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


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.


In this answer a hydrocarbon anion called Kuhn's anion has the formula $\ce{C_{67}H_{39}^-}$ has 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

  • $\begingroup$ Ehhm, but the question is about neutral hydrocarbons… $\endgroup$
    – andselisk
    Nov 29 '19 at 13:56
  • 2
    $\begingroup$ But the neutral hydrocarbon is the acid. It's mentioned here as the conjugate of the anion. $\endgroup$ Nov 29 '19 at 14:25
  • $\begingroup$ 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? $\endgroup$
    – andselisk
    Nov 29 '19 at 14:41
  • $\begingroup$ 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. $\endgroup$ Nov 29 '19 at 14:51
  • $\begingroup$ 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. $\endgroup$ Nov 29 '19 at 16:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.