NMR of Alanine



Figure 1. The Entire Spectrum

Figure 2.
Figure 2. The H-atoms on the methyl group at the end form a doublet at around 1.450-1.4800 because they are H-atoms from a regular alkyl. In addition, the three H-atoms are identical so there is only a doublet formed due to the H-atom on the adjacent Carbon atom.

Figure 3.
Figure 3. The H-atom on the second Carbon atom forms a quartet due to the three H-atoms on the adjacent Carbon atom. In addition, the H-atom is less shielded, and thereby between 3.730 and 3.800, due to the nitrogen directed bonded to its Carbon atom.

Figure 4.
Figure 4. Because the solvent for this molecule is Deuterium Oxide, the H-atoms are quickly exchanged by the heavy water. As a result, no splitting or coupling occurs with the H-atom. In addition, the H-atom is less shielded resulting in a singlet at around 4.800. [Yes this is the spectrum of HDO JCB]

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