Aliphatic hydrocarbon alkanes

1
Late Ku. Durga K. Banmeru Science College,
LONAR DIST. BULDANA (Maharashtra), India.
Section –II
Unit-3
C) AliphaticHydrocarbon
“ALKANES”
B. Sc. Ist year Sem-Ist
Subject:- Chemistry

2
Dr. Suryakant B. Borul
(M.Sc., M.Phil., Ph.D.)
Head Of Department
Department of Chemistry
Late Ku. Durga K. Banmeru Science College,
Lonar
Teacher Profile

C) Aliphatic Hydrocarbons
Aliphatic hydrocabons are the main class of organic compounds.
Which involved basic or parent compounds of organic.
I) Alkanes
II) Alkenes
III) Alkynes
IV) Alkadienes

Q.- What are alkanes?
Ans-
“The aliphatic saturated hydrocarbon containing carbon-carbon single bond
they have general formula CnH2n+2 are called as alkanes.”
“ALKANES”
Some members of alkanes as follows-
1. CH4 -Methane
2. CH3-CH3 -Ethane
3. CH3-CH2-CH3 -Propane
4. C4H10 n-Butane

Methods of Formation of Alkanes
1. Wurtz Synthesis-
Q. Write a note on Wurtz synthesis.
Ans-
Alkyl halide reacts with sodium metal in presence of dry ether
then it gives higher alkanes, this reaction is called as Wurtz reaction.

ii) When ethyl bromide reacts with sodium metal in presence of dry ether then it
gives n-butane.
Examples- i) When methyl bromide reacts with sodium metal in presence of dry
ether then it gives ethane.

2. Corey House Alkane Synthesis-
Q. Write a note on Corey House Alkane Synthesis.
Ans- When alkyl halide reacts with lithium metal in presence of ether then it
gives alkyl lithium, which on treated with CuI (copper iodide) to from
lithium dialkyl cuprate , then alkyl halide react with dialkyl cuprate to
from alkane.

Q. How will you prepare propane from ethyl bromide?
Ans- When ethyl bromide reacts with lithium metal in presence of ether then it
gives ethyl lithium, which on treated with CuI (copper iodide) to from
lithium diethyl cuprate , then methyl bromide react with lithium diethyl
cuprate to from Propane.

Chemical Reactions of Alkanes
1. Halogenation-
“The process in which replacement of hydrogen atom from alkane by
halogen is called as halogenation.”
 Chlorination-
“When alkanes react with chlorine in presence of UV light or mild sunlight
or at high temp (300-400oC) it gives mixture of haloalkane i.e.
chloroalkane.”
Ex.- Methane react with Cl2 in UV light then it gives methyl chloride,
methylene chloride, trchloromethane carbon tetrachloride.

CH4 + Cl Cl
UV light
C
H3 Cl + Cl
H
Methane Chlorine Methyl chloride
C
H3 Cl Cl Cl
+ CH2Cl2 + Cl
H
CH2Cl2 Cl2
+ CHCl 3 + Cl
H
CHCl 3 + Cl2 CCl 4 Cl
H
+
Methyl chloride
Methylene dichloride
Methylene dichloride Trichloromethane
Trichloromethane Carbon tetrachloride
If excess of methane is used then it gives methyl chloride is major product.

Q. Explain the mechanism of chlorination of methane.
Ans-
When methane reacts with chlorine in presence of ultra violet
light then it gives methyl chloride.
Step-1: Chain Initiation-
Chlorine molecule undergoes homolytic bond fission to give chlorine radicals
Cl Cl Cl Cl
+
UV light
Chlorine Chlorine f ree radicals
Free radical mechanism involved following steps-

Step-2: Chain Propagation-
Chlorine free radical attacks on methane to form HCl and methyl free
radical.
Cl CH3
H
+ Cl
H + CH3
CH3 + Cl Cl C
H3 Cl + Cl
Methy l Chloride
Step-3: Chain Termination-
Chain propagation steps when any two free radical combine.
Cl Cl
Cl
Cl
Cl
+
CH3
CH3 +
C
H3 Cl
C
H3 CH3
CH3

