# Why do Ketones Have Lower Wavenumbers than Esters?

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?

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.
• 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 $\mu$. Aug 19, 2019 at 12:26