Chemical Shift Assignments:

¹H NMR: δ 5.72 (1H, brd. s), 1.75 (3H, s), 1.74 (3H, s) and 1.49 (3H, s)

¹³C NMR: 197.48, 154.12, 123.70, 30.98, 26.85, 19.88 ppm.

The degree of unsaturation is 2. The weak singlet at 197.48 ppm indicates a carbonyl group of a ketone. If you check the ¹³C shift table, you will see that amides and esters, which have similar chemical shifts, are excluded by the formula. The carbon is not that of an aldehyde because there is no low field aldehyde hydrogen in the ¹H NMR. Removing C=O from the formula gives C₅H₁₀. The ¹H NMR reveals three methyl singlets, two of which are quite similar (δ 1.74 and 1.75 ). Subtracting 3 x CH₃ from the residual formula gives C₂H, which must contain 1 degree of unsaturation. That is, C₂H is C=CH to which three methyls (uncoupled) and C=O must be attached. The solution below is the only possibility. If one of the methyl groups attached to the double bond were switched with the vinyl hydrogen, then the hydrogen and the remaining methyl group would be coupled to one another. Return to Menu.