoms);
to form alkene is known as dehydrohalogenation reaction
or elimination reaction or α-β elimination reaction or
β elimination reaction.
 Alkyl halides undergo dehydrohalogenation when heated
with alcoholic KOH or NaOH (alkali); to form alkenes.
 The decreasing order of reactivity of alkyl halides in
dehydrohalogenation is,
 Tertiary (3o) > Secondary (2o) > Primary(1o)

LOGO
From 1o Alkyl halide:-
Q.1) How will you convert Ethyl bromide to ethylene. (S-11, 1 Mark)
Q.2) What happens when- Ethyl bromide is heated with alcoholic KOH? (W-11 & W-12, 1 Mark)
Q.3) What happen when, Ethyl bromide is heated with alcoholic KOH? (S-14, 2 Mark)
Q.4) Ethyl bromide when treated with alcoholic KOH gives _____________. (W-14, ½ Mark)
Q.5) Give the preparation of ethylene from ethyl bromide and discuss its mechanism. (S-16, 4 M)
Q.6) Complete the following reaction: (S-17, 2 Mark)
H3C CH2 Br Alcohol
KOH H2C CH2
Ethylene
Ethyl bromide
+ + KBr + H2O

LOGO
From 2o Alkyl halide:
e.g. (i) From iso-propyl chloride:
CH3CHCH3
Cl
KOH CH3CH=CH2 KCl H2 O
+
Alcohol
+ +
iso-propyl chloride
Propylene
Q.1) Explain Saytzeff rule. (S-18, 2 Mark)
Saytzeff's rule state that, “In elimination reaction; hydrogen atom from
β- carbon atoms is preferentially eliminated from that C-atom having less no. of
H-atoms & to form major product is a more substituted alkene”.
This is because the more substituted alkene is more stable and it is formed
more easily.
C C
H3C
H
H Br
H
C
H
H
H
H3C CH CH CH3
H3C CH2 CH CH2
1-butene
(minor product)
Less substituted product
2-butene
(major product)
More substituted product
2-Butyl bromide
or 2-Bromobutane
(- H2O)
alcoholic
KOH
(- KBr)
 


pramodpadole@gmail.com By Dr. Pramod R. Padole
Mechanism of
Dehydrohalogenation:
E1
Mechanism:
(Unimolecular
Elimination
reaction):
Mechanism
of Dehydrohalogenation:
E2
Mechanism:
(Bimolecular
Elimination
reaction)
There are two mechanisms of dehydrohalogenation of alkyl halides through,

What Do The E1 and E2 Reactions
Have In Common?
Here’s what each of these two reactions has in
common:
in both cases, we form a new C-C π bond,
and break a C-H bond and a C–(leaving
group) bond
in both reactions, a species acts as a base to
remove a proton, forming the new π bond
both reactions follow Zaitsev’s
rule (where possible)
both reactions are favored by heat.

How Are The E1 and E2 Reactions
Different?
 Now, let’s also look at how these two mechanisms
are different.
 Let’s look at this handy dandy chart:

E1 Mechanism:
(Unimolecular Elimination reaction):
Unimolecular Elimination (E1) is
a reaction in which the removal of an HX
substituent results in the formation of a
double bond.
It is similar to
(i)a unimolecular nucleophilic substitution
reaction (SN1) in various ways. One being the
formation of a carbocation intermediate.
(ii) Also, the only rate determining (slow) step is
the dissociation of the leaving group to form
a carbocation, hence the name unimolecular.

E1 Mechanism:
 Tertiary alkyl halides undergo dehydrohalogenation in a
solution of low base concentration by E1 mechanism.
 For example,
Q.1) Explain E1 mechanism with suitable example. (W-16 & W-17, 4 Mark)
Q.2) E1 mechanism, _______ reactants involved in rate determining step.
(a) 1 (b) 2 (c) 3 (d) 4 (S-18, ½ Mark)
Q.3) Explain the E1 mechanism. (S-19, 4 Mark)
C CH3
CH3
H3C
Br
OH
C CH2
CH3
H3C
2-bromo-2-methyl-propane
or tert-butyl bromide
2-methyl-propene
or isobutylene
or
CH3O
H2O
or
CH3-OH




