Methylcyclopentane xrightarrow{Br_2/hv} matho | vedclass

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(D) $1$. Step $1$: Free radical bromination of methylcyclopentane with $Br_2/hv$ occurs preferentially at the tertiary carbon to form $1-$bromo$-1-$me

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$1-$Bromo$-2-$methylcyclopentane

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(D) $1$. Step $1$: Free radical bromination of methylcyclopentane with $Br_2/hv$ occurs preferentially at the tertiary carbon to form $1-$bromo$-1-$methylcyclopentane $(X)$ as the major product.$2$. Step $2$: Dehydrohalogenation of $(X)$ with alcoholic $KOH/\Delta$ follows the $E2$ mechanism to form the more substituted alkene,$1$-methylcyclopentene $(Y)$,as the major product.$3$. Step $3$: Anti-Markovnikov addition of $HBr$ to $1-$methylcyclopentene $(Y)$ in the presence of peroxide yields $1-$bromo$-2-$methylcyclopentane $(Z)$ as the major product.

Methylcyclopentane $\xrightarrow{Br_2/hv} \mathop{Major}\limits_{(X)}$ $\xrightarrow[KOH/\Delta]{Alcoholic} \mathop{Major}\limits_{(Y)}$ $\xrightarrow[Peroxide]{HBr} \mathop{Major}\limits_{(Z)}$
Major final product $(Z)$ is

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Methylcyclopentane $\xrightarrow{Br_2/hv} \mathop{Major}\limits_{(X)}$ $\xrightarrow[KOH/\Delta]{Alcoholic} \mathop{Major}\limits_{(Y)}$ $\xrightarrow[Peroxide]{HBr} \mathop{Major}\limits_{(Z)}$Major final product $(Z)$ is