The compound $C_7H_8$ undergoes the following reactions:
$$ C_7H_8 \xrightarrow{3Cl_2/\Delta} A \xrightarrow{B

Question

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:

Accepted Answer

1. **Identify Reactant:** The compound $C_7H_8$ is Toluene ($C_6H_5CH_3$).
2. **Step 1 (Side Chain Halogenation):** Reaction with $3Cl_2$ in the presence of heat/light ($h
u$) is a free radical substitution reaction. All three hydrogen atoms of the methyl group are replaced by chlorine atoms [NCERT 12th, Ch 10, Sec 10.6].
   $$ C_6H_5CH_3 + 3Cl_2 \xrightarrow{h
u} C_6H_5CCl_3 	ext{ (A: Benzotrichloride)} $$
3. **Step 2 (Electrophilic Aromatic Substitution):** Reaction of 'A' with $Br_2/Fe$ is a ring substitution (halogenation). The $-CCl_3$ group is a strong electron-withdrawing group (due to -I effect of three Cl atoms) and acts as a **meta-directing group** [NCERT 11th, Ch 13, Sec 13.5.6].
   $$ C_6H_5CCl_3 + Br_2 \xrightarrow{Fe} m	ext{-Bromo-benzotrichloride (B)} $$
4. **Step 3 (Reduction):** Reaction with $Zn/HCl$ reduces the trichloromethyl group ($-CCl_3$) back to the methyl group ($-CH_3$). The aromatic C-Br bond is strong and generally resists this reduction condition.
   $$ m	ext{-Br-}C_6H_4CCl_3 \xrightarrow{Zn/HCl} m	ext{-Br-}C_6H_4CH_3 	ext{ (C: m-Bromotoluene)} $$

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