¹H NMR: δ 4.17 (6H, m), 2.97 (2H, d, J = 20 Hz), 1.35 (6H, t, J = 7 Hz), and 1.29 (3H, q, J = 7 Hz)

¹³C NMR: 165.7 (d, J = 6 Hz), 62.6 (d, J = 6 Hz), 61.4, 34.3 (d, J = 134 Hz), 16.4 (d, J = 7 Hz), and 14.1.

Phosphorus is present in this compound and coupling between ³¹P (abundance = 100%) should be anticipated. The large J value of 20 Hz in the ¹H NMR spectrum is indicative of ¹H-³¹P coupling as are the couplings in the ¹³C NMR spectrum telltale signs of ¹³C-³¹P coupling. The 3H triplets at δ 1.35 and 1.29 are methyl groups attached to three unresolved methylene groups at δ 4.17. [Note: Of the six hydrogens colored violet four are of a lighter shade and two are a darker shade. The matched pair in the ester group are enantiotopic. The two mismatched pairs are diastereotopic, a fact that may add to the complexity of the signal.] The shift of the methylene groups indicate that they are attached to oxygen atoms. Two of the C₂H₅O- groups are identical from the integration. C₆H₁₅O₃ has been determined; C₂H₂PO₂ remains. The remaining two equivalent hydrogens are at δ 2.97 in the ¹H NMR spectrum and they are coupled by 20 Hz to phosphorus. This means that the CH₂ group is attached to phosphorus and that the carbon is the one at 34.3 ppm (J = 134 Hz). The CH₂ is also attached to a carbon atom based on its chemical shift. The unknown assignments are CPO₂. The doublet (J = 6 Hz) at 165.7 ppm is in agreement with an ester carbonyl carbon. All that remains to be assigned is PO. It is possible that the phosphorus is trivalent and that the remaining oxygen is part of a perester, or a P=O bond is present. A ³¹P NMR would distinguish these two options. Working on the assumption that a P=O bond is present, we have P(=O)CH₂CO to which the three C₂H₅O groups must be attached. Return to Menu.