Markovnikov's Rule - Markovnikov's addition – Addition of hydrogen halides - Hydrohalogenation Anti-Markovnikov's rule - Anti-Markovnikov's addition – Addition of hydrogen bromide – Peroxide effect - Kharasch effect

1. Markovnikov’s addition
&
Anti-Markovnikov's addition
Prof.Dr.P.Venkatesh
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2. Markovnikov's Rule Markovnikov's addition
Addition of hydrogen halide Hydrohalogenation
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3. Russian chemist Markovnikov studied a number of addition reactions.
He suggested a rule to predict the course of reaction for addition of hydrogen halides
and other unsymmetrical reagents (HCl, HBr, HI, HOCl, H2SO4, etc) to unsymmetrical
alkenes.
The rule named as Markovnikov Rule
Markovnikov Rule stated as: In the ionic addition of an unsymmetrical reagent (H-G),
the hydrogen or positive end of the reagent becomes attached to the carbon atom of the
double bond bearing the larger number of hydrogen atoms.
C C
H
H
R
H
+ H G
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4. Alkenes reacts with hydrogen halides (HX) by addition across the carbon-
carbon double bond to form alkyl halides or haloalkanes.
Hydrohalogenation:
Reactions involving the addition of a hydrogen atom and a halogen atom
to the carbons of a carbon-carbon double bond.
C C + H X C C
H X
Alkene Hydrogen
halide
Alkyl halide
[Haloalkane]
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5. Addition to symmetrical alkenes
Addition of a hydrogen halide to a symmetrical alkene gives only one product.
It doesn’t matter to on which carbon of the double bond the halogen atom is
placed.
H2C CH2 + HBr H2C CH2
H Br
(Or) H2C CH2
H Br
Ethylene
Ethyl bromide
(Bromoethane)
H3CHC CHCH 3 + HBr H3CHC CHCH 3
H Br
(Or) H3CHC CHCH 3
H Br
2-butene
2-bromobutane
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6. Mechanism
Hydrohalogenation of a symmetrical alkene follows a carbonium ion mechanism
Hydrogen halide forms a π complex which rearranges to give a carbonium ion.
Nucleophilic attack of bromide ion on the carbonium ion gives alkyl halide.
Only one type of carbonium ion is possible from a symmetrical alkene
Hence a single product formed
H2C CH2 + HBr
Ethylene
H3C CH2
+
Carboniumion
+ Br
H3C CH2
Carboniumion
+ Br H3C CH2 Br
Ethyl bromide
+
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7. Addition to unsymmetrical alkenes
Hydrogen halide when reacts with an unsymmetrical alkene – two addition products
possible.
Hydrogen atom can be placed on one or the other carbon of the double bond.
Propylene with HBr can form n-propyl bromide & isopropyl bromide.
[But experimentally isopropyl bromide obtained predominantly]
H2C CH CH3
Propylene
+ HBr
H3C CH CH3
H2C CH2 CH3
Br
Isopropyl bromide
(Major Product)
Br
n-propyl bromide
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8. Reaction of Isobutene with HBr
[The addition reactions of an alkene which follow Markovnikov Rule are
referred to as Markovnikov additions]
+ HBr
C CH2
H3C
CH3
2-Methylpropene
(Isobutene)
C CH3
H3C
CH3
Br
2-bromo- 2-methylpropane
HC CH2
H3C
CH3
Br
1-bromo- 2-methylpropane
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9. Mechanism
Addition of hydrogen halides to unsymmetrical alkenes involves the formation of
carbonium ions. Ex: Addition of hydrogen bromide to propylene.
(i) Formation of a π complex, giving cyclic intermediate and bromide ion.
(ii) The cyclic intermediate is capable of forming two carbonium ions.
H2C CH CH3 + HBr H2C CH CH3
H
+
+ Br
Propene Cyclic intermediate
H2C CH CH3
H
+
H3C CH CH3
H3C CH2 CH2
+
[more stable]
+
1° Carbonium ion
2° Carbonium ion
[less stable)
I
II
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10. (iii) Carbonium ions I & II react with nucleophile bromide ion from step (i) to
give two possible alkyl bromides.
Stabilities of carbonium ions are in order 3°>2°>1°
So, carbonium ion I dominates in step (ii)
In step (iii), I reacts with bromide ion to form isopropyl bromide as the major product.
