Which of the following amine forms a stable diazonium salt?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Primary aromatic amines
Primary amines react with nitrous acid, made in the reaction mixture from sodium nitrite and hydrochloric acid, to give diazonium salts. Whether the salt survives depends on the group attached to nitrogen.
With a primary aromatic amine at 273–278 K, the arenediazonium salt is formed and stays in solution for a short time. (it is stabilised by resonance with the ring):
\( \displaystyle \ce{C6H5NH2 + NaNO2 + 2HCl} \xrightarrow{273-278\ \text{K}} \ce{C6H5N2+Cl- + NaCl + 2H2O} \)
A primary aliphatic amine gives an alkyldiazonium salt that at once loses nitrogen:
\( \displaystyle \ce{R-NH2 + HNO2} \xrightarrow{\ce{NaNO2 + HCl}} [\ce{R-N2+Cl-}] \xrightarrow{\ce{H2O}} \ce{R-OH + N2 ^ + HCl} \)
So a stable diazonium salt is formed only by primary aromatic amines.
Why the other options are incorrect
Option (1) Primary aliphatic amines — Incorrect. They form alkyldiazonium salts that decompose at once with the evolution of nitrogen gas.
Option (2) Secondary aliphatic amines — Incorrect. A diazonium ion needs two hydrogens on nitrogen; a secondary amine has only one and gives an N-nitrosamine (a yellow oily compound) instead.
Option (4) Secondary aromatic amines — Incorrect. For the same reason, N-methylaniline and similar amines give N-nitroso compounds, not diazonium salts.
NCERT (2026-27 edition):
Chemical Reactions – …
