Methylcyclopentane xrightarrow{Br_2/hv} matho | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$1-$Bromo$-2-$methylcyclopentane

Vedclass · Answer

(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

Similar Questions

The product $(A)$ of the given reaction is:

$A$ molecule $P$ on treatment with acid undergoes rearrangement and gives $Q$. $Q$ on ozonolysis followed by reflux under alkaline condition gives $R$. The structure of $R$ is given below:
[Image of $R$]
The structure of $P$ is

What is the organic product obtained when ethylene glycol is heated with acidified potassium permanganate?

Which of the following reactions does not represent the major product correctly?

Which of the following is not a member of the same homologous series?

Vedclass Test Series

Exam Paper Generator

Online Exam Module

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