2019 H2 Chem P1 Q23 — Nitrogen Compound Solution

The answer is A.

• W, \(\text{C}_5\text{H}_{10}\text{O}_2\), is a carboxylic acid \(\text{C}_4\text{H}_9\text{CO}_2\text{H}\) and is chiral. Of the four such acids (pentanoic, 2-methylbutanoic, 3-methylbutanoic and 2,2-dimethylpropanoic acid), only 2-methylbutanoic acid, \(\text{CH}_3\text{CH}_2\text{CH(CH}_3\text{)CO}_2\text{H}\), has a chiral carbon (marked in the figure).
• X reacts with an amine to form an amide, so X is the acyl chloride \(\text{CH}_3\text{CH}_2\text{CH(CH}_3\text{)COCl}\) (made from W with, for example, \(\text{SOCl}_2\) or \(\text{PCl}_5\)).
• The amine is piperidine, a secondary amine (N-H in a six-membered ring). Its N atom bonds to the acyl carbon and loses its H, giving a tertiary amide with N in the ring. So Y is A: the 2-methylbutanoyl group joined to the ring N of piperidine.

• B: a piperidine amide, but its acyl group is 3-methylbutanoyl, \((\text{CH}_3)_2\text{CHCH}_2\text{CO}\), from 3-methylbutanoic acid, which is achiral, so it cannot come from W.
• C: the correct 2-methylbutanoyl group, but it is an N-cyclohexyl secondary amide (N-H), which would come from cyclohexylamine, not piperidine.
• D: the wrong acyl group (3-methylbutanoyl) and the wrong amine (cyclohexylamine).