Formula: C₄H₈O

Answer: Ethyl vinyl ether

Chemical Shift Assignments:

¹H NMR: δ 1.27 (t, 3H, J = 7.2 Hz), 3.72 (quartet, 2H, J = 7.2 Hz), 3.95 (dd, 1H, J = 6.8 and 1.6 Hz), 4.14 (dd, 1H, J = 14.4 and 1.6 Hz), and 6.43 (dd, 1H, J = 14.4 and 6.8 Hz)

¹³C NMR: 151.8, 86.2, 63.5 and 14.6 ppm

The formula indicates one degree of unsaturation. The signals at δ 1.27 and 3.72 in the ¹H NMR spectrum comprise an ethyl pattern with an attachment to a heteroatom-- namely, oxygen. All that remains of the formula after C₂H₅O- is C₂H₃. The chemical shifts (δ 3.95, 4.14, and 6.37 ) as well as their mutual pairs of coupling constants require of the remaining three proton signals constitute a vinyl pattern. Thus, C₂H₅O- and CH₂=CH- gives ethyl vinyl ether. The J values of the hydrogens of the double bond are: trans: 14.4 Hz, cis: 6.8 Hz, geminal: 1.6 Hz. The more shielded of the two vinyl carbons (86.2 ppm) is the terminal one. Consider the effect of resonance on electron density. The ¹H NMR spectra of ethyl acrylate and ethyl vinyl ether are quite similar. In particular, the chemical shifts of the terminal, geminal hydrogens of the respective vinyl groups are quite different. Use resonance concepts to rationalize these differences.

For an overlay ¹H NMR spectum of ethyl acrylate and ethyl vinyl ether, click here. Return to Menu.

For an overlay ¹³C NMR spectum of ethyl acrylate and ethyl vinyl ether, click here. Return to Menu.