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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?

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    $\begingroup$ 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 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. $\endgroup$ – Safdar Faisal Apr 12 at 9:22
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    $\begingroup$ Triphenyl methane is some kind of propeller shaped, so it has less conjugation, making it very less acidic than what we except. $\endgroup$ – 0.00 May 26 at 14:32
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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.

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