The compound C_{7}H_{8} undergoes the followi | vedclass

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(A) Step $1$: Toluene $(C_{7}H_{8})$ reacts with $3Cl_{2}$ in the presence of heat to undergo side-chain chlorination,forming benzotrichloride $(C_{6}

Vedclass · Accepted Answer

$m-$bromotoluene

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(A) Step $1$: Toluene $(C_{7}H_{8})$ reacts with $3Cl_{2}$ in the presence of heat to undergo side-chain chlorination,forming benzotrichloride $(C_{6}H_{5}CCl_{3})$ as product '$A$'.Step $2$: The $-CCl_{3}$ group is strongly electron-withdrawing and meta-directing. Therefore,electrophilic substitution with $Br_{2}/Fe$ occurs at the meta position,forming $m-$bromobenzotrichloride as product '$B$'.Step $3$: Reduction of the $-CCl_{3}$ group with $Zn/HCl$ converts it back into a methyl group $(-CH_{3})$,yielding $m-$bromotoluene as the final product '$C$'.Thus,the correct option is $A$.

The compound $C_{7}H_{8}$ undergoes the following reactions:
$C_{7}H_{8}$ $\xrightarrow{3Cl_{2} / \Delta} A$ $\xrightarrow{Br_{2} / Fe} B$ $\xrightarrow{Zn / HCl} C$
The product '$C$' is

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The compound $C_{7}H_{8}$ undergoes the following reactions:$C_{7}H_{8}$ $\xrightarrow{3Cl_{2} / \Delta} A$ $\xrightarrow{Br_{2} / Fe} B$ $\xrightarrow{Zn / HCl} C$The product '$C$' is