Arrange the three types of hydrogens in 1‐ethynyl‐3‐methylbenzene in the order from the most acidic to the least.


From my understanding, the hydrogen on methyl would be the most acidic as the negative charge after removing the H would be in resonance.

The next one would be the the ethyne hydrogen because the negative charge after removing it would stay on a sp-carbon. The least acidic H would be the one from the benzene ring as the negative charge would fall on a sp2-carbon.

But the answer says the correct order is: ethyne's H > methyl's H > benzene ring's H. Where am I going wrong?

  • 3
    $\begingroup$ Your approach is fine. It is just that resonance is a thing, and $sp$ acidity is a thing too, and you can't easily tell which of these things is bigger. $\endgroup$ Commented Mar 15, 2023 at 13:32
  • 2
    $\begingroup$ People who pose these question seriously should go to school and learn about chemistry. $\endgroup$ Commented Mar 15, 2023 at 21:32
  • $\begingroup$ Well, obvious conclusion from the answer is that sp-carbon effect is bigger than single ring effect, and it's indeed true. $\endgroup$
    – Mithoron
    Commented Mar 16, 2023 at 0:19
  • 1
    $\begingroup$ @Satya One approach could be to 1) compute the heat of formation of the structure above, 2) compute/look up the heat of formation for a proton, 3) compute the heat of formation of the deprotonated structure of the above so that one can apply Hess' law. It would be an iteration for all positions with a proton attached. But solvation of the molecules and ions has to be considered, too. Though there still is a need to predict pKa, C-H acidities are not always considered in models based on empiric data like MolGpKa. $\endgroup$
    – Buttonwood
    Commented Mar 16, 2023 at 9:08
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    $\begingroup$ @Buttonwood I've seen more pKa values in my life than any reasonable person should ;p In a question like that you either know ethynyl is more acidic than benzyl, or not - the difference in pKa is huge. $\endgroup$
    – Mithoron
    Commented Mar 16, 2023 at 13:56

1 Answer 1


Based on the papers I've found, the answer is correct. According to Bouchoux,[1] ethynylbenzene – an analogue to the compound in question – the $\Delta G$ of deprotonation for terminal $\ce{H}$ of ethyne group is $\pu{1522.2 kJ mol-1}$ while that for ring $\ce{H}$s of benzene at o-, m-, p-positions are $1612, 1614,$ and $\pu{1614.1 kJ mol-1}$, respectively.

According to Koppel,[2] $\Delta G$ of methyl $\ce{H}$ for toluene is $\pu{373.7 kcal mol-1}$ (equivalent to $\pu{1564.23 kJ mol-1}$).

According to Hansch,[3] the Hammett constant for methyl substitution is $-0.07$ and $-0.17$ when it is at m- and p-position, respectively, which means that being substituted by a methyl of phenyl ring would almost not affect its proton acidity.

So, the sequence is ethyne's $\ce{H} \gt$ methyl's $\ce{H} \gt$ benzene ring's $\ce{H}$.


  1. Bouchoux, G. Gas phase acidity of substituted benzenes. Chem. Phys. Lett. 2011, 506, 167-174. DOI 10.1016/j.cplett.2011.03.032
  2. Koppel, I. A.; Koppel, J.; Pihl, V.; Leito, I.; Mishima, M.; Vlasov, V. M; Yagupolskiid, L. M.; Taft, R. W. Comparison of Brønsted acidities of neutral CH acids in gas phase and dimethyl sulfoxide. J. Chem. Soc., Perkin Trans. 2 2000, 6, 1125-1133. DOI: 10.1039/B001792M
  3. Hansch, C.; Leo, A.; Taft, R. W. A survey of Hammett substituent constants and resonance and field parameters. Chem. Rev. 1991, 91, 165-195. DOI 10.1021/cr00002a004

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