The hydrogen atom has two quantum-mechanically allowed orientations in a magnetic field. The consequence of this is that as soon as you turn the magnetic field on, any protons around are forced to point their "intrinsic magnetic spin vectors" either along the direction of the applied magnetic field or opposite to the field. Pointing along the field is a more stable, lower energy position (state). As the field gets bigger, so does the energetic advantage of pointing with the field.

Every chemically different proton in a molecule will a different splitting of its "with and against" states in an applied magnetic field. An applied field might affect some protons a lot, so that their states split very far apart in energy and make a big fat sideways V on the diagram above. Other protons might only be affected a small amount, so that their energy diagram has a very narrow splitting. We can figure out how big the splitting is for a particular proton in a molecule by holding the magnetic field constant (say at 3) and shining photons of various frequencies onto a sample. The photons that are absorbed must have an energy difference just exactly equal to the splitting of the states as depicted in the diagram above. When a photon is absorbed to make a proton jump up into the higher energy state, we say that the proton "flipped its spin." From the diagram, you can see that it will take photons of larger energy to make protons with larger splitting perform a spin flip.

In another kind of NMR experiment, we shine photons of constant frequency onto a sample but change the magnetic field. From the diagram, you can see that it will take a larger magnetic field to make protons with smaller splitting perform a spin flip in this "constant frequency" experiment.

The reason we want to know the splitting of the magnetic spin states of protons in a molecule is that believe it or not, that splitting can help us figure out the identity of an unknown molecule! Read the following vinette to see how.

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Tricia is one of the hydrogens in methane. Fred is the hydrogen in chloroform. When a magnetic field is turned on, both Fred and Tricia have to make the Choice.

At 5T, who would absorb the bigger photon, Fred or Tricia? Why? Note that this is a constant field, varying frequencies experiment.

Answer: Electronegative substituents such as chlorine pull electron density away from nucleii, which makes them more vulnerable to the applied magnetic fields. For this reason, we say that a proton near an electronegative group is "deshielded" (i.e. deprived of its electron density). For the same applied field, Fred’s energy splitting is bigger than Tricia’s, so he will resonate at a higher frequency. On the other hand, if you set the frequency to 3 and started sweeping the field from 0T to 7T, Fred would perform the spin flip at a lower applied field.

No matter how we actually take the NMR spectrum, we plot the data as if we had set the frequency and swept the field. "Downfield" spin-flip peaks (like Fred’s) are plotted to the left. They are said to have a higher "chemical shift" than upfield peaks. Downfield = deshielded = high chemical shift = to the left on the NMR spectrum.

Both Fred and Tricia have only one chemical type of hydrogen. If a molecule has more than one chemical type of hydrogen, then there will be a different signal for each type, and that signal may be "split." Here is a hypothetical NMR spectrum for 1-chloro, 3,3-dimethylbutane:

There are three groups of "chemically equivalent" hydrogens in this molecule: 9 methyl protons, two protons in the middle, and two on the chlorine-containing carbon. As we get closer to the chlorine, the protons are more deshielded and therefore show up to the left on the NMR spectrum.

Notice that two of the peaks are actually clusters of peaks. This is because the magnetic fields of the protons on adjacent carbons are "coupling." Take a close-up look at the part of the molecule that is responsible for this coupling.

Each H is "coupled" with the two H's on the neighboring C. The chemical shift of each of the H's depends on the spins of the other H's to which it is coupled. There are four possibilities, but two of them are equivalent, so really there are only three possibilities:

The distance between two of the peaks in a cluster is the "coupling constant J." If two clusters have the same coupling constant, then they probably represent adjacent groups of protons, because two groups of protons coupled to each other have the same coupling constant J.

Chemically equivalent protons do not couple. (That's why protons on the same C do not split each other.) Here’s an example of a molecule whose protons are all chemically equivalent. The NMR absorption spectrum will have one peak.

Look at the difference between the spectrum above and the spectrum of the molecule below. The difference is only one atom, but that’s enough to destroy the chemical equivalency.

The above examples are illustrations of the "n+1" rule. The signal from a given proton is split into n+1 signals, where n is the number of chemically inequivalent protons on an adjacent carbon. This rule, together with a knowledge of how electronegative substituents affect peak positions, is invaluable for deducing the structure of an unknown molecule from its NMR spectrum. Following are a few practical considerations about reading NMR spectra.

If the signals overlap, as in toluene, then the spectrum is ugly. It may be extremely difficult to tell what is splitting what.

If there is more than one adjacent carbon with protons attached (e.g. trans-cinnamaldehyde), then the splitting is also more complex, because both adjacent carbon protons will split the signal.

For H₂, there are 4 possibilities instead of the usual three because the last two aren't equivilant anymore:

Therefore, we get a quartet (a "doublet of doublets) where before we would have predicted a triplet.

