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Spectroscopy Tutorial: Examples

Example 3

C7H16O

MW 116

First calculate the degree of unsaturation: the result is 0. The compound will not have a ring or a doublebond.

IR Spectrum

The compound has an oxygen, but no double bonds, so it must either be an alcohol (­OH) or an ether (­O­). Alcohols are quite distinctive in IR, showing up with a strong, broad band in the region 3500­3250.Note the broad band at 3391: this indicates that the compounds is indeed an alcohol. You can also lookfor the C­OH stretch at 1260­1050.

Think of possible structures

Now we know that the compound has an ­OH group, but there are lot of ways that this seven­carbonmolecule could be put together. It could be straight­chained or it could be branched; the ­OH group couldbe (theoretically) anywhere in the molecule. Let's look at the NMR to get an idea of how many differenthydrogens are in the molecule so that we can narrow down the number of possible structures.

Proton NMR Spectrum

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The NMR of this compound is below:

There are only four different kinds of protons, and 12 of them (at 0.9 ppm) are next to a carbon that hasonly one hydrogen (because it is a doublet). This indicates that the molecule has two of the sub­structuresbelow:

These two sub­structures account for 6 of the 7 carbons in the molecule. Therefore the "something" ineach must be the same carbon. Let's put them end­to­end and replace the "something" with a carbon, thensketch in the proper number of bonds:

Now we need to place an ­OH group in the molecule. It can't be on either of the carbons that areconnected to the two methyl groups, because the position of each hydrogen in blue is dictated by the factthat it splits the 12 hydrogens at 0.9 ppm into a doublet. Therefore, it must be on the middle carbon, likethis:

In the NMR, protons on the same carbon as an ­OH group ­ the hydrogen in purple above ­ will show upfrom 2­4 ppm. Sure enough, there is a peak constituting one proton at 3.1 ppm; furthermore, this peak is

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a triplet indicating 2 protons (blue) on adjacent carbons.

The peak at 1.7­1.9 ppm has two protons and is an octet (although it's kind of hard to see all 8 peaks);this peak corresponds to the blue protons, which are split by both the red and purple protons (a total of 7hydrogens on neighboring carbons).

The hydrogen on the oxygen (green) shows up as a singlet at 1.4 ppm. Although hydroxyl protonsusually show up from 2­4 ppm, the exact position is not always predictable. Hydroxyl protons are usuallynot split.

Summary

Example 3 is 2,4­dimethyl­3­pentanol: