In the following reaction,the compound C is:{ | vedclass

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(C) Step $1$: Toluene $({C_7}{H_8})$ reacts with $3Cl_2$ in the presence of $\Delta$ (heat) to undergo side-chain chlorination,forming benzotrichlorid

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$m-$bromotoluene

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(C) Step $1$: Toluene $({C_7}{H_8})$ reacts with $3Cl_2$ in the presence of $\Delta$ (heat) to undergo side-chain chlorination,forming benzotrichloride $({C_6}{H_5}CCl_3)$ as compound $A$.Step $2$: Benzotrichloride $({C_6}{H_5}CCl_3)$ contains a $-CCl_3$ group,which is a strongly deactivating and meta-directing group. Reaction with $Br_2/Fe$ leads to electrophilic aromatic substitution at the meta position,forming $m-$bromobenzotrichloride as compound $B$.Step $3$: Reduction of the $-CCl_3$ group using $Zn/HCl$ converts it back into a methyl $(-CH_3)$ group. Thus,$m-$bromobenzotrichloride is reduced to $m-$bromotoluene,which is compound $C$.

In the following reaction,the compound $C$ is:
${C_7}{H_8}$ $\xrightarrow[\Delta]{3Cl_2} A$ $\xrightarrow{Br_2/Fe} B$ $\xrightarrow{Zn/HCl} C$

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In the following reaction,the compound $C$ is:${C_7}{H_8}$ $\xrightarrow[\Delta]{3Cl_2} A$ $\xrightarrow{Br_2/Fe} B$ $\xrightarrow{Zn/HCl} C$