Answer to Question #96012 in Organic Chemistry for Krina

Question

Why is the pKa of pyridine lower than that of piperidine?
Why is the pKa of pyrrole smaller than that of pyridine?

Answer 1

According to general equation of dissociation of protonated base

$HB^+ \rightleftharpoons H^+ + B$

equilibrium constant is as follows:

$K_a = \frac{[H^+][B]}{[HB^+]}$

$pK_a = -lgK_a = -lg \frac{[H^+][B]}{[HB^+]}=lg[HB^+] - lg[H^+][B]$

Comparing the basicity of three nitrogen containing heterocycles (piperidine, pyridine and pyrrole) it is clear from structures that pyrrole is the weakest base as the electron pair of the nitrogen is included in the aromatic system. Piperidine is an aliphatic secondary cyclic amine and its basicity is highest. Pyridine is weaker base than piperidine because its electron pair is in an sp²-hybridized orbital, and the electron pair is more tightly held by the atom.

According to the qualitative analysis it is clear that [HB⁺] is highest in the case of piperidine, then follows pyridine and the smallest value is of pyrrole. Thus according to the last equation, pK_a of piperidine is higher that of pyridine and pK_a of pyridine is higher that of pyrrole.