13.6 Interpreting Proton NMR Spectra The ability of an NMR spectrometer to separate signals that have similar chemi- cal shifts is termed its resolving power and is directly related to the magnetic field strength of the instrument. Two closely spaced signals at 60 MHz become well separated if a 300-MHz instrument is used. (Remember, though, that the chemical shift 8, cited in parts per million, is independent of the field strength. 13.6 INTERPRETING PROTON NMR SPECTRA Analyzing an NMR spectrum in terms of a unique molecular structure begins with the information contained in Table 13. 1. By knowing the chemical shifts characteristic of various proton environments, the presence of a particular structural unit in an unknown ompound may be inferred. An NMR spectrum also provides other useful information, including: l. The number of signals, which tells us how many different kinds of protons there 2. The intensity of the signals as measured by the area under each peak, which tells us the relative ratios of the different kinds 3. The multiplicity, or splitting of each signal, which tells us how many protons are vicinal to the one giving the signal Protons that have different chemical shifts are said to be chemical- shift-non equivalent (or chemically nonequivalent). A separate NMR signal is given for each hemical-shift ivalent proton in a substance. Figure 13.9 shows the 200-MHZ H NMR spectrum of methoxyacetonitrile( CH3OCH2CN), a molecule with protons in two different environments. The three protons in the CH3o group constitute one set, the twe 3H N=CCH,OCH3 CH CH 10.0 6.0 5.0 Chemical shift(8, ppm) FIGURE 13.9 The 200-MHZ H NMR spectrum of methoxyacetonitrile(CH3OCH, CN) Back Forward Main MenuToc Study Guide ToC Student o MHHE Website
The ability of an NMR spectrometer to separate signals that have similar chemical shifts is termed its resolving power and is directly related to the magnetic field strength of the instrument. Two closely spaced signals at 60 MHz become well separated if a 300-MHz instrument is used. (Remember, though, that the chemical shift , cited in parts per million, is independent of the field strength.) 13.6 INTERPRETING PROTON NMR SPECTRA Analyzing an NMR spectrum in terms of a unique molecular structure begins with the information contained in Table 13.1. By knowing the chemical shifts characteristic of various proton environments, the presence of a particular structural unit in an unknown compound may be inferred. An NMR spectrum also provides other useful information, including: 1. The number of signals, which tells us how many different kinds of protons there are. 2. The intensity of the signals as measured by the area under each peak, which tells us the relative ratios of the different kinds of protons. 3. The multiplicity, or splitting, of each signal, which tells us how many protons are vicinal to the one giving the signal. Protons that have different chemical shifts are said to be chemical-shift-nonequivalent (or chemically nonequivalent). A separate NMR signal is given for each chemical-shift-nonequivalent proton in a substance. Figure 13.9 shows the 200-MHz 1 H NMR spectrum of methoxyacetonitrile (CH3OCH2CN), a molecule with protons in two different environments. The three protons in the CH3O group constitute one set, the two 13.6 Interpreting Proton NMR Spectra 497 4.0 3.0 2.0 1.0 0.0 Chemical shift (δ, ppm) 10.0 5.0 9.0 8.0 7.0 6.0 3H 2H CH3 CH2 NPCCH2OCH3 FIGURE 13.9 The 200-MHz 1 H NMR spectrum of methoxyacetonitrile (CH3OCH2CN). Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
