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Question
Account for the following:
pKb of aniline is more than that of methylamine.
Solution 1
pKb of aniline is more than that of methylamine:
Aniline undergoes resonance, and as a result, the electrons on the \[\ce{N}\]-atom are delocalized over the benzene ring. Therefore, the electrons on the N-atom are less available to donate.
On the other hand, in the case of methylamine (due to the +I effect of the methyl group), the electron density on the \[\ce{N}\]-atom is increased. As a result, aniline is less basic than methylamine. Thus, pKb of aniline is more than that of methylamine.
Solution 2
In aniline, the lone pair of electrons on the \[\ce{N}\]-atom is delocalised over the benzene ring.
As a result, electron density on the nitrogen atom decreases. Whereas in \[\ce{CH3NH2 + I}\] effect of \[\ce{- CH3}\] group increases the electron density on the \[\ce{N}\]-atom. Therefore, aniline is a weaker base than methylamine, and hence its pKb value is higher than that of methylamine.
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