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The compound C7H8C_7H_8 undergoes the following reactions: C7H8→3Cl2/ΔA→Br2/FeB→Zn/HClCC_7H_8 \xrightarrow{3Cl_2/\Delta} A \xrightarrow{Br_2/Fe} B \xrightarrow{Zn/HCl} C The product 'C' is:

A

m–Bromotoluene

B

o–Bromotoluene

C

3–Bromo–2,4,6–trichlorotoluene

D

p–Bromotoluene

Step-by-Step Solution

  1. Identify Reactant: The compound C7H8C_7H_8 is Toluene (C6H5CH3C_6H_5CH_3).
  2. Step 1 (Side Chain Halogenation): Reaction with 3Cl23Cl_2 in the presence of heat/light (hνh\nu) 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]. C6H5CH3+3Cl2→hνC6H5CCl3 (A: Benzotrichloride)C_6H_5CH_3 + 3Cl_2 \xrightarrow{h\nu} C_6H_5CCl_3 \text{ (A: Benzotrichloride)}
  3. Step 2 (Electrophilic Aromatic Substitution): Reaction of 'A' with Br2/FeBr_2/Fe is a ring substitution (halogenation). The −CCl3-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]. C6H5CCl3+Br2→Fem-Bromo-benzotrichloride (B)C_6H_5CCl_3 + Br_2 \xrightarrow{Fe} m\text{-Bromo-benzotrichloride (B)}
  4. Step 3 (Reduction): Reaction with Zn/HClZn/HCl reduces the trichloromethyl group (−CCl3-CCl_3) back to the methyl group (−CH3-CH_3). The aromatic C-Br bond is strong and generally resists this reduction condition. m-Br-C6H4CCl3→Zn/HClm-Br-C6H4CH3 (C: m-Bromotoluene)m\text{-Br-}C_6H_4CCl_3 \xrightarrow{Zn/HCl} m\text{-Br-}C_6H_4CH_3 \text{ (C: m-Bromotoluene)}
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