Q. What is action of Chlorine in presence of UV light on
1) Ethane 2) Propane.
Ans- 1) Ethane with chlorine- When ethane reacts with chlorine in presence
of ultra violet light then it gives ethyl chloride.
+
UV light
Chlorine
Cl
H
Cl Cl
C
H3 CH3 + C
H3 Cl
Ethane Ethyl chloride
2) Propane with chlorine- When propane reacts with chlorine in presence of
ultra violet light then it gives ethyl chloride.
C
H3 CH2 CH3 + Cl Cl
UV light
C
H3 C
H CH3
Cl
C
H3 CH2 CH2 Cl
2 - Chloro Propane
1 - Chloro Propane
Propane Chlorine

2. Aromatization-
“The process in which formation of aromatic ring from alkenes is called as
Aromatization of Alkane.”
Q. How will you convert 1) n-Hexane to Benzene
2) n-Heptane to Toluene.
Ans- 1) n-Hexane to Benzene –
When n-hexane heated with Cr2O3-Al2O3 at 600oC and10
atmospheric pressure then it undergoes aromatization to form benzene.
C
H3 CH2 CH2 CH2 CH2 CH3
Cr2O3 - Al 2O3
600oC / 10atm + H2
4
n - Hexane
Benzene

2) n-Heptane to Toluene –
When n-heptane heated with Pt-Al2O3 at 600oC and10 atmospheric pressure
then it undergoes aromatization to form toluene.
CH2 CH2 CH2 CH2 CH2 CH3
C
H3
Pt - Al 2O3
600oC / 10atm
CH3
+ H2
4
n - Heptane
Toluene

Q.- What are alkenes?
Ans-
“The aliphatic unsaturated hydrocarbon containing carbon-carbon double
bond they have general formula CnH2n are called as alkenes or olefins.”
“ALKENES”
Some members of alkenes as follows-
1. C2H4 -Ethene / Ethylene
2. C3H6 -Propene / Propylene
3. C4H10- Butene / Butylene
CH2=CH2
CH3-CH=CH2
CH3-CH=CH-CH3
• They have general formula CnH2n, where n=number of carbon atom
• They have Functional group -C=C- Ethylene linkage
• It is also called as Olefins.

Methods of Formation of Alkenes
I. Dehydrohalogenation of Alkyl halides Or Elimination Reaction
Q. Write a note on Elimination Reaction .
Ans-
When alkyl halide reacts with alcoholic KOH or NaOH to form
alkene.
C
H3 CH2 Br Alcoholic KOH C
H2 CH2 + KOH H2O
+
ethyl bromide ethylene
Two types of Elimination reactions
1. E1 reaction-
2. E2 reaction-

 E1 stand for; 1- stand for Unimolecular,
E- stand for Elimination reaction
Mechanism of Elimination reactions
1. E1 Reaction Mechanism-
Q. Explain E1 Reaction Mechanism with example.
Ans-
When 2-Bromo 2-methyl Propane (t- butyl bromide) reacts with alcoholic
NaOH to form 2-methyl propane.
C
CH3
CH3
C
H3
Br
+ C
H3 O
- CH3 - OH / Na
C
CH3
CH2
C
H3 + C
H3 OH+ Br
-
2 - bromo 2 - methylpropane
2 - methyl propane
Rate of reaction (R) = K C
CH3
CH3
C
H3
Br

Mechanism:-
Step-I – Formation of Carbocation-
In this step breaking C-Br bond take place and this is slow step hence it is
rate determine step (RDS).
C
CH3
CH3
C
H3
Br
slow
C
+
CH3
CH3
C
H3 + Br
-
2 - bromo 2 - methylpropane
Carbocation
Step-II – Elimination of Proton-
In this step breaking C-H bond and formation of carbon carbon double
bond take place. This is fast step.
2 - methyl propane
Carbocation
C
H3 O
-
+ C
+
CH3
CH3
CH2
H Fast C
CH3
CH2
C
H3 + C
H3 OH
Rate of this reaction depends on only one reactant molecule hence it is
unimolecular i.e. E1

 E2 stand for; 2- stand for bimolecular,
E- stand for Elimination reaction
2. E2 Reaction Mechanism-
Q. Explain E2 Reaction Mechanism with example.
Ans-
When 1-Bromo Propane (n- Propyl bromide) reacts with sodium ethoxide
in presence of alcoholic NaOH it gives Propane and ethanol.
C
H3 CH2 CH2 Br + C
H3 CH2 O Na C
H3 CH CH2 C
H3 CH2 OH
+
1 - Bromopropane sod. ethoxide
1 - Propane Ethanol

