The carbonyl stretching frequencies for an acid anhydride are approximately $1820~\mathrm{cm^{-1}}$ and $1760~\mathrm{cm^{-1}}$. These are both higher frequencies than a simple ketone owing to the electron withdrawing capability of the adjacent group. However, why are there two distinct infrared stretching frequencies?


2 Answers 2


The two observed C=O frequencies are due to the symmetric and asymmetric stretching modes of the anhydride.

enter image description here
Source: Introduction to Spectroscopy, Pavia and Lampman

You can see that the lower frequency symmetric stretch occurs where both C=O bonds are lengthening and shortening in tandem, whilst the higher frequency asymmetric stretch occurs when one C=O group is lengthening as the other is shortening. You can see an example of this kind of spectral pattern in the example below.

enter image description here
Source: Introduction to Spectroscopy, Pavia and Lampman

This explains why you get two frequencies for C=O bond in acid anhydrides, whilst ketones/aldehydes etc. only give a single IR frequency for the C=O bond.

  • $\begingroup$ I'm just curious that if both C=O are shortening, would that give 1760 cm^-1 ? $\endgroup$
    – Viv
    Dec 29, 2015 at 18:38
  • 2
    $\begingroup$ Its a continous motion of shortening and lengthening, so yes $\endgroup$
    – NotEvans.
    Dec 29, 2015 at 18:44
  • 1
    $\begingroup$ Is there any simple explanation why the asymmetric vibration corresponds to a higher energy, I would have expected it to correspond to a lower energy. $\endgroup$ Sep 13, 2017 at 19:02

Just to add on to NotEvans's answer:

The stretching frequencies of compounds can theoretically be predicted by group theory and symmetry considerations - some introduction can be found here. In general, for "simple" molecules such as $\ce{CO2}$ ($D_{\infty\mathrm{h}}$) and $\ce{XeF4}$ ($D_{4\mathrm{h}}$), the bonds do not stretch individually, but rather their vibrations are coupled. A diagram of the stretching vibrations for the two compounds mentioned above will illustrate this (there are other non-stretching vibrational modes which are omitted):

CO2 Dinftyh

XeF4 D4h

(source: Orbital Interactions in Chemistry 2nd ed., Albright, Burdett & Whangbo)

In general this arises because the $\ce{C=O}$, or the $\ce{Xe-F}$, bonds are related by symmetry. In ketones, there aren't any other $\ce{C=O}$ groups next to it, so the $\ce{C=O}$ bond tends to vibrate individually, which allows us to speak specifically of "the $\ce{C=O}$ stretch" in IR spectroscopy.

However, in anhydrides, there are two carbonyl groups next door to each other, and while they are not always exactly equivalent by symmetry, they still display coupling to each other, as @NotEvans described. It's hardly limited to anhydrides, though. In acetylacetone, where the two $\ce{C=O}$ groups are equivalent:

Acetylacetone IR
(source: Introduction to Spectroscopy 5th ed., Pavia et al.)

Another extremely easy-to-spot example of coupling is in primary amines $\ce{RNH2}$ or primary amides $\ce{RCONH2}$ - you will observe two distinct "$\ce{N-H}$" stretches, which are actually symmetric and antisymmetric stretches. Here's an example of a primary amine, $\ce{BuNH2}$:


Compare that with a secondary amine, $\ce{Bu2NH}$, which only has one $\ce{N-H}$ bond:

(source: Pavia et al.)


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