4
$\begingroup$

Fisher Esterification

Is the O of the C=O group more electronegative or is it for the sake of getting from product A to B that the carbonyl is protonated rather than the hydroxyl?

$\endgroup$
1
  • $\begingroup$ I suppose it because $\ce{R-C^{+}(OH)2}$ is more preferred. I think protonation of both oxygens is possible, but the symmetric structures is more stable. $\endgroup$
    – Poutnik
    Apr 15, 2019 at 7:27

1 Answer 1

6
$\begingroup$

Because the protonated acid is resonance-stabilized when the carbonyl is protonated, but isn't if the hydroxyl is protonated:

protonation_resonance

So the first protonated form is much more energetically favourable than the second.

The same resonance in the non-protonated acid means that the lone pair on the alkoxide has reduced basicity, meaning that the carbonyl lone pair is the more basic since it isn't involved in any energy-lowering orbital interactions(/resonances).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.