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

(A) The addition of $HBr$ to propene in the absence of peroxide follows Markovnikov's rule,which is an electrophilic addition reaction.
$HBr$ dissociates to provide an electrophile,$H^+$. This electrophile attacks the double bond to form $1^{\circ}$ and $2^{\circ}$ carbocations:
$CH_3-CH=CH_2 + H^+ \to CH_3-CH_2-CH_2^+$ (Primary,less stable)
$CH_3-CH=CH_2 + H^+ \to CH_3-CH^+-CH_3$ (Secondary,more stable)
Since the secondary carbocation is more stable,it forms faster and reacts with $Br^-$ to yield $2$-bromopropane as the major product.
In the presence of benzoyl peroxide,the reaction follows a free radical mechanism (peroxide effect or Kharasch effect),which is anti-Markovnikov addition:
$1$. Initiation: Benzoyl peroxide undergoes homolysis to form phenyl radicals,which react with $HBr$ to generate $Br^{\centerdot}$.
$2$. Propagation: $Br^{\centerdot}$ attacks propene to form a secondary free radical $(CH_3-overset{\centerdot}{C}H-CH_2Br)$ which is more stable than the primary radical $(CH_3-CH(Br)-overset{\centerdot}{C}H_2)$.
$3$. The secondary radical reacts with $HBr$ to form $1$-bromopropane as the major product and regenerates $Br^{\centerdot}$.