Chlorobenzene can be prepared by reacting aniline with

The correct answer is Option (4) → nitrous acid, followed by treatment with cuprous chloride

The –NH₂ group of aniline cannot be replaced by chlorine directly. It is first changed into a diazonium group, which Cu(I) then replaces by –Cl (Sandmeyer’s reaction).

• Diazotisation: nitrous acid, made in the mixture from sodium nitrite and hydrochloric acid, converts aniline into benzenediazonium chloride at 273–278 K.

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

• Sandmeyer’s reaction: cuprous chloride replaces the diazonium group by –Cl and nitrogen gas is given off.

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

Why the other options are incorrect

Option (1) hydrochloric acid — Incorrect. Aniline is a base; with HCl alone it only forms the salt anilinium chloride, and the C–N bond is not broken.

Option (2) cuprous chloride — Incorrect. Cu(I) replaces a diazonium group, not an amino group; without diazotisation first, no chlorobenzene forms.

Option (3) chlorine in the presence of anhydrous AlCl₃ — Incorrect. This puts chlorine on the ring and does not remove the –NH₂ group; besides, aniline forms a salt with the Lewis acid AlCl₃.

NCERT (2026-27 edition):
6.5 Preparation of Haloarenes – “(ii) From amines by Sandmeyer’s reaction
When a primary aromatic amine, dissolved or suspended in cold aqueous mineral acid, is treated with sodium nitrite, a diazonium salt is formed. Mixing the solution of freshly prepared diazonium salt with cuprous chloride or cuprous bromide results in the replacement of the diazonium group by –Cl or –Br.”