1.10. Extra Considerations and Complications
Nuclear Magnetic Resonance (NMR) is incredibly complex. Although including all of the advanced details would be helpful for structure determination, at an introductory level this would be excessive. However, some additional concepts and details are very helpful for analysis.
Topicity is a sub-section of stereochemistry. With topicity the stereochemical relationship between substituents within a molecule is described (as opposed to the stereochemical relationship between two different molecules). Although this seems specific and niche, topicity relationships are important in enzymatic reactions and for NMR spectroscopy.
Recall that “substitution tests” are performed by separately replacing two atoms with a hypothetical group and compare the resulting structures. Many sources use “Z” as it does not correspond to any element or functional group. Previous substitution tests (see Section 1.6.2.3) were used to see if hydrogen groups were chemically equivalent. However, this is only a simplified application of substitution tests. Defining the relationship between molecules generated by substitution tests establishes the topicity relationship between the groups that were substituted.
If the resulting two structures are chemically identical then the topicity relationship between the two groups that were replaced is homotopic (Figure 1.36). In 1H NMR spectroscopy homotopic hydrogens are chemically equivalent and will contribute to the same signal in the spectrum.
Figure 1.36 – Example of a Substitution Test on Two Hydrogens of Propan-2-one Determining a Homotopic Relationship Between Them.
If the resulting two structures are chemically different (constitutional isomers) then the topicity relationship between the two groups that were replaced is heterotopic (Figure 1.37). In 1H NMR spectroscopy heterotopic hydrogens are NOT chemically equivalent and will NOT contribute to the same signal in the spectrum.
Figure 1.37 – Example of a Substitution Test on Two Hydrogens of Propan-2-one Determining a Heterotopic Relationship Between Them.
If the resulting two structures are a pair of enantiomers then the topicity relationship between the two groups that were replaced is enantiotopic (Figure 1.38). In 1H NMR spectroscopy enantiotopic hydrogens are chemically equivalent and will contribute to the same signal in the spectrum.
Figure 1.38 – Example of a Substitution Test on Two Hydrogens of Propan-2-one Determining an Enantiotopic Relationship Between Them.
Although enantiotopic hydrogens are equivalent in NMR spectroscopy they are NOT equivalent in enzymatic reactions. This has far reaching consequences and is the primary reason for exploring topicity relationships in biological systems.
When there are pre-existing stereogenic centres or stereogenic elements then the results are more complicated.
If the resulting two structures are a pair of diastereomers then the topicity relationship between the two groups that were replaced is diastereotopic (Figure 1.39). In 1H NMR spectroscopy diastereotopic hydrogens are NOT chemically equivalent and will NOT contribute to the same signal in the spectrum.
Figure 1.39 – Example of a Substitution Test on Two Hydrogens of (S)-1-Bromo-1-chloro-1-fluoropentan-3-one Determining a Diastereotopic Relationship Between Them.
1.10.1.4.1. Advanced Coupling – Diastereotopic Couplings
Because diastereotopic hydrogens are not chemically equivalent in 1H NMR spectroscopy, they will have (slightly) different chemical shifts. Because diastereotopic hydrogens are not chemically equivalent in 1H NMR spectroscopy, they will also spin couple to one another if close enough (within three bonds). Both of these effects result in much more complicated spectra than might otherwise be predicted.
In simple cases involving diastereotopic hydrogens the outcome is more intuitive. Consider the three hydrogens of 3,3,3-tribromoprop-1-ene (Figure 1.40). The hydrogen at position 2 (blue) is definitely not chemically equivalent to either of the other hydrogens. A substitution test shows that the two hydrogens at position 1 (red and green) are diastereotopic to each other; the results of substitution tests with these hydrogens produce a pair of E/Z diastereomers.
Figure 1.40 – Substitution Test on the Two Hydrogens at Position 1 of 3,3,3-Tribromoprop-1-ene Determining a Diastereotopic Relationship Between Them.
