# Formation of Tropylium carbocation

I've read that Tropylium carbocation is formed by the ring expansion of benzyl carbocation. But in benzyl carbocation, the $\ce{C}$ atoms in the benzene ring are $sp^2$ hybridized i.e. $120^\circ$ is the required angle for stability. And in the benzene ring of the benzyl carbocation, the angle is already 120 degrees. Therefore benzyl ring need not expand for stability. Therefore why does benzyl rearrange to give tropylium?

• Note that such rearrangement is typical for mass spectrometry, not "normal" reactions. – Mithoron Apr 18 '17 at 19:29

More stable products are lower in energy. At first glance the bonding arrangements might seem equivalent in both cases in terms of number of $\ce{C-C}$ and $\ce{C-H}$ bonds from $\ce{sp^2}$ carbons, which would result in no net change. However, the change in molecular structure leads to a change in the molecular orbitals and the resulting energies of the molecule for the aromatic/resonance component to the energy. Using the Frost diagram technique we want to compare the orbital energies, nicely pictured here:
• if we are using Frost circles then we can generalize it to the Huckel model. The pi electron stabilization I get from that model = $8.00\beta$ (vs nonbonding $p$ orbitals) for benzene but $8.72\beta$ for benzyl cation, versus $8.99\beta$ for tropylium. The predicted driving force for rearrangement is still positive but much less than if you ignored the conjugation of the benzylic carbon. So, I would expect the benzylic conjugation to be fairly important in more advanced models too. – Oscar Lanzi May 5 '17 at 0:03
• That's what the more general model is for. For an all carbon system you just set up the connectivity matrix and get the eigenvalues in units of $\beta$, and those are your energy levels (positive= bonding, negative =antibonding). The Frost circle is just the pattern of eigenvalues specific to a single ring with no pendant pi bonds. – Oscar Lanzi May 5 '17 at 18:05