This question came through answering a question that talked about the acidic order of four compounds.

Upon researching the actual $\mathrm p K_\mathrm a$ values I noticed that even though there was a resonance destabilization of electrons (albeit partial) in triphenyl methane, the $\mathrm p K_\mathrm a$ values of acetylene were less than that of triphenyl methane? I was unable to find a solid reason for why this is the case.

Does anyone in the community have the reason?

  • 1
    $\begingroup$ Basically you are seeking confirmation for your answer there, which is unusual. Anyway, I just upvoted it because is it at least in the right direction. $\endgroup$
    – Alchimista
    Apr 12, 2021 at 9:21
  • $\begingroup$ @Alchimista yeah. I would convert my answer there into a community post if I can get a proper explanation for this one. Since the first part answers the actual question but not the question I pose. $\endgroup$ Apr 12, 2021 at 9:22
  • 1
    $\begingroup$ Triphenyl methane is some kind of propeller shaped, so it has less conjugation, making it very less acidic than what we except. $\endgroup$
    – Adithya
    May 26, 2021 at 14:32

1 Answer 1


As mentioned by @orthocresol, just comparing methane and acetylene shows a difference of $\approx 20$ $\mathrm pK_\mathrm a$ units and the amount of resonance required to overcome this difference is very high.

Comparing $\mathrm p K_\mathrm a$ of phenol($10.0$) and methanol($15.5$), we see that the presence of one resonating ring only produces a 5.5 $\mathrm p K_\mathrm a$ difference. Resonance in triphenyl system is effectively only one phenyl group at a time due to the steric effects due to the bulkiness of the phenyl groups.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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