How many of the following amines liberate nitrogen gas when treated with nitrous acid?
Ethanamine, Propan-1-amine, N-Methylethanamine, Propan-2-amine, N,N-Dimethylmethanamine, 2-Methylpropan-2-amine, N-Methylmethanamine
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 4
A primary aliphatic amine forms an aliphatic diazonium salt with nitrous acid. The salt is unstable and breaks down at once, giving off nitrogen gas and leaving an alcohol. Secondary and tertiary amines react in a different manner and do not give this nitrogen.
\( \displaystyle \ce{R-NH2} \xrightarrow{\ce{NaNO2 + HCl}} \ce{[R-N2^+Cl^-]} \xrightarrow{\ce{H2O}} \ce{R-OH + N2 ^ + HCl} \)
Liberate nitrogen (primary amines):
Ethanamine (gives ethanol)
Propan-1-amine (gives propan-1-ol)
Propan-2-amine (gives propan-2-ol)
2-Methylpropan-2-amine (gives 2-methylpropan-2-ol)
No nitrogen gas:
N-Methylethanamine — secondary amine.
N,N-Dimethylmethanamine — tertiary amine.
N-Methylmethanamine — secondary amine.
Liberate nitrogen: 4 amines.
NCERT (2026-27 edition):
9.6 Chemical Reactions – “Three classes of amines react differently with nitrous acid which is prepared in situ from a mineral acid and sodium nitrite.”
“(a) Primary aliphatic amines react with nitrous acid to form aliphatic diazonium salts which being unstable, liberate nitrogen gas quantitatively and alcohols.” … “Secondary and tertiary amines react with nitrous acid in a different manner.”