Which reactive intermediate is involved in th | vedclass

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

Question

(D) The reaction of an alkane with $Br_2$ in the presence of $hv$ (ultraviolet light) proceeds via a free radical substitution mechanism.In the first

Vedclass · Accepted Answer

Secondary free radical

Vedclass · Answer

(D) The reaction of an alkane with $Br_2$ in the presence of $hv$ (ultraviolet light) proceeds via a free radical substitution mechanism.In the first step,the bromine molecule undergoes homolytic cleavage to form bromine radicals $(Br^{\bullet})$.These bromine radicals then abstract a hydrogen atom from the alkane to form an alkyl radical intermediate.For $2-	ext{methylbutane}$ $(CH_3-CH(CH_3)-CH_2-CH_3)$,the abstraction of a hydrogen atom from the $C-2$ position leads to a tertiary free radical,while abstraction from the $C-3$ position leads to a secondary free radical.The formation of $2-	ext{bromo}-3-	ext{methylbutane}$ involves the secondary free radical intermediate formed at the $C-3$ position.

Which reactive intermediate is involved in the following reaction?
$2-\text{methylbutane} \xrightarrow{Br_2, hv} 2-\text{bromo}-3-\text{methylbutane}$ (not the major product)

Similar Questions

What will be the major product when $2-$methylbutane undergoes bromination in the presence of light?

$CH_4$ is formed when:

$A$ fuel contains $25 \%$ $n-$heptane and $75 \%$ iso-octane. Its octane number is

The reaction ${C_2H_5Cl} \xrightarrow{[H]} x$ produces $x$ as:

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Which reactive intermediate is involved in the following reaction?$2-\text{methylbutane} \xrightarrow{Br_2, hv} 2-\text{bromo}-3-\text{methylbutane}$ (not the major product)