2024 H2 Chem P1 Q25 — Nitrogen Compound Solution

The answer is D (\(\text{HCO}_2^-\text{Na}^+\) and \((\text{CH}_3)_2\text{NH}\)).
Dimethyl formamide is the amide \(\text{HCON(CH}_3)_2\). Heating with aqueous NaOH hydrolyses the amide C-N bond:
\(\text{HCON(CH}_3)_2 + \text{NaOH} \rightarrow \text{HCO}_2^-\text{Na}^+ + (\text{CH}_3)_2\text{NH}\)
The acyl part, HCO, has only one carbon, so it becomes methanoate; in alkaline solution it is present as the sodium salt, sodium methanoate. The nitrogen keeps both of its methyl groups, so the amine is dimethylamine, \((\text{CH}_3)_2\text{NH}\).
• A is wrong: the acyl group in the amide is HCO (one carbon), not \(\text{CH}_3\text{CO}\), so ethanoic acid cannot form; the N carries two methyl groups, so the amine is \((\text{CH}_3)_2\text{NH}\), not \(\text{CH}_3\text{NH}_2\); and in NaOH the acid would be present as its salt.
• B is wrong: the salt is sodium methanoate, not sodium ethanoate, and the amine is dimethylamine, not methylamine.
• C is wrong: the organic products are right, but in excess aqueous NaOH methanoic acid is neutralised, so it is present as \(\text{HCO}_2^-\text{Na}^+\), not \(\text{HCO}_2\text{H}\).