In a question asking to compare the basicity of pyrrole and aniline, the explanation given for aniline being more basic than pyrrole was as follows:

Basicity depends on the availability of lone pair of $e^-$ on Nitrogen atom.

In pyrrole, lone pair of $e^–$ present on $\ce{N}$ are involved in resonance with the ring which provide aromatic character to the ring.

On protonation of pyrrole, the electron will not be above to participate in resonance, While anilines aromatic even though the involvement of line pair of nitrogen of $\ce{—NH2}$ group is not there and lone pair of aniline is delocalized in benzene ring but due to deocalization benzene loss its aromatic character so, the delocalization is not that strong as in pyrole molecule.

Lone pair is more available of protonation. Hence, aniline is more basic than pyrrole.

But, why should the aromatic character get destroyed? It is still a planar structure and should be aromatic. What is an intuitive explanation?


1 Answer 1


There would be no aromaticity in pyrrole after protonation since there is no lone pair to allow for Hückel's rule for aromaticity (the $4n+2$ rule). So, upon protonation, you are destroying aromaticity in pyrrole. Also the lone pair is completely delocalised in pyrrole and it cannot easily participate as a base.

However, in aniline, there is a partial loss of the lone pair due to resonance but the lone pair of nitrogen is still available for protonation. Aromaticity is not lost in the resonance structures of aniline.

Therefore aniline is more basic than pyrrole but is less basic compared to pyridine (where the lone pair is out of plane of resonance).


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