Question

Deduce the structure of an unknown compound using the data below: Molecular Formula: C6H12O IR: 1705 cm-1 1H NMR: no absorptions greater than δ 3 ppm 13C NMR: δ 24.4, δ 26.4, δ 44.2, and δ 212.6 ppm. Resonances at δ 44.2 and 212.6 have very low intensity.

Answer 1

C6H12O

degree of unsaturation = ( nC x 2+ 2-nH) /2 = ( 2 x 6 + 2 -12) /2 = 1         ( where O is considered 0H)

IR peak at 1705 cm-1 indicates C=O

now HNMr peaks all less than 3ppm idnicates we have simple alkanes and CH groups adjacent to C=O

there is no aldehyde H peak which lies near 9 ,hence our C=O is ketone

C-NMR showed 4 peaks which indicates 4 differnt types of carbons , low intensity means Carbon having no Hydrogens attached.

CNMR peak at 212.6 ppm indicates C=O

Hence based on this