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(A) The reaction of methylenecyclopentane with $HBr$ in the presence of a peroxide $(R_2O_2)$ follows the anti-Markovnikov addition mechanism (peroxid

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Cyclopentylmethyl bromide

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(A) The reaction of methylenecyclopentane with $HBr$ in the presence of a peroxide $(R_2O_2)$ follows the anti-Markovnikov addition mechanism (peroxide effect or Kharasch effect).In this reaction,the $Br^\bullet$ radical attacks the less substituted carbon of the double bond to form a more stable radical intermediate.The terminal carbon $(CH_2)$ of the exocyclic double bond is less substituted than the ring carbon,so the bromine atom attaches to the terminal $CH_2$ group.Thus,the product formed is $1$-(bromomethyl)cyclopentane,which is also known as cyclopentylmethyl bromide.

What is the product of the following reaction?
Methylenecyclopentane $\xrightarrow{HBr, R_2O_2}$ Product is

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What is the product of the following reaction?Methylenecyclopentane $\xrightarrow{HBr, R_2O_2}$ Product is