Alkanes

Now let's take a look at how these generalities I've been talking about correspond to the spectra of some real compounds.

Hexane

First, let's consider hexane (in spectrum #1 in your workbook). Where does it have absorptions?

Hexane IR spectrum.[61irsp01.JPG (10672 bytes)]

There are three absorptions close to the right of 3000. That corresponds to hydrogen attached to a carbon atom which has only single bonds. The fact that there are three absorptions does not mean that there are three hydrogen atoms. Instead, it corresponds to three ways that the hydrogen in the compound can absorb energy. The rest of the absorptions are in the fingerprint region that we won't worry about. Altogether, what does this spectrum show? It indicated that hexane contains carbon and hydrogen and single bonds. In short, hexane is an alkane.

Octane

Next, the spectrum for octane (spectrum #2 in your workbook).

Octane IR spectrum.[61irsp02.JPG (10352 bytes)]

As with hexane the C-H bonds show up to the right of 3000 and nothing shows up in the multiple bond region. Again, this compound is an alkane. Notice that there is not much in the spectra that would allow us to distinguish between hexane and octane. That is characteristic of IR spectra, they will help identify the class of compound, but not the specific compound. You would have to use something like the molecular weight to do that.

Cyclohexane

Next, cyclohexane (spectrum #3 in your workbook). Again, it is an alkane as indicated by the location of the absorptions.

Cyclohexane IR spectrum.[61irsp03.JPG (10351 bytes)]

There is a difference in fingerprint region, however, and also there are only two peaks to the right of 3000 rather than three. A skilled interpreter of these spectra would pick up on that and figure out that this is a cyclic compound. (That is because the third peak comes from vibrations caused by having three hydrogen atoms on the carbon atom at the end of the molecule. The absence of the third peak indicates the absence of three hydrogen atoms at the end of the molecule. In this case the molecule does not have an end.)

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E-mail instructor: Eden Francis

Clackamas Community College
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