(a) Explain how pyrrole is isoelectronic with the cyclopentadienyl anion.
(b) Specifically, what is the difference between the cyclopentadienyl anion and pyrrole?
(c) Draw resonance forms to show the charge distribution on the pyrrole structure.
(a) The pyrrole is an aromatic five-membered heterocyclic, with one nitrogen atom and two double bonds. Although it may seem that pyrrole has only four pi electrons, the nitrogen atom has alone pair of electrons.
The cyclopentadienyl anion is an aromatic five-membered ring, with two double bonds and negative charged of lone pair of electrons. So, it is also contain 6 pi electron. So, it is isoelectronic with cyclopentadienyl anion.
(b) These two compounds are aromatic compounds. The cyclopentadiene is unusually acidic because loss of a proton converts the non-aromatic diene to the aromatic cyclopentadienyl anion. The pyrrole has amino group. So, it has a basic nature. The pyrrole has 4 pi electrons and the nitrogen atom has a lone pair of electrons. So, it has 6
Pi electrons. The cyclopentadienyl anion has 4 pi electrons and the negatively charged of lone pair of electrons of methylene group. So, i is also 6 pi electrons
The cyclopentadienyl anion does not have any heteroatom but pyrrole has nitrogen atom (hetero atom)
(c) The lone pair of electrons on the nitrogen atom undergoes resonance to form four resonance contributors. They are shown below.