I was working on nucleophilic addition reactions to the carbonyl group (namely, hemiacetal formation) when I came across the following explanation for the use of acid catalyst and how it increases reaction rate:

In acid (dilute HCl, say), the mechanism is different in detail. The first step is now protonation of the carbonyl group’s lone pair: the positive charge makes it much more electrophilic so the addition reaction is faster.

The reaction at hand is the following:

Hemiacetal formation mechanism

Ignoring the presence of the positive formal charge on the oxygen, I am trying to understand on a molecular orbital theory level why this makes sense. I'm suspecting that this somehow lowers carbon's LUMO (${π}^{*}$ orbital), which would enhance the electrophilicity of carbon.

Would someone be able to give an explanation to the enhanced electrophilicity of the carbonyl group based on molecular orbital theory?


Clayden, J., Greeves, N., Warren, S. Organic chemistry, 2nd ed.; Oxford University Press: New York, 2012.

  • 1
    $\begingroup$ I think title should reflect you wanna talk about MOs. $\endgroup$
    – Mithoron
    Dec 27, 2018 at 17:18

1 Answer 1


Ignoring the presence of the positive formal charge on the oxygen

I think this is exactly what is responsible, even from the MO viewpoint. From the MO perspective, the electrophilicity of the carbonyl group can be viewed as how energetically-accessible is the $\ce {C=O \pi ^*}$ MO to the nucleophile's HOMO. The protonation of the oxygen atom effectively decreases the energies of all MOs of the carbonyl, including the $\ce {C=O \pi ^*}$ MO. Why? The proton is positively-charged. By attaching to the oxygen atom, there is a net decrease in the potential energy of the molecule due to there being more electron-nuclei attractive forces in the resultant entity. Hence, protonation makes the $\pi ^*$ LUMO more accessible to the attacking nucleophile, increasing electrophilicity.

  • $\begingroup$ Wouldn't that interaction be only between the oxygen and hydrogen atom? Or does that interaction affect the electrophilicity of the C atom? $\endgroup$
    – Ethiopius
    Dec 28, 2018 at 1:50
  • $\begingroup$ That is not true. The positive charge is more delocalised than that. The protonation will cause all the electrons in the molecule to redistribute themselves. $\endgroup$ Dec 28, 2018 at 3:10

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.