SolutionsFive-membered aromatic heterocycles: pyrrole, furan and thiophene
Exercise 1
Why is pyrrole (pKaH = −3.8) a much weaker base than pyrrolidine (pKaH = 11.3) and pyridine (5.2)? Where is it protonated and what happens to it in concentrated acid?
The lone pair of the pyrrole nitrogen is part of the aromatic sextet; protonating it on nitrogen would destroy the aromaticity without compensation. In pyrrolidine and pyridine the pair is free (in an sp3 or sp2 orbital, outside the π system). Pyrrole is protonated at C2, giving a resonance-stabilized but non-aromatic cation; this cation is an electrophile that attacks another pyrrole molecule, and the chain reaction produces polymers (“pyrrole red”).
Exercise 2
Give the main product: a) furan with acetic anhydride and BF3·OEt2; b) thiophene with n-BuLi and then CO2 and H3O+; c) 1-methylpyrrole with DMF and POCl3, then water. Why is AlCl3 not used in a)?
a) 2-Acetylfuran. b) Thiophene-2-carboxylic acid: butyllithium deprotonates C2 and 2-thienyllithium is carboxylated. c) 1-Methylpyrrole-2-carbaldehyde. AlCl3 is too strong a Lewis acid: it opens or polymerizes the weakly aromatic furan; a mild catalyst is enough, because furan is much more nucleophilic than benzene.