It is said that the smaller the ring size, the greater the wavelength of the carbonyl group, but why does cyclobutanone have a higher wavelength than cyclopropanone?


1 Answer 1


Why does cyclobutanone have a higher wavelength than cyclopropanone?

Seemingly, you were informed wrong. According to MSU website, following wave numbers listed for different cyclic ketones:

Wave numbers for different cyclic ketones

The increase in frequency ranges from $30$ to $\pu{45 cm-1}$ for a 5-membered ring, to $50$ to $\pu{60 cm-1}$ for a 4-membered ring, and nearly $\pu{130 cm-1}$ for a 3-membered ring. According to this, the increase in stretching frequency of exo-carbonyl group is because small rings have more $\mathrm{p}$-character in their ring bonds, which means the bonds outside the ring (exo-bonds) have more $\mathrm{s}$-character and are stronger.

Edit: There is a relation between stretching frequency and corresponding wavelengths:

$$ \begin{array}{|c|c|} \hline \text{Stretching frequency, }\pu{cm-1} & \text{corresponding wavelength, }\pu{nm} \\ \hline 1716 & 5820 \\ 1748 & 5720 \\ 1783 & 5600 \\ 1850 & 5400 \\ \hline \end{array} $$

Also, you can find this convertor here.

  • 1
    $\begingroup$ The questioner seems unsure about wavenumber[frequency] and wavelength $\endgroup$
    – jimchmst
    May 26 at 22:30

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