So far proton NMR has been not unlike carbon NMR on a smaller scale. However, we have yet to discuss the real strength of proton NMR, something more important than chemical shifts and something that allows us to look not just at individual atoms but also at the way the C–H skeleton is joined together. This is the result of the interaction between nearby protons, known as coupling. An example we could have chosen in the last section is the nucleic acid component cytosine, which has exchanging NH₂ and NH protons giving a peak for HOD at 4.5 ppm. We didn’t choose this example because the other two peaks would have puzzled you. Instead of giving just one line for each proton, they give two lines each—doublets as you will learn to call them—and it is time to discuss the origin of this ‘coupling’.

You might have expected a spectrum like that of the heterocycle below, which like cytosine is also a pyrimidine. It too has exchanging NH₂ protons and two protons on the hetero cyclic ring. But these two protons give the expected two lines instead of the four lines in the cytosine spectrum. It is easy to assign the spectrum: the green proton labelled HA is attached to an aldehyde-like C=N and so comes at lowest fi eld. The red proton labelled HX is ortho to two electron-donating NH₂ groups and so comes at high field for an aromatic proton (p. 272). These protons do not couple with each other because they are too far apart. They are separated by five bonds whereas the ring protons in cytosine are separated by just three bonds.

Understanding this phenomenon is so important that we are going to explain it in three different ways—you choose which appeals to you most. Each method offers a different insight. The diaminopyrimidine spectrum you have just seen has two single lines (singlets we shall call them from now on) because each proton, H^A or H^X, can be aligned either with or against the applied magnetic field. The cytosine spectrum is different because each proton, say HA, is near enough to experience the small magnetic fi eld of the other proton HX as well as the fi eld of the magnet itself. The diagram shows the result.

If each proton interacted only with the applied field we would get two singlets. But pro ton H^A actually experiences two slightly different fields: the applied field plus the field of H^X or the applied fi eld minus the field of HX. HX acts either to increase or decrease the field experienced by HA. The position of a resonance depends on the fi eld experienced by the proton so these two situations give rise to two slightly different peaks—a doublet as we shall call it. And whatever happens to HA happens to HX as well, so the spectrum has two doublets, one for each proton. Each couples with the other. The field of a proton is a very small indeed in comparison with the field of the magnet and the separation between the lines of a doublet is very small. We shall discuss the size of the coupling later (pp. 294–300). The second explanation takes into account the energy levels of the nucleus. In Chapter 4, when we discussed chemical bonds, we imagined electronic energy levels on neighbouring atoms interacting with each other and splitting to produce new molecular energy levels, some higher in energy and some lower in energy than the original atomic energy levels. When hydrogen nuclei are near each other in a molecule, the nuclear energy levels also interact and split to produce new energy levels. If a single hydrogen nucleus interacts with a magnetic field, we have the picture on p. 270 of this chapter: there are two energy levels as the nucleus can be aligned with or against the applied magnetics field, there is one energy jump possible, and there is a resonance at one frequency. This you have now seen many times and it can be sum marized as shown below.

The spectrum of the pyrimidine on p. 286 shows exactly this situation: two protons well separated in the molecules and each behaving independently. Each has two energy levels, each gives a singlet, and there are two lines in the spectrum. But in cytosine, whose spectrum is shown on p. 285, the situation is different: each hydrogen atom has another hydrogen nucleus nearby and there are now four energy levels. Each nucleus HA and HX can be aligned with or against the applied fi eld. There is one (lower) energy level where they are both aligned with the fi eld and one (higher) level where they are both aligned against. In between there are two different energy levels in which one nucleus is aligned with the fi eld and one against. Exciting H from alignment with to alignment against the applied fi eld can be done in two slightly different ways, shown as A1 and A2 on the diagram. The result is two resonances very close together in the spectrum.

