Question

The proton NMR spectrum for a compound with the formula C6H12O2 is shown below along with carbon-13 spectral data in tabular form. (It may be necessary to expand (zoom) some of the 1H signals to view spin-spin splitting details.)

This compound exhibits strong infrared absorption at 1738 and 1245 cm^-1.

¹³C NMR Data

Normal Carbon	DEPT-135	DEPT-90	Proton Shift	Relative Area
19.1	Positive	No peak	0.93	6.0
20.9	Positive	No peak	1.92	0.9
27.7	Positive	Positive	2.06	3.0
70.6	Negative	No peak	3.85	1.9
171.1	No peak	No peak

Answer 1

C6H12O2 ,

degree of unsaturation = ( nC x2 +2- nH) /2 = ( 2 x 6+2 -12) /2 = 1

hence 1 double bond which si C=O as we have IR peak 1738 cm-1

DEPT 135shows peak for CH and CH3 only , we had peak at 0.93 with relative area 6

hence we had 2 CH3 groups ( i.e 6 H)

peak at 1.92 ppm would be CH ,

peak at 2.06 ppm would be CH3 adjacent to C=O

peak at 3.85 ppm is OCH2   as   DEPT 135 showed negative peak

hence we had C=O and also C-O

based on this structure is