The resonance energy of pyrole furran and pyridne heterocycles is
Answers
have been taught that the order of aromaticity is: benzene > thiophene > pyrrole > furan. However, I have been unable to deduce any logical explanation for that order. Also, I wish to compare pyridine's aromaticity with these, but my professor does not know about it.
According to me, all the rings are aromatic and satisfy the Huckel's rule of aromaticity with 6π resonating electrons and a planar ring. So, that point is useless for doing any comparison of aromaticity.
But, we note that 2pπ-2pπ overlap is most favorable in conjugated systems. Hence, thiophene should be the least aromatic (since it has a 3pπ-2pπ overlap instead).
I agree that benzene should be the most aromatic because both its resonating structures are equivalent and have no charge separation. Using exactly the same reasons, pyridine should be the second best aromatic ring (lesser than benzene because the N atom is an electron withdrawing group and is reducing the electron density in the ring)
That said, every resonating structure (except the uncharged one) for pyrrole and furan will create a formal positive charge on the hetero-atom. Since N is less electronegative than O, it will be slightly more stable than O with that positive charge. Hence, pyrrole will be more aromatic than furan.
Therefore, according to me, the aromaticity order should be: benzene > pyridine > pyrrole > furan > thiophene. But it is obviously wrong.
Explanation: