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(B) The reaction of an alkene with $HBr$ in the presence of a peroxide $(R_2O_2)$ and light $(hv)$ proceeds via a free radical mechanism,which follows

Vedclass · Accepted Answer

$2-$bromo$-1-$methylbicyclo[$4.2$.$0$]octane

Vedclass · Answer

(B) The reaction of an alkene with $HBr$ in the presence of a peroxide $(R_2O_2)$ and light $(hv)$ proceeds via a free radical mechanism,which follows the Anti-Markovnikov rule.In this mechanism,the bromine radical $(Br^{\bullet})$ attacks the double bond to form the more stable radical intermediate.For the given substrate,the $Br^{\bullet}$ radical attacks the less substituted carbon of the double bond,leading to a more stable tertiary $(3^{\circ})$ radical intermediate.Subsequently,this radical abstracts a hydrogen atom from $HBr$ to form the final product,which is $1$-methyl-$2$-bromobicyclo$[4.2.0]$octane.

The major product $(A)$ is:

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