Rate of Reaction- Rate of reaction depends on two molecules hence it is
biomolecular reaction.
Rate = K C
H3 CH2 CH2 Br C
H3 CH2 O Na
Mechanism:-
Rate This is one step mechanism hence breaking of C-Br, C-H bond and
making of carbon-carbon double bond take place simultaneously.
C
H3 CH2 O
-
+
C
H3
CH CH2 Br
H C
H3 CH CH2 C
H3 CH2 OH
1 - Propane Ethanol
+

Saytzeff rule -
If dehydrohalogenation of alkyl halides gives more than one alkenes, then
the most highly substituted alkene is major product.
C
H3 CH CH2 CH3
Br C
H3 CH CH CH3
C
H3 CH2 CH CH2
2 - Butene ( Major)
1 - Butene ( Minor)
2 - bromo Butane

II. Dehydration of Alcohols-
When alcohol treated with strong acid like H2SO4 or p-toluene Sulphonic acid
then it undergoes dehydration to form alkenes.
C
H3 CH2 CH2 OH
Conc.H2SO4
C
H3 CH CH2+ H2O
1 - Propanol 1 - Propene
• Mechanism-
Step-I- Protonation of Alcohol-
C
H3 CH2 CH2 OH C
H3 CH2 CH2 OH2
+
n - Propyl alcohol
+ H2SO4
Protonated Alcohol
Step-II- Loss of H+ (proton) & H2O to form Alkene-
+H2O
1 - Propene
+ HSO 4
-
Protonated Alcohol
C
H3 CH2 CH2 OH2
+
C
H3 CH CH2 + H2SO4

Alkenes-“The aliphatic unsaturated hydrocarbon containing carbon-carbon
double bond they have general formula CnH2n are called as alkenes or olefins.”
“Reactions of Alkenes”
• C4H8- Butene / Butylene
CH3-CH=CH-CH3, or CH3-CH2-CH=CH2
• C5H10- Pentene/ Pentylene
CH3-CH2-CH2-CH=CH2 or CH3-CH2-CH=CH-CH3
From this example it is clear that there are Two Types of Alkenes
A) Symmetrical Alkenes-
B) Unsymmetrical Alkenes-

A) Symmetrical Alkenes- “The alkenes which have same or equal number of
atoms on the both sides of double bonds such alkenes are called as
symmetrical alkenes”
• Ex- C4H8- Butene- CH3-CH=CH-CH3 2-Butene
C6H12- Butene- CH3-CH2-CH=CH-CH2-CH3 3-Hexene
A) Unsymmetrical Alkenes- “The alkenes which have different or un-equal number of
atoms on the both sides of double bonds such alkenes are called as unsymmetrical
alkenes”
• Ex- C4H8- Butene CH3--CH2-CH=CH2 1-Butene
C5H10- Pentene CH3-CH2-CH2-CH=CH2 or 1-Pentene
CH3-CH2-CH=CH-CH3 2-Pentene

Reactions of Alkenes
1. Hydrohalogenation or Addition of HX (Halo acid)-
A) Addition of HX in Symmetrical alkenes- The symmetrical alkenes react with
HX then it gives alkyl halides”
Ex- i) When Ethene react with HBr then it gives Ethyl bromide.
C
H3 CH CH CH3+ C
H3 CH2 CH CH3
Br
Br
H
2 - Butene 2 - bromo butane
C
H2 CH2+ C
H3 CH2
Br
Br
H
Ethene
Ethyl bromide
ii) When 2-butene react with HBr then it gives 2-bromo butane.

B) Addition of HX in Unsymmetrical alkenes- The unsymmetrical alkenes react
with HX then it gives mixture of alkyl halides”
Ex- i) Propene react with HBr then it gives 1-bromo propene and 2-bromo propene.
CH CH2
C
H3 + CH2 CH2 Br
C
H3
Br
H
Propene 1 - bromo propene
+ CH CH3
C
H3
Br
2 - bromo propene
Here mixture of 1-bromo propene and 2-bromo propene obtained but
which is major product. According to Markownikoff addition rule 2- bromo
propene is the major product.

