# Identifying the molecular structure of compound with sum formula C8H14O3 formula using NMR and IR spectrometry data

I am looking to identify the molecular structure of a compound with the sum formula $$\ce{C8H14O3}$$ using the NMR and IR spectrometry data below.

I was able to figure out the molecule as

My friend asked me why I didn't consider the following molecule:

I couldn't answer that. What inferences from the IR & NMR data can I observe, to explain why the structure my friend suggested is wrong?

Source for these IR & NMR Spectra: https://www.orgchemboulder.com/Spectroscopy/Problems/7.shtml

• You have a singlet with 2, so there must be a methylene group coupling with no other protons. Jan 10, 2021 at 12:45
• @Martin-マーチン, I drew the structure(That my friend suggested) incorrectly. Could you be kind enough to look at the question again? I appreciate any help that you can provide. Jan 10, 2021 at 15:28
• Look at the two peaks at around 4.2 and 3.5 ppm. The quartet at 4.2 is more deshielded than the singlet at 3.5. If the structure your friend suggested is right, then the 4.2 signal must come from the $\ce{CH3C\textit{H}_2O -}$, and the 3.5 signal would come from $\ce{-O-C\textit{H}_2-C=O}$. This would not make sense, as the CH2 in between two electron withdrawing groups should be more deshielded than the one that is next to one electron-withdrawing group. Jan 10, 2021 at 17:30
• @xavitop Instead of adding "(ANSWERED)" to the title, please feel free to post an answer to your question if you think it has been solved. Jan 11, 2021 at 5:55

## 1 Answer

Compare the quartet at $$\pu{4.2ppm}$$ and the singlet at $$\pu{3.5ppm}$$. These quite obviously would, in your friend’s structure, correspond to the protons on either side of the ether-oxygen. However, we should expect the methylene singlet to be more deshielded than the ethyl group’s methylene: It is squashed between an electron-withdrawing $$\ce{OR}$$ and an electron-withdrawing $$\ce{C=O}$$. The ethyl group’s methylene is between an electron-withdrawing $$\ce{OR}$$ and a mostly electron-neutral $$\ce{CH3}$$ group. Therefore, the order of peaks we observe in the spectrum does not make sense when we consider your friend’s proposal.

Thereafter, one might also argue that a quartet at $$\pu{4.2ppm}$$ is so obviously an ethyl ester as likely every organic chemist has had way to many NMR spectra after a column that contained a peak exactly there. An ethyl-ether is more shielded (diethyl ether’s methylene proton has a chemical shift of $$\pu{3.5ppm}$$.

• I’m sure someone else can provide an answer that finds ester signals in the IR and a lack of a 1,2-diketone but I won’t use IR if I can avoid it ;)
– Jan
Jan 12, 2021 at 15:59