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Consider the cleavage of the indicated hydrogen in cyclopentadiene:
The hydrogen indicated will be easily removed as $\ce{H^.}$ or $\ce{H^+}$. Assume a nonpolar medium.
My question was that would aromaticity give enough stabilization for hydrogen to be more easily removed as $\ce{H+}$?
The cleavage of the C-H bond could be envisioned to take place in two steps:
Ionization of the hydrogen atom requires 313.8 kcal/mol. Aromatic stabilization, calculated from unexpected differences between calculated and experimental heats of hydrogenation, yield an "extraordinary 36 kcal/mol" greater stability than expected. Well, 36 kcal/mol could be extraordinary under certain circumstances, but compared to an ionization energy, it's not very large.
The molecular orbital situation for the cyclopentadienyl radical isn't exactly non-aromatic, it's just not as fully aromatic as it could be. The five pi electrons of the radical are all bonding, but there's room for one more bonding electron. The question is whether the extra stabilization of that electron is sufficient to cause the ionization of the hydrogen atom. A typical carbon-carbon single bond has a dissociation energy of about 100 kcal/mol and a pi bond would be a bit less. A bond would involve two electrons, so one electron would be responsible for about half that, or less than 50 kcal/mol.
So between the "extraordinary" stabilization of benzene (36 kcal/mol) and the extra bond energy of the sixth electron (50 kcal/mol), I don't think there is enough energy available to ionize the hydrogen atom. Homolytic cleavage would be easier than heterolytic. However, if you provide an ionic solvent that stabilizes the proton and cyclopentadienyl anion, circumstances could change completely.
