# Predicting IR frequencies of esters / vinylogous esters

Here's the problem...

Three isomeric structures with formula $$\ce{C10H8O2}$$ are shown:

Identify the correct ordering of carbonyl IR stretching frequencies.

I understand that conjugation in the systems will affect the carbonyl stretching frequencies in their IR spectra. How do they order in terms of absorption wavelength though?

1 is a conjugated ester with $$\nu \approx \pu{1715 cm^-1}$$.

However, 2 is the one I'm most confused about. It seems to have two sources of conjugation: the phenyl ring as well as the oxygen lone pair. In general, is conjugation weaker from a different ring?

3 is a conjugated ketone. I predict a wavenumber of around $$\mathrm{1695~cm^{-1}}$$.

SO, my answer is A>C>B in this order.

Can someone explain the conjugative effects of the middle compound in particular since there seems a lot going on? Also, what is the correct order of wavelengths of these three compounds?

• I have formatted your question using $\LaTeX$ to improve its readability. For more information on how to do this yourself please see here. – bon May 18 '15 at 17:31
• C is conjugated with phenyl only, so it likely has highest frequency, typical for methylphenylketones . A & B should have close IR shifts, as both have strong OC ... O interaction (double bond is a good conductor of such interaction). However, which one has larger shift is unclear. – permeakra May 18 '15 at 20:16
• @permeakra is C not also conjugated with O lone pair since the carbonyl is separated with a double bond in between? – justbehappy May 19 '15 at 8:27
• @justbehappy Surprisingly, benzere ring is a rather poor conductor of conjugation. It may be rationalized, assuming that to conduct conjugation, it must partially loose its aromaticity. Also, see carefully, and try to draw resonance structure for C. The way it includes double bond is quite unfavorable to conduct +/-M effects. – permeakra May 19 '15 at 12:00