I did an undergraduate experiment and could use some help in interpreting the $\ce{^1H}$ NMR spectra.

This was the task: About $\pu{5-10mg}$ (1-2 drops) distilled amine is weighed in a NMR tube. Put the top on the tube immediately to prevent reaction between the amine and $\ce{CO2}$ from the air. Weigh exactly 1.1 equivalent of (S)-alpha-acetoxyphenylacetic acid. Dissolve the acid in about $\pu{0.4 ml }\ce{CDCl3}$ and transfer the solution to the NMR tube.

This here is the $\ce{^1H}$ NMR:


Could I get affirmation of the choices drawn in red and what does the shift at $\pu{4.2ppm}$ correspond to? Could it be because $\ce{CO2}$ reacted with the amine?

(ChemdDraw was used to get the molecular figure and chemical shifts on the figure.)

  • $\begingroup$ Looks like your NH3+ group $\endgroup$
    – Waylander
    Sep 29, 2018 at 17:38
  • $\begingroup$ but it says 7.0 for the chemical shift at that one. $\endgroup$
    – user43537
    Sep 29, 2018 at 17:39
  • 1
    $\begingroup$ I doubt it - the signal at ~7 looks to sharp to contain NH3+ . Shame that signal is not integrated. To be sure you need to do a D2O shake with the sample and rerun. Note that several of your other signals are not where they are predicted to be. Heteroatom-H signals are notoriously difficult to predict. $\endgroup$
    – Waylander
    Sep 29, 2018 at 17:42
  • $\begingroup$ The integration came at 12.0 when not including the little top beside it, with the little top it came at 14 $\endgroup$
    – user43537
    Sep 29, 2018 at 17:48
  • $\begingroup$ and you have 10 aromatic protons to account for. The rest of the integrations of your phenethylamine molecule are off so I would not be too concerned. If that hump at 4.2 is not your NH3+ then where is it? $\endgroup$
    – Waylander
    Sep 29, 2018 at 17:55

1 Answer 1


I think some of your assignments are not correct. The doublet at $\pu{1.3 ppm}$ is correctly assigned. But close inspect of that signal suggests $\pu{12 Hz}$ coupling constant. Therefore, the signal given by the single proton causing that splitting should be a visual quartet with $\pu{12 Hz}$ coupling constant. But, your assignment for that proton is broad singlet at $\pu{4.75 ppm}$ with 2.7 integration (keep in mind that ChemDraw suggestion is not always true). I think that signal at $\pu{4.75 ppm}$ belongs to $\ce{NH3+}$. The small signal at $\pu{4.1 ppm}$, which is seemingly a quartet with large coupling constant is your $\ce{H-C(Ph)(CH3)NH3+}$ resonance. My best bet for the broad singlet next to it at $\pu{4.2 ppm}$ is crystalline water from (S)-alpha-acetoxyphenylacetic acid ((+)-O-Acetyl-L-mandelic acid), which is crystalline solid.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.