Objectives After completing this section, you should be able to explain the spin-spin splitting pattern observed in the 1 H NMR spectrum of a simple organic compound, such as chloroethane or 2-bromopropane. interpret the splitting pattern of a given 1 H NMR spectrum. determine the structure of a relatively simple organic compound, given its 1 H NMR spectrum and other relevant information. use coupling constants to determine which groups of protons are coupling with one another in a 1 H NMR spectrum. predict the splitting pattern which should be observed in the 1 H NMR spectrum of a given organic compound. Key Terms Make certain that you can define, and use in context, the key terms below. coupling constant multiplet quartet triplet doublet Study Notes From what we have learned about 1 H NMR spectra so far, we might predict that the spectrum of 1,1,2-trichloroethane, CHCl 2 CH 2 Cl, would consist of two peaks—one, at about 2.5-4.0 δ , expected for CH 2 -halogen compounds and one shifted downfield because of the presence of an additional electronegative chlorine atom on the second carbon. However, when we look at the spectrum it appears to be much more complex. True, we see absorptions in the regions we predicted, but these absorptions appear as a group of two peaks (a doublet ) and a group of three peaks (a triplet ). This complication, which may be disturbing to a student who longs for the simple life, is in fact very useful to the organic chemist, and adds greatly to the power of NMR spectroscopy as a tool for the elucidation of chemical structures. The split peaks ( multiplets ) arise because the magnetic field experienced by the protons of one group is influenced by the spin arrangements of the protons in an adjacent group. Spin-spin coupling is often one of the more challenging topics for organic chemistry students to master. Remember the n + 1 rule and the associated coupling patterns. The source of spin-spin coupling The 1 H-NMR spectra that we have seen so far (of methyl acetate and para -xylene) are somewhat unusual in the sense that in both of these molecules, each set of protons generates a single NMR signal. In fact, the 1 H-NMR spectra of most organic molecules contain proton signals that are 'split' into two or more sub-peaks. Rather than being a complication, however, this splitting behavior actually provides us with more information about our sample molecule. Consider the spectrum for 1,1,2-trichloroethane. In this and in many spectra to follow, we show enlargements of individual signals so that the signal splitting patterns are recognizable. The signal at 3.96 ppm, corresponding to the two H a protons, is split into two subpeaks of equal height (and area) – this is referred to as a doublet . The H b signal at 5.76 ppm, on the other hand, is split into three sub-peaks, with the middle peak higher than the two outside peaks - if we were to integrate each subpeak, we would see that the area under the middle peak is twice that of each of the outside peaks. This is called a triplet . The source of signal splitting is a phenomenon called spin-spin coupling , a term that describes the magnetic interactions between neighboring, non-equivalent NMR-active nuclei. In our 1,1,2 trichloromethane example, the H a and H b protons are spin-coupled to each other. Here's how it works, looking first at the H a signal: in addition to being shielded by nearby valence electrons, each of the H a protons is also influenced by the small magnetic field generated by H b next door (remember, each spinning proton is like a tiny magnet). The magnetic moment of H b will be aligned with B 0 in (slightly more than) half of the molecules in the sample, while in the remaining half of the molecules it will be opposed to B 0 . The B eff ‘felt’ by H a is a slightly weaker if H b is aligned against B 0 , or slightly stronger if H b is aligned with B 0 . In other words, in half of the molecules H a is shielded by H b (thus the NMR signal is shifted slightly upfield) and in the other half H a is deshielded by H b (and the NMR signal shifted slightly downfield). What would otherwise be a single H a peak has been split into two sub-peaks (a doublet), one upfield and one downfield of the original signal. These ideas an be illustrated by a splitting diagram , as shown below. Now, let's think about the H b signal. The magnetic environment experienced by H b is influenced by the fields of both neighboring H a protons, which we will call H a1 and H a2 . There are four possibilities here, each of which is equally probable. First, the magnetic fields of both H a1 and H a2 could be aligned with B 0 , which would deshield H b , shifting its NMR signal slightly downfield. Second, both the H a1 and H a2 magnetic fields could be aligned opposed to B 0 , which would shield H b , shifting its resonance signal slightly upfield. Third and fourth, H a1 could be with B 0 and H a2 opposed, or H a1 opposed to B 0 and H a2 with B 0 . In each of the last two cases, the shielding effect of one H a proton would cancel the deshielding effect of the other, and the chemical shift of H b would be unchanged. So in the end, the signal for H b is a triplet , with the middle peak twice as large as the two outer peaks because there are two ways that H a1 and H a2 can cancel each other out. Now, consider the spectrum for ethyl acetate: We see an unsplit 'singlet' peak at 1.833 ppm that corresponds to the acetyl (H a ) hydrogens – this is similar to the signal for the acetate hydrogens in methyl acetate that we considered earlier. This signal is unsplit because there are no adjacent hydrogens on the molecule. The signal at 1.055 ppm for the H c hydrogens is split into a triplet by the two H b hydrogens next door. The explanation here is the same as the explanation for the triplet peak we saw previously for 1,1,2-trichloroethane. The H b hydrogens give rise to a quartet signal at 3.915 ppm – notice that the two middle peaks are taller then the two outside peaks. This splitting pattern results from the spin-coupling effect of the three H c hydrogens next door, and can be explained by an analysis similar to that which we used to explain the doublet and triplet patterns. Example 13.11.1 Explain, using left and right arrows to illustrate the possible combinations of nuclear spin states for the H c hydrogens, why the H