Question

Reaction of $HBr$ with propene in the presence of peroxide gives

• A. isopropyi bromide
• B. 3-bromo propane
• C. allyl bromide
• D. n-propyl bromide

Answer 1

Answer: (D)

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.)
$C{H}_3-CH=C{H}_2+HBr\stackrel{\text{Peroxide}}{\to }\underset{\text{n - propyl bromide}}{C{H}_3-C{H}_2-C{H}_{2}Br}$
Mechanism of this reaction is represented as follows
Step 1 . formation of free radical of peroxide by means of decomposition.
$\underset{\text{Benzoyl peroxide}}{C_6{H}_5-\underset{\stackrel{||}{O}}{C}-O-O-\underset{\stackrel{||}{O}}{C}-C_6{H}_5}\stackrel{\Delta }{\to }$
$\underset{\text{Benzoate free radical}}{2C_6{H}_5-CO\stackrel{\bullet }{O}}$
Step 2 . Benzoate free radical forms bromine free radical with $HBr$.
$C_6{H}_{5}CO\stackrel{\bullet }{O}+HBr\to C_6{H}_{5}COOH+\stackrel{\bullet }{\mathrm{Br}}$
Step 3 . Bromine free radical attacks on $C=C$ of propene to form intermediate free radical.

Hence, $C{H}_3-\stackrel{\bullet }{C}H-C{H}_{2}Br$ 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.
$C{H}_3-\stackrel{\bullet }{C}H-C{H}_{2}Br+C_6{H}_{5}COOH\to \underset{\text{n-propyl bromide}}{C{H}_3-C{H}_2-C{H}_{2}Br+C_6{H}_{5}CO\stackrel{\bullet }{O}}$
Step 5. Benzoate free radicals are changed into benzoyl peroxide for the termination of free radical chain.
$C_6{H}_{5}CO\stackrel{\bullet }{O}+C_6{H}_{5}CO\stackrel{\bullet }{O}\to (C_6{H}_{5}CO{)}_2{O}_2$