Because all three hydrogens are chemically distinct they will each give rise to their own signal in 1H NMR spectroscopy. They are also close enough (within three bonds) to spin couple to one another. It is possible to predict the coupling pattern for each signal (Figure 1.41). Each hydrogen will couple to two sets of 1 chemically equivalent hydrogen and form a doublet of doublets. Recall that the actual shape of multi-coupled signals can vary significantly; the doublet of doublets for the hydrogen at position 2 looks different from the doublet of doublets for either of the hydrogens at position 1 but all three are still a “doublet of doublets”.
Figure 1.41 – Predicted and Actual Multiplicities of 1H NMR Signals of 3,3,3-Tribromoprop-1-ene.
It is also possible to predict that the chemical shifts for the two hydrogens at position 1 will be only slightly different (because they are diastereotopic). Note that while it is possible to predict that the two signals will have similar chemical shifts it is not possible at an introductory level to predict which of the two will have the higher/lower shift.
The exact same approach is used when stereocentres are involved. For simplicity the example will have a very limited number of hydrogens. Consider the three hydrogens of (R)-1,1,2,2,3,3,4-heptachlorocyclopentane (Figure 1.42). The hydrogen at position 4 (blue) is definitely not chemically equivalent to either of the other hydrogens. A substitution test shows that the two hydrogens at position 5 (red and green) are diastereotopic to each other; the results of substitution tests with these hydrogens produce a pair of diastereomers.
Figure 1.42 – Substitution Test on the Two Hydrogens at Position 5 of (R)-1,1,2,2,3,3,4-Heptachlorocyclopentane Determining a Diastereotopic Relationship Between Them.
Because all three hydrogens are chemically distinct they will each give rise to their own signal in 1H NMR spectroscopy. They are also close enough (within three bonds) to spin couple to one another. It is possible to predict the coupling pattern for each signal (Figure 1.43). Each hydrogen will couple to two sets of 1 chemically equivalent hydrogen and form a doublet of doublets. As before, it is also possible to predict that the chemical shifts for the two hydrogens at position 5 will be only slightly different but not possible to predict which of the two will have the higher/lower chemical shift.
Figure 1.43 – Predicted and Actual Multiplicities of 1H NMR Signals of (R)-1,1,2,2,3,3,4-Heptachlorocyclopentane Omitting the Signal for Hydrogen at Position 4.
Recall that the actual shape of multi-coupled signals can vary significantly. This example highlights a problem with this. The signal for the hydrogen at position 4 (blue) is a doublet of doublets. However, by coincidence the peaks overlap and look like a triplet (Figure 1.44). This is the same outcome that would have been predicted if the diastereotopic hydrogens were treated as chemically equivalent. This results from the two coupling constants (JblueH-redH and Jblue-H-greenH) being VERY similar.
Figure 1.44 – Predicted and Actual Multiplicities of the 1H NMR Signal for Hydrogen at Position 4 of (R)-1,1,2,2,3,3,4-Heptachlorocyclopentane.
This type of coincidental overlap is possible but rare for multi-coupled splitting patterns except when one of the couplings is to a diastereotopic partner. In these cases it is somewhat common; coupling constants to diastereotopic groups are often VERY similar, sometimes coincidentally identical. As a result, the theoretical coupling pattern (e.g. dd) may be more complex than the actual signal shape (e.g. apparent triplet).
When predicting multiplicities treat diastereotopic hydrogens as chemically distinct and describe the theoretical coupling pattern. However, when interpreting spectra be aware that it is possible for diastereotopic hydrogens to coincidentally “act like” they are chemically equivalent in some or all spin couplings. Often this is indicated in some way to assist students in structure determination.
The consequences of stereochemistry and diastereotopic hydrogens being chemically distinct can rapidly add insurmountable amounts of complexity to NMR spectra. This example is included only to show how easily spectra can be complicated by the presence of a stereocentre/stereogenic element. Students would not be expected to analyze spectra of this complexity beyond commenting that the complexity could arise from stereochemistry.
Consider the 1H NMR spectrum for 1,1,3,3-tetrachlorocyclopentane (Figure 1.45). The two hydrogens at position 2 (blue) are chemically equivalent to each other (homotopic). The four hydrogens at positions 4 and 5 (red) are chemically equivalent to each other (internal mirror plane; checking topicity will show enantiotopic relationships between each). No sets of chemically distinct hydrogens can spin couple to each other. The resulting spectrum is very simple.
