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?

  • $\begingroup$ 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
    Commented Apr 15, 2019 at 7:27

1 Answer 1


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


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).


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