CHAPTER THIRTEEN Spectroscopy protons in the OCH,CN group the other. These two sets of protons give rise to the two peaks that we see in the NMR spectrum and can be assigned on the basis of their chem ical shifts. The protons in the OCH2CN group are connected to a carbon that bears two lectronegative substituents (O and C=N) and are less shielded than those of the CH3O group, which are attached to a carbon that bears only one electronegative atom(O). The signal for the protons in the OCH,CN group appears at 84.1 ppm; the signal corre- sponding to the CH3o protons is at 83.3 ppm. Another way to assign the peaks is by comparing their intensities. The three equiv alent protons of the CH3o group give rise to a more intense peak than the two equiva- lent protons of the OCH_CN group. This is clear by simply comparing the heights of the peaks in the spectrum. It is better, though, to compare peak areas by a process called integration. This is done electronically at the time the NMr spectrum is recorded, and the integrated areas are displayed on the computer screen or printed out. Peak areas are proportional to the number of equivalent protons responsible for that signal It is important to remember that integration of peak areas gives relative, not absolute, proton counts. Thus, a 3: 2 ratio of areas can, as in the case of CH3OCH,CN, correspond to a 3: 2 ratio of protons. But in some other compound a 3: 2 ratio of areas might correspond to a 6: 4 or 9: 6 ratio of protons PROBLEM 13.5 The 200-MHZ H NMR spectrum of 1, 4-dimethylbenzene looks exactly like that of CH3 OCH2 CN except the chemical shifts of the two peaks are 8 2.2 ppm and 87.0 ppm. Assign the peaks to the appropriate protons of 1,4- dimethylbenzene Protons are equivalent to one another and have the same chemical shift when they are in equivalent environments. Often it is an easy matter to decide, simply by inspec tion, when protons are equivalent or not. In more difficult cases, mentally replacing a proton in a molecule by a"test group"can help. We'll illustrate the procedure for a sim- ple case-the protons of propane. To see if they have the same chemical shift, replace one of the methyl protons at C-1 by chlorine, then do the same thing for a proton at C-3. Both replacements give the same molecule, 1-chloropropane. Therefore the methyl protons at C-1 are equivalent to those at C-3 CH CHCH CICHCH.CH CHCHCHCI Propane 1-Chloropropane 1-Chloropropane If the two structures produced by mental replacement of two different hydrogens in a molecule by a test group are the same, the hydrogens are chemically equivalent. Thus, the six methyl protons of propane are all chemically equivalent to one another and have he same chemical shift Replacement of either one of the methylene protons of propane generates 2-chloro- propane. Both methylene protons are equivalent. Neither of them is equivalent to any of the methyl protons The H NMR spectrum of propane contains two signals: one for the six equiva lent methyl protons, the other for the pair of equivalent methylene protons PROBLEM 13.6 How many signals would you expect to find in the H NMr spec trum of each of the following compounds? (a)1-Bromobutane (c) Butane (b)1-Butanol d)1, 4-Dibromobutane Back Forward Main MenuToc Study Guide ToC Student o MHHE Website
protons in the OCH2CN group the other. These two sets of protons give rise to the two peaks that we see in the NMR spectrum and can be assigned on the basis of their chemical shifts. The protons in the OCH2CN group are connected to a carbon that bears two electronegative substituents (O and CPN) and are less shielded than those of the CH3O group, which are attached to a carbon that bears only one electronegative atom (O). The signal for the protons in the OCH2CN group appears at 4.1 ppm; the signal corresponding to the CH3O protons is at 3.3 ppm. Another way to assign the peaks is by comparing their intensities. The three equivalent protons of the CH3O group give rise to a more intense peak than the two equivalent protons of the OCH2CN group. This is clear by simply comparing the heights of the peaks in the spectrum. It is better, though, to compare peak areas by a process called integration. This is done electronically at the time the NMR spectrum is recorded, and the integrated areas are displayed on the computer screen or printed out. Peak areas are proportional to the number of equivalent protons responsible for that signal. It is important to remember that integration of peak areas gives relative, not absolute, proton counts. Thus, a 3:2 ratio of areas can, as in the case of CH3OCH2CN, correspond to a 3:2 ratio of protons. But in some other compound a 3:2 ratio of areas might correspond to a 6:4 or 9:6 ratio of protons. PROBLEM 13.5 The 