Figure 1.45 – 1H NMR Spectrum of 1,1,3,3-Tetrachlorocyclopentane.
Consider the 1H NMR spectrum for (S)-1-bromo-1,3,3-trichlorocyclopentane (Figure 1.46). Replacing one of the chlorine atoms with a bromine forms a stereogenic centre. The total number of hydrogens stays the same, but now they are all chemically distinct by being heterotopic or diastereotopic. Many spin couplings split each signal. Because they have similar chemical shifts some of the signals from different hydrogens overlap with each other. The resulting spectrum is VERY complex.
Figure 1.46 – 1H NMR Spectrum of (S)-1-Bromo-1,3,3-trichlorocyclopentane.
In most cases, molecules with one or more stereogenic centres result in the hydrogen atoms of CH2 groups being diastereotopic and giving rise to two (slightly) different signals in NMR spectra. If it is not obvious a substitution test is still recommended. Conversely, the hydrogen atoms of methyl (CH3) groups will always be homotopic and give rise to the same signal.
Recall that all O-H and N-H bonds are polar, with an excess of electron density on the O/N and a deficiency of electron density on the H. Functional groups with O-H or N-H bonds (alcohols, carboxylic acids, hydrates, hemiacetals, amines, amides, etc.) can participate in hydrogen bonding (Figure 1.47). The hydrogen of the group can act as a hydrogen-donor. The oxygen or nitrogen can act as a hydrogen-acceptor. Because these groups can act as both donors and acceptors they can always participate in hydrogen bonding with another equivalent of themselves. As a result, there is always some hydrogen bonding occurring in samples with O-H or N-H bonds.
Figure 1.47 – Representation of Hydrogen Bonding in Functional Groups with O-H or N-H Bonds.
When hydrogen bonding is occurring the actual distance between the hydrogen and the oxygen/nitrogen fluctuates rapidly. This affects how much electron density is experienced by the hydrogen nucleus. At any point in time some molecules will have a short bond distance, some will be middling, and in some it will be long. As a result, signals for hydrogens attached to oxygen or nitrogen tend to be broad (a “hill” rather than a sharp peak) because the signal is arising from many slightly different electron density environments (Figure 1.48).
Figure 1.48 – Example of a Broad O-H or N-H 1H NMR Peak.
When hydrogen bonding is occurring the individual hydrogens are swapping between molecules (acid-base chemistry). This inhibits spin coupling. As a result, signals for hydrogens attached to oxygen or nitrogen typically do not split nearby signals nor are split themselves (Figure 1.49). It is possible for spin coupling to occur, but this typically requires specific conditions be met during the sample preparation. As a secondary consequence, signals for hydrogens attached to oxygen or nitrogen may not always integrate properly (the integral value is occasionally less than would be expected). The signal may be significantly “smaller” than usual. Often, but not always, this is indicated in some way to assist students in structure determination.
Figure 1.49 – 1H NMR Spectrum of Methanol.
The exact amount of hydrogen bonding occurring will affect the electron density around the hydrogen nuclei in O-H and N-H bonds. This is heavily influenced by the concentration of the sample in the spectrometer. This is why the standard chemical shift ranges for hydrogens in O-H and N-H bonds are so large (see Figure 1.24). The same compound analyzed twice at different concentrations will often have a different chemical shift for the signals of hydrogens in O-H and N-H bonds.
1.10.2.1. Extra Experiments – D2O Exchange
It is possible to take advantage of the above effects to determine which, if any, signals in a 1H NMR spectrum come from O-H and/or N-H groups. The simplest way this is done is through a deuterium exchange experiment. The sample is placed in the spectrometer and the 1H NMR spectrum is collected as usual. Then, deuterated water (D2O) is added to the sample and the mixture is allowed to equilibrate (Figure 1.50). Because of the acid-base chemistry occurring this replaces the protium nuclei in O-H and N-H groups with deuterium nuclei. Deuterium is a different isotope of hydrogen and has a different spin quantum number. It does not produce signals in 1H NMR spectra. The sample is then analyzed again and any signals which disappeared correspond to those from O-H and/or N-H groups.
