Question

Deduce the structure of an unknown compound using the data.

Molecular Formula: C6H12O${\text{C}}_6^{}{\text{H}}_{12}^{}\text{O}$

IR: 1705 cm−1

H1${}^1\text{H}$ NMR: no absorptions greater than ? 3 ppm$\delta 3\text{ ppm}$

C13${}^{13}\text{C}$ NMR: ? 24.4,$\delta 24.4\text{,}$ ? 26.4,$\delta 26.4\text{,}$ ? 44.2,$\delta 44.2\text{,}$ and ? 212.6 ppm.$\delta 212.6\text{ ppm.}$ Resonances at ? 44.2$\delta 44.2$ 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