In lectures, we were taught that ketones and aldehydes tend to have chemical shifts of around 200 ppm, while acid derivatives (carboxylic acids esters etc) tend to have values around 160–170 ppm.

The reasoning given by my supervisor was that despite acid derivatives having more electron withdrawing groups, due to conjugation, the net effect was that these additional groups donated electron density to the carbonyl carbon, hence shielding it more than the carbonyl carbons in ketones/aldehydes (i.e. the conjugative effect dominates over the inductive effect). This all makes sense for me except for acyl chlorides.

The reason why acyl chlorides are not adding up for me is that in the very next topic, we were taught that the CO double bond of an acyl chloride is stronger than that of a ketone/aldehyde! Since carbonyl bonds are strengthened by anything that withdraws electrons, I can only conclude that the Cl has a net electron withdrawing effect (i.e. its inductive effect dominates the conjugative effect).

Since the Cl group appears to actually withdraw electron density, going back to NMR, shouldn't the carbonyl carbon in acyl chlorides resonate at a higher frequency than the carbonyl carbon in a ketone/aldehyde?

  • 2
    $\begingroup$ Relevant, but I'm too tired to read thoroughly now: doi.org/10.1063/1.1703231 The issue is more complicated than just electron density, as 13C usually doesn't play by the same rules as 1H (see chemistry.stackexchange.com/a/62565/16683). $\endgroup$ Commented Dec 17, 2018 at 23:00
  • $\begingroup$ @HunnyBunch Actually, I am not actually convinced by the reasoning presented in your 2nd para regarding resonance donation outweighing inductive withdrawal. If we were to draw the resonance structures showing delocalisation of the lone pair into the carbonyl group, we necessarily would have to show the C-O $\pi$ bond breaking. Thus, even though there is resonance donation of the lone pair, increasing electron density around the carbonyl carbon, the effect of C-O $\pi$ bond breaking actually reduces electron density around the carbon atom. Thus, the reasoning fails to explain the phenomena. $\endgroup$ Commented May 18, 2019 at 5:09
  • $\begingroup$ @orthocresol I am not entirely convinced by the explanation they have suggested in their discussion. The reason for my disagreement with it is given in my previous comment. $\endgroup$ Commented May 18, 2019 at 5:19

1 Answer 1


Halogens $(\ce{Cl},$ $\ce{Br},$ $\ce{I})$ also act as conjugative donors. Not very good ones but you still have to take that into account especially because, as you said, conjugative effect is larger than inductive effect (and it is very different in $\ce{^1H}$- or $\ce{^13C}$-NMR).

The $\ce{^13C}$ chemical shift in $\ce{CH2Cl2}$ is $\pu{53 ppm}$ but the one for $\ce{CH2I2}$ is $\pu{–62 ppm}$ (yes, with a minus sign).

  • $\begingroup$ Thank you for your response, but how would you rationalise the second part of my post? $\endgroup$
    – HunnyBunch
    Commented Dec 20, 2018 at 14:47
  • $\begingroup$ Well in the 2nd part of your post, you actually consider only one bond (C=O) out of the 3 possible bonds (C=O, C-Cl, C-C). $\endgroup$
    – SteffX
    Commented Dec 20, 2018 at 15:58
  • 2
    $\begingroup$ Hmm, I'm not altogether convinced... Indeed 1H and 13C effects are very different, but I don't see how conjugation applies to 13C shielding in CH2I2? There is a known shielding effect with heavy atoms, see e.g. chemistry.stackexchange.com/a/73685/16683 onlinelibrary.wiley.com/doi/full/10.1002/mrc.4720 $\endgroup$ Commented Feb 15, 2021 at 18:06
  • 1
    $\begingroup$ I guess conjugation effects are usually how the relative IR stretches are rationalised, but I'm not sure if NMR is the same: the picture is rather complicated as there are both diamagnetic and paramagnetic shielding effects. $\endgroup$ Commented Feb 15, 2021 at 18:08

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.