The average wavenumber for a ketone is about $\pu{1720 cm-1}$ and the average wavenumber for an ester is about $\pu{1740 cm-1}$. This, however, does not make sense, as the carbonyl group of an ester should have a greater single bond character than the ketone due to resonance from the adjacent oxygen atom. This greater single bond character should thus result in a lower wavenumber for the ester, but it does not. Is there an explanation for this?


1 Answer 1


For an undergraduate starts to learn IR spectroscopy, the stretching frequency of any $A-B$ chemical bond, $\over\nu$ (in $\pu{cm-1}$) can be calculated by using following equation: $${\bar\nu} = \frac{1}{2\pi c}\sqrt{\frac{k}{\mu}}= \frac{1}{2\pi c}\sqrt{\frac{k(m_A+m_B)}{(m_Am_B}}$$ where $k=\text{the force constant of the bond}=\text{bond strength}$, $m_A= \text{mass of atom }A$, $m_B= \text{mass of atom }B$, and $c= \text{speed of light}$. The equation shows $\bar\nu$ is at least depends on two factors, reduced mass $\mu$ and the force constant of the bond $k$.

The dependency on $\mu$ would be explained by the difference in stretching frequency of $\ce{C-H}$ ($\pu{\approx 3000 cm-1}$) and $\ce{C-D}$ ($\pu{\approx 2200 cm-1}$). The force constants of $\ce{C-H}$ and $\ce{C-D}$ are approximately equal.

The most important fact when compared to different carbonyl bond stretching frequencies is force constants of the bonds of interest. For example, ring strain in a cyclic ketone usually increases the $\ce{C=O}$ stretching frequency. That of cycloheptanone is $\pu{\approx 1702 cm-1}$, cyclohexanone is $\pu{\approx 1714 cm-1}$, cyclopentanone is $\pu{\approx 1747 cm-1}$, and cyclobutanone is $\pu{\approx 1783 cm-1}$. Therefore, we can generally conclude that the stretching frequency of the bond increase with the increase of the reactivity of carbonyl bond. Likewise, reactivity of carbonyl compounds in increasing order is: acid chlorides ($\pu{1780-1820 cm-1}$) > acid anhydrides ($\pu{\approx 1760 cm-1}$ and $\pu{\approx 1810 cm-1}$) > esters ($\pu{1730-1750 cm-1}$) > aldehydes ($\pu{1720-1740 cm-1}$) > ketones ($\pu{1705-1725 cm-1}$) > carboxylic acids ($\pu{1700-17205 cm-1}$) > acid amides ($\pu{1630-1680 cm-1}$). The reduction of double bond character in ester (average bond length of methyl acetate is $\pu{\approx 123.2 pm}$) is minimal but has a better leaving group than in ketones (average bond length of acetaldehyde is $\pu{\approx 123.1 pm}$) when subjected to react.

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    $\begingroup$ Using the equation above when isotopes are concerned, the force constant is equal between CD and CH because only a change in mass is involved so any change in frequency is solely due to changes in $\mu$. $\endgroup$
    – porphyrin
    Aug 19, 2019 at 12:26
  • $\begingroup$ But why does the leaving group ability explain the stretch frequency and the resonance effects? $\endgroup$
    – S R Maiti
    Mar 2, 2021 at 11:48

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