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)

n-propyl bromide

Answer 2

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.) $CH_3 - CH = CH_2 + HBr {->[\text{Peroxide}]} \underset{\text{n - propyl bromide}}{CH_3 - CH_2 - CH_2Br}$ Mechanism of this reaction is represented as follows Step 1 . formation of free radical of peroxide by means of decomposition. $\underset{\text{Benzoyl peroxide}}{C_6H_5 - \underset{\overset{||}{O}}{C} - O -O - \underset{\overset{||}{O}}{C} - C_6H_5} {->[\Delta]}$ $\underset{\text{Benzoate free radical}}{2C_6H_5 - CO\overset{\bullet}{O}}$ 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. $C_6H_5CO\overset{\bullet}{O}+ C_6H_5CO\overset{\bullet}{O} \to (C_6H_5CO)_2O_2$