C_7H_8 xrightarrow{3Cl_2 / text{heat}} A xrig | vedclass

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Question

(D) Step $1$: Toluene $(C_7H_8)$ reacts with $3Cl_2$ in the presence of heat (free radical substitution) to form benzotrichloride $(C_6H_5CCl_3)$,whic

Vedclass · Accepted Answer

$m-	ext{bromotoluene}$

Vedclass · Answer

(D) Step $1$: Toluene $(C_7H_8)$ reacts with $3Cl_2$ in the presence of heat (free radical substitution) to form benzotrichloride $(C_6H_5CCl_3)$,which is compound $A$.Step $2$: The $-CCl_3$ group is a strong deactivating and meta-directing group. Therefore,reaction with $Fe/Br_2$ (electrophilic aromatic substitution) yields $m-	ext{bromobenzotrichloride}$ $(C_6H_4(Br)CCl_3)$,which is compound $B$.Step $3$: Reduction of the $-CCl_3$ group with $Zn/HCl$ converts it back to a methyl group $(-CH_3)$. Thus,the product $C$ is $m-	ext{bromotoluene}$.

$C_7H_8$ $\xrightarrow{3Cl_2 / \text{heat}} A$ $\xrightarrow{Fe/Br_2} B$ $\xrightarrow{Zn/HCl} C$. What is compound $C$?

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