+ Br
NaOH + CH3-OH CH3O--Na + H2O
(ii)
(i) NaOH Na + OH
Alcohol
OR

E1 Mechanism:
 The mechanism of involves following two steps -
Step-1) Dissociation of alkyl halide; to form carbocation and
halide ion (bromide ion).
 This is slow step hence called as rate determining step
(only one reactant is involved).
Step-2) Loss of proton from carbocation; to form alkene.
OH
2-methyl-propene
or isobutylene
or
CH3O
H2O
or
CH3-OH
C CH3
CH3
H2
C
H
Fast
C CH2
CH3
H3C
Abstraction of proton by base

LOGO
E2 Mechanism:
(Bimolecular Elimination reaction)
Q.1) Explain E2 mechanism and state Saytzeff rule. (S-18, 4 Mark)
Q.2) Explain E2 mechanism with example. (W-19, 4 Mark)
E2 stands for bimolecular elimination.
The reaction involves a one-step mechanism in which carbon-hydrogen and carbon-halogen
bonds break to form a double bond (C=C Pi bond).
E2 is a single step elimination, with a single transition state.
E2, bimolecular elimination, was proposed in the 1920s by British chemist Christopher Kelk Ingold.
Unlike E1 reactions, E2 reactions remove two subsituents with the addition of a strong base, resulting
in an alkene.

E2 Mechanism:
 Primary alkyl halides undergo dehydrohalogenation in a
solution of high base concentration by E2 mechanism.
 For example,
OH
or
C2H5O
H2O
or
C2H5OH
+ Br
NaOH + C2H5OH C2H5O--Na + H2O
(ii)
(i) NaOH Na + OH
Alcohol
OR
CH3-CH2-Br + CH2 CH2
Ethene
or Ethylene
Ethyl bromide

E2 Mechanism:
 The mechanism involves only one step (called concerted
mechanism).
 The removal of halide ion from α(alpha) carbon atom and
abstraction of proton from β(beta) carbon atom by OH- take
place simultaneously.
C H
H
C
Br


H
H
H
OH
or
C2H5O
H2O
or
C2H5OH
+ Br
+
CH2 CH2
Ethene
or Ethylene
Ethyl bromide
Abstraction of proton by base
RDS
Note:-
Secondary alkyl halides undergo dehydrohalogenation by both E1 and E2 mechanisms.
A low concentration of base favours E1 mechanism while its high concentration favours E2 mechanism.

Chemical Reactions of Alkenes:
Halogenation
(Addition of Halogen):
Hydrohalogenation
( Addition
of
Halogen acid,
HX):
Chemical Reactions:
a) Chlorination b) Bromination:
in presence of CCl4
(inert / organic solvent)
Symmetrical
alkene
Unsymmetrical
alkene
Geminal
Dihalides
(1,1-dihalides) Vicinal dihalides
(1,2-dihalides)
X

Addition of halogen atoms is known as Halogenation
i) Reaction with
Chlorine
in presence of
CCl4
(inert / organic solvent):
Halogenation
Of
Alkanes
ii) Reaction with
Bromine
in presence of
CCl4
(inert / organic solvent) :

1) Halogenation (Addition of Halogens):
Or Preparation of Vicinal dihalide(1,2-dihalide)
 Alkenes undergo an addition reaction with halogens;
the halogen atoms partially break the carbon-carbon double
bond in the alkene to a single bond and add across it. ...
 Addition of halogens to Alkene:
When alkene is treated or reacted with halogens (chlorine
and bromine only) in presence of CCl4 (inert /organic solvent);
to form vicinal dihalides (or 1,2-dihalides).

a) Chlorination (Addition of Chlorine):
e.g. i) From ethylene:
When ethylene (ethene) is treated or reacted with chlorine
in presence of CCl4 (inert / organic solvent) ; to form ethylene
dichloride (or 1,2-dichloroethane).
Q.1) How will you bring the following conversions? (S-09 & W-09, 1 Mark)
i) Ethylene dichloride from ethylene (ethene).
Q.2) How will you bring the following conversions? (S-10, 1 Mark)
i) Propylene dichloride from propylene (propene).
Q.3) Discuss the (Electrophilic or Free Radical) mechanism of addition of chlorine to ethylene (or propylene).
Q.5) What happens when: Propane on treatment with chlorine in presence of U.V. light? (W-18, 2 Mark)
CH2=CH2
Cl2
CCl4
CH2
- CH2
Cl Cl
+ (Inert solvent)
Ethylene Ethylene dichloride (1,2-dichloroethane)

