Spectroscopy Tutorial: Examples

Example 16


MW 153

Calculate the degree of unsaturation: the answer is 1. The molecule has one double bonds or rings.

IR Spectrum

Since the molecule has an oxygen and one degree of unsaturation, look for a carbonyl band - there is one at about 1700. There is also a wide O-H stretch band centering about 3000; this is characteristic of a carboxylic acid.

NMR Spectrum

The singlet at 11.6 ppm supports the IR suggestion that the compound is a carboxylic aci, R-CO2H. The triplets of two hyrogens each at 3 and 3.6 ppm indicate two different methylene groups, both adjacent to a carbon that has two hydrogens. Since the molecule only has 3 carbons, these must be next to each other. A methylene group which has a bromine on the same molecule will be from about 2.5-4 ppm and a methylene group adjacent to a carboxylic acid will be from about 2-2.7 ppm. The following molecule satisfies these criteria:


Example 16 is 3-bromopropionic acid: