I've been analyzing a compound "propiophenone", using NMR and IR, and I can't tell which of the nmr spectrums relates to which isomer of C9H10O2 is which.

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At first I thought the two isomers were ethyl benzoate and phenyl propanoate, which would make the 1.5 peak from the CH3 group and the 2.7 and 4.2 peaks from the CH2, and the CO stretch in ethyl benzoate is about 1730. But then I didn't know why the stretch in the other isomer would be 1770 and I thought that the ethyl benzoatate would be the major isomer, but I was told that the isomer D (with a stretch at 1730) is the minor isomer.

  • $\begingroup$ Someone can freely expand this hint into an answer - conjugation lowers the frequency of IR absorption for carbonyl compounds. $\endgroup$
    – Ben Norris
    Mar 30, 2016 at 21:28

2 Answers 2


As this is clearly a homework type question, this answer will just give you some helpful pointers to head you in the right direction, and leave you to piece together the answer.

My usual approach for these types of problems is to have a look at the NMR data first, and ignore all the IR data. We might use that later for double checking. From both spectra, it is possible to identify the following spin systems:

  • -CH2CH3
  • -C6H5 (monosubstituted benzene).

We also know we must have a -OC(O)- group in there also. That satisfies our molecular formula.

Looking at Isomer C first:

  • CH2 group at 2.7, indicative of it being adjacent to the carbonyl of an ester
  • Ortho aromatics at 7.0 and upfield of the meta, indicative of substitution with an electron donating (or aromatic activating) group, Ar-O- I'll let you piece together the structure of C.

Looking at Isomer D:

  • CH2 group at 4.4, indicative of being adjacent to a heteroatom, so -O-CH2
  • Ortho protons at 8.0 and downfield of the meta , indicative of substitution with an electron withdrawing (or aromatic deactivating) group; Ar-C(O)- I'll let you put this structure together also.

And what about the IR data? We know the following, which is consistent with what we propose from NMR:

  • Al-C(O)-O-Ar 1740-1770cm-1
  • Ar-C(O)-O-Al 1715-1730cm-1

The general rule with IR absorptions is that conjugation lowers the frequency - I'll leave you to rationalise this with some fancy arrow-pushing from the aromatic ring to the carbonyl group.


It is very obvious on NMR that C is the benzoate since the 2.7 2H is very typical for protons on the carbon next to the carbonyl. D is the propanoate since the 4.4 2H is for protons on the carbon with C-O bond on it.


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