I have an unknown ester with the chemical formula $\ce{C9H10O2}$ that is used as a flavoring agent in candies. It exhibits the following H-NMR and C-NMR (Attached file: unknown ester NMR)

Based on the NMR peaks I believe the unknown is benzyl acetate, as the C-NMR and the first two peaks of H-NMR are identical to the unknown NMR (Attached file: benzyl acetate NMR).

Knowing that aromatic protons are not equivalent, as shown in the unknown's H-NMR at 7.4 and 7.5 ppm, I do not know why the aromatic protons for benzyl acetate are integrated and shown as a singlet.

Is this just due to the sensitivity of the prediction algorithm (I am using Chembiodraw)? Or benzyl acetate is not the unknown?

Your help is greatly appreciated!

enter image description here enter image description here

  • 3
    $\begingroup$ Benzyl acetate seems very plausible to me, although if this is supposed to be your homework, using ChemDraw to predict NMR is probably not encouraged. Better to use chemical reasoning to rationalise why benzyl acetate fits the spectrum you're given. As for ChemDraw's prediction, did you take a look at whatever was below the spectrum? As far as I know, the software should tell you how it calculates the chemical shifts. i.stack.imgur.com/WBVRA.png $\endgroup$
    – orthocresol
    Feb 20 '17 at 20:29
  • $\begingroup$ Thank you for the quick response! Yes, I have looked at Chemdraw's numerical output, which was the same as the picture you've linked. And that is source of my confusion as in why the program is considering the aromatic protons to be equivalent :s $\endgroup$
    – AMahder
    Feb 20 '17 at 20:45

Prediction software always has its limitations, and there is always a degree of error in the calculation. For the ChemDraw predictions, you will see that for the 3 aromatic environments, it has done 3 independent calculations, and has happened to arrive at the same chemical shift. This simply means that these shifts are coincident, not equivalent.

Remember, prediction software is like any tool — only as good as the person using it, and should not replace a proper assessment of the data.

Your assessment here should first look at your proton environments; all 10 protons are accounted for. We have a $\ce{CH3}$, a $\ce{CH2}$, and a mono-substituted benzene. The $\ce{CH3}$ and $\ce{CH2}$ are not directly connected to any other group that is causing obvious splitting. Secondly, the $\ce{^13C}$ spectrum shows 9 peaks, consistent with our $\ce{CH3, CH2}$ and a mono-substituted benzene. We also have a peak at ~δ170, which is a $\ce{-C(O)-{}}$

There are only a few ways you could put these groups together. Rather than using ChemDraw to predict the spectrum and see which is identical to your question, you should rationalise why each possibility is or isn't the correct answer.

Four possible structures: benzyl acetate, 1-phenyloxyacetone, 2-methoxyacetophenone, methyl 2-phenylacetate

Trying to rationalise a bunch of peaks around δ7.4–7.5 is the of minimal interest, and remember that a real spectrum will almost always look different to a simulation. The peaks you should focus on willjustify the chemical shifts for the $\ce{-CH3}$ group and the $\ce{-CH2}$ group in the proton spectrum, and the $\ce{-CH3, -CH2}$ and $\ce{-C(O)-{}}$ in the carbon spectrum. There is only one possible correct answer.

  • $\begingroup$ LOL - As I have heard it put "A fool with a tool is still a fool." $\endgroup$
    – MaxW
    Feb 20 '17 at 21:55
  • $\begingroup$ I would like to argue that $170~\mathrm{ppm}$ clearly shows a carboxy group or amide group and a pure ketone would have something closer to $200~\mathrm{ppm}$. That would cut down the number of possibilities to two from four. Otherwise have my full approval and upvote. $\endgroup$
    – Jan
    Feb 20 '17 at 22:22
  • $\begingroup$ @Jan - I think that is exactly Long's point. The OP shouldn't just rely on pattern matching but use some knowledge of spectral interpretation to solve the problem. $\endgroup$
    – MaxW
    Feb 20 '17 at 22:34
  • $\begingroup$ @Jan - precisely. Leaving this deduction to OP. Similar arguments can be made for the methyl group - it is clearly not oxygen-bound. And so on... $\endgroup$
    – long
    Feb 20 '17 at 22:34
  • $\begingroup$ I am just learning about NMR specteoscopy. But one thing I don't realize is why the measurementss are in unit ppm. Isn't it should be a unit of magnetic field? $\endgroup$ Feb 20 '17 at 22:49

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