[Rutherford]

In esters 1 and 4 there is conjugation and the IR frequencies should have lower values, so I would assign them the A spectrum. 2 and 3 would have spectrum B. Correct?

[discodermolide]

Ester 1 is a vinyl ester and is not conjugated. The only one conjugated with the carbonyl is ester 4.

[Rutherford]

But the lone pair on oxygen produces the conjugation from C=C to C=O.

[discodermolide]

Surely that is a resonance form  and not an actual molecule?

[Rutherford]

But the resonance from would contribute to the IR shift.

[orgopete]

In esters 1 and 4 there is conjugation and the IR frequencies should have lower values, so I would assign them the A spectrum. 2 and 3 would have spectrum B. Correct?

This makes no sense. Only one structure gives Spectrum A and one for B.

[Rutherford]

That was my original thinking, but I don't see a way how to do it. How to conclude which one?

[CrazyAssasin]

First of all, we can throw away the second one because it has no C=C group. The only one conjugated is 4 and it matches A spectra. While speaking about 1 and 3, they both have approximately the same shifts for C=C and C=O groups, however the intensity of C=C shift varies. The intensity in IR spectra shows the polarity of the functional group. The polarity of C=C is much more greater in 1 than in 3, because former has one carbon atom attached to oxygen. Therefore the intensity of C=C shift is larger of the first one. So only 3 should match B spectra.

[Rutherford]

I still don't understand why everyone is excluding 1. I think it has conjugation through the lone pair on oxygen, if it didn't, then we could also say that pyrrole isn't aromatic.

[discodermolide]

That resonance structure does not look at all likely to me..
If you want to do this then do not use the C=C.

[Rutherford]

Okay, I thought it could contribute.

[Rutherford]

The polarity of C=C is much more greater in 1 than in 3, because former has one carbon atom attached to oxygen. Therefore the intensity of C=C shift is larger of the first one. So only 3 should match B spectra.

I don't understand your conclusion. Why can't 1 match spectrum B? Both 1 and 3 have bigger IR shifts than A.

[orgopete]

The purported resonance structures are incorrect. There should be a positive charge on the carbon in which electrons were removed and not on the oxygen. The oxygen now has 10 electrons and no charge. However, carbonyl frequencies respond to donating and withdrawing groups. The C=C-C=O would be shifted to lower frequencies. Structure 1 would match spectrum B. I guess you can get the rest.

This is how I reason structure 1. The oxygen is attached to an electron withdrawing C=C. Although oxygen could provide some inductive withdrawal and resonance donation (but weaker than nitrogen), the C=C group negates that property. It is now a stronger withdrawing group and shifts the carbonyl to higher frequencies.

[Rutherford]

Yes, this is the part I can't solve, but I don't understand your thoughts completely. Could you elaborate it more deeply?

[orgopete]

Okay, the spectra cannot correspond with structure 2, because it doesn't have a double bond nor 4 because it would be shifted to lower frequencies. Allyl ester 3 is spectrum B because the double bond is isolated from the carbonyl group. The carbonyl is like an aliphatic ester. This leaves 1 and A. The carbonyl of an acid chloride is shifted to higher frequency because of the electron withdrawing effect of the chlorine. A nitrogen might be expected to also shift the carbonyl group to higher frequency, however the non-bonded electrons can be donated to the carbonyl group. This shifts the carbonyl frequency lower. This donation can be seen in the NMR spectra by a reduced rotation about the C-N bond and a compound like DMF will show two methyl peaks.

This brings us to oxygen. Will the non-bonded electrons shift the carbonyl group to lower frequencies as the nitrogen or to higher frequencies as a chlorine? In regular esters, the effects seem to nearly cancel each other and an ester frequency is quite similar to a ketone. If a C=O is attached to the oxygen, the carbonyl will shift to higher frequencies (look up an anhydride). Although a C=C is not as withdrawing as a carbonyl group, it is withdrawing none the less. You may check pKa tables to realize this effect. The vinyl ester, 1, corresponds with spectrum A.