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The reason given for the question is that in aniline the lone pair on N atom is delocalized throughout the benzene ring, making it less available for donation.

But the conjugate acid of aniline should be more stable given the fact that the positive charge on N atom is also delocalized throughout the ring, hence making it more stable than the conjugate acid of methylamine which has +I effect. On that account, shouldn't aniline be more basic?

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    $\begingroup$ Are you sure the positive charge is delocalised throughout the ring? Have you tried writing the resonant forms of $\ce{C6H5NH3+}$ that are involved in that? $\endgroup$
    – user41033
    Jun 12, 2017 at 14:17
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    $\begingroup$ Okay. Now I got it. The positive charge cannot be delocalised because N can't form 5 covalent bonds. Since the C-N double bond can't be formed, there would be no charge dispersion. $\endgroup$
    – Siddharth Garg
    Jun 12, 2017 at 14:49
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    $\begingroup$ Exactly. And the basic form, aniline, is actually stabilised by resonant structures in a way that alkyl amines can't. $\endgroup$
    – user41033
    Jun 12, 2017 at 15:05
  • $\begingroup$ The question premise is itself wrong. $\endgroup$
    – tryst with freedom
    Jan 19, 2021 at 8:50

2 Answers 2

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N in aniline is forced into an $\ce{sp^2}$ configuration because only then can its p orbital interact to form aromatic MO’s. By contrast, methylamine has its N in the usual $\ce{sp^3}$ configuration. An $\ce{sp^2}$ lone pair has a great contribution from an s AO than an $\ce{sp^3}$ lone pair (33% vs 25%), and because an s AO is lower in energy than a p AO (due to a region of electronic density closer to the nucleus), it is also lower in energy. Therefore, the HOMO of aniline has a poorer energy match with the H 1s LUMO than the HOMO of methylamine, and the resonance energy stabilizing the protonated adduct is smaller. Hence, aniline is less basic.

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  • $\begingroup$ It’s worthwhile to note that acidity is a thermodynamic, not kinetic effect. Yet, I can still consider HOMO and LUMO interactions here because the interaction between N and H is largely local, leaving all other MO’s largely unaffected, so the interaction between the FMO’s determine the stabilization of the product and not just the TS. $\endgroup$
    – GingerBadger
    Aug 6, 2018 at 22:41
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You are wrong when you say that positive charge on the anilinium ion is delocalised in the benzene ring because that would mean nitrogen having $5$ bonds (which is not possible since nitrogen cannot expand its octet). Hence, the aliphatic amines are generally more basic than aromatic amines.

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