Please notice carefully that we cannot have this discussion about H^A without discussing HX in the same way. If there are two slightly different energy jumps to excite H^A, there must also be two slightly different energy jumps to excite H^X. A₁, A₂, X₁, and X2 are all different, but the difference between A1 and A2 is exactly the same as the difference between X₁ and X₂. Each proton now gives two lines (a doublet) in the NMR spectrum and the splitting of the two doublets is exactly the same. We describe this situation as coupling. We say ‘A and X are coupled’ or ‘X is coupled to A’ (and vice versa, of course). We shall be using this language from now on and so must you. Now look back at the spectrum of cytosine at the beginning of this section. You can see the two doublets, one for each of the protons on the aromatic ring. Each is split by the same amount (this is easy to check with a ruler). The separation of the lines is the coupling constant and is called J. In this case J = 4 Hz. Why do we measure J in hertz and not in ppm? We pointed out on p. 55 (Chapter 3) that we measure chemical shifts in ppm because we get the same number regardless of the rating of the NMR machine in MHz. We measure J in Hz because we also get the same number regardless of the machine. The spectra below show ¹H NMR spectra of the same compound run on two different NMR machines—one a 90 MHz spectrometer and one a 300 MHz spectrometer (these are at the lower and upper ends of the range of fi eld strengths in common use). Notice that the peaks stay in the same place on the chemical shift scale (ppm) but the size of the coupling appears to change because 1 ppm is worth 90 Hz in the top spectrum but 300 Hz in the bottom.

Measuring coupling constants in hertz

 To measure a coupling constant, it is essential to know the rating of the NMR machine in MHz (megahertz). This is why you are told that each illustrated spectrum is, say, a ‘400 MHz 1H NMR spectrum’. Couplings may be marked on the spectrum, electronically, but if not then to measure the coupling, measure the distance between the lines by ruler or dividers and use the horizontal scale to find out the separation in ppm. The conversion is then easy—to turn parts per million of megahertz into hertz you just leave out the million! So, 1 ppm on a 300 MHz machine is 300 Hz. On a 500 MHz machine, a 10 Hz coupling is a splitting of 0.02 ppm.

● Spectra from different machines

 When you change from one machine to another, say, from a 200 MHz to a 500 MHz NMR machine, chemical shifts (δ) stay the same in ppm and coupling constants (J ) stay the same in Hz.

Now for the third way to describe coupling. If you look again at what the spectrum would be like without interaction between H^A and H^X you will see the pattern on the right, with the chemical shift of each proton clearly obvious. But you don’t see this because each proton couples with the other and splits its signal by an equal amount either side of the true chemical shift. The true spectrum has a pair of doublets each split by an identical amount. Note that no line appears at the true chemical shift, but it is easy to measure the chemical shift by taking the midpoint of the doublet. So, this spectrum would be described as δH 7.5 (1H, d, J 4 Hz, H^A) and 5.8 (1H, d, J 4 Hz, H^X). The main number gives the chemical shift in ppm and then, in brackets, comes the integra tion as the number of Hs, the shape of the signal (here ‘d’ for doublet), the size of coupling constants in Hz, and the assignment, usually related to a diagram. The integration refers to the combined area under both peaks in the doublet. If the doublet is exactly symmetrical, each peak integrates to half a proton. The combined signal, however complicated, integrates to the right number of protons. We have described these protons as A and X with a purpose in mind. A spectrum of two equal doublets is called an AX spectrum. A is always the proton you are discussing and X is another proton with a different chemical shift. The alphabet is used as a ruler: nearby protons (on the chemical shift scale—not necessarily nearby in the structure!) are called B, C, etc. and distant ones are called X, Y, etc. You will see the reason for this soon. If there are more protons involved, the splitting process continues. Here is the NMR spectrum of a famous perfumery compound supposed to have the smell of ‘green leaf lilac’. The compound is an acetal with five nearly identical aromatic protons at the normal benzene position (7.2–7.3 ppm) and six protons on two identical OMe groups.

and a ¹H triplet at 4.6 ppm. In NMR talk, triplet means three equally spaced lines in the ratio 1:2:1. The triplet arises from the three possible states of the two identical protons in the CH₂ group. If one proton H^A interacts with two protons H^X, it can experience protons H^X in three different possible states. Both protons H^X can be aligned with the magnet or both against. These states will increase or decrease the applied field just as before. But if one proton H^X is aligned with and one against the applied field, there is no net change to the field experienced by HA. There are two arrangements for this (see diagram overleaf). We’ll therefore see a signal of double intensity for H^A at the correct chemical shift, one signal at higher field and one at lower field. In other words, a 1:2:1 triplet.