Markownikoff’s Addition Rule
Q.- State and Explain the Markownikoff’s Addition Rule
Ans- Statement- “When an addition of unsymmetrical reagent (HX) in an unsymmetrical
alkenes then the negative part of reagent (X-) get attached to the double bonded carbon
atom which carries less number of hydrogen atom”.
Ex- i) When Propene react with HBr then it gives 2-bromo propene.
CH CH2
C
H3 + Br
H
Propene
CH CH3
C
H3
Br
2- bromopropene
Here Br- ion get attached to middle double bonded carbon atom of propene
because according to M.K. rule which carries less number of hydrogen atom.

Example ii) When 1-butene react with HBr then it gives 2-bromo butene.
C
H3 CH2 CH CH2+ C
H3 CH2 CH CH3
Br
Br
H
1 - Butene 2 - bromo butane

Q.-Explain the Mechanism of addition of HBr to Propylene.
Ans- Propene react with HBr then it gives 2-bromo propene.
CH CH2
C
H3 + Br
H
Propene
CH CH3
C
H3
Br
2- bromopropene
Mechanism-
Mechanism of addition of HBr in Propylene in three steps are as follows-
Step-I:- Formation of Electrophile- In this HBr undergoes heterolytic fission.
Br
H Br
-
+
H
+
Heterolytic fission

Step-II:- Formation of Carbocation- Attack of Proton on propylene to from
primary (less stable) carbocation and secondary (more stable) carbocation.
H
+
C
H3 CH CH2
C
H3 CH
+
CH3
C
H3 CH2 CH2
+
Propylene
Secondary Carbocation
PrimaryCarbocation
More Stable
Less Stable
Step-III:- Formation of Product- Attack of Br- ion on more stable carbocation to
from 2- bromo Propylene (iso propyl bromide).
Br
-
+
C
H3 CH
+
CH3
SecondaryCarbocation
More Stable
C
H3 CH CH3
Br
2 - Bromo Propane
iso propyl bromide

Peroxide Effect or Anti-Markownikoff’s Addition Rule
Q.- Explain the Peroxide effect or Anti-Markownikoff’s Addition Rule
Ans-“When an addition of unsymmetrical reagent (HX) in an unsymmetrical alkenes
in presence of peroxide then the negative part of reagent (X-) get attached to
the double bonded carbon atom which carries more number of hydrogen
atom” is called as Peroxide effect or Anti Markownikoff’s addition rule.
Br
H
+
C
H3 CH CH2
Propylene
Peroxide
R - O - O - R
C
H3 CH2 CH2 Br
1 - Bromo Propane
n propyl bromide
For Example – When Propylene or Propene react with HBr in presence of peroxide (R-O-O-
R) then it gives n-propyl brmode or 1-bromo propene.

Q.-Explain the Mechanism of addition of HBr to Propylene in presence
of peroxide.
Ans- When Propylene or Propene react with HBr in presence of peroxide (R-O-O-R) then it
gives n-propyl brmode or 1-bromo propene.
Mechanism-
Mechanism of addition of HBr to Propylene in presence of peroxide occur in
three steps, these are as follows-
Step-I:- Chain Initiation-
Step-II:- Chain Propagation-
Step-III:- Chain Termination-
Br
H
+
C
H3 CH CH2
Propylene
Peroxide
R - O - O - R
C
H3 CH2 CH2 Br
1 - Bromo Propane
n propyl bromide

Step-I:- Chain Initiation-
In this step peroxide undergoes homolysis to form free radicals of alkoxy
group, which attack on HBr to from alcohol and bromine free radical.
Br
H +
Homolytic fission
R O O R R O R O
+
Alkyl Peroxide alkoxy f ree radicals
R O + R OH Br
Alkoxy Free radical
Alcohol Bromine free radical

Step-II:- Chain Propagation-
In this step bromine free radical attack on propylene to form iso-propyl
bromide primary (less stable) and n-propyl bromide secondary free radical (more
stable).
Then n-propyl bromide secondary free radical attack on HBr to form n-
propyl bromide and bromine free radical.
+ Br
Bromine Free radical
C
H3 CH CH2
Propylene
C
H3 CH CH2
Br
C
H3 CH CH2 Br
Primary free radical
Secondary free radical
less stable
more stable

Step-III:- Chain Termination-
In this step bromine free radicals attack one another to form stable
bromine molecule.
+ Br
Bromine free radicals
Br Br Br
Bromine molecule
Br
H +
+ Br
C
H3 CH CH2 Br
Secondaryfree radical
C
H3 CH2 CH2 Br
1 - Bromo Propane
n propyl bromide

