Question

Does $CH_3NH_3^+$ give a diazonium salt?

Answer 1

No

Answer 2

Diazonium salts are formed from the reaction of primary aromatic amines with nitrous acid ($HNO_2$, generated in situ from $NaNO_2$ and a strong acid like $HCl$) at low temperatures (0-5 °C).

Methylamine ($CH_3NH_2$) is a primary aliphatic amine. When primary aliphatic amines react with nitrous acid, the intermediate diazonium ion formed is highly unstable. In aqueous solution, it rapidly decomposes to form an alcohol and nitrogen gas, rather than a stable diazonium salt.

The reactions are:

1. Formation of nitrous acid: $NaNO_2 + HCl \rightarrow HNO_2 + NaCl$
2. Reaction of methylamine with nitrous acid: $CH_3NH_2 + HNO_2 \xrightarrow{0-5^\circ C} [CH_3N_2^+] + 2H_2O$
3. Decomposition of the unstable aliphatic diazonium ion: $[CH_3N_2^+] + H_2O \rightarrow CH_3OH + N_2 \uparrow + H^+$

Since the diazonium ion derived from methylamine is unstable and decomposes, methylamine does not yield a stable diazonium salt. The methylammonium ion ($CH_3NH_3^+$) is the protonated form of methylamine. In acidic conditions required for the formation of nitrous acid, methylamine exists predominantly as $CH_3NH_3^+$. However, the reaction with nitrous acid requires the free amine ($CH_3NH_2$), and even if formed, the resulting diazonium ion is unstable.