2017 H2 Chem P1 Q21 — Nitrogen Compound Solution

The answer is A (\(\text{CH}_3\text{CH}_2\text{CONHCH}_2\text{CH}_3\) \(\lt\) N-ethylphenylamine \(\lt\) \((\text{CH}_3\text{CH}_2)_2\text{NH}\)).
Basicity depends on how available the lone pair on N is to accept a proton.
• Amide, \(\text{CH}_3\text{CH}_2\text{CONHCH}_2\text{CH}_3\), is the least basic (effectively neutral): the lone pair on N is delocalised into the adjacent C=O group, so it is not available to accept a proton.
• N-ethylphenylamine, \(\text{C}_6\text{H}_5\text{NHCH}_2\text{CH}_3\), is weakly basic: the p orbital containing the N lone pair overlaps with the \(\pi\) electron cloud of the benzene ring, so the lone pair is partly delocalised into the ring and less available.
• Diethylamine, \((\text{CH}_3\text{CH}_2)_2\text{NH}\), is the most basic: the lone pair is not delocalised, and the two electron-donating ethyl groups increase the electron density on N, making the lone pair more available.
• B is wrong: it places N-ethylphenylamine below the amide. The phenylamine is more basic than the amide, whose lone pair is fully delocalised into C=O.
• C is wrong: it reverses the order, placing diethylamine as least basic and the amide as most basic.
• D is wrong: it places N-ethylphenylamine as the most basic, above diethylamine. The delocalisation of its lone pair into the ring makes it less basic than diethylamine.