Four isomeric amines of molecular formula C4H11N are given below. Which of them liberates nitrogen gas when treated with nitrous acid?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2-Methylpropan-2-amine
Only a primary aliphatic amine (–NH2 group, one alkyl group on nitrogen) forms an aliphatic diazonium salt with nitrous acid. The salt is unstable and gives off nitrogen gas and an alcohol. Secondary and tertiary amines react with nitrous acid in a different manner.
The class of an amine depends on the number of groups on the nitrogen, not on the carbon that carries it.
N-Ethylethanamine, (C2H5)2NH — two alkyl groups on nitrogen: secondary.
2-Methylpropan-2-amine, (CH3)3C–NH2 — one alkyl group on nitrogen: primary, although the carbon bonded to nitrogen is tertiary.
N,N-Dimethylethanamine, C2H5N(CH3)2 — three alkyl groups on nitrogen: tertiary.
N-Methylpropan-2-amine, (CH3)2CH–NH–CH3 — two alkyl groups on nitrogen: secondary.
\( \displaystyle \ce{(CH3)3C-NH2 + NaNO2 + 2HCl -> [(CH3)3C-N2^+Cl^-] + NaCl + 2H2O} \)
\( \displaystyle \ce{[(CH3)3C-N2^+Cl^-] + H2O -> (CH3)3C-OH + N2 ^ + HCl} \)
NCERT (2026-27 edition):
9.6 Chemical Reactions – “(a) Primary aliphatic amines react with nitrous acid to form aliphatic diazonium salts which being unstable, liberate nitrogen gas quantitatively and alcohols. Quantitative evolution of nitrogen is used in estimation of amino acids and proteins.”
“Secondary and tertiary amines react with nitrous acid in a different manner.”