Past exam of the chemistry A course of the University of Cambridge / 2022 / Paper 1 / A4 / c

Fill four spatial molecular orbitals with the eight $\pi$ electrons. For negative ${\beta }_S,{\beta }_L$ and $|{\beta }_S|>|{\beta }_L|$, the occupied energies are $\alpha +{\beta }_S+{\beta }_L$, $\alpha +{\beta }_S-{\beta }_L$, and the pair at $\alpha -\sqrt{{\beta }_S^2+{\beta }_L^2}$. ThereforeFor equal integrals, ${\beta }_S={\beta }_L=\beta <0$, this becomesAt equal bond lengths, the last two electrons occupy the frontier degeneracy at $E=\alpha$, so their distribution within that degenerate space does not change the bare Hückel method energy.

For ${\beta }_L=0$, four isolated short-bond dimers each contribute $2(\alpha +{\beta }_S)$:These are different coupling limits. Keeping ${\beta }_S=\beta$ while simply removing long-bond coupling does not represent a small distortion at fixed mean bond strength.

The uniform planar eight-electron ring has a partially filled frontier degeneracy. Bond-length alternation splits those levels, permitting the two frontier electrons to occupy the lowered orbital and opening a gap to the empty partner. This is the electronic mechanism for destabilizing the equal-bond planar geometry.

A controlled comparison keeps the mean resonance integral fixed. PutUsing part (i), the change from the uniform ring isThus a small alternation lowers the electronic energy, with a leading gain linear in $\delta$. An ordinary elastic cost starts quadratically, so the equal-bond planar state is unstable in this model. The equilibrium distortion requires those geometric and sigma-bond costs as well.

The eight $\pi$ electrons constitute a $4n$ cyclic count rather than the closed-shell $4n+2$ case associated with aromaticity. The result concerns the imposed planar geometry; it does not establish the unconstrained three-dimensional equilibrium shape of cyclooctatetraene.