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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:

NMR OF DISTILLED AMINE

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.)

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  • $\begingroup$ 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
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    $\begingroup$ 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

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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.

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