24.4 Nuclear Magnetic Resonance Spectroscopy 1019
molecules, almost all of the carbon atoms are^12 C , which have no magnetic dipole.
Almost all of the oxygen atoms are^16 O, which also have no magnetic dipole. In an
organic molecule that contains only carbon, hydrogen, oxygen, nitrogen, and chlorine,
only other protons provide noticeable spin–spin coupling for protons.
If there are more than two protons in a molecule, multiple splittings can occur. For
example, consider a proton (number 1) on a carbon adjacent to a methyl group. Call
the protons on the methyl group 2, 3, and 4. The three protons on the methyl group
have the same chemical shift, so they do not give multiplet splitting with each other.
Each of them will have the same splitting constant with the proton number 1.J 12 J 13 J 14 (24.4-15)The protons on the methyl group might all have spins up, two might have spin up,
one might have spin up, or none might have spin up. Different molecules will occupy
these states with relative populations in the ratios 1 : 3 : 3 : 1, since there are three
ways to divide a set of three objects into a subset of two and a subset of one. Pro-
ton 1 will produce four lines (a quartet) with splitting equal toJ 12 and with inten-
sities in the ratios 1 : 3 : 3 : 1. This pattern can be derived with the same kind of
mnemonic device as was shown in Figure 24.2 for the ESR lines. This mnemonic
device is shown in Figure 24.4, and gives the intensities in the ratios 1 : 3 : 3 : 1. If
there is a single proton on the adjacent carbon, the protons on the methyl group will
produce two lines with splitting equal toJ 12 , because of the splitting from the first pro-
ton. The total intensity of this doublet will equal three times the total intensity of the
quartet.Figure 24.4 Mnemonic Device for
the Spin–Spin Splitting for Three
Protons.If a proton exhibits spin–spin coupling withnother protons that have equal cou-
pling constants with the first proton, there aren+1 possible values for the sum in
Eq. (24.4-10). There can benprotons with spin up, there can ben−1 with spin up,
and so on down tonprotons with no spins up. This means that the spectral line of pro-
tons on the first carbon atom will be split inton+1 lines. Since the energy differences
are much smaller thankBT, all of these states will be nearly equally populated, and
the intensities of the lines will be proportional to the degeneracies of the levels. The
degeneracy of the level withmprotons having spin up out of a set ofnprotons is the
number of ways of choosing a subset ofmmembers and a subset ofn−mmembers
out of a set ofnmembers:Degeneracyn!
m!(n−m)!(24.4-16)
which is the formula for binomial coefficients. For example, protons on the middle
carbon of propane will be coupled to six protons on the two outer carbons, and will
give seven lines with intensities in the ratios 1 : 6 : 15 : 20 : 15 : 6 : 1. Since protons
on the outer carbons will be coupled to two protons on the center carton, they will give
three lines with intensities in the ratios 1 : 2 : 1. This triplet will have three times the
total area of the total area of the septet.
If a proton exhibits spin–spin coupling with two or more other protons that have
different chemical shifts, the spin–spin coupling constants will have different values,
since the coupling constants depend on the electronic environment of both protons.
One way to predict the effect of the spin–spin coupling in such a case is to divide the
other protons into sets of equal coupling constants. First determine the splitting due to
the protons in one set. Then split each resulting line according to the splittings of the
protons in the next set.