1H NMR: Intermediate Level, Spectrum 19

Formula: C2H3F3O

Answer: 2,2,2-Trifluoroethanol

 

Chemical Shift Assignment:

1H NMR: δ 4.38 (1H, m) and 7.28 (2H, q, J = 9.0 Hz)

13C NMR: 124.6 (q, J = 275 Hz) and 61.0 (q, J = 37 Hz) ppm

The degree of unsaturation is 0. The presence of fluorine (19F, abundance 100%, s = 1/2) in this compound suggests vicinal coupling between fluorine and hydrogen. There are only three hydrogens in the compound of two different chemical shifts, one of which is a quartet . This signal at δ 7.28 is coupled (J = 9.0 Hz) to three equivalent fluorine atoms. Thus, we have F3CCH2-, which leaves only the hydroxyl group to be added to the structure. The broadband, proton decoupled 13C NMR spectrum reveals two quartets from 19F-13C coupling. The signal at 124.6 ppm, which is the trifluoromethyl carbon having one bond coupling, has J = 275 Hz (do the measurement). The two bond coupling of fluorine to the carbon centered at 61.0 ppm has a smaller coupling constant (J = 37 Hz). 13C NMR spectrum. Return to Menu.