Past exam of the chemistry A course of the University of Cambridge / 2023 / Paper 1 / A5 / b / iv / Solution

The six $\pi$ electrons give the Hückel method energyThe appropriate localized reference contains two C=N $\pi$ bonds and one C=C $\pi$ bond. Three C=C bonds would incorrectly replace the nitrogen one-centre parameters. For one isolated C=N pair, the dimensionless matrix iswhose eigenvalues are $(1\pm \sqrt{5})/2$. Since $\beta <0$, the occupied bonding molecular orbital has energy $\alpha +(1+\sqrt{5})\beta /2$. Two doubly occupied C=N bonds and one doubly occupied C=C bond consequently giveTherefore the signed delocalization energy isThe positive stabilization is $1.82039|\beta |$, smaller than the benzene value $2|\beta |$ in the stated model but still stabilizing.

Pyrazine is aromatic: its planar cyclic conjugated system contains six $\pi$ electrons and has a filled set of three bonding $\pi$ molecular orbitals. Each ring nitrogen contributes one $p_z$ electron; its in-plane lone pair is outside this $\pi$ system. The heteroatoms change orbital energies and coefficients without interrupting cyclic conjugation or the $4n+2$ count. Aromaticity does not require the stabilization to equal benzene's.