CHEM 2319
Quiz #8
Dr. Bachrach
April 14, 2008

The proton and ¹³C NMR spectra for a molecule with molecular formula C₅H₁₁Cl. Identify the structure of the compound and assign every peak in ¹H NMR spectra to specific proton(s). Indicate what information you gained from the ¹³C NMR.

¹³C NMR:

Ok, start with the ¹³C NMR. It has four peaks, indicating that there are four unique carbons in the molecule. Since the molecule contains five carbons – two of them MUST BE EQUIVALENT. Now the degree of unsaturation is zero, so there are no double bonds or rings. The ¹³C NMR has no signals above 80 ppm, so that confirms no double bonds. The peak at 72 ppm is the carbon that has the chlorine on it.

So let’s look at the ¹H MNR. There are three signals, indicating three different types of hydrogen. There is a triplet a 1.0 ppm, a singlet at 1.5 ppm and a quartet at 1.75 ppm. Since there is no singal between 2.5-4 ppm, there CAN BE NO HYDROGENS ON THE CARBON THAT ALSO IS BONDED OT THE CHLORINE. So we must have on carbon that is bonded to three carbons and the chlorine:

The singlet at 1.5 means that these hydrogens have no neighbors. The chemical shift is at the high end for methyl group, but if these methyl groups are attached to a carbon with a chlorine, they will be shifted downfield. So that makes our fragment

We are left with a triplet at 1.0 ppm – clearly a methyl group with a neighboring carbon having 2 hydrogens. And the quartet at 1.75 ppm must have three neighbors – this just accounted for methyl group. So this is an ethyl group, and we need to attach it to the fragment above – and there is only one way to do that!