Addition of $HBr$ to propene yields $2-$bromopropane,while in the presence of benzoyl peroxide,the same reaction yields $1-$bromopropane. Explain and give the mechanism.

(N/A) $1$. Addition of $HBr$ to propene in the absence of peroxide follows Markovnikov's rule. The reaction proceeds via an electrophilic addition mechanism involving a carbocation intermediate. The secondary carbocation $(CH_3-CH^+-CH_3)$ is more stable than the primary carbocation $(CH_3-CH_2-CH_2^+)$,leading to the formation of $2-$bromopropane.
$2$. In the presence of benzoyl peroxide,the reaction follows the anti-Markovnikov rule (peroxide effect or Kharasch effect). This proceeds via a free radical mechanism. The benzoyl peroxide generates bromine radicals $(Br^\bullet)$. The bromine radical attacks the double bond to form the more stable secondary free radical $(CH_3-CH^\bullet-CH_2Br)$,which then abstracts a hydrogen atom from $HBr$ to yield $1-$bromopropane.