Question

The reaction of HBr with $ \left [ CH_3-CCH_3=CH_2 \right ]$ in the presence of peroxide will give
A. $CH_3-CCH_3Br-CH_3$
B. $CH_3-CH_2-CH_2-CH_2Br$
C. $CH_3-CHCH_3-CH_2Br$
D. $ CH_3-CH_2-CHCH_3-CH_3$

Answer 1

Hint: Not all alkene addition reactions follow markovnikov’s rule. And in presence of peroxide addition HX in alkene follows a radical mechanism. The product formed in this reaction is anti-markovnikov's product and this addition reaction is called anti-markovnikov addition reaction.

Complete Step by Step Answer:
According to the addition mechanism of the alkene as the formation of carbocation takes place first the electrophile gets attached to the carbon which contains more no. of hydrogen atoms attached to the carbon and then the nucleophile gets attached to the carbon that contains less number of hydrogen atoms because according to markovnikov’s rule the nucleophile will attach with the carbon which contains less number of hydrogen.

But when peroxide is added to the reaction, the reaction follows a different mechanism and these reactions are called radical reactions of alkene. In these reactions, free radicals are formed. Free radicals are formed due to homolytic cleavage in hydrogen peroxide. The reaction is:
$CH_3-CCH_3=CH_2 + \ HBr/H_2O_2\ \ \ \rightarrow CH_3-CHCH_3-CH_2Br$

The mechanism of the reaction is:
$H-O-O-H \longrightarrow OH \bullet + \bullet OH$
$OH\bullet+H-Br \longrightarrow H_2O+Br\bullet$
$CH_3-CCH_3=CH_2+Br \bullet \ \longrightarrow CH_3-\bullet CCH_3-CH_2Br \longrightarrow \ CH_3-CHCH_3-CH_2Br$
As we know, the stability of 2˚ radical is higher than the stability of 1˚ radical so 2˚ radical will be formed as a major product thus the major product formed is $ \left [ CH_3-CHCH_3-CH_2Br \right ]$.
Thus, Option (C) is correct

Note: This reaction was first observed by M. S. Kharash and this reaction is also known as Kharash effect. Attack of any radical on any substrate to form a new radical is called hemolysis. Homolytic bond cleavage means breaking of bond when electron pairs split evenly between the products and form radicals.