200-MHz 1 H NMR spectrum of 1,4-dimethylbenzene looks exactly like that of CH3OCH2CN except the chemical shifts of the two peaks are 2.2 ppm and 7.0 ppm. Assign the peaks to the appropriate protons of 1,4- dimethylbenzene. Protons are equivalent to one another and have the same chemical shift when they are in equivalent environments. Often it is an easy matter to decide, simply by inspection, when protons are equivalent or not. In more difficult cases, mentally replacing a proton in a molecule by a “test group” can help. We’ll illustrate the procedure for a simple case—the protons of propane. To see if they have the same chemical shift, replace one of the methyl protons at C-1 by chlorine, then do the same thing for a proton at C-3. Both replacements give the same molecule, 1-chloropropane. Therefore the methyl protons at C-1 are equivalent to those at C-3. If the two structures produced by mental replacement of two different hydrogens in a molecule by a test group are the same, the hydrogens are chemically equivalent. Thus, the six methyl protons of propane are all chemically equivalent to one another and have the same chemical shift. Replacement of either one of the methylene protons of propane generates 2-chloropropane. Both methylene protons are equivalent. Neither of them is equivalent to any of the methyl protons. The 1 H NMR spectrum of propane contains two signals: one for the six equivalent methyl protons, the other for the pair of equivalent methylene protons. PROBLEM 13.6 How many signals would you expect to find in the 1 H NMR spectrum of each of the following compounds? (a) 1-Bromobutane (c) Butane (b) 1-Butanol (d) 1,4-Dibromobutane CH3CH2CH3 Propane ClCH2CH2CH3 1-Chloropropane CH3CH2CH2Cl 1-Chloropropane 498 CHAPTER THIRTEEN Spectroscopy Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
13.6 Interpreting Proton NMR Spectra 2.2-Dibromobutane (g)1, 1, 4-Tribromobutane 23, 3-Tetrabromobutane (h )1,1, 1-Tribromobutane SAMPLE SOLUTION (a)To test for chemical-shift equivalence, replace the pro- tons at c-1, C-2, C-3, and C-4 of 1-bromobutane by some test group such as chlo- rine. Four constitutional isomers result. CH3 CH2CH2 CHBr CHa CH2 CHCH2 Br CH3 CHCH2 CH2 Br CICH2 CH2CH2 2 Br 1-Bromo-1 1-Bromo-2 1-Bromo-3- 1-Bromo-4- hlorobutane chlorobutane Thus, separate signals will be seen for the protons at C-1, C-2, C-3, and C-4. Bar ring any accidental overlap, we expect to find four signals in the NMR spectrum of 1-bromobutane Chemical-shift nonequivalence can occur when two environments are stereochem- ically different. The two vinyl protons of 2-bromopropene have different chemical shifts H3C H85.5 One of the vinyl protons is cis to bromine; the other trans. Replacing one of the vinyl protons by some test group, say, chlorine, gives the Z isomer of 2-bromo-1-chloro- propene; replacing the other gives the E stereoisomer. The E and Z forms of 2-bromo- I-chloropropene are stereoisomers that are not enantiomers; they are diastereomers. Pro- tons that yield diastereomers on being replaced by some test group are described as diastereotopic. The vinyl protons of 2-bromopropene are diastereotopic. Diastereotopic protons can have different chemical shifts. Because their environments are similar, how ever, this difference in chemical shift is usually small, and it sometimes happens that two diastereotopic protons accidentally have the same chemical shift. Recording the spec trum on a higher field NMR spectrometer is often helpful in resolving signals with sim- lar chemical shifts trum of each of the following compounds s ou expect to PROBLEM 13.7 How many signals would you expect to find in the H NMr spec- (a)Vinyl bromide (d) trans-1, 2-Dibromoethene D1,1-Dibromoethene (e)Allyl bromide (c) cis-1, 2-Dibromoethene (f)2-Methyl-2-butene SAMPLE SOLUTION (a) Each proton of vinyl bromide is unique and has a chem ical shift different from the other two. The least shielded proton is attached to the carbon that bears the bromine. The pair of protons at C-2 are diastereotopic with respect to each other; one is cis to bromine while the other is trans to bromine. there are three proton signals in the Nmr spectrum of vinyl bromide Their observed chemical shifts are as indicated 86.4 ppm H H 85.8 ppm Back Forward Main MenuToc Study Guide ToC Student o MHHE Website
