Solution

In the presence of a peroxide, HBr adds to an unsymmetrical alkene against Markovnikov's rule (Kharasch effect or peroxide effect). The reaction goes by a free radical mechanism.
The peroxide gives a bromine radical, which adds to the terminal carbon of propene so that the more stable secondary radical forms:
$CH_3-CH=CH_2 + \dot{B}r \rightarrow CH_3-\dot{C}H-CH_2Br$
$CH_3-\dot{C}H-CH_2Br + HBr \rightarrow CH_3CH_2CH_2Br + \dot{B}r$
Overall: $CH_3CH=CH_2 + HBr \xrightarrow{peroxide} CH_3CH_2CH_2Br$
The product is 1-bromopropane, i.e. n-propyl bromide. (Without peroxide, Markovnikov addition would give isopropyl bromide.)

Concept Kharasch peroxide effect

Hydrocarbons in past papers

75 questions from this chapter have appeared across 26 exam years.