a) Chlorination (Addition of Chlorine):
 e.g. ii) From Propylene:-
When propylene (propene) is treated or reacted with chlorine
in presence of CCl4 (Inert / organic solvent); to form
propylene dichloride (or 1,2-dichloropropane).
Q.1) How will you bring the following conversions? (S-09 & W-09, 1 Mark)
i) Ethylene dichloride from ethylene (ethene).
Q.2) How will you bring the following conversions? (S-10, 1 Mark)
i) Propylene dichloride from propylene (propene).
Q.3) Discuss the (Electrophilic or Free Radical) mechanism of addition of chlorine to ethylene (or propylene).
Q.5) What happens when: Propane on treatment with chlorine in presence of U.V. light? (W-18, 2 Mark)
CH3CH=CH2 Cl2
CCl4
CH3
CH - CH2
Cl Cl
+ Inert Solvent
Propylene dichloride(1,2-dichloropropane)

b) Bromination (Addition of Bromine)

https://www.footprints-science.co.uk/index.php?quiz=Reaction_of_alkenes_and_hydrogen

b) Bromination (Addition of Bromine)
 e.g. i) From ethylene:
When ethylene reacts with bromine in presence of CCl4 (inert / organic
solvent); to form ethylene dibromide (or 1,2-dibromoethane).
 For example,
Q.1) How will you bring the following conversions? (S-10 & S-12, 1 Mark)
Ethylene dibromide from ethylene (ethene).
Q.2) What happens when ethylene is treated with - (S-04, W-13 & W-14, 2 Mark)
Bromine in presence of Carbon tetrachloride (CCl4)?
Q.3) Discuss the (Electrophilic or Free Radical) mechanism of addition of bromine to ethylene (or
propylene).
Q.4) Complete the following reaction. (S-13, 2 Mark)
Q.5) How will you convert: Ethylene into ethyl bromide? (S-16, 2 Mark)

b) Bromination (Addition of Bromine):
 e.g. ii) From Propylene:-
When propylene (propene) is treated or reacted with bromine in
presence of CCl4 (Inert / organic solvent); to form propylene dibromide
(or 1,2-dibromopropane).
Q.1) How will you bring the following conversions? (S-10 & S-12, 1 Mark)
Ethylene dibromide from ethylene (ethene).
Q.2) What happens when ethylene is treated with - (S-04, W-13 & W-14, 2 Mark)
Bromine in presence of Carbon tetrachloride (CCl4)?
Q.3) Discuss the (Electrophilic or Free Radical) mechanism of addition of bromine to ethylene (or
propylene).
Q.4) Complete the following reaction. (S-13, 2 Mark)
Q.5) How will you convert: Ethylene into ethyl bromide? (S-16, 2 Mark)

LOGO Confirmatory Test: On adding to an alkene, the brown colour of bromine in CCl4 disappears.
Mechanism of
halogenations:
Halogenation of an alkene takes place by electrophilic addition
mechanism as well as Free radical mechanism.

LOGO
Mechanism of halogenations:
Free Radical
Addition
Mechanism
of
Br2
to Ethylene:
Electrophilic
(or Ionic)
addition
Mechanism
of
Br2 / Cl2 :
Mechanism
Of Halogenations

LOGO
Electrophilic (or Ionic) addition Mechanism of Br2 / Cl2 :
 It is an ionic reaction initiated by the electrophile (Br+ ) released
from Br2 .
 The mechanism of addition of Br2 / Cl2 to the ethylene (ethene)
reaction involves three steps.
Step-1) Formation of an electrophile( Br+) :-
When Br2 undergoes heterolytic fission (ionizes on interaction
with π – electron cloud) ; to form Br+ ion (bromonium ion) as an
electrophile & Br- ion(bromide ion) as a nucleophile.
Q.1) Explain the mechanism of addition of bromine to ethylene. (S-12, 4 Mark)

LOGO
Step -2) Formation of Carbonium ion
Or Attack of an electrophile ( Br+) :-
When Br+ ion, as an electrophile, attacks on the C=C bond
of the ethylene; to form carbonium ion

