Explain the mechanism when a secondary amine reacts with nitrous acid. Include the formation of the N‑nitrosamine intermediate and why the reaction stops without forming a diazonium ion. State the final product.

Step-by-step solution

1. Write the generation of nitrous acid in situ: $\mathrm{NaN}{\mathrm{O}}_2+\mathrm{HCl}\to \mathrm{HN}{\mathrm{O}}_2+\mathrm{NaCl}$.
2. Protonate the secondary amine (${\mathrm{R}}_2\mathrm{NH}$) to give ${\mathrm{R}}_2\mathrm{N}{\mathrm{H}}_2^{+}$, making the nitrogen more nucleophilic toward $\mathrm{N}{\mathrm{O}}^{+}$ (the active species from $\mathrm{HN}{\mathrm{O}}_2$).
3. The $\mathrm{N}{\mathrm{O}}^{+}$ attacks the lone pair on nitrogen, forming an $N–N$ bond and giving the N‑nitrosamine intermediate ${\mathrm{R}}_{2}N–N=\mathrm{O}$ after loss of water.
4. Because the nitrogen of a secondary amine already bears two alkyl groups, the resulting diazonium ion (${\mathrm{R}}_2{\mathrm{N}}^{+}={\mathrm{N}}_2$) would be highly unstable and rapidly loses ${\mathrm{N}}_2$, so the reaction does not proceed beyond the nitroso stage.
5. Thus the stable product isolated is the N‑nitrosoamine ${\mathrm{R}}_{2}N–N=\mathrm{O}$.

Concept tested: Reaction of secondary amines with nitrous acid (formation of N‑nitrosamines)

Published 5 September 2026 · How this solution was prepared and checked