My notes states this:

An electronegative element tends to draw in the electrons between the carbon and oxygen atoms through its electron-withdrawing effect. This strengthens the C=O bond.

An electron-donating group, on the other hand, supplies electrons to the C=O bond, weakening the bond.

I just can't rationalise this. How does the electron-withdrawing group strengthen (and not weaken) the C=O bond and how does the electron-donating group weaken (but not strengthen) the C=O bond?

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    $\begingroup$ That depends if effect is mesomeric or inductive. $\endgroup$
    – Mithoron
    Apr 8, 2022 at 13:16

1 Answer 1


This is easiest to see when a mesomerically electron-donating group is attached to the carbonyl, as in the case of carboxylate: enter image description here

The O atom attached to the carbonyl donates a lone pair of electrons via resonance. It does so by overlapping the p orbital containing the lone pair with the C=O $\pi^*$ LUMO. This donation of electrons into the $\pi^*$ LUMO weakens the C=O bond $\pi$ bond. This can be observed on the diagram above; in the lewis structure on the right the C=O $\pi$ bond is shown to be completely broken as a result of the conjugation of lone pair electrons. In reality the C=O $\pi$ bond is not completely broken but simply weakened, which is shown on the rightmost portion of the diagram.

This concept can be extended to inductively electron donating groups as well. For example, a CH3 group would donate electron density from its C-H $\sigma$ bonds via hyperconjugation into the C=O $\pi^*$ LUMO, weakening the $\pi$ bond as well.

For electron withdrawing groups, the the logic is the same just that the reasoning is opposite. If a halogen X was attached to the carbonyl then it would exert an electron withdrawing effect on the C=O C atom, giving it a delta positive charge. This makes breaking the C=O $\pi$ bond difficult as it would then intensify the positive charge on the C atom.

Side note

As a side note, perhaps you were confused because of the conflicting observation where the electron withdrawing groups that strengthen the C=O $\pi$ bond actually make the C=O more reactive, while electron donating groups that weaken it make it less reactive. This is because bond strength does not tell the whole picture of a molecule's reactivity.

The electron donating groups do weaken the C=O $\pi$ bond (as measured by FTIR) but at the same time reduce the kinetics of nucleophilic addition. The electron donating effect reduces the delta positive of the C=O C atom, decreasing the electrostatic attraction to electron rich nucleophiles. The overlap of electrons with the LUMO stabilises the lone pair of electrons but also increases the LUMO energy, increasing the HOMO-LUMO gap for any incoming nucleophiles. There are thermodynamic considerations as well: the electron donating and delocalisation of charge is generally stabilising. Overall this means that while the C=O bond is weakened, the reactivity of the molecule as a whole is also decreased.

  • $\begingroup$ Oh thanks, this is a comprehensive answer! Then if we link this back to IR spectroscopy, is it true that the absorption wavelength for, for instance, an ester tend to be higher than an aldehyde, due to mesomeric electron-donating effect? $\endgroup$
    – warren
    Apr 8, 2022 at 13:29

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