Spectroscopy Tutorial: Examples

Example 15


MW 70

Calculate the degree of unsaturation: the answer is 2. The molecule has 2 double bonds and/or rings.

IR Spectrum

Since the molecule has an oxygen and two double bonds, look for a carbonyl band. There is one at 1693, and the compound is probably an aldehyde because of the C-H stretch bands from 2830-2695. Also look for a C=C stretch at 1680-1640: there is a band there, at 1641, indicating an alkene.

NMR Spectrum

From the IR and degree of unsaturation, we suspect both an aldehyde and a carbon-carbon double bond. The peak of one proton at 9.9 ppm supports the presence of an aldehyde. The peak at 2 ppm is a methyl group next to a carbon that has one hydrogen.

There are only 4 carbons in the molecule - these groups "tie up" 3 of them. One of two ways the molecule can be put together is shown below (the molecule is drawn trans, the other way would be the cis isomer).

The discussion above does not cover coupling constants or the explanation for the number and type of peaks seen in the NMR spectrum; knowledge of these details allow you to designate whether the molecule is cis or trans. If you want to learn more about this, ask your TA or instructor or consult one of the references.


Example 14 is trans-2-butenal: