2025 H2 Chem P1 Q24 — Nitrogen Compound Solution

The answer is B (Y \(\rightarrow\) Z \(\rightarrow\) X).
Base strength depends on how available the lone pair on N is for donation to a Lewis acid. In the gaseous state there is no solvation, so only electronic effects matter.
• Y (phenylamine): the lone pair on N is delocalised into the benzene ring, so it is the least available. Y is the weakest base.
• Z (cyclohexylmethanamine, a primary amine): one electron-donating alkyl group on N increases the electron density on N.
• X (piperidine, a secondary amine): two electron-donating alkyl groups on N, so its lone pair is the most available. X is the strongest base.
Base strength increases: Y \(<\) Z \(<\) X.
• A is wrong: it places Z (one alkyl group) above X (two alkyl groups).
• C is wrong: it makes Y the strongest base, but its lone pair is delocalised into the ring.
• D is wrong: it places Y above Z, but Z has an electron-donating alkyl group while the lone pair of Y is delocalised.