Q.49. - Organic Chemistry

Question

Question: Draw the conjugate bases of pyrrole and cyclopentadiene. Explain why the $\mathit{s}{\mathit{p}}^{\mathbf{3}}$ hybridized C-H bond of cyclopentadiene is more acidic than the N-H bond of pyrrole.

Step-by-Step Solution

Answer

Answer

The $s{p}^3$ hybridized C-H bond of cyclopentadiene is more acidic than the N-H bond of pyrrole as the conjugate base formed by cyclopentadiene is more stable than the conjugate base formed from pyrrole.

Step 1: Pyrrole

There are several heterocyclic compounds, and one among them is pyrrole. The heteroatom found in pyrrole is nitrogen.

Step 2: Cyclopentadiene

Cyclopentadiene is a compound consisting of five carbons in the form of a ring and two double bonds. The dimerization of cyclopentadiene leads to the formation of dicyclopentadiene, which occurs via the Diels Alder reaction.

Step 3: Conjugate bases of pyrrole and cyclopentadiene

The conjugate base of pyrrole can be given as:

The conjugate base of pyrrole

 

The count of pi electrons in pyrrole and the conjugate base of pyrrole is $6\mathrm{\pi }$ electrons leading them to be aromatic. The deprotonation of pyrrole does not lead to the gain of aromaticity as the starting material is aromatic.

The conjugate base of cyclopentadiene can be given as:

The conjugate base of cyclopentadiene

 

Cyclopentadiene is not aromatic, but its conjugate base is aromatic and contains $6\mathrm{\pi }$ electrons.

This makes the C-H bond in cyclopentadiene more acidic than the N-H bond in pyrrole, as the deprotonation of cyclopentadiene leads to the formation of an aromatic conjugate base.