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Q.
Reaction of $HBr$ with propene in the presence of peroxide gives
ManipalManipal 2007Hydrocarbons
Solution:
Reaction of $HBr$ with propene in the presence of peroxide gives $n$-propyl bromide. This addition reaction is an example of anti-Markownikoff addition reaction.
(ie, it is completed in form of free radical addition.)
$CH_3 - CH = CH_2 + HBr \xrightarrow{\text{Peroxide}} \underset{\text{n - propyl bromide}}{CH_3 - CH_2 - CH_2Br}$
Mechanism of this reaction is represented as follows
Step 1 . formation of free radical of peroxide by means of decomposition.
$\underset{\text{Benzoyl peroxide}}{C_6H_5 - \underset{\overset{||}{O}}{C} - O -O - \underset{\overset{||}{O}}{C} - C_6H_5} \xrightarrow{\Delta} $
$\underset{\text{Benzoate free radical}}{2C_6H_5 - CO\overset{\bullet}{O}}$
Step 2 . Benzoate free radical forms bromine free radical with $HBr$.
$C_6H_5CO\overset{\bullet}{O} + HBr \to C_6H_5COOH + \overset{\bullet}{Br}$
Step 3 . Bromine free radical attacks on $C =C$ of propene to form intermediate free radical.
Hence, $CH_3 - \overset{\bullet}{C}H-CH_2Br$ is the major product of this step.
Step 4 . More stable free radical accept hydrogen free radical from benzoic acid and give final product of reaction.
$CH_3 - \overset{\bullet}{C}H - CH_2Br+C_6H_5COOH \to \underset{\text{n-propyl bromide}}{CH_3-CH_2-CH_2Br+C_6H_5CO\overset{\bullet}{O}}$
Step 5. Benzoate free radicals are changed into benzoyl peroxide for the termination of free radical chain.
$C_6H_5CO\overset{\bullet}{O}+ C_6H_5CO\overset{\bullet}{O} \to (C_6H_5CO)_2O_2$
