CH_3-CH=CH_2 xrightarrow[{(low , conc.)}]{Br_ | vedclass

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Question

(B) The reaction of propene with $Br_2$ at low concentration in the presence of light $(h
u)$ or high temperature follows a free radical substitution

Vedclass · Accepted Answer

$CH_2=CH-CH_2-Br$

Vedclass · Answer

(B) The reaction of propene with $Br_2$ at low concentration in the presence of light $(h
u)$ or high temperature follows a free radical substitution mechanism at the allylic position.This is known as allylic bromination.The hydrogen atom attached to the allylic carbon ($CH_3$ group in propene) is replaced by a bromine atom.The product $(A)$ is allyl bromide: $CH_2=CH-CH_2-Br$.

$CH_3-CH=CH_2 \xrightarrow[{(low \, conc.)}]{Br_2/h\nu} (A)$; Product $(A)$ of the reaction is

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