Doing an exchange term so I've missed some classes that the others have attended, and unfortunately we're getting contradictionary explanations, so I thought I'd try and ask you.

Would be extremely thankful if someone could explain the effect of conjugation of a carbonyl group on the resonance frequency in IR spectroscopy.

Our 'assignment' is basically to figure out if it's supposed to be lower or higher than an unconjugated carbonyl.

Thanks in advance!

  • 1
    $\begingroup$ As a short answer: conjugation (with a $\ce{C=C}$ bond) moves the IR peak of a carbonyl group to lower wave numbers. $\endgroup$
    – Philipp
    Commented Dec 5, 2014 at 1:07

1 Answer 1


It would shift the IR peak to lower wavenumbers. You can rationalise this by understanding the bonding in the carbonyl group; in essence, by conjugating the carbonyl, the $\ce{C=O}$ bond is weakened, which shifts the IR peak down. You can think of it in terms of, if there is little or no conjugation, the electrons are firmly based in the $\ce{C=O}$ bond. As soon as we introduce some resonance structure, or conjugation, now those electrons can delocalise over 3-4 atoms or even more. This weakens the $\ce{C=O}$ bond, moving the IR peak down to lower wavenumbers.


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