2. Addition of Halogen (Halogenation) or Alkenes-
When alkenes reacts with halogens (chlorine & Bromine) then it gives
vicinal dihalidess.
R CH CH R + X2 R CH CH R
X X
Alkene halogen
Vicinal dihalide
Q.- What is action of i) Br2 in CCl4 ii) Cl2 in CCl4 on ethylene.
Ans- i) Action of Br2 in CCl4 on ethylene / ethene-
ii) Action of Cl2 in CCl4 on ethylene / ethene-
C
H2 CH2 + Br2 C
H2 CH2
Br Br
Ethene Bromine
1, 2 dibromoethane
CCl4
C
H2 CH2 + Cl2 C
H2 CH2
Cl Cl
Ethene Chlorine
1, 2 dichloroethane
CCl4

Electrophilic Addition Mechanism
Mechanism of Halogenation of Alkenes
Q- Explain electrophilic addition mechanism of bromine to propylene.
CH CH2
C
H3 + CH CH2
Br Br
C
H3
Ethene
1, 2 dibromoethane
CCl4
Br2
Bromine
Mechanism-
Step- I- Electrophilic attack forms bromonium ion and bromide ion.
Ans-
+ Br Br
Bromine molecule
C
H3
C
H
H
H
C
H3
C
H
H
H
Br
+
+ +Br
Bromonium ion

Step- II- Attack of Bromide on bromonium ion-
Br
-
C
H3
C
H
H
H
Br
+
+
Bromoniumion
C
H3 CH CH2 Br
Br
1, 2 - dibromopropane

Free Radical Addition Mechanism
Mechanism of Halogenation of Alkenes
Q- Explain Free radical addition mechanism of bromine to propylene.
C
H2 CH2 + Cl Cl
UV light
Homolysis
C
H2 CH2
Cl Cl
Ethene
1, 2 dichloropropane
Ans- Stewart in 1935 proposed this mechanism, When ethene react with chlorine in
presence of UV light then it undergoes homolysis to form 1, 2 dichloroethane.
Cl Cl
UV light
Homolysis
Cl Cl
+
Chlorine Chlorine free radicals
Step-I- Chain Initiation- In this step chlorine undergoes homolysis in presence of
UV light then it gives chlorine free radicals.
Mechanism-

Cl
Chlorine free radicals
C
H2 CH2 + UV light
Homolysis
C
H2 CH2
Cl
Ethene
ethylene chloride free radical
+ UV light
Homolysis
C
H2 CH2
Cl
ethylene chloride free radical
Cl Cl C
H CH
Cl Cl
Ehtylene dichloride
Step-II- Chain Propagation- In this step chlorine free radical attack on ethylene to
form ethylene chloride free radical first then it react with chlorine molecule to form
ethylene dichloride.
Step-II- Chain Termination- In this step chlorine free radical combine with
chloride free radical to form chlorine molecule.
Chlorine
Cl Cl
Cl
+
Cl

Q.- What are alkynes?
Ans-
“The aliphatic unsaturated hydrocarbon containing carbon-carbon triple
bond they have general formula CnH2n-2 are called as alkynes or
acetylenes.”
“ALKYNES”
Some members of alkynes as follows-
• They have general formula CnH2n-2, where n=number of carbon atom
• They have Functional group acetylene linkage
• It is also called as Acetylenes.

Methods of Formation of Alkynes
I. Dehydrohalogenation of Vicinal dihalides-
When vicinal dihalides reacts with alcoholic KOH or NaOH then it gives
vinyl halide which on react with sodium amide it gives acetylene.
General reaction-
R C C H
H
X
X
H
Alcohol
KOH
R C C H
H
X
Vinyl halide
Vicinal dihalide
R C C H
H
X
Vinyl halide
Alcohol
NaNH2
R C C H
Alkyne
+ +
NaX NH3

Q. How will you convert ethylene dibromide to Acetylene.
Ans-
When ethylene dibromide first reacts with alcoholic KOH or NaOH then it
gives vinyl bromide, which on react with sodium amide it gives acetylene.
H C C H
H
Br
Br
H
Alcohol
KOH
H C C H
H
Br
Vinyl bromide
Ethylene dibromide
H C C H
H
Br
Vinyl bromide
Alcohol
NaNH2
H C C H
Acetylene
+ +
NaBr NH3