LOGO
Step-3) Formation of Product Or Attack of a nucleophile( Br -):
When Br – ion, as a nucleophile, then attacks to the
carbonium ion from opposite side(back side) ; to form
ethylene dibromide(vicinal dibromide).
C - C
H
H
H H
Br
Carbonium ion
Br C
C
H
H
H
H
Br
Br
+ Back side attack
Nucleophile
( )
Bromide ion 1,2-dibromoethane
(Ethylene dibromide)
Above mechanism of addition of bromine to ethylene / propylene
in another way as below,

Mechanism (Electrophilic addition):
 Above mechanism of addition of bromine to ethylene /
propylene in another way as below,
 Consider the addition of bromine to propene, it involves
following steps:

Mechanism (Electrophilic addition):
Step -1: Electrophilic attack forms bromonium ion and bromide ion.
Step - 2: The bromide ion attacks on bromonium ion from back side; to form
1, 2-dibromopropane.

LOGO
“ Add your company slogan ”
Free Radical
Addition Mechanism
of Br2 to Ethylene:

Free Radical Addition Mechanism of Br2 to Ethylene:
 In the presence of sun light (h‫)ע‬ / peroxide / heat at high temp., the addition
of Br2 to ethylene / propylene takes place; to from vicinal dihalide.
Mechanism:
It is a free radical addition reaction initiated by bromine free radical (Br.).
 The mechanism involves three steps.
Step-1) Chain Initiation: When bromine molecule undergoes homolytic
fission; to form bromine free radicals (Br*).
Br Br Br Br
Homolysis
+
* *
Bromine
molecule
Bromine Bromine
free radical free radical

Mechanism:
Step-2) Chain Propagation:
a) When bromine free radical attacks on ethylene; to form the ethylene free
radical.
b) When ethylene free radical attacks on Bromine molecule; to form ethylene
dibromide & bromine free radical again.
Br
CH2 = CH2
CH2--CH2--Br
*
Bromine
free radical
+
Ethylene
*
ethylene free radical
In step-2) The reactions (a) & (b) are repeated over and over again, called as propagation step.

Mechanism:
Step -3) ChainTermination –
The above chain reaction comes to an end (or stopped / terminated) by
combination of two free radicals.
(For Home work)
Q.1) Discuss the Electrophilic or ionic mechanism of addition of chlorine to
ethylene / propylene.
-----*****-----
Q.2) Explain the mechanism for photochemical addition of chlorine to ethylene /
propylene. (S-19, 4 Mark)
-----*****-----
Q.3) Discuss the Free radical mechanism of addition of chlorine to ethylene /
propylene.
-----*****-----

LOGO
( Addition of Halogen acid, HX):
Hydrohalogenation:

LOGO
Hydrohalogenation:
Unsymmetrical
alkene
e.g.
CH3-CH=CH2
Symmetrical
alkene
e.g.
CH2=CH2
H-X
Addition of
Halogen acid,

LOGO
Symmetrical alkene ( e.g. CH2=CH2 ):
a) Reaction with HBr:
 When ethylene is treated or reacted with strong aq. solution
of HBr ; to form ethyl bromide.
b) Reaction with HCl:
 When ethylene is treated or reacted with strong aq. solution
of HCl ; to form ethyl chloride.
Q.1) What happens when ethylene is treated / reacted with strong aqueous solution of HBr.
Q.2) What happens when ethylene is treated with HBr? (W-11 & S-16, 2 Mark)
Q.3) Complete the following reaction: (S-14 & W-17, 2 Mark)
CH2
= CH2 H Cl CH3
CH2
-Cl
+
Ethylene Ethyl Chloride
The decreasing order of ease of addition of hydrogen halide is,
HI> HBr >HCl

Mechanism of HBr addition:
HBr molecule is polarized
because there is a electronegativity difference between hydrogen and bromine atoms.
Also, electrons density around the double bond is higher.
(due to higher electrons density), those places can be attracted by positive charges.

LOGO
On Unsymmetrical alkene ( e.g. CH3-CH=CH2 ):
Hydrohalogenation

Hydrohalogenation on
Unsymmetrical alkene:
Markownikoff’s
Rule:
or
Markovnikov
Rule:
(Anti Peroxide
Effect)
Alkene
Unsymmetrical
Anti-
Markownikoff’s
Rule :
(Peroxide Effect
Or
Kharasch effect):

LOGO
Aliphatic Hydrocarbons by Dr Pramod R. Padole
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Semester - I C) Aliphatic Hydrocarbons by Dr Pramod R Padole

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