Though I found links about ring strain and other things, I actually fail to understand the exact reason for this difference in acidity.

  • $\begingroup$ If you think of both compounds as bases and find their conjugate acids, then apply hyperconjugation to find the stronger conjugate acid, you would find the stronger base(which will be the weaker acid). $\endgroup$ Commented Feb 4, 2017 at 7:26

1 Answer 1


Indeed, according to the Evans $\mathrm{p}K_\mathrm{a}$ table the cyclopropane $\ce{C-H}$ bond $(\mathrm{p}K_\mathrm{a} \approx 46)$ is more acidic than the $\ce{C-H}$ bond on the central carbon in propane $(\mathrm{p}K_\mathrm{a} \approx 51)$.

We know that, in general, the acidity of $\ce{C-H}$ bonds follows the order $\mathrm{sp \gt sp^2 \gt sp^3}$. This is because the more $\mathrm{s}$-character in an orbital, the more stable (lower energy) electrons are in the orbital.

The $\ce{C-H}$ bonds around the central carbon in propane are close to being $\mathrm{sp^3}$ hybridized, whereas the $\ce{C-H}$ bonds in cyclopropane are roughly $\mathrm{sp^{2.46}}$ (see this earlier answer to see how the hybridization in cyclopropane was determined).

Now, when we compare the hybridization of the $\ce{C-H}$ bonds in cyclopropane and propane we see that there is more s character in the cyclopropane $\ce{C-H}$ bonds, this explains the increased acidity.


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