Question

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

• A. iso-propyl 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-Markownikoffs
addition reaction. (i.e. it is completed in form of free radical addition) $CH_3-CH =CH_2 +HBr \xrightarrow{Peroxide}$ \hspace50mmCH_3-CH_2-CH_2Br \hspace50mm _{}^{\, \, \, \, \, \, \, n-propyl\, bromide} Mechanism of this reaction is represented as follows:

Step II Benzoate free radical forms bromine free radical with HBr.
$C_6 H_5 CO \mathring{O}+H-Br \longrightarrow C_6 H_5 COOH +B \mathring{r}$ free radical from benzoic acid and give final product of reaction.
$CH_3 - \mathring{C}H-CH_2 Br + C_6 H_5 COOH \longrightarrow$
\hspace20mm CH_3 - CH_2-CH_2 Br + C_6 H_5 CO \mathring{O}
\hspace20mm _{}^{n-propyl\, bromide}

Step V Benzoate free radicals are changed into benzoyl peroxide for the termination of free radical chain.

$C_6 H_5 CO \mathring{O}+C_6 H_5 CO \mathring{O} \longrightarrow (CC_6 H_5 CO)_2 O_2$

**Read more from the chapter: ****Haloalkanes and Haloarenes **