Identify A and B in the reaction sequence given below (X = Cl or Br).

The correct answer is Option (1) → \(A = C_6H_5N_2Cl, B = C_6H_5X\)

Forming A (diazotisation): aniline reacts with nitrous acid, made in the mixture from \( \ce{NaNO2} \) and \( \ce{HCl} \), in the cold (0 °C) to give benzenediazonium chloride:

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

Forming B (Sandmeyer reaction): in the presence of Cu(I) halide, the –N₂⁺ group is replaced by one halogen atom and nitrogen gas is given off, so B is the halobenzene \( \ce{C6H5X} \):

    \( \displaystyle \ce{C6H5N2^+X^-} \xrightarrow{\ce{CuX}} \ce{C6H5X + N2} \)

Why the other options are incorrect

Option (2) \(A = C_6H_5N_2Cl, B = C_6H_4X_2\) — Incorrect. A is right, but the diazonium group is replaced by only one halogen atom; no second ring hydrogen is substituted.

Option (3) \(A = C_6H_5NCl, B = C_6H_5X\) — Incorrect. The diazonium salt carries two nitrogen atoms (–N₂⁺); C₆H₅NCl is not benzenediazonium chloride.

Option (4) \(A = C_6H_5NCl, B = C_6H_4X_2\) — Incorrect. Both are wrong: A must contain –N₂⁺, and B carries only one halogen atom.

NCERT (2026-27 edition):
9.7 Method of Preparation of Diazonium Salts – “Benzenediazonium chloride is prepared by the reaction of aniline with nitrous acid at 273-278K. Nitrous acid is produced in the reaction mixture by the reaction of sodium nitrite with hydrochloric acid.”

9.9 Chemical Reactions – “1. Replacement by halide or cyanide ion: The Cl^–, Br^– and CN^– nucleophiles can easily be introduced in the benzene ring in the presence of Cu(I) ion. This reaction is called Sandmeyer reaction.”