Other problems to work: 3-Chloropentane (diastereotopic H s), 1- chloropentane.

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Let s look at some specific examples. Dichloroacetaldehyde, l 2 HHO, has two inequivalent toms, H1 and H2. We expect to see two resonances, one at around δ 10.5 ppm and one around δ 5.5 ppm. (The H2 resonance is so far downfield because there are two l atoms attached to 2.) One tom contributes to one resonance, and one tom contributes to the other resonance, so we expect them to integrate to a 1:1 ratio. The 1 H resonating at 10 ppm, H1, has one magnetically active neighbor, H2. These atoms are separated by three bonds, so we expect to see a coupling constant. The resonance at δ 10 ppm is split into two peaks of equal intensity, a doublet. Likewise, H2 has one magnetically active neighbor, so its resonance is split into a doublet too. The spectrum consists of two doublets. There is one H1- H2 coupling constant, so the J values for one resonance, obtained by measuring the distance between the peaks in ppm and multiplying by the spectrometer frequency, is the same as the J value for the other resonance. hloroacetaldehyde, lh 2 HO, has two kinds of inequivalent toms, one H1 and two H2. We expect to see two resonances, one at around δ 10.0 ppm and one around δ 4.5 ppm. One tom contributes to the downfield resonance, and two toms contribute to the upfield resonance, so we expect them to integrate to a 1:2 ratio. This time, H1 has two equivalent magnetically active neighbors, so the downfield resonance is split into a triplet with 1:2:1 intensity. One the other hand, each H2 has one magnetically active and inequivalent neighbor, so the upfield resonance is split into a doublet of 1:1 intensity. Acetaldehyde, H 3 HO, has two kinds of inequivalent toms, one H1 and three H2. We expect to see two resonances, one at around δ 9.5 ppm and one around δ 2.2 ppm. One tom contributes to the downfield resonance, and three toms contribute to the upfield resonance, so we expect them to integrate to a 1:3 ratio. H1 has three equivalent magnetically active neighbors, so the downfield resonance is split into a quartet with 1:3:3:1 intensity. One the other hand, each H2 has one magnetically active and inequivalent neighbor, so the upfield resonance is split into a doublet of 1:1 intensity. Iodoethane, H 3 H 2 I, has two kinds of inequivalent toms, one H1 and two H2. We expect to see two resonances, one at around δ 3.2 ppm and one around δ 1.85 ppm. (The H 3 peak is a little more downfield of the normal saturated alkyl range because of the effect of the electronegative I atom.) Two toms contribute to the downfield resonance, and three toms contribute to the upfield resonance, so we expect them to integrate to a 2:3 ratio. Each H1 has three equivalent magnetically active and inequivalent neighbors, so the downfield resonance is split into a quartet with 1:3:3:1 intensity. One the other hand, each H2 has two magnetically active and inequivalent neighbors, so the upfield resonance is split into a triplet of 1:2:1 intensity. A downfield quartet and an upfield triplet with identical coupling constants and in a 2:3 ratio is diagnostic for the presence of an ethyl group attached to an electronegative atom. 14

Isopropanol, H 3 H(OH)