b signal in ethyl acetate is split into a quartet. The integration ratio of doublets is 1:1, and of triplets is 1:2:1. What is the integration ratio of the H b quartet in ethyl acetate? (Hint – use the illustration that you drew in part a to answer this question.) Solution By now, you probably have recognized the pattern which is usually referred to as the n + 1 rule : if a set of hydrogens has n neighboring, non-equivalent hydrogens, it will be split into n + 1 subpeaks. Thus the two H b hydrogens in ethyl acetate split the H c signal into a triplet, and the three H c hydrogens split the H b signal into a quartet. This is very useful information if we are trying to determine the structure of an unknown molecule: if we see a triplet signal, we know that the corresponding hydrogen or set of hydrogens has two `neighbors`. When we begin to determine structures of unknown compounds using 1 H-NMR spectral data, it will become more apparent how this kind of information can be used. Three important points need to be emphasized here. First, signal splitting only occurs between non-equivalent hydrogens – in other words, H a1 in 1,1,2-trichloroethane is not split by H a2 , and vice-versa. Second, splitting occurs primarily between hydrogens that are separated by three bonds. This is why the H a hydrogens in ethyl acetate form a singlet– the nearest hydrogen neighbors are five bonds away, too far for coupling to occur. Occasionally we will see four-bond and even 5-bond splitting, but in these cases the magnetic influence of one set of hydrogens on the other set is much more subtle than what we typically see in three-bond splitting (more details about how we quantify coupling interactions is provided in section 5.5B). Finally, splitting is most noticeable with hydrogens bonded to carbon. Hydrogens that are bonded to heteroatoms (alcohol or amino hydrogens, for example) are coupled weakly - or not at all - to their neighbors. This has to do with the fact that these protons exchange rapidly with solvent or other sample molecules. Below are a few more examples of chemical shift and splitting pattern information for some relatively simple organic molecules. Multiplicity in Proton NMR The number of lines in a peak is always one more (n+1) than the number of hydrogens on the neighboring carbon. This table summarizes coupling patterns that arise when protons have different numbers of neighbors. # of lines ratio of lines term for peak # of neighbors 1 - singlet 0 2 1:1 doublet 1 3 1:2:1 triplet 2 4 1:3:3:1 quartet 3 5 1:4:6:4:1 quintet 4 6 1:5:10:10:5:1 sextet 5 7 1:6:15:20:15:6:1 septet 6 8 1:7:21:35:35:21:7:1 octet 7 9 1:8:28:56:70:56:28:8:1 nonet 8 Example 13.11.2 How many proton signals would you expect to see in the 1 H-NMR spectrum of the structure shown? For each of the proton signals, predict the splitting pattern. Assume that you see only 3-bond coupling. Answer Because of the symmetry in the molecule, there are only four proton signals. Predicted splitting is indicated. Example 13.11.3 Predict the splitting pattern for the 1 H-NMR signals corresponding to the protons at the locations indicated by arrows (the structure is that of the neurotransmitter serotonin). Solutions Coupling constants Chemists quantify the spin-spin coupling effect using something called the coupling constant , which is abbreviated with the capital letter J . The coupling constant is simply the difference, expressed in Hz, between two adjacent sub-peaks in a split signal. For our doublet in the 1,1,2-trichloroethane spectrum, for example, the two subpeaks are separated by 6.1 Hz, and thus we write 3 J a -b = 6.1 Hz. The superscript 3 tells us that this is a three-bond coupling interaction, and the a-b subscript tells us that we are talking about coupling between H a and H b . Unlike the chemical shift value, the coupling constant, expressed in Hz, is the same regardless of the applied field strength of the NMR magnet . This is because the strength of the magnetic moment of a neighboring proton, which is the source of the spin-spin coupling phenomenon, does not depend on the applied field strength. When we look closely at the triplet signal in 1,1,2-trichloroethane, we see that the coupling constant - the `gap` between subpeaks - is 6.1 Hz, the same as for the doublet. This is an important concept! The coupling constant 3 J a -b quantifies the magnetic interaction between the H a and H b hydrogen sets, and this interaction is of the same magnitude in either direction . In other words, H a influences H b to the same extent that H b influences H a . When looking at more complex NMR spectra, this idea of reciprocal coupling constants can be very helpful in identifying the coupling relationships between proton sets. Coupling constants between proton sets on neighboring sp 3 -hybridized carbons is typically in the region of 6-8 Hz. With protons bound to sp 2 -hybridized carbons, coupling constants can range from 0 Hz (no coupling at all) to 18 Hz, depending on the bonding arrangement. For vinylic hydrogens in a trans configuration, we see coupling constants in the range of 3 J = 11-18 Hz, while cis hydrogens couple in the 3 J = 6-15 Hz range. The 2-bond coupling between hydrogens bound to the same alkene carbon (referred to as geminal hydrogens) is very fine, generally 5 Hz or lower. Ortho hydrogens on a benzene ring couple at 6-10 Hz, while 4-bond coupling of up to 4 Hz is sometimes seen between meta hydrogens. Fine (2-3 Hz) coupling is often seen between an aldehyde proton and a three-bond neighbor. Table 4 lists typical constant values. Remember, chemically equivalent protons do not couple with one another to give spin-spin splitting. Exercise \(\PageIndex{1}\) Predict the splitting patterns of the following molecules: Answer H : Doublet. H : Septet H: Doublet, H : Triplet H : Singlet, H : Quartet, H : Triplet Exercise \(\PageIndex{2}\) Draw the following according to the criteria given. C 3 H 5 O; two triplet, 1 doublet C 4 H 8 O 2 ; three singlets C 5 H 12 ; one singlet Answer These are just some drawings, more may be possible. Exercise \(\PageIndex{3}\) The following spectrum is for C 3 H 8 O. Determine the structure. A triplet; B singlet; C sextet; D triplet Source: SDBSWeb : http://sdbs.db.aist.go.jp (National Institute of Advanced Industrial Science and Technology, 3 December 2016) Answer