# NMR of phenylethane amine

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.)

• Looks like your NH3+ group – Waylander Sep 29 '18 at 17:38
• but it says 7.0 for the chemical shift at that one. – user43537 Sep 29 '18 at 17:39
• 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. – Waylander Sep 29 '18 at 17:42
• The integration came at 12.0 when not including the little top beside it, with the little top it came at 14 – user43537 Sep 29 '18 at 17:48
• 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? – Waylander Sep 29 '18 at 17:55

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.