Q.Aniline reacts with nitrous acid at $0$–$5^\circ\text{C}$ to give a stable diazonium salt, whereas $N$-methylaniline (a secondary aromatic amine) gives an $N$-nitrosamine under the same conditions. Explain this difference in terms of the number of hydrogens on nitrogen available for substitution.

Nitrous acid (HNO2) converts a primary aliphatic amine into the corresponding alcohol with liberation of nitrogen gas.

A primary aliphatic amine reacts with nitrous acid (generated in situ from NaNO2 + HCl) to form an unstable diazonium salt, which immediately decomposes to give the alcohol, releasing N2 and brisk effervescence:

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Only secondary amines form stable N-nitrosamines with nitrous acid, so the $C_3H_9N$ amine described must be the secondary amine CH₃-NH-C₂H₅.

The reaction of an amine with nitrous acid ($NaNO_2/HCl$) depends on the amine's class:

• 1° amines give unstable diazonium salts (aliphatic ones decompose instantly to alcohols + $N_2$).
• 2° amines give a stable, yellow oily N-nitrosamine, $R_2N-N=O$ — these are known carcinogens.
• 3° amines simply form an unstable salt with the acid, with no further reaction of the nitrogen.

Since the question specifies a carcinogenic nitrosamine is formed, the amine must be secondary. With molecular formula $C_3H_9N$ and a secondary ($R_2NH$) structure, the amine is:

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