Question

If we replace one of the carbon atoms (and its associated H atom) in benzene by a nitrogen atom, we obtain the molecule pyridine which is also an aromatic six membered ring. Sketch a qualitative MO diagram for pyridine and describe how you expect the energy levels of pyridine to differ from those of benzene.

Answer 1

This MO diagrams were extracted from the BOOOK:

ORBITAL INTERACTION THEORY OF ORGANIC CHEMISTRY . ARVI RAUK

Here we have the MOs of the both compounds

Since nitrogen is trivalent, it can also be doubly bonded to carbon, i.e., it can form a C=N bond in the context of an aromatic system. In this case, nitrogen is using its only 2p_z AO to form a double bond in the conjugated system, so that the unshared pair is in an sp² AO in the trigonal plane. Consequently, the latter pair is not counted as part of the aromatic system. Nitrogen, then, only contributes one electron to the conjugated system, as does any of the carbon atoms of benzene

An sp orbital of a particular element is lower in energy than an sp2 orbital of the same element, which is lower in energy than an sp3 orbital of the same element. A pair of electrons residing in, say, an N(sp) orbital will therefore be less reactive than a pair of electrons in an N(sp2) orbital, which will be less reactive than a pair of electrons in an N(sp3) orbital.

Hybridization also affects energy by affecting geometry. If the shape of a compound forces an atom to assume bond angles or shapes that diverge from the optimum bond angles or shapes as determined by that atom’s hybridization, then the compound is higher in energy than one might expect.

In contrast to benzene, the electron density of the pyridine is not evenly distributed over the ring, reflecting the negative inductive effect of the nitrogen atom. For this reason, pyridine has a dipole moment and a weaker resonant stabilization than benzene, and a higher energy that the benzene.