Nmr  (Page 4/6)

We can also use integration to find the relative amounts of substances present. For example, if we have a mixture of iodomethane and chloromethane dissolved in solution, the ratio of the two peak areas will give the relative amounts of these two compounds. This is useful for determining purity and product yield. We can introduce a known amount of one sample and use its concentration to determine the concentration of another species in solution. The one caution here is that the relative peak areas need to be normalized for the number of protons that produce the signal. In the example of iodomethane and chloromethane, this is easy since they both have methyl groups and the signals are both due to three protons. However, if we were to determine the concentration of dichloromethane versus chloromethane, we would need to take into account that one molecule has three H atoms whereas the other only has two.

Butane

Pentane

The two methyl groups are equivalent to each other, as are two of the two methylene groups. The middle methylene group is unique. In principle, we would expect to see three signals in a 6:4:2 or 3:2:1 ratio.

Ethyl acetate

Ethyl acetate has three types of H atoms. There are two methyl groups and one methylene group. We would expect to see three signals in a 3:3:2 area ratio.

Coupling

Coupling is another very important aspect of NMR spectroscopy that gives a great deal of information about structure. Coupling results in a signal being split into more signals due to neighboring atoms which also have a nuclear spin. This arises frequently between protons on adjacent carbon atoms. One important aspect of this phenomenon is that equivalent nuclei do not couple to each other.

The reason coupling occurs can be described as follows. For example, let us consider the case of a molecule containing two carbon atoms, each with a single proton and other atoms which do not have a nuclear spin coupling to the protons. One such molecule is 1,1-dibromo-2,2-dichloroethane.