We could look at this result by our other methods too. There is one way in which both nuclei can be aligned with and one way in which both can be aligned against the applied field, but two ways in which they can be aligned one with and one against. Proton HA interacts with each of these states. The result is a 1:2:1 triplet.

Using our third way of seeing coupling to see how the triplet arises, we can just make the peaks split in successive stages:

COUPLING IN THE PROTON NMR SPECTRUM

If there are more protons involved, we continue to get more complex systems, but the intensities can all be deduced simply from Pascal’s triangle, which gives the coefficients in a binomial expansion. If you are unfamiliar with this simple device, here it is.

You can read off from the triangle what pattern you may expect when a proton is coupled to n equivalent neighbours. There are always n + 1 peaks with the intensities shown by the triangle. So far, you’ve seen 1:1 doublets (line 2 of the triangle) from coupling to 1 proton, and 1:2:1 triplets (line 3) from coupling to 2. You will often meet ethyl groups (CH3CH2X), where the CH2 group couples to three identical protons and appears as a 1:3:3:1 quartet and the methyl group as a 1:2:1 triplet. In isopropyl groups, (CH3)2CHX, the methyl groups appear as a 6H doublet and the CH group as a septuplet. Here is a simple example: the four-membered cyclic ether oxetane. Its NMR spectrum has a 4H triplet for the two identical CH₂ groups next to oxygen and a 2H quintet for the CH₂ in the middle. Each proton HX ‘sees’ four identical neighbours (HA) and is split equally by them all to give a 1:4:6:4:1 quintet. Each proton HA ‘sees’ two identical neighbours HX and is split into a 1:2:1 triplet. The combined integral of all the lines in the quintet together is 2 and of all the lines in the triplet is 4.

A slightly more complicated example is the diethyl acetal below. It has a simple AX pair of doublets for the two protons on the ‘backbone’ (red and green) and a typical ethyl group (2H quartet and 3H triplet). An ethyl group is attached to only one substituent through its CH₂ group, so the chemical shift of that CH₂ group tells us what it is joined to. Here the peak at 3.76 ppm can only be an OEt group. There are, of course, two identical CH₂ groups in this molecule.

In all of these molecules, a proton may have had several neighbours, but all those neigh bours have been the same. And therefore, all the coupling constants have been the same. What happens when coupling constants differ? Chrysanthemic acid, the structural core of the insecticides produced by pyrethrum flowers, gives an example of the simplest situation— where a proton has two different neighbours.

Chrysanthemic acid has a carboxylic acid, an alkene, and two methyl groups on the three membered ring. Proton HA has two neighbours, H^X and H^M. The coupling constant to HX is 8 Hz, and that to H^M is 5.5 Hz. We can construct the splitting pattern as shown on the right. The result is four lines of equal intensity called a double doublet (or sometimes a doublet of doublets), abbreviation dd. The smaller coupling constant can be read off from the separation between lines 1 and 2 or between lines 3 and 4, while the larger coupling constant is between lines 1 and 3 or between lines 2 and 4. The separation between the middle two lines is not a coupling constant. You could view a double doublet as an imperfect triplet where the second coupling is too small to bring the central lines together: alternatively, look at a triplet as a special case of a double doublet where the two couplings are identical and the two middle lines coincide.

مواضيع ذات صلة

Some compounds can exist as a pair of mirror-image forms

Exchange of acidic protons is revealed in proton NMR spectra

Protons on heteroatoms have more variable shifts than protons on carbon

Non-aldehyde protons in the aldehyde region: pyridines

The aldehyde region: unsaturated carbon bonded to oxygen

Structural information from the alkene region

Electron-rich and electron-deficient alkenes

How electron donation and withdrawal change chemical shifts

Uneven electron distribution in aromatic rings

The benzene ring current causes large shifts for aromatic protons

Summary chart of proton NMR shifts

Chemical shifts of CH groups