So, the original question was which compound is more basic.

Compound A Compound B

Compound A and Compound B respectively.

My reasoning is that on protonation, Compound B would be more stable than Compound A as well, Compound B is an allylic nitrogen cation. Thus, compound B should be more basic.

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Why don't nitrogen cations act similar to carbo cations in this case?

But the given answer is that compound A is more basic, because of being connected to a sp2 hybridized carbon. Another given reason was that the first cation was stabilized by water while the second one wasn't?

Where am I wrong?

  • $\begingroup$ Being connected to a sp2 hybridised carbon makes it weaker not stronger. Still, enamines can be even weaker because of mesomeric effects. $\endgroup$
    – Mithoron
    Sep 2, 2021 at 15:24
  • $\begingroup$ @Mithoron, I know right! I was like, that sp2 reasoning doesn't make much sense. $\endgroup$ Sep 2, 2021 at 15:30
  • $\begingroup$ The lone pair in B is conjugated with the C=C bond. Similar to the nitrogen lone pair in an amide, which is also not very basic $\endgroup$
    – Andrew
    Sep 2, 2021 at 19:55

2 Answers 2


What may be happening is that the A cation has a resonance structure where the $\pi$ bond polarizes to the nitrogen and thus turns the $sp^2$ carbon into a carbocation center (leaving the nitrogen formally uncharged). This carbocation contribution is enhanced by the $sp^2$ carbon also hyperconjugating with the carbon atom next to it in the ring. Also the B cation cannot form an allylic structure because protonation of the nitrogen atom in that compound saturates it.


The A ammonium ion is sp2 hybridized so less basic[really the ion is more acidic] than an sp3 nitrogen. However, the enamine structure in B delocalizes the electron pair over the nitrogen and the double bound reducing basicity even more. There is no resonance of an ammonium ion with a double bond; there are no orbitals available on the nitrogen to accept electrons to reduce the charge[except antibonding].


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