Identify the compound Y in the following reaction :

The correct answer is Option (1) → chlorobenzene

Aniline is first diazotised: with \( \ce{NaNO2} \) and HCl at 273–278 K it gives benzenediazonium chloride:

    \( \displaystyle \ce{C6H5NH2 + NaNO2 + 2HCl} \xrightarrow{273-278\ \mathrm{K}} \ce{C6H5N2+Cl- + NaCl + 2H2O} \)

In the Sandmeyer reaction, the diazonium group is replaced by \( \ce{Cl^{-}} \) in the presence of Cu(I), and nitrogen gas escapes:

    \( \displaystyle \ce{C6H5N2+Cl-} \xrightarrow{\ce{Cu2Cl2}} \ce{C6H5Cl + N2} \)

So Y is chlorobenzene.

The related replacements of the diazonium group with copper are:

• Sandmeyer reaction — Cu(I) salt with its acid: CuCl/HCl gives ArCl, CuBr/HBr gives ArBr, CuCN/KCN gives ArCN.

      \( \ce{ArN2+X^-} \xrightarrow{\ce{CuCN/KCN}} \ce{ArCN + N2} \)

• Gattermann reaction — copper powder with the halogen acid; it also gives ArCl or ArBr, but the yield is lower than in the Sandmeyer reaction.

      \( \ce{ArN2+X^-} \xrightarrow{\ce{Cu/HX}} \ce{ArX + N2} \)

Why the other options are incorrect

• Option (2) benzene — Incorrect. It is formed when the diazonium salt is reduced with hypophosphorous acid (\( \ce{H3PO2} \)) or ethanol, not with \( \ce{Cu2Cl2} \).

• Option (3) 1,3-dichlorobenzene — Incorrect. Only one diazonium group is present, so only one Cl can be introduced.

• Option (4) 1,4-dichlorobenzene — Incorrect. For the same reason, a second Cl cannot be introduced.

NCERT (2026-27 edition):
Replacement by Halide or Cyanide Ion