H3C CH CH3
+
+ Br
I
H3C CH CH3
Br
Isopropyl bromide
(major product)
H3C CH2 CH2
+
II
+ Br H3C CH2 CH2
Br
n-propyl bromide
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11. Ex: Addition of hydrogen bromide to 2-methylpropylene
 3° & 1° carbonium ions produced, in which 3° carbonium ion is much more
stable than 1° carbonium ion.
 So, 3° carbonium ion reacts with bromide ion to give almost exclusively 2-
bromo-2-methyl propane.
3° Carbonium ion
C CH2
H3C
CH3
2-methyl propylene
+ HBr
C CH3
H3C
CH3
+ C CH3
H3C
CH3
Br
2-bromo- 2-methyl propane
(major product)
H3C CH CH2
+
CH3
1° Carbonium ion
Br
Br
H3C CH CH2
CH3
Br
1-bromo- 2-methyl propane
(small amount)
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12. When the two carbonium ions produced are about the same stability,
almost equal amounts of the products obtained.
HC CH
H3C CH2CH3 + HBr
HC CH2
H3C CH2CH3
+
2° Carbonium ion
2° Carbonium ion
Br
HC CH2
H3C CH2CH3
Br
2-bromopentane
Br
H2C CH
H3C CH2CH3
H2C CH
H3C CH2CH3
+
Br
3-bromopentane
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13. Anti-Markovnikov's rule
Anti-Markovnikov's addition
Addition of hydrogen bromide
Kharasch effect
Peroxide effect
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14. Kharash discovered that the addition of hydrogen bromide to unsymmetrical
alkenes in the presence of organic peroxides (R-O-O-R) takes a course opposite
to that predicted by Markovnikov Rule.
The phenomenon of anti-Markovnikov addition caused by the presence of
peroxide, is called Kharash Peroxide effect.
Ex: Propylene when reacts with hydrogen bromide
In the presence of peroxide – Mainly n-propyl bromide is the product
In the absence of peroxide – Main product is isopropyl bromide
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15. Anti-Markovnikov addition of hydrogen bromide to alkenes also takes place
when the reactants are exposed to UV light.
This mode of addition is not the reverse process of Markovnikov addition.
Markovnikov addition follows Ionic mechanism
Anti-Markovnikov addition proceeds by free radical mechanism.
HC CH2
H3C + HBr
Peroxide present
Peroxide absent
H3C CH2 CH2
Br
Propylene
n-propyl bromide
(anti-Markovnikov product)
HC CH3
H3C
Br
Isopropyl bromide
(Markovnikov product)
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16. Mechanism
Peroxides as well as UV light are able to initiate the chain reaction which results
in the formation of anti-Markovnikov product.
Ex: Reaction of Propylene with hydrogen bromide to give n-propyl bromide.
(i) Chain initiating step
 Peroxide dissociates to give two free alkoxy radicals which attack hydrogen bromide
to form bromine free radical.
 Hydrogen bromide on photochemical dissociation (UV light) gives two free radicals.
RO : OR 2RO
.
RO
. + H : Br ROH + Br
.
H : Br
.
H
.
+ Br
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17. (ii) Bromine free radical attacks the alkene molecule gives two possible bromoalkyl free
radicals. [Order of stability of free radicals is 3°>2°>1°]
(iii) The more stable radical from step (ii) reacts with hydrogen bromide forming n-
propyl bromide.
Br
. + H2C CH CH3
Propylene
. . Br CH2 CH CH3
.
2° free radical
(more stable)
Br
.
+
H2C CH CH3
Propylene
. .
H2C CH CH3
.
Br
1° free radical
(less stable)
.
Br CH2 CH CH3
.
2° free radical
+ H : Br Br-CH 2-CH2-CH3 + Br
n-propyl bromide
anti-Markovnikov product
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18. Note:
HCl and HI do not give anti-Markovnikov products in the presence of
peroxides. Because
 The H–Cl bond is stronger than H–Br bond. It is not broken by the
alkoxy free radicals obtained from peroxides.
 The H–I bond is weaker than H–Br. It’s broken by the alkoxy free
radicals obtained from peroxides. But the iodine atoms so formed
readily combine with each other to give iodine molecules, rather than
attack the double bond alkenes.
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Presented by Prof.Dr.P.Venkatesh
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