Identify the product of the given reaction.
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → o-Bromoanisole
o-Anisidine (2-methoxyaniline) is a primary aromatic amine. In step (i), nitrous acid formed from \( \ce{NaNO2} \) and HCl converts its –NH2 group into a diazonium group (diazotisation):
\( \displaystyle \ce{CH3O-C6H4-NH2 + NaNO2 + 2HCl} \xrightarrow{273-278\,\mathrm{K}} \ce{CH3O-C6H4-N2^+Cl^- + NaCl + 2H2O} \)
In step (ii), CuBr (a Cu(I) salt) replaces the diazonium group by Br, and nitrogen gas escapes (Sandmeyer reaction):
\( \displaystyle \ce{CH3O-C6H4-N2^+Cl^- + CuBr -> CH3O-C6H4-Br + N2 ^ + CuCl} \)
Br takes the same ring position that the –NH2 group held, next to the –OCH3 group. The product is 2-bromoanisole, o-bromoanisole.
Why the other options are incorrect
Option (2) m-Bromoanisole — Incorrect. Br enters only where the diazonium group was; it does not move to the meta position.
Option (3) p-Bromoanisole — Incorrect. This would need an amino group para to –OCH3, as in p-anisidine; in o-anisidine it is ortho.
Option (4) No product — Incorrect. A primary aromatic amine is diazotised readily, and the diazonium group is a very good leaving group that Br− replaces in the presence of Cu(I).
NCERT (2026-27 edition):
9.7 Method of Preparation of Diazonium Salts – “… The conversion of primary aromatic amines into diazonium salts is known as diazotisation.”
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.”