CH-CH 2 - e-CTLT

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MARKOVNIKOV’S RULE
 The compounds which contains only carbon and hydrogen are
called hydrocarbons. Hydrocarbons are classified into three
categories according to c-c bond present.

Hydrocarbon
Saturated
Unsaturated
alkane

alkene
Symmetrical
CH2=CH2
Ethene
alkyne
Unsymmetrical
CH3-CH=CH2
Propene
Aromatic
 In case of symmetrical alkene when HBr is react with the
product will be the same.
 ie
H
H
H
H
H
H


C=C
+ HBr
H C C
H OR H C
C
H


H
H

SimilarlyH
H
H Br
I-Bromoethane
Br H
I-Bromoethane
H
CH3-CH(Br)-CH2-CH3
H
C
H
C
=C
CH3 + HBr
2-Bromobutane
CH3-CH2-CH(Br)CH
2-Bromobutane
 But when Hbr is added to the unsymmetrical alkene then we




will get two different product i.e.
CH3-CH2-CH2-Br
CH3-CH=CH2+HBr
I-Bromopropane
CH3-CH-CH3
Br
In this case the addition governed by Markovnikov Rule who was a Russian
chemist.
 The rules goes like this
 During the addition acroos unsymmetrical multiple bond the negatively part
attacking molecule joins the carbon atom which carries small number of
hydrogen atoms while the positive part goes to carbon atom with more
hydrogen atoms.
 So in this above case 2-Bromopropoane is the major product i.e.
CH3-CH=CH2+HBr
CH3-CH-CH3+ CH3-CH2-CH2
Br
(90%)
Br
(10%)
 Mechanism In order to understand Markovnikov’s rule.
 Let consider the attack of electrophile(H+) on the
unsymmetrical alkene molecule (Propene). Here two
intermediate carbocation are formed.

(+)
CH3-CH-CH3
CH3-CH=CH2+ H+
2 Degree(carbocation)
(+)
CH3-CH2-CH2
1 Degree(carbo cation)
 As we know that carbocation is stabilises by no-bond
resonance structure. More the resonating structure more in
the stable.

(+)
H (+) H
H
 CH3-CH-CH3
or H-C-CH-C-H
CH3-CH=CH
H
 2 Degree(carbonation)

H
 H + C=CH-CH3

H
H
H-C=CH-CH3
H+

H
H
CH3-CH=C H +
H
H
H
CH3-CH=CH
H+
H+
H-C=CH-CH3
H
Total no: 6(six)
 Where as in case of 1degree carbon cation it has 3 resonating structure.
 So 2 dgree carbon cation is more stable than 1 degree carbon cation.
 This 2 degree carbon cation further react with Br- to form the addition product.

(+)
(-)
 CH3-CH-CH3+Br

CH3-CH-CH3
Br
 So we can describe Markovnikov ‘s Rule as follows.
 The electrophilic addition to an unsymmetrical alkene always occur through the
formation of more stable carbo cation intermediate.
Thus the complete mechanism for the addition of HBr to
propane as follow.
HBr
H+ + BrCH3-CH=CH2+H+
CH -CH-CH +Br
3
3
CH3-CH-CH3
2◦ Carbocation
CH3-CH-CH3
Br
Anti Markovnikov Rule or peroxide effect
Observed by Kharasch that when HBr is added to alkene in presence of
O
O
organic peroxides provide (C6H5-C-O-O-C-C6H5) the reaction takes place opposite
to Markovnikov rule. This is known as anti Markovnikov ‘srule or peroxide effect.
According to this rule the Br atom will join to that carbon atom which carries more
hydrogen atoms. While the hydrogen (H)-atom go to the other C- atom.
Thus CH3-CH=CH2+HBr
Peroxide
CH3-CH2CH2Br
1-Bromopropane
Mechanism/ Explanation for anti- Markovnikov‘s Rule:The addition of HBr to alkene in the presence of Peroxide occurs by free radical
mechanism.
◦
Here the first two steps result into generation of free radical Br. Then it adds to
double bond in such a way to give more stable radical. Then this free radical
abstracts a H – free radical form to complete the addition.
The various steps of mechanism are:-O
O
O
homolysis
1. C6H5----C-O-O---C-C6H5
.
2C6H5----C----O
2C6H5+2CO2
Free radical
.
2. C6H5+H----Br
(phenyl free radical).
C6H6+Br (Free radical).
.
.
CH 3CH-CH2Br
3. Br+CH3-CH
CH2
2 degree free radical(more stable)(I)
.
CH3-CH-CH2 (less stable) (II)
Br
1 degree free radical
2 degree free radical is more stable then 1 degree free radical. It will be preferably formed.
.
4. CH3-CH-CH2Br + H---Br
CH3-CH2-CH2Br + Br
Exceptional behaviour of HBr :- It is interesting to note that
peroxide effect is applicable only to HBr and not to HF,HCl or
HI. This can be understand on tha basis of energy changes in
the prpogation step using different halogen atom.
.
Step iii
X+CH3-CH=CH2
CH3-CH-CH2X
X=F
ΔH= -209 KJ/MOL
X=Cl ΔH= -101 KJ/MOL
X=Br ΔH= -42 KJ/MOL
X=I
ΔH= +12 KJ/MOL
.
.
Step iv
H-X+CH3-CH-CH2X
CH3-CH2-CH2X+X
X=F
ΔH= +159 KJ/MOL
X=Cl ΔH= +27 KJ/MOL
X=Br ΔH= -37 KJ/MOL
X=I
ΔH= - 106 KJ/MOL
From the above data it has been observed that only with HBr both steps are
exothermic where as others do not posses both steps as exothermic.
In other words homolysis of HF or HCl does not take place. In case of HI, the
homolysis of HI is possible as evident from step 4 but the attack of I to propene
does not take place (step III is exothermic). So the free radical I will combine
with similar free radical from I2 molecule. Thus peroxide effect is observed
incase of HBr.
Thank You

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