Compound (A) on reaction with NaNO2 and HCl at low temperature forms…

The correct answer is Option (2) → A: Aniline, B: Benzenediazonium chloride, C: p-hydroxyazobenzene

Two reactions are hidden in this question:

• Diazotisation: a primary aromatic amine reacts with nitrous acid (made in the mixture from \( \ce{NaNO2} \) and HCl) at 273–278 K to give an arenediazonium salt. The low temperature is needed because the salt decomposes when warm.

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

• Coupling: the diazonium ion is a weak electrophile. It attacks the para position of phenol in a mildly alkaline medium (where phenol is present as the more reactive phenoxide ion) and forms an azo compound, –N=N–, which is coloured.

      \( \displaystyle \ce{C6H5N2^+Cl^- + C6H5OH} \xrightarrow{\ce{OH^-}} \ce{C6H5-N=N-C6H4-OH + Cl^- + H2O} \)

Working backwards: the orange dye C is p-hydroxyazobenzene, so B must be benzenediazonium chloride, and A, the compound diazotised by \( \ce{NaNO2} \)/HCl, is aniline.

Why the other options are incorrect

• Option (1) A: Phenol, B: Aniline, C: p-aminoazobenzene — Incorrect. Phenol is not diazotised by \( \ce{NaNO2} \)/HCl, and aniline does not form from it. p-Aminoazobenzene is the yellow dye formed when the diazonium salt couples with aniline in a mildly acidic medium, not with phenol.

• Option (3) A: p-nitroaniline, B: Hydroxyazobenzene, C: p-aminoazobenzene — Incorrect. Hydroxyazobenzene is the final coupling product, not the intermediate B, and coupling with phenol cannot give an amino-substituted dye.

• Option (4) A: Nitrobenzene, B: Aniline, and C: Phenol — Incorrect. \( \ce{NaNO2} \)/HCl does not reduce nitrobenzene to aniline (that needs Sn/HCl or Fe/HCl), and phenol is not a dye.

NCERT (2026-27 edition):
Reactions Involving Retention of Diazo Group