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Question

(A) The reaction of propene with $HBr$ in the presence of benzoyl peroxide $((C_6H_5CO)_2O_2)$ proceeds via a free radical mechanism.This is known as

Vedclass · Accepted Answer

$CH_3 - CH_2 - CH_2 - Br$

Vedclass · Answer

(A) The reaction of propene with $HBr$ in the presence of benzoyl peroxide $((C_6H_5CO)_2O_2)$ proceeds via a free radical mechanism.This is known as the Kharasch effect or Peroxide effect.According to this effect,the addition of $HBr$ to an unsymmetrical alkene follows Anti-Markovnikov's rule.Therefore,the bromine atom attaches to the carbon atom with the greater number of hydrogen atoms.The reaction is: $CH_3 - CH = CH_2 + HBr \xrightarrow{Peroxide} CH_3 - CH_2 - CH_2 - Br$ ($1$-bromopropane).Thus,the correct option is $A$.

The major product of the following reaction is $CH_3 - CH = CH_2 + HBr \xrightarrow{(C_6H_5CO)_2O_2}$

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