Figure 1.50 – Deuterium Exchange Experiment for 3-Amino-2-hydroxypropanoic acid.
This is particularly helpful for establishing what functional groups are, or are not, present when trying to determine an unknown structure from spectral data. For example, if the structures of the compounds in Figure 1.50 were not given then the deuterium exchange experiment demonstrated that there are 3 chemically distinct hydrogens attached to oxygen and/or nitrogen atoms in the molecule.
To generate data for an NMR spectrum a sample is placed in the spectrometer and then analyzed. Theoretically the sample could be “neat”, with only the pure compound in a vessel. In practice this does not work well (detector saturation) and/or is not feasible (if only a small amount of the compound is available). Instead, samples are dissolved in a solvent and then analyzed.
There are MANY more molecules of solvent than the compound being analyzed in the vessel. As a result, if a solvent with protium nuclei were used the signal(s) from the solvent would be so large that any signals from the sample would be reduced to “noise”. Deuterium is a different isotope of hydrogen and has a different spin quantum number. It does not produce signals in 1H NMR spectra. So-called deuterated solvents, where the solvent was synthesized to contain deuterium instead of protium for its hydrogens, are used for NMR sample preparation.
By far the most common solvent used is deuterated chloroform (CDCl3; Figure 1.51). Other frequently used solvents are deuterated dimethylsulfoxide (commonly written as DMSO-D6), deuterated water (D2O), deuterated tetrahydrofuran (commonly written as THF-D8), and deuterated benzene (C6D6). There are many other deuterated solvents that are less commonly used. The variety is for solubility: some compounds are only soluble in certain solvents.
Figure 1.51 – Common Deuterated Solvents for NMR Spectroscopy.
Chemical shift is dependent on local electron density. This is heavily influenced by the electron density and polarity of the solvent. As a result, spectra collected in different solvents will have slightly different chemical shifts. For example, the chemical shifts for the signals of hydrogens in a sample of methanol vary depending on the solvent (Table 1.3).
Table 1.3 – 1H NMR Chemical Shifts of C-H’s of Methanol in Deuterated Solvents.
Because shifts will depend on the solvent used, the solvent is normally included in the description when reporting NMR data. This text avoids this requirement by always using values and ranges for signals in deuterated chloroform. However, this is only possible by judicious selection of which compounds to discuss. In practical applications such as laboratory settings the solvent must be explicitly mentioned.
Modern synthetic techniques are not flawless. For example, even using best practices a synthetic solvent such as CDCl3 will still have a small (~0.2%) amount of non-deuterated molecules (CHCl3) contaminating it. When collecting data signals will also be collected for the small amount of non-deuterated solvent. In “real” spectra researchers must identify and know to ignore these signals (Figure 1.52; see Organometallics 2010, 29, 2176–2179 for a useful resource for this).
Figure 1.52 – 1H NMR Spectrum of Methanol with Residual Solvent Signal Not Omitted.
For simplicity this text omits these signals (all spectra in this text are simulated by the computer). However, this is only considered acceptable for teaching purposes.
1.10.4. Degree of Unsaturation // Index of Hydrogen Deficiency
When attempting to determine an unknown structure any additional information can greatly simplify the process. It is possible to use a relatively simple mathematical formula and the molecular formula to determine how many rings and/or pi (π) bonds are in the structure. This is sometimes referred to as the molecule’s Degree of Unsaturation (DoU; Figure 1.53). Some sources may use the phrase Index of Hydrogen Deficiency (IHD) instead.
Figure 1.53 – Formula for Degree of Unsaturation.
The degree of unsaturation number indicates the total number of rings and/or π bonds there are in the structure, but does not differentiate between them. For example, if the DoU is 3 there could be: 3 rings and 0 π bonds, 2 rings and 1 π bond, 1 ring and 2 π bonds, or 0 rings and 3 π bonds.
Using the degree of unsaturation formula can make structure determination much faster and/or simpler. Most often this is done by using this number and any obvious functional groups to rule out other options. It is also helpful for double checking proposed structures (see Section 1.12).