Diastereotopic Protons: Say we took an NMR spectrum of this compound:

From what we’ve said about NMR so far, you would expect to see four signals: a doublet integral 3 from the methyl on the left, a quartet of triplets (i.e. a multiplet: 12 lines if we could see them all) integral 1 for the proton on the chlorine-containing carbon, a quartet of doublets (i.e. a multiplet: 8 lines if we could see them all) integral 2 from the C with two protons, and a triplet integral 3 from the methyl on the right. In deriving this result, we’ve assumed that H₁ and H₂ are in the same physical environment; i.e. that they are chemically equivalent.

However, H₁ and H₂ are not chemically equivalent, because they are not in the same physical environment. You can see this most clearly by mentally removing H₁ (replacing it with something random like X) and comparing the result to what you get if you mentally remove H₂ (replacing it with X). The two compounds that result are diastereomers of each other.

If we take one of the H’s away, we get something different from what we get if we take the other H away. Therefore the two H’s are not in the same physical environment, so they are not chemically equivalent, so they do split each other. Most of the time, this splitting is so small that we can’t really distinguish it on the NMR spectrum unless our resolution is really really good. But theoretically anyway, the spectrum of this compound would have a doublet integral 3 from the methyl on the left, a quartet of doublets of doublets (i.e. a multiplet: 16 lines if we could see them all) integral 1 for the proton on the chlorine-containing carbon, a quartet of doublets of doublets for H₁, a quartet of doublets of doublets for H₂, and a doublet of doublets integral 3 for the methyl on the left. That would have to be a really fancy NMR spectrometer, though. In reality, the spectrum would probably be very messy, but you would be able to distinguish five different signals instead of the four that you might have predicted before you learned about diastereotopy.

See if you can figure out why in order for two protons to be diastereotopic: 1. they have to be the only two protons on the carbon in question and 2. there has to be another chiral center somewhere in the molecule.

Recall from our studies of chirality that two enantiomers have the exact same physical properties. That’s why they’re so difficult to separate. (Remember that in order to separate enantiomers we have to make diastereomers out of them by crystallizing them with an enantiomerically pure chiral salt . . .)

Say we took an NMR spectrum of this compound:

If I remove H₁ and replace it with X, I get the enantiomer of what I would get if I removed H₂ and replaced it with X. We might call the two protons "enantiotopic." Why don’t they split each other?

Enantiomers, as we’ve said above, have the same physical properties. Therefore, the two protons are in the same physical environment. As such, they are chemically equivalent, and thus they do not split each other. (If there were another chiral center in the molecule, then H₁ and H₂ would be diastereotopic, and not chemically equivalent.)

Remember, although enantiomers rotate light in different directions, they absorb light in equal amounts, which causes them to give the same spectroscopic signals. Therefore, from an absorption or an NMR spectrum, it is impossible to tell whether you are dealing with one enantiomer, the other enantiomer, or a racemic mixture of the two enantiomers. If you were dealing with two diastereomers, would it be possible to tell them apart using NMR?

So far, we’ve only dealt with protons splitting each others’ signal. ¹²C and ¹⁶O don’t split the signal from protons because those nucleii have the special property of not being affected by a magnetic field. (Nucleii containing an even number of protons and an even number of neutrons never respond to magnetic fields.) Other atoms, however, do affect the NMR signal of protons to which they are adjacent. Some examples of other atoms that do this are deuterium and fluorine. Just like ¹H, ¹⁷F has two allowed spin states in a magnetic field, and therefore will split proton NMR signals. Deuterium, on the other hand, has three allowed spin states in a magnetic field. The ¹H NMR signal of HF, therefore, will consist of a doublet, and the ¹H NMR signal of HD will consist of a triplet. (The signals from the F and the D won’t show up in ¹H NMR; they F and D flip at an entirely different frequency. It is instructive to work through the expected ¹H NMR signals for molecules containing heteroatoms with different spin states like F and D.

Most C is 12C, which has 6 protons and 6 neutrons. Since it's an even-even nucleus, it doesn't react to a magnetic field. But a few 13C atoms are mixed in. There are only a few of these, so you can see that it would be extremely rare for 2 13C atoms to end up side-by-side in a molecule. In fact, we can assume that this never happens, so we will never see splitting between 2 adjacent 13C atoms.

When we take a 13C NMR spectrum, we don’t see proton NMR peaks, because the protons perform spin flips at a frequency very different from the frequency at which carbon NMR is taken. As expected, carbons bonded to electronegative substituents resonate downfield, just like in proton NMR. In "proton-decoupled" NMR spectra, the carbon signals are not split by the protons to which they are bonded. A useful rule of thumb is that sp2 C's typically resonante downfield from sp3 C's. In carbon NMR as in proton NMR, we have to watch out for equivalent carbons. Symmetry planes can make C’s equivalent.