Spectroscopy Tutorial: Examples

Example 2


MW 114

First calculate the degree of unsaturation: the result is 1. The compound has four carbons and an oxygen, thus it can have a carbon-carbon double bond, a carbon-oxygen double bond - a carbonyl, or a ring.

IR Spectrum

Since the degree of unsaturation indicates that the compound could have a carbonyl, let's look for that first, since carbonyl bands are strong and distinct. Carbonyls show up in the region 1760-1665, and specifically, saturated aliphatic ketone close to 1715. Sure enough, there is a band at 1713 indicating a saturated aliphatic ketone.

Think of possible structures

Now we know that the compound has a carbon-oxygen double bond, but there are lot of ways that this seven-carbon molecule could be put together. It could be straight-chain with the carbonyl at any one of the carbons, it could be branched with the carbonyl at one of several different carbons. Some examples are shown below:

It is not likely that the compound is an aldehyde, because the carbonyl of saturated aliphatic aldehydes show up at 1740-1720. While in the above IR spectrum the band at 1713 might be close to the range of saturated aliphatic aldehydes, an aldehyde would also show a distinct band for H-C=O stretch in the region 2830-2695, so it is not likely that it is an aldehyde. The molecule is most likely a ketone, and we're going to have to look at the NMR to determine how it's put together.

Proton NMR Spectrum

The NMR of Example 2 is below:

The integral values 2, 2, 3 add up to 7, and there are 14 hydrogens in the molecular formula. Thus, we have to multiply each by two. The true integral values are:

This molecule must be symmetrical, since there are 7 carbons but only three different types of protons. The protons at 2.3-2.4 ppm are closest to the carbonyl because they are further downfield, and since they are split to a triplet, they are next to a carbon that has two protons:

The protons at 1.5-1.6 ppm are further from the carbonyl because they are not as far downfield; they are next to carbons with a total of 5 hydrogens, since the peak is a sextuplet. The methylene group in blue below satisfies this stipulation.

Finally, there is a the triplet at 0.8-0.9 ppm. This means that the hydrogens are next to a carbon that has 2 hydrogens. The methyl group in green satisfies this stipulation:

Since the molecule has 7 carbons, the pattern above is repeated on each side of the carbonyl to give the proper number of hydrogens:

and these correlate with the NMR like this:


Example 2 is 4-heptanone: