Spectroscopy Tutorial: Examples

Example 1


MW 72

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

IR Spectrum

A table of characteristic IR absorptions is available online: click on the link below. Note that this chart is also linked to in the frame to the left.

The IR spectrum for Example 1 is below. 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 1718 indicating a saturated aliphatic ketone.

Think of possible structures

Now we know that the compound has a carbon-oxygen double bond, but there are still a few ways that this four-carbon molecule could be put together. Examples are below:

The second and third structures above are saturated aliphatic aldehydes, which show up at 1740-1720. While in the above IR spectrum the band at 1715 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 one of these structures. That leaves the first compound, which is 2-butanone.

Proton NMR Spectrum

A table of characteristic NMR shifts is available online: click on the link below. Note that this chart is also linked to in the frame to the left.

Before you look at the NMR spectrum, think about what the spectrum of 2-butanone should look like. There are three different types of protons:

The 3 protons in green will be a singlet and show up from 2-2.7 ppm. The 2 protons in blue will be split to a quartet by the protons in red; they will show up from 2-2.7 ppm. They will be further downfield (have a higher ppm value) than the protons in green because they are shielded both by the carbonyl and by the red methyl group. The 3 red protons are the farthest from the carbonyl and are split into a triplet by the blue protons. Let's look at the NMR and see if this is what we see.

Sure enough, here is how they correlate with the structure:

Note the pattern of the ethyl group -CH2CH3 in the above NMR spectrum. Whenever you suspect an ethyl group in a molecule, look for a quartet of 3 protons and a triplet of 2 protons, with the methylene (-CH2-) group further downfield than the methyl group (-CH3).


Example 1 is 2-butanone: