The compound C7H8 undergoes the following reactions:C7H8 → (3Cl2/hν) → A → (Br2/Fe) → B → (Zn/HCl) →

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Question

The compound C7H8 undergoes the following reactions:C7H8 → (3Cl2/hν) → A → (Br2/Fe) → B → (Zn/HCl) → CThe product 'C' is

Accepted Answer

Correct Answer: A — m-bromotolueneStep-by-step Solution:C7H8 is toluene (methylbenzene).Step 1: Toluene + 3Cl2/hν → A (side-chain chlorination under light/heat). The methyl group undergoes free radical chlorination: –CH3 → –CCl3. Product A = benzotrichloride (C6H5CCl3).Step 2: A + Br2/Fe → B (electrophilic aromatic substitution on the ring). The –CCl3 group is a strong electron-withdrawing group (–I and –M effects), making it a meta-director. Therefore, Br enters the meta position. Product B = m-bromobenzotrichloride.Step 3: B + Zn/HCl → C. This is a reduction step that removes the three chlorine atoms from the side chain (–CCl3 → –CH3) while retaining the ring bromine. Product C = m-bromotoluene.Why other options are incorrect:Options B and C (ortho and para products) would require the substituent to be an ortho/para director, but –CCl3 is a meta-director due to its strong electron-withdrawing nature.Option D is an intermediate (B), not the final product after Zn/HCl reduction.Key Concept: Free radical halogenation occurs on the side chain (not the ring) when Cl2 is used in the presence of light (hν) or heat. Electrophilic aromatic substitution occurs on the ring in the presence of a Lewis acid catalyst (Fe). –CCl3 is a meta-director due to its strong –I effect.

The compound C7H8 undergoes the following reactions:C7H8 → (3Cl2/hν) → A → (Br2/Fe) → B → (Zn/HCl) → CThe product 'C' is

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The compound C7H8 undergoes the following reactions:C7H8 → (3Cl2/hν) → A → (Br2/Fe) → B → (Zn/HCl) → CThe product 'C' is

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