# Why is aniline less basic than methylamine?

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?

• 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? – user41033 Jun 12 '17 at 14:17
• 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. – Siddharth Garg Jun 12 '17 at 14:49
• Exactly. And the basic form, aniline, is actually stabilised by resonant structures in a way that alkyl amines can't. – user41033 Jun 12 '17 at 15:05

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.
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.