Looking for classes? Ksquare Career Institute, Bengaluru →
The compound $C_7H_8$ undergoes the following reactions:
$C_7H_8 \xrightarrow{3Cl_2/\Delta} A \xrightarrow{Br_2/Fe} B \xrightarrow{Zn/HCl} C$.
The product 'C' is
$C_7H_8 \xrightarrow{3Cl_2/\Delta} A \xrightarrow{Br_2/Fe} B \xrightarrow{Zn/HCl} C$.
The product 'C' is
Explanation
The $-CCl_3$ group directs Br to the meta position before reduction back to $-CH_3$.
Detailed Solution
$C_6H_5CH_3\ (C_7H_8) \xrightarrow{3Cl_2/\Delta} C_6H_5CCl_3\ (A)$ (side-chain chlorination)
$-CCl_3$ is meta-directing: $C_6H_5CCl_3 \xrightarrow{Br_2/Fe} m-BrC_6H_4CCl_3\ (B)$
$m-BrC_6H_4CCl_3 \xrightarrow{Zn/HCl} m-BrC_6H_4CH_3\ (C)$
So C is m-bromotoluene.
$-CCl_3$ is meta-directing: $C_6H_5CCl_3 \xrightarrow{Br_2/Fe} m-BrC_6H_4CCl_3\ (B)$
$m-BrC_6H_4CCl_3 \xrightarrow{Zn/HCl} m-BrC_6H_4CH_3\ (C)$
So C is m-bromotoluene.