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This is a question I've been thinking about quite frequently. I had seen that this hydrocarbon has a low $\mathrm pK_\mathrm a$ with reference to the highlighted H-atom

c

Loss of the central gives aromatic character to the three 5-membered rings in the compound, contributing to its acidity.

Are there any other neutral hydrocarbons which are stronger acids? How do their $\mathrm pK_\mathrm a$'s stack up against, say another organic acid such as acetic acid?

EDIT: As pointed out in the comments, $\ce{CH_5^+}$ (methanium) is a superacid, but is not a neutral hydrocarbon.

Aniruddha Deb
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    $\ce{CH5+}$ is cheating? – Karsten Nov 28 '19 at 03:39
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    See also https://en.wikipedia.org/wiki/Methanium. But the OP may mean a neutral molecule. – Poutnik Nov 28 '19 at 05:55
  • @KarstenTheis yep ;) Nice try though. I will read about methanium. Also edited the question accordingly. – Aniruddha Deb Nov 28 '19 at 10:09
  • I think you can get much further by drawing a neutral hydrocarbon which is initially non-aromatic, but becomes aromatic when deprotonated. Cyclopentadiene is extraordinarily acidic for a hydrocarbon, with a pKa comparable to alcohols. With some imagination, it's likely possible to set up a structure with cascading aromatization upon deprotonation. – Nicolau Saker Neto Nov 28 '19 at 11:13
  • @NicolauSakerNeto "cascading aromatization" . Nice to imagine in my head. – Beyond Zero Nov 28 '19 at 11:46
  • @NicolauSakerNeto That's exactly what I have drawn! The question is: are there any compounds that are more acidic than what I've drawn. – Aniruddha Deb Nov 28 '19 at 16:48
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    @AniruddhaDeb Well, what I had in mind was for the initial structure to be completely non-aromatic. For reference, cyclopentadiene is approximately as acidic as 9-phenyl-9H-fluorene, according to Hans Reich's pKa tables in DMSO, even though the latter has far more extended conjugation. It's better for aromaticity to suddently appear, instead of existing aromatic systems merely being conjugated. – Nicolau Saker Neto Nov 28 '19 at 21:13
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    related https://chemistry.stackexchange.com/questions/9212/can-a-organic-compounds-such-as-hydrocarbons-contain-an-ionic-bond https://chemistry.stackexchange.com/questions/20283/whats-the-strongest-known-organic-acid – Mithoron Nov 28 '19 at 22:15

1 Answers1

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In this answer a hydrocarbon anion called Kuhn's anion (1) has the formula $\ce{C67H_{39}^-}$ and reported $pK_b=8.1$, which would correspond to the neutral hydrocarbon having $pK_a=5.9$. The anion, together with several hydrocarbon cations with which it forms stable salts, is shown below (taken from the answer referenced above; primary reference J. Org. Chem. 1990, 55 (3), 996–1002):

enter image description here

Oscar Lanzi
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  • Ehhm, but the question is about neutral hydrocarbons… – andselisk Nov 29 '19 at 13:56
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    But the neutral hydrocarbon is the acid. It's mentioned here as the conjugate of the anion. – Oscar Lanzi Nov 29 '19 at 14:25
  • Right, but the illustration showing only ions looks confusing to me. Why not to draw alongside the actual compound you are posting $\mathrm{p}K_\mathrm{a}$ value for? – andselisk Nov 29 '19 at 14:41
  • Not sure how to modify the anion, which would involve using CH in the center instead of the carbanion. Would need a desktop I think and I cannot get to one this weekend. – Oscar Lanzi Nov 29 '19 at 14:51
  • How is this hydrocarbon so acidic? I understand that the $pK_a$ was derived via a practical experiment but how would one go about explaining it theoretically? Is it merely because of $\pi$-resonance? I don't see any aromatic rings forming/breaking to contribute to such a pKa. – Aniruddha Deb Nov 29 '19 at 16:54
  • Conjugation through the central carbon of the anion. – Oscar Lanzi Nov 29 '19 at 17:07
  • @AniruddhaDeb It does aromatise three cyclopentadienyl rings, only partially though. – Mithoron Nov 30 '19 at 00:02
  • When we put the proton on the central carbon, saturating it, we fix a pendant pi bond to one of the atoms in each five-membered ring, so aromatic coupling cannot extend to those rings. With deprotonation to form the anion, the central carbon can donate a pi electron pair, which delocalizes those double bonds and thus allows aromatic coupling through all the rings. – Oscar Lanzi Aug 29 '23 at 20:34