In the following reaction sequence, [C] is :

Solution

Chemistry ma’am · 4 steps

Follow the chlorine: the ring Cl stays put, and only the side-chain Cl reacts with sodium.

1. Sandmeyer gives A. NaNO₂ + HCl at 0 to 5 °C turns p-toluidine into the diazonium salt. Cu₂Cl₂ + HCl then replaces -N₂⁺ with Cl. So [A] is p-chlorotoluene, Cl-C₆H₄-CH₃.
2. Light chlorinates CH3. Cl₂ with hν is a free-radical reaction. It attacks the benzylic CH₃, not the ring. So [B] is p-chlorobenzyl chloride, Cl-C₆H₄-CH₂Cl.
3. Which Cl reacts. [B] now has two C–Cl bonds. The ring C–Cl has partial double-bond character from resonance, so Na in dry ether leaves it alone. The benzylic C–Cl is reactive, so only that end takes part in the Wurtz coupling.

   2 Cl-C₆H₄-CH₂Cl + 2Na →(dry ether) Cl-C₆H₄-CH₂-CH₂-C₆H₄-Cl + 2NaCl

4. So. Two benzylic carbons join through a new CH₂-CH₂ bridge and both ring chlorines stay in place. [C] is 1,2-bis(4-chlorophenyl)ethane, option (c).

The idea to keep. When a molecule carries both an aryl Cl and an alkyl or benzylic Cl, Na in dry ether couples only the alkyl one. The ring C–Cl survives because resonance gives it partial double-bond character.

Answer from the NTA final answer key. Solution by PrepWiser.