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The $\mathrm{p}K_\mathrm{a}$ values of methylene protons in cycloheptatriene vs cyclopropene are found to be 60 and 36, respectively.

From what I understand - both conjugate bases are non-aromatic (not anti-aromatic) as a result of some sort of Jahn-Teller distortion. I can't pin down the nature of the distortions but they appear to be a pyramidal geometry for the cyclopropenyl anion and either $\mathrm{sp^{2}}$ hybridisation or unequal bond lengths for the 'tropylium' anion.

In either case, if both species and both conjugate bases are non-aromatic what is the cause of the massive $\mathrm{p}K_\mathrm{a}$ difference? Is it something to do with the $\mathrm{sp^{2.68}}$ hybridisation in the three membered ring?

Mathew Mahindaratne
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Quark
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  • pKa = 60 looks like very rough estimate. Wouldn't be surprising if another source said 55 or even 50. – Mithoron Apr 09 '22 at 10:58
  • Would you not agree that in either case the difference seems strange? – Quark Apr 09 '22 at 11:07
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    Hmm, gonna check out the values, but it may be that tropyl is still stabilised by resonance without antiaromatic contribution and cyclopropenyl can't do the same. (Big ring could be only partially flat.) – Mithoron Apr 09 '22 at 11:13
  • I was thinking that too, but couldn't quite rationalise how it could be stabilised by resonance but not destabilised by the resulting antiaromaticity. Maybe unequal bond lengths remove the degeneracy (and thus the antiaromaticity) but it's still flat enough for p-orbital overlap and thus resonance? – Quark Apr 09 '22 at 11:30
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    I mean, tropyl is obviously not flat overall - not antiaromatic, but could be locally flat - just one allyl would be enough for such effect. – Mithoron Apr 09 '22 at 11:39
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    Plot thickens! While for methylene protons it might be about 60, cyclopropene's likely deprotonated on vinylic position and might be actually more acidic https://chemistry.stackexchange.com/questions/105637/acidity-comparison-between-cyclopropene-and-cyclopropane – Mithoron Apr 09 '22 at 12:48
  • There might not be antiaromaticity. But very harsh to remove allylic hydrogen. See here. – Mathew Mahindaratne Apr 10 '22 at 22:04
  • @mithoron I am thinking the same thing about "partial" conjugation. I suggest below a valence-bond contribution that could do that. – Oscar Lanzi Apr 11 '22 at 18:38

1 Answers1

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Not all "antiaromatic" rings are created equal.

Assuming the cyclopropenide anion is formed by deprotonating a methylene proton, forming a cyclopropen-3-ide ion (in reality, a vinylic hydrigen appears to be deprotonated instead, forming cyclopropen-1-ide ion; see Poutnik's comment to the question), it can remove the unfavorable antiaromatic coupling only by fully localizing the negative charge and pyramidalizing the bonding at the selected carbon, giving up any possibility of stabilizing the charge via delocalization.

Not so with the cycloheptatrien-7-ide ion. While delocalizing the charge over the full ring results in an antiaromatic coupling, a nonaromatic and resonance-stabilized alternative is available by forming an allyl-anion-butadiene contributing structure (see below). Thus with the seven-carbon anion, some stabilization is possible while still circumventing antiaromatic coupling. With multiple positions possible for the allyl-anion component, the $pK_a$ reported here for cycloheptatriene actually beats that for the simple allyl-anion former propene ($36$ versus $43$).

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Oscar Lanzi
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