The following example involves a much more complex structure than would be typical at an introductory level. It is included only to show the potential amount of information that can be gained from applying the Degree of Unsaturation formula.
Determining structures from spectral data is a logic puzzle. Imagine attempting to determine the structure of an unknown molecule using spectral data. The molecular formula is given as C18H29BrCl2FN3O3. There are MANY possible structures that match this formula. The degree of unsaturation is calculated as [2+(2*18)+3-(1+2+1)-29]/2 = [8]/2 = 4. This means the molecule has a total of 4 rings and/or π bonds. Checking the 1H NMR spectrum (Figure 1.54) shows a very complicated set of signals and patterns. The amount and complexity of information is daunting.
Figure 1.54 – 1H NMR Spectrum of an Unknown Compound with Molecular Formula C18H29BrCl2FN3O3.
Focusing on the less crowded regions might simplify things. One could look at the data and gather that there are probably several O-H and/or N-H bonds in the molecule (multiple broad peaks; Figure 1.55). However, this still leaves many candidate structures containing many possible functional groups. Examination shows that there are multiple non-broad peaks (i.e. not O-H or N-H) around 7 ppm. Consulting a table of standard chemical shifts (see Figure 1.24) indicates that these are likely for hydrogens attached to an aromatic ring. On its own this piece of information is only slightly helpful.
Figure 1.55 – Analysis of Some Signals in a 1H NMR Spectrum of an Unknown Compound with Molecular Formula C18H29BrCl2FN3O3.
We can combine this with the DoU value to know more. The structure of a typical aromatic ring (Figure 1.56) has one ring and three π bonds. 1+3 = 4 and the DoU is 4, so all unsaturations are accounted for.
Figure 1.56 – Generic Aromatic Ring Highlighting Unsaturations.
This immediately means any other possible structures with more rings are incorrect. Any other possible structures with functional groups with π bonds (alkenes, alkynes, any functional group with a carbonyl, etc.) are incorrect. The O-H/N-H signals can only come from a limited number of functional groups. Although the spectrum remains too challenging to fully analyze it would be much easier to select a corresponding structure from a set of potential options.
Other nuclei than protium are also analyzed with NMR spectroscopy. An isotope of carbon, 13C, is the second most common type of nucleus studied. So-called 13C NMR, spoken aloud simply as “carbon NMR”, is again very useful for structure determination. Fortunately, interpretation of 13C NMR spectra is typically simpler.
Just as with 1H NMR, the number of signals in a 13C NMR spectrum is equal to the number of chemically distinct carbon nuclei. As before, chemical equivalence by symmetry and/or rotation is possible (Figure 1.57).
Figure 1.57 – 13C NMR Spectra of Benzene and 2,2-Dimethylpropionaldehyde Highlighting Chemical Equivalence by Symmetry and Rotation.
Just as with 1H NMR, the chemical shift of signals in a 13C NMR spectrum is determined by the chemical environment (electron density) and can be used to determine functional groups (Figure 1.58). The ppm range for 13C NMR is significantly larger than 1H NMR. The chemical shift ranges for carbon nuclei attached to halogens is highly variable, with potential values descending well into the negative ppm region.
Figure 1.58 – Typical 13C NMR Chemical Shifts for Common Functional Groups.
Just as with 1H NMR, there are usually solvent signals in a 13C NMR spectrum that should be ignored. As before, this text omits these signals but they may be present in other sources.
Unlike with 1H NMR, the integration (area under the curve) in a 13C NMR spectrum is NOT equal to the number of carbon nuclei contributing to that signal. Integrations and signal intensities are not meaningful in carbon NMR. The reasons for this are complicated and involve the specifics of how the sample is pulsed to increase the amount and resolution of the signals.
Unlike with 1H NMR, there is no C-C spin coupling (splitting) observed and all signals are singlets in a 13C NMR spectrum. The reason for this is the low natural abundance (~1.1%) of the 13C isotope. Because of the low natural abundance very few (less than half a percent) molecules will have a 13C nucleus adjacent to another 13C nucleus. The net result is a coupling that is too weak to observe. Although this makes interpreting spectra easier, it also means that connectivity information is not included.
1.10.5.1. Extra Experiments – DEPT
There are three special types of 13C NMR experiments that are exceptionally useful for structural determination. Exactly how these experiments work is extremely complicated and typically only explored in advanced courses dedicated to NMR spectroscopy. However, the practical consequences are simple to understand and use.
All three experiments involve specific pulsing of both the carbon and hydrogen (protium) nuclei and then observing the results of a transfer of energy from one type of nucleus to another. As a result, they are referred to as “Distortionless Enhancement by Polarization Transfer” (DEPT) experiments. They are classed with a number in the name, referring to how the pulse sequence for that experiment works. Understanding the specifics is not important; do not worry about what the numbers mean, simply view each as a differently named experiment.
Because of how the nuclei are pulsed the phase (positive, zero, negative) of the 13C NMR signal in the resulting spectrum is determined by the number of hydrogen atoms directly attached to the carbon(s) of the signal (Table 1.4).
Table 1.4 – Phase Results in DEPT Spectra.
The most commonly used of the three is DEPT-135. This allows an educated guess at assignment of signals based on the phase. Below is an example of using only a 13C NMR spectrum and a DEPT-135 spectrum to generate a set of fragments of an unknown compound.
Consider a standard 13C NMR spectrum for an unknown compound with molecular formula C7H12O2 (Figure 1.59). There are two signals around 205 ppm. These could be aldehydes and/or ketones. There are four signals in the 20-50 ppm range. These could be CH3’s, CH2’s, CH’s, or carbons with no hydrogens attached.
Figure 1.59 – 13C NMR Spectrum of an Unknown Compound and Interpretation of Information.
The DEPT-135 spectrum is collected and added to the data (Figure 1.60).
Figure 1.60 – DEPT-135 Spectrum of an Unknown Compound.
It is possible to analyze the data using the chemical shift and phase of each signal (Figure 1.61).
Figure 1.61 – 13C NMR and DEPT-135 Spectra of an Unknown Compound with Interpretation of Information.
The phase of the downfield/deshielded/higher ppm signal around 205 ppm became zero. Because of the chemical shift and the phase, this must be a ketone.
The phase of the upfield/shielded/lower ppm signal around 205 ppm stayed positive. Because of the chemical shift and the phase, this must be an aldehyde.
The phase of the signal around 50 ppm became negative. Because of the chemical shift and the phase, this must be a CH2 group.
The phase of the signal around 45 ppm became zero. Because of the chemical shift and the phase, this must be a carbon with no hydrogen atoms attached to it. Because both oxygens are already accounted for, this cannot be a carbon attached to an oxygen. Therefore, this must be a carbon attached to four other carbons.
The phase of the signal around 30 ppm stayed positive. Because of the chemical shift and the phase, this could be a CH or a CH3.
The phase of the signal around 20 ppm stayed positive. Because of the chemical shift and the phase, this must be a CH3.
There are only six signals but seven carbon atoms in the formula. Therefore, one of the signals must represent a pair of chemically equivalent carbons.
Theoretically, it is possible to go further using the molecular formula because only a few atoms are not already accounted for. Considering all possible combinations of CH vs. CH3 and all possible options for which signal must be two equivalent carbons takes surprisingly little time. This would show that the fragment must be a CH3 rather than a CH and that it must be one of the signals for a CH3 that represents two chemically equivalent carbons. With all elements accounted for and a list of all fragments making up the molecule, it is actually possible to rule out options such that only four structures are candidates (Figure 1.62) using ONLY this 13C NMR and DEPT data.
Figure 1.62 – Candidate Molecular Structures of an Unknown Compound Determined Using Only 13C NMR and DEPT-135 Spectra .
This is feasible but requires a significant amount of rationalizing (and practice). This is shown only to highlight how much could be determined in principle. Normally only the simple analysis (Figure 1.61) would be expected. A deep analysis relying on incomplete data like this would not be expected at an introductory level. For contrast, using the integrations from the 1H NMR data would make figuring out which fragment is a pair of chemically groups trivial. It would also make generating possible structures much easier (see Section 1.12) and reduce the number of possible structures to only two.