(e) 2,2-Dibromobutane (g) 1,1,4-Tribromobutane (f) 2,2,3,3-Tetrabromobutane (h) 1,1,1-Tribromobutane SAMPLE SOLUTION (a) To test for chemical-shift equivalence, replace the protons at C-1, C-2, C-3, and C-4 of 1-bromobutane by some test group such as chlorine. Four constitutional isomers result: Thus, separate signals will be seen for the protons at C-1, C-2, C-3, and C-4. Barring any accidental overlap, we expect to find four signals in the NMR spectrum of 1-bromobutane. Chemical-shift nonequivalence can occur when two environments are stereochemically different. The two vinyl protons of 2-bromopropene have different chemical shifts. One of the vinyl protons is cis to bromine; the other trans. Replacing one of the vinyl protons by some test group, say, chlorine, gives the Z isomer of 2-bromo-1-chloropropene; replacing the other gives the E stereoisomer. The E and Z forms of 2-bromo- 1-chloropropene are stereoisomers that are not enantiomers; they are diastereomers. Protons that yield diastereomers on being replaced by some test group are described as diastereotopic. The vinyl protons of 2-bromopropene are diastereotopic. Diastereotopic protons can have different chemical shifts. Because their environments are similar, however, this difference in chemical shift is usually small, and it sometimes happens that two diastereotopic protons accidentally have the same chemical shift. Recording the spectrum on a higher field NMR spectrometer is often helpful in resolving signals with similar chemical shifts. PROBLEM 13.7 How many signals would you expect to find in the 1 H NMR spectrum of each of the following compounds? (a) Vinyl bromide (d) trans-1,2-Dibromoethene (b) 1,1-Dibromoethene (e) Allyl bromide (c) cis-1,2-Dibromoethene (f) 2-Methyl-2-butene SAMPLE SOLUTION (a) Each proton of vinyl bromide is unique and has a chemical shift different from the other two. The least shielded proton is attached to the carbon that bears the bromine. The pair of protons at C-2 are diastereotopic with respect to each other; one is cis to bromine while the other is trans to bromine. There are three proton signals in the NMR spectrum of vinyl bromide. Their observed chemical shifts are as indicated. 5.7 ppm 6.4 ppm 5.8 ppm C Br H H H C 5.3 ppm 5.5 ppm C Br H H3C H C 2-Bromopropene CH3CH2CH2CHBr W Cl 1-Bromo-1- chlorobutane CH3CH2CHCH2Br W Cl 1-Bromo-2- chlorobutane CH3CHCH2CH2Br W Cl 1-Bromo-3- chlorobutane ClCH2CH2CH2CH2Br 1-Bromo-4- chlorobutane 13.6 Interpreting Proton NMR Spectra 499 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
CHAPTER THIRTEEN Spectroscopy When enantiomers are generated by replacing first one proton and then another by a test group, the pair of protons are enantiotopic with respect to one another. The meth ylene protons at C-2 of 1-propanol, for example, are enantiotopic. CH CH CH,OH CHOH H CHOH Enantiotopic 1-Propanol (R)-2-Chloro-l-propanol (S)-2-Chloro-1-propanol have different chemica chemical shift, regardless field strength of the NMR spectro shifts in a chiral solvent. Be. cause the customary solvent At the beginning of this section we noted that an NMR spectrum provides struc CDCl3)used in NMR mea- tural information based on chemical shift, the number of peaks, their relative areas, and the multiplicity, or splitting, of the peaks. We have discussed the first three of these fea- phenomenon is not observed tures of H NMR spectroscopy. Lets now turn our attention to peak splitting to see what kind of information it offers 13.7 SPIN-SPIN SPLITTING IN NMR SPECTROSCOPY The H NMR spectrum of CH3OCH2CN(see Figure 13.9)discussed in the preceding section is relatively simple because both signals are singlets; that is, each one consists of a single peak. It is quite common though to see a signal for a particular proton appear not as a singlet, but as a collection of peaks. The signal may be split into two peaks(a doublet), three peaks(a triplet), four peaks(a quartet), or even more. Figure 13.10 nows the H NMR spectrum of 1, 1-dichloroethane(CH; CHCI), which