Exercise 1. Predict approximate chemical shifts for all the carbon and hydrogen atoms which are explicitly shown in the following molecules:

If two or more protons (or indeed with two or more carbons) are in an equivalent environment, then they will have the same chemical shift and appear as one signal. A common example of this is a CH₃ group, where all three protons are always equivalent. In a CH₂ group, the two protons are also equivalent, unless adjacent to a chiral center. Look for any plane of symmetry in the molecule, which will render the two halves equivalent. In the examples below, the different types of atom are given a subscript; atoms which are equivalent will have the same subscript.

Exercise 2. Predict how many peaks you would expect to see in the 1H and 13C NMR spectra of the following molecules:

4. Integration

For example:

Exercise 3: What is the ratio of the hydrogens in the following ¹H spectrum?

As mentioned in section 2 above, the chemical shift is determined by B_eff, which is affected by nearby nuclei and electrons. However, the nearby nuclei are themselves being excited to the higher energy state by the radio waves. It turns out that any particular ¹H will spend about 50% of the time in the lower energy state, and about 50% of the time in the upper energy state. Consider two neighboring protons in the following system:

If we consider B_eff for H_a, we will find that 50% of the time H_b will be increasing B_eff for H_a (because H_b is aligned with the field), and 50% of the time H_b will be decreasing B_eff for H_a (see above). The net result is that 50% of the signal for H_a will be shifted slightly downfield by H_b, while 50% of the signal for H_a will be shifted upfield by H_b. This process is called coupling, and it leads to a splitting of the signal into a doublet.

If now we turn to H_b, we will find that H_a has the exact same effect on H_b that H_b had on H_a. Thus the signal for H_b will be split by the same amount as H_a.

Now let us consider a more complicated system. If H_a has two neighboring H_b nuclei which are equivalent, the effect of these H_b nuclei may cancel out or not, as shown in the diagram below. This leads to the H_a signal being split into a triplet, with the three parts of the peak having areas in the ratio 1 : 2 : 1

(i.e., 25% : 50% : 25%). [However H_b will only be a simple doublet, as in the previous situation, because there is only one neighboring H_a affecting it.]

How do things work out in practice?

1. Hydrogens which are equivalent do not couple to one another (though they may couple to other nearby protons). A common effect of this is that 2 or 3 hydrogens attached to the same carbon do not usually split one another- thus an isolated methyl group always shows up as a singlet.
2. Only hydrogens which are attached to neighboring carbons usually couple to one another. Since only 1% of carbon is ¹³C, coupling of carbon is not seen in ¹H spectra.
3. The "n + 1 rule", which says that if a proton H_a has n equivalent protons on neighboring carbons, then the signal for H_a will be split into n + 1 peaks. For example, a CH₃ peak will split any "next door" proton signals into 4 peaks, called a quartet.
4. In a variant of the above rule, if H_a has m equivalent protons of type Hb and n equivalent protons of type Hc, all on neighboring carbons, then Ha will be split into (m + 1)(n + 1) peaks. Thus we see patterns such as a doublet of triplets, etc.

Consider the following examples:

A.

B.

Exercise 4. Predict the coupling patterns in the following molecules:

6. Decoupling a ¹³C spectrum

Since ¹³C makes up only 1% of natural carbon, coupling between carbon atoms is rarely observed. However coupling by nearby hydrogens would very often make ¹³C spectra very hard to read, so we routinely use a technique called decoupling to eliminate all coupling effects from all hydrogens. This is done by blasting the sample with radio waves which excite in the ¹H region (this scrambles all of the hydrogens), while observing in the ¹³C region. There is a price to pay: integration can no longer be done accurately. However there is also a benefit: the hydrogens transfer some of their energy to the carbons and this improves the otherwise feeble absorption of energy by the ¹³C nuclei. A side effect of this is that carbons which have no hydrogens attached to them tend to be considerably smaller than the other carbons, and such carbons can easily be identified in a ¹³C spectrum.

Therefore in a decoupled 13C spectrum we see no coupling (except in the CDCl3 solvent, which is split into three peaks by the deuterium). This means that each carbon gives rise to a single sharp peak, and in a clear 13C spectrum the total number of such peaks (excluding TMS and solvent) is equal to the number of types of carbon in the molecule.

7. Putting it all together: determining a structure from an NMR spectrum

Interpreting an NMR spectrum is a skill acquired through much practice, and there are few hard and fast rules since every spectrum is different. However there are a few general guidlines which are usually helpful. I personally follow this procedure:

1. If you have a molecular formula, determine the number of double bonds or rings in the molecule. This is given (if no nitrogens present) by the formula 1 + # carbons - ¹/₂(# hydrogens + halogens)
2. Alternatively, if the starting material from which the sample was made is known, use this as a starting point- look to see which groups are unchanged.
3. Use the integration in the 1H spectrum to find out the approximate number of hydrogens giving rise to each set of peaks.
4. Look for distinctive groups in the spectrum. For example, a sharp singlet (integration of 3H) at around 2 ppm is likely to be a CH₃ which is next to a C=O, C=C or aromatic ring. A sharp singlet at around 0.8 ppm (integration of 9H) arises from a tert-butyl group, making this group a useful marker in chemical reactions. Learn to recognize other simple alkyl groups up to 4 carbons, particularly a simple ethyl (triplet for CH₃ + quartet for CH₂). The isopropyl group (doublet for the two CH₃ s+ tiny peak for CH, split into 7) can be tricky, since the two equivalent but split methyl groups may easily be confused with two slightly different methyl groups. Peaks at around 3.5 - 4 ppm are likely to be adjacent to oxygen. Peaks above 7 are probably aromatic, but may possibly be vinylic if C=O groups (or similar) are nearby. If you see a pair of doublets between 7 and 8, this is most likely to be a para disubstituted aromatic ring (the only disubstitution pattern which is obvious). In 13C spectra, look for peaks above 160 ppm (C=O) and peaks around 60 ppm (C-O). Peaks above 100 ppm are sp² carbons.
5. Write down all of the fragments of the molecule that you can, then (if possible) consider what "barrier groups" are present (e.g. ketone C=O, ester or ether C-O). These groups act as barriers preventing coupling between protons on carbons either side of them. Remember too that a group attached to an aromatic ring will not be coupled to the aromatic protons. Taking into account all of these barrier groups, and the chemical shifts & coupling patterns, you should be able to assemble all of your fragments into a complete structure.

Exercise 5.  Try it out on these NMR spectra:

All spectra from the Sigma-Aldrich Handbook of FT-NMR Spectra.
Go to the answers to NMR exercises.
Back to the Chemistry 29b home page
Back to the Chemistry 25b home page