To illustrate the power of NMR, consider these three alcohols of formula C4H10O, each of which has a quite different ¹³C NMR spectrum. Peaks from the spectra are shown in the table below.

Each alcohol has a saturated carbon atom next to oxygen, all appearing in the region typical of saturated carbon atoms next to oxygen (p. 56). Then there are carbons next door but one to oxygen: they are back in the 0–50 ppm region but at its low fi eld end—about 30–35 ppm— because they are still deshielded by the nearby oxygen atom. Two of the alcohols have carbon(s) one further away still at yet smaller chemical shift (further upfi eld, more shielded) at about 20 ppm, but only the n-butanol has a more remote carbon still at 15.2. The number and the chemical shift of the signals identify the molecules very clearly. A common situation chemists fi nd themselves in is that they have some idea about a molec ular formula—from high-resolution mass spectrometry, for example—and need to match a structure to NMR data. Here’s an example: the formula C3H6O is represented by seven reason able structures, as shown in the margin. The three 13C NMR spectra below represent three of these compounds. The challenge is to identify which three. We will give you some clues, and then we suggest you try to work out the answer for yourself before turning the page. Simple symmetry can distinguish structures A, C, and E from the rest as these three have only two types of carbon atom. The two carbonyl compounds, D and E, will have one peak in the 150–200 ppm region but D has two different saturated carbon atoms while E has only one. The two alkenes, F and G, both have two unsaturated carbon atoms (100–200 ppm) but in ether G one of them is joined to oxygen—you would expect it therefore to be deshielded and to appear between 150 and 200 ppm.

The three saturated compounds (A, B, and C) present the greatest problem. The epoxide, B, has two different carbon atoms next to oxygen (50–100 ppm) and one normal saturated carbon atom (0–50 ppm). The remaining two both have one signal in the 0–50 ppm region and one in the 50–100 ppm region, and only the more powerful techniques of ¹H NMR and, to a certain extent, infrared spectroscopy (which we will move on to shortly) will distinguish them reliably. Here are NMR spectra of three of these molecules. Before reading further see if you can assign them to the structures on the previous page. Try also to suggest which signals belong to which carbon atoms.

We hope these didn’t give you too much trouble. The only carbonyl compound with two identical carbons is acetone (E) so spectrum 1 must be that one. Notice the very low fi eld signal (206.6 ppm) typical of a simple ketone C=O carbon atom. Spectrum 2 has two unsatu rated carbons and a saturated carbon next to oxygen so it must be F or G. In fact, it has to be F as both unsaturated carbons are similar (137 and 116 ppm) and neither is next to oxygen (>150 ppm). This leaves spectrum 3, which appears to have no carbon atoms next to oxygen as all chemical shifts are less than 50 ppm. No compound fits that description and the two signals at 48.0 and 48.2 ppm are suspiciously close to the arbitrary 50 ppm borderline. They are, of course, both next to oxygen and this is compound B.