In the following reaction sequence, [C] is : (Sequence: p-toluidine, (i) \(\mathrm{NaNO_2 + HCl}\), \(0-5^\circ\mathrm{C}\); (ii) \(\mathrm{Cu_2Cl_2 + HCl}\) gives [A]; [A] with \(\mathrm{Cl_2}\), \(h\nu\) gives [B]; [B] with Na + dry ether gives [C] (Major Product).)
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Sandmeyer reaction replaces the diazonium group by Cl; \(\mathrm{Cl_2}\)/\(h\nu\) chlorinates the benzylic \(\mathrm{CH_3}\); Na/dry ether then couples two benzylic chlorides (Wurtz-type).
Step 1: p-Toluidine with \(\mathrm{NaNO_2 + HCl}\) at 0-5 \(^\circ\)C gives p-methylbenzenediazonium chloride; \(\mathrm{Cu_2Cl_2 + HCl}\) (Sandmeyer) replaces \(\mathrm{N_2^+}\) by Cl. So [A] = 4-chlorotoluene. Step 2: \(\mathrm{Cl_2/h\nu}\) is free-radical chlorination, which occurs at the benzylic side chain, not on the ring. [B] = 4-chlorobenzyl chloride, \(\mathrm{Cl-C_6H_4-CH_2Cl}\). Step 3: Na in dry ether (Wurtz reaction) joins two benzylic carbons by removing the benzylic Cl (the aryl Cl is inert): \(\mathrm{2\,Cl-C_6H_4-CH_2Cl + 2Na \to Cl-C_6H_4-CH_2-CH_2-C_6H_4-Cl + 2NaCl}\). So [C] = 1,2-bis(4-chlorophenyl)ethane, which is option (3). Answer: (3) 1,2-bis(4-chlorophenyl)ethane, Cl-C6H4-CH2-CH2-C6H4-Cl
