Name the following reaction.
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Gattermann reaction
An arenediazonium salt, \( \ce{ArN2+X-} \), easily loses nitrogen gas, so its –\( \ce{N2^{+}} \) group can be replaced by other groups. Two named reactions introduce Cl, Br or CN in this way and differ only in the copper reagent:
Sandmeyer reaction: the diazonium salt is warmed with a copper(I) salt, \( \ce{Cu2Cl2} \)/HCl, \( \ce{Cu2Br2} \)/HBr or CuCN/KCN.
Gattermann reaction: the diazonium salt is treated with the halogen acid in the presence of copper powder. The yield is usually lower than in the Sandmeyer reaction.
The reaction shown uses Cu/HBr, copper powder with hydrobromic acid, so it is the Gattermann reaction.
Why the other options are incorrect
Option (1) Gabriel synthesis — Incorrect. It makes primary alkylamines from phthalimide: potassium phthalimide is alkylated by an alkyl halide and then hydrolysed. No diazonium salt is involved.
\( \displaystyle \ce{C6H4(CO)2NK + R-X -> C6H4(CO)2N-R + KX} \)
\( \displaystyle \ce{C6H4(CO)2N-R} \xrightarrow{\ce{NaOH(aq)}} \ce{R-NH2 + C6H4(COONa)2} \)
Option (2) Sandmeyer reaction — Incorrect. It gives the same aryl bromide, but with a copper(I) salt (\( \ce{Cu2Br2} \)), not with copper powder.
Option (3) Carbylamine reaction — Incorrect. It is a test for primary amines: heating with chloroform and alcoholic KOH gives a foul-smelling isocyanide.
\( \displaystyle \ce{R-NH2 + CHCl3 + 3KOH} \xrightarrow{\Delta} \ce{R-NC + 3KCl + 3H2O} \)
NCERT (2026-27 edition):
Replacement by Halide or Cyanide Ion
