Question

Predict the molecule's structure giving the following Proton- NMR

C14H22O

triplet at 7.14 ppm , doublet at 6.63 ppm, a singlet at 5.70 ppm exchangeable with D2O, and a singlet at 1.4 ppm

integration in order 1: 2 : 1 : 18

I'm unsure what the structure is supposed to look like. I assumed there would be an alcohol and the index of unsaturation is 4 but I don't think an aromatic is possible with the give H-NMR.

*I couldn't upload a picture

Answer 1

degree of unsaturation for C14H22O

U = (C+1) - (H)/2

= (14+1) - 22/2

= 15-11 = 4

As the mlecule contans a singlet at 5.70 ppm exchangeable with D2O indicates the presence of OH group attached to aromatic ring. The presence of aromatic ring is evidenced from the signals in aromatic region.

The intensity ratio 1: 2 : 1 : 18 indictes that molecule is symmetrical as there are only 4 signals.

protons for 18H indicates two sets of C(CH3)3 units (this will only provide 18H as a singlet).

Based on these observations structure can be given as: