There many sources on the internet saying that cyclopropane is more acidic than cyclopropene because the conjugate base in latter would have antiaromaticity which would destabilize it.

But in this explanation we are assuming that the central $\ce{C-H}$ bond in cyclopropene ionizes. I argue that the $\ce{C(sp^2)-H(s)}$ bond should ionize because of two factors:

  1. The $\ce{sp^2}$ carbon would better accommodate the negative charge than the central $\ce{sp^3}$ carbon.

  2. Consequent conjugate base would not contain anti aromaticity.

My logic can also be rationalized by considering acidity order between straight chain alkanes and alkenes.

This would mean inversion of the order which is incorrect according to experimental data. What is the flaw in my reasoning?


1 Answer 1


This reference does in fact indicate that cyclopropene ($pK_a=29$) has a much higher a acidity than cyclopropane ($pK_a=44$). Deprotonation at the vinylic hydrogen would account for this. That cyclopropene is deprotonated at a vinylic carbon is briefly summarized here with a primary reference given.

  • $\begingroup$ Many sources say that pKa of cyclopropene is 61. $\endgroup$ Dec 15, 2018 at 12:16
  • $\begingroup$ Say you have a molecule of citric acid. If you could magically prevent the carboxyl protons from dissociating you'd find that there is a certain $pK_a$ for the remaining hydroxyl (alcohol) hydrogen, which in practice chemists could get from extrapolating data for citrate esters. But in the real acid the carboxyl protons beat the alcohol protons to the punch and you see only the former dissociations. Similarly, $pK_a=61$ for cyclopropene picks the methylene hydrogens but the molecule really deprotonates at the vinylic hydrogens with $pK_a=29$. $\endgroup$ Dec 16, 2018 at 11:09

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