2022 H2 Chem P1 Q27 — Nitrogen Compound Solution

The answer is A (\([\text{CH}_3\text{NH}_2\text{CH}_2\text{CONH}_2]^+\text{Cl}^-\)).

The molecule has a secondary amine group (\(\text{CH}_3\text{NH}-\)) and a primary amide group (\(-\text{CONH}_2\)).
• The secondary amine N has a lone pair available to accept \(\text{H}^+\), so it is basic and is protonated by HCl, giving \([\text{CH}_3\text{NH}_2\text{CH}_2\text{CONH}_2]^+\text{Cl}^-\).
• In the amide, the lone pair on N is delocalised into the C=O (the p orbital on N overlaps with the \(\pi\) electron cloud of C=O), so it is not available to accept \(\text{H}^+\). The amide is neutral and is not protonated.
• Amide hydrolysis needs heating with acid, so it does not happen in cold HCl(aq).

• B is wrong: it protonates the amide N, which is not basic, and leaves the basic amine N unprotonated.
• C is wrong: even with excess acid, the amide N is not protonated, so only one \(\text{H}^+\) is added.
• D is wrong: cold HCl(aq) does not hydrolyse the amide. Hydrolysis on heating would give a carboxylic acid and \(\text{NH}_4^+\), never an acyl chloride, which is itself hydrolysed by water.