II. Dehydrohalogenation of Geminal dihalides-
When geminal dihalides reacts with alcoholic KOH or NaOH then it gives
vinyl halide which on react with sodium amide it gives acetylene.
General reaction-
R C C H
X
X
H
H
Alcohol
KOH
R C C H
X
H
Vinyl halide
Geminal dihalide
R C C H
X
H
Vinyl halide
Alcohol
NaNH2
R C C H
Alkyne
+ +
NaX NH3

Q. How will you convert ethylidene dibromide to Acetylene.
Ans-
When ethylidene dibromide first reacts with alcoholic KOH or NaOH then it
gives vinyl bromide, which on react with sodium amide it gives acetylene.
H C C H
Br
Br
H
H
Alcohol
KOH
H C C H
Br
H
Vinyl bromide
Ethylidene dibromide
H C C H
Br
H
Vinyl bromide
Alcohol
NaNH2
H C C H
Acetylene
+ +
NaBr NH3

Reactions of Alkynes
I. Hydrogenation of Alkynes-
A. Catalytic hydrogenation- When ethyne reacts with hydrogen in presence of
catalyst as Ni or Pt or Pd then it first gives ethylene and finally ethane.
C
H CH + H2 C
H2 CH2 + H2 C
H3 CH3
Acetylene
Ni
Catalyst
Ni
Catalyst
Ethylene Ethane
B. Controlled hydrogenation- When ethyne reacts with hydrogen in Pd-BaSO4 in
quinoline then it gives ethylene.
C
H CH + H2 C
H2 CH2
Acetylene
Pb - BaSO 4
Quilnoline
Ethylene
Palladium in Barium Sulphate in quinoline medium (Pd-BaSO4 in
quinoline) is called as “Lindlar’s Catalyst”

Q.- What are alkadienes?
Ans-
“The aliphatic unsaturated hydrocarbon containing carbon-carbon two
double bond they have general formula CnH2n-2 are called as alkadienes.”
“ALKADIENES”
Some members of alkynes as follows-
• They have general formula CnH2n-2, where n=number of carbon atom
• They have Functional group -diene-

Q.- How alkadienes are classified?
Ans-
On the basis of position of double bond alkadienes are classified into three
classes.
1. Conjugated dienes-
2. Isolated dienes-
3. Cumulated dienes-
1. Conjugated dienes-
The alkadienes in which two double bonds are separated by one single
bond are called as conjugated dienes.
ex.-
C
H2 CH CH CH2
C
H3 CH CH CH CH2
1, 3 Butadiene
1, 3 Pentadiene

2. Isolated dienes-
The alkadienes in which two double bonds are separated by more than one
single bond are called as isolated dienes.
ex.-
1, 4 Pentadiene
C
H2 CH CH2 CH CH2
3. Cumulated dienes-
The alkadienes in which two double bonds are adjacent to each other are
called as cumulated dienes.
ex.-
C
H3 CH C CH2
C
H3 CH2 CH C CH2
1, 2 Butadiene
1, 2 Pentadiene

Synthesis of 1, 3 Butadiene
A) From Cyclohexene-
Q. How will you prepared 1, 3 butadiene from cyclohexene?
Ans-
When cyclohexene heated with 873k in presence of Ni-Cr alloy then it
gives 1,3 butadiene.
C
H2 CH CH CH2
1, 3 Butadiene
873k
Ni - Cr Alloy
+ C
H2 CH2
Cyclohexene
Ethene

Reactions of 1, 3 Butadiene
1) Addition of Hydrogen-
Q. What is the action of H2 in presence of Ni/Pt on 1, 3 butadiene.
Ans-When 1, 3 butadiene react with hydrogen gas in presence of Ni or Pt, then it
undergoes 1, 2 addition & 1, 4 addition then it gives 1-butene & 2-butene.

2) Addition of Halogen-
Q. What is the action of chlorine in presence of CCl4 on 1, 3 butadiene.
Ans-When 1, 3 butadiene react with halogen gas in presence of CCl4 then it
undergoes 1, 2 addition & 1, 4 addition then it gives 3, 4 dichloro-1-butene &
1, 4 dichloro-2-butene .

3) Addition of Halogen acid (HX)-
Q. What is the action of HBr on 1, 3 butadiene.
Ans-When 1, 3 butadiene react with halogen acid then it undergoes 1, 2 addition &
1, 4 addition then it gives 3-dibromo-1-butene & 1-bromo-2-butene .

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