Problems and solutions in Heterocyclic Chemistry

Answer
I. (a) 2,3-Dihydro- 1 H-pyrrole; (b) 2,5-dihydro-1 H-pyrrole; (c)
2H-pyran; (d) 4H-pyran; (e) benzo[b]thiophene; (0 benzo[c]thiophene.


1.2. (a) Antiaromatic and unstable 4n type; (b) aromatic, planar
(4n + 2) type; lone pair formally on N delocalized into ring; (c) aromatic,
planar and ‘benzene-like constitution’; (d) Non-aromatic,
non-planar aminotriene; (e) Non-aromatic, non-planar.


1.3. Normally an N-substituent of a piperidine larger than a proton
assumes an equatorial orientation. Thus, in the isomers A and
B the trunsoid conformation A is more stable, and hence more highly
populated, than B.