is characterized by a doublet centered at 8 2 1 ppm for the methyl protons and a quartet at 8 5.9 ppm for the methine proton The number of peaks into which the signal for a particular proton is split is called its multiplicity. For simple cases the rule that allows us to predict splitting in H NMR More complicated splitting Multiplicity of signal for Ha = n+1 patterns conform to an ex- tension of the"n+1"rule where n is equal to the number of equivalent protons that are vicinal to Ha. Two pro- and will be discussed in se tons are vicinal to each other when they are bonded to adjacent atoms. Protons vicinal tion 13.11 to Ha are separated from Ha by three bonds. The three methyl protons of 1, 1 dichloroethane are vicinal to the methine proton and split its signal into a quartet. The ingle methine proton, in turn, splits the methyl protons' signal into a doublet This proton splits the signal for the methyl protons into H c-CI These three protons split the signal for the me proton into a quartet. The physical basis for peak splitting in 1, I-dichloroethane can be explained with the aid of which examines how the chemical shift of the methyl protons is affected by the spin of the methine proton. There are two magnetic environments for the methyl protons: one in which the magnetic moment of the methine proton is paral lel to the applied field, and the other in which it is antiparallel to it. When the magnetic Back Forward Main MenuToc Study Guide ToC Student o MHHE Website
When enantiomers are generated by replacing first one proton and then another by a test group, the pair of protons are enantiotopic with respect to one another. The methylene protons at C-2 of 1-propanol, for example, are enantiotopic. Replacing one of these protons by chlorine as a test group gives (R)-2-chloro-1-propanol; replacing the other gives (S)-2-chloro-1-propanol. Enantiotopic protons have the same chemical shift, regardless of the field strength of the NMR spectrometer. At the beginning of this section we noted that an NMR spectrum provides structural information based on chemical shift, the number of peaks, their relative areas, and the multiplicity, or splitting, of the peaks. We have discussed the first three of these features of 1 H NMR spectroscopy. Let’s now turn our attention to peak splitting to see what kind of information it offers. 13.7 SPIN–SPIN SPLITTING IN NMR SPECTROSCOPY The 1 H NMR spectrum of CH3OCH2CN (see Figure 13.9) discussed in the preceding section is relatively simple because both signals are singlets; that is, each one consists of a single peak. It is quite common though to see a signal for a particular proton appear not as a singlet, but as a collection of peaks. The signal may be split into two peaks (a doublet), three peaks (a triplet), four peaks (a quartet), or even more. Figure 13.10 shows the 1 H NMR spectrum of 1,1-dichloroethane (CH3CHCl2), which is characterized by a doublet centered at 2.1 ppm for the methyl protons and a quartet at 5.9 ppm for the methine proton. The number of peaks into which the signal for a particular proton is split is called its multiplicity. For simple cases the rule that allows us to predict splitting in 1 H NMR spectroscopy is Multiplicity of signal for Ha n 1 where n is equal to the number of equivalent protons that are vicinal to Ha. Two protons are vicinal to each other when they are bonded to adjacent atoms. Protons vicinal to Ha are separated from Ha by three bonds. The three methyl protons of 1,1- dichloroethane are vicinal to the methine proton and split its signal into a quartet. The single methine proton, in turn, splits the methyl protons’ signal into a doublet. The physical basis for peak splitting in 1,1-dichloroethane can be explained with the aid of Figure 13.11, which examines how the chemical shift of the methyl protons is affected by the spin of the methine proton. There are two magnetic environments for the methyl protons: one in which the magnetic moment of the methine proton is parallel to the applied field, and the other in which it is antiparallel to it. When the magnetic This proton splits the signal for the methyl protons into a doublet. These three protons split the signal for the methine proton into a quartet. Cl H CH3 C Cl Enantiotopic protons C CH3 CH2OH H H H C CH2OH C CH2OH 1-Propanol CH3 Cl H (R)-2-Chloro-1-propanol CH3 Cl (S)-2-Chloro-1-propanol 500 CHAPTER THIRTEEN Spectroscopy Enantiotopic protons can have different chemical shifts in a chiral solvent. Because the customary solvent (CDCl3) used in NMR measurements is achiral, this phenomenon is not observed in routine work. More complicated splitting patterns conform to an extension of the “n 1” rule and will be discussed in Section 13.11. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
3.7 Spin-Spin Splitting in NMR Spectroscopy H3 Cl2CHCH3 8.0 5.0 Chemical shift(8, ppm) FIGURE 13. 10 The 200-MHz H NMR spectrum of 1, 1-dichloroethane, showing the methine roton as a quartet and the methyl protons as a doublet. the peak multiplicities are seen more moment of the methine proton is parallel to the applied field, it reinforces it. This decreases the shielding of the methyl protons and causes their signal to appear at slightly lower field strength. Conversely, when the magnetic moment of the methine proton is antiparallel to the applied field, it opposes it and increases the shielding of the methyl protons. Instead of a single peak for the methyl protons, there are two of approximately qual intensity: one at slightly higher field than the"true"chemical shift, the other at slightly lower field Turning now to the methine proton, its signal is split by the methyl protons into a quartet. The same kind of analysis applies here and is outlined in Figure 13.12. The methine proton"sees"eight different combinations of nuclear spins for the methyl FIGURE 13. 11 The magnetic moments (blue arrows) of the two possible spin states of the methine proton affect the chemical shift of the H-C-Cl dichloroethane. When the magnetic moment is parallel the external field external field and a smalle geo is needed for resonance When it is antiparallel to the of methin reinforces t hine proton shields external field, it subtracts a weaker lo is needed for resonance. methyl protons from Co- Methyl signal appears a Methyl signal appears at higher field. methyl protons. Back Forward Main MenuToc Study Guide ToC Student o MHHE Website
moment of the methine proton is parallel to the applied field, it reinforces it. This decreases the shielding of the methyl protons and causes their signal to appear at slightly lower field strength. Conversely, when the magnetic moment of the methine proton is antiparallel to the applied field, it opposes it and increases the shielding of the methyl protons. Instead of a single peak for the methyl protons, there are two of approximately equal intensity: one at slightly higher field than the “true” chemical shift, the other at slightly lower field. Turning now to the methine proton, its signal is split by the methyl protons into a quartet. The same kind of analysis applies here and is outlined in Figure 13.12. The methine proton “sees” eight different combinations of nuclear spins for the methyl 13.7 Spin–Spin Splitting in NMR Spectroscopy 501 4.0 3.0 2.0 1.0 0.0 Chemical shift (δ, ppm) 8.0 7.0 6.0 6.4 6.0 5.6 2.4 2.0 9.0 5.0 CH3 Cl2CH Cl2CHCH3 FIGURE 13.10 The 200-MHz 1 H NMR spectrum of 1,1-dichloroethane, showing the methine proton as a quartet and the methyl protons as a doublet. The peak multiplicities are seen more clearly in the scale-expanded insets. Cl W W CH3 0 H±C±Cl Spin of methine proton reinforces 0; a weaker 0 is needed for resonance. Methyl signal appears at lower field. Spin of methine proton shields methyl protons from 0. Methyl signal appears at higher field. Cl W W CH3 H±C±Cl FIGURE 13.11 The magnetic moments (blue arrows) of the two possible spin states of the methine proton affect the chemical shift of the methyl protons in 1,1- dichloroethane. When the magnetic moment is parallel to the external field 0 (green arrow), it adds to the external field and a smaller 0 is needed for resonance. When it is antiparallel to the external field, it subtracts from it and shields the methyl protons. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website