I guess the ester is a weaker nucleophile because it does have an additional oxygen atom, unlike the ketone, that is pulling electrons from the C-O double bond towards the carbon atom (this happens vice versa too of course with the ester oxygen). This lowers the ionic character of the C-O double bond (smaller $\delta^{+}$ and $\delta^{-}$ charge on the carbon and oxygen atoms) and thus increases the energy of the C-O $\pi^{*}$ bond, making it less nucleophilic. Is this correct?


2 Answers 2


The ester carbonyl carbon is a stronger nucleophile and less prone to nucleophilic attack than the carbonyl carbon in a ketone. I think you are trying to understand why the carbonyl in a ketone typically reacts faster with a nucleophile than the carbonyl in an ester. Look at the resonance structures drawn below. Both the ketone and ester have a resonance structure that places positive charge on the carbonyl carbon, however you can also draw a third resonance structure for the ester that removes some of the positive charge from the carbonyl carbon and places it on the ester oxygen. This makes the ester carbonyl carbon less positive (more negative, more nucleophilic, less prone to attack by a nucleophile) than the ketone carbonyl carbon (more positive, less nucleophilic, more prone to attack by a nucleophile). A nucleophile has electrons it would like to share and will react preferentially with centers of low electron density (more positively charged). Therefor a nucleophile will react faster with a ketone carbonyl than an ester carbonyl.

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I like Martin's posts, they make me think - and that is always a good thing. Here is where my thinking has taken me:

  • I believe Martin's approach (basically frontier MO) is fundamentally correct. I would be more convinced that his actual numbers are meaningful if he were to calculate 1) the carbonyl carbon LUMO coefficient and 2) the HOMO-LUMO separation for the following series of compounds in order to see if his calculated values correlated with this order of reactivity of carbonyl compounds towards nucleophilic attack (we could use water as the nucleophile) - but Martin, I think such calculations might be a poor use of your time and resources.

acyl halide > acid anhydride > aldehyde > ketone > ester ~ carboxylic acid > amide > carboxylate ion

  • My argument based on how resonance affects the charge density at the carbonyl carbon is one approach used to explain ("teach") the above order of reactivity. See for example page 4 on these UCLA notes. I was just trying to keep it simple.
  • The issue I have with Martin's approach is, what should the reader do next week when he wants to know if an acid chloride reacts faster or slower than an ester. While the resonance approach may not be rigorous, it does yield the correct order of reactivity and allows the user to make predictions while only using pencil and paper.
  • I am considering further editing of this post to use the resonance approach to explain differences in ground state stabilization of the carbonyl series. Then, since they all go (more or less) to the same tetrahedral intermediate, one could explain the reactivity order based on differences in Eact. This is another textbook method used to explain the carbonyl reactivity pattern.
  • Finally, what is the "correct" answer? Is it the real truth as nature knows it, but in a form that is difficult for the common individual to use and extend; or is it a rough approximation of the truth that can be easily used? I think the answer to that question depends upon who is asking the question, who is the audience.
  • 1
    $\begingroup$ However, the enolate of a ketone will be more nucleophilic than the enolate of an ester (for almost all of the same reasons). $\endgroup$
    – Ben Norris
    Commented Jul 18, 2014 at 1:19
  • $\begingroup$ Thank-you Ron. I completely forgot about this question until I saw it bumped up today. I think the original problem I had with this explanation is that the resonance structure where the positive charge is on the ester oxygen does not block the mechanism of attack. On nucleophilic attack the negative charge would just "go" to the ester oxygen instead of the carbonyl oxygen. Instead of the I guess you can only explain it then with that the LUMO is more delocalized in the ester, increasing its energy. $\endgroup$
    – Jori
    Commented Aug 11, 2015 at 10:57
  • $\begingroup$ @ron Will an amide be more nucleophilic than ester? Also, please explain electrophilic order for an ester, 0° amide, 1° - amide, 2* amide. Please reply. $\endgroup$ Commented Aug 24, 2022 at 12:34

Looking at the partial charges of esters and ketones, unfortunately ron's answer is only half true. For a simple model I have chosen 3-pentanone and ethyl acetate. You can see, that the carbonyl carbon in the ketone has a smaller positive charge ($q=0.6$) than in the ester ($q=0.8$), so the latter should be more prone to nucleophilic attacks.

This is not the case and the explanation lies within the molecular orbitals. The LUMO of the ketone has a slightly lower ($-1.57~\mathrm{eV}$) energy than that of the ester ($-0.87~\mathrm{eV}$). More important, the coefficient in this orbital is larger at the carbonyl carbon in the ketone compared to the ester. And this orbital will be attacked by the nucleophile, which usually has a high lying HOMO. So the LUMO of the ketone will be more accessible compared to the ester (smaller energy difference).

Another important fact is, that the ester is much more rigid than the ketone as it has one low lying MO that spans the $\ce{O=C-O}$ moiety (MO20), which is obviously not present (as it cannot be) in the ketone (MO18). You can also see, that the $\pi$ contribution of carbon in the ketone is lower.


  • $\begingroup$ Thank-you Martin. I completely forgot about this question until I saw it bumped up today. Why is the ketone LUMO lower in energy than the ester LUMO? Because of the delocalization I suppose. The energies you list are relative to what precisely? $\endgroup$
    – Jori
    Commented Aug 11, 2015 at 10:52
  • $\begingroup$ Martin, I read this thread once before and then came back to it after some time. I believe that you had posted pictures of the MO's from your simulation, now it looks they are no longer there? $\endgroup$ Commented Aug 23, 2017 at 14:24
  • $\begingroup$ @MikeMurphy I don't know, at my end it displays fine. Maybe there was some issue with imgur. In any case, there is nothing I could do about it. Did you experience it anywhere else in the site? You can also check with other users in Chemistry Chat. I hope it works again for you now. $\endgroup$ Commented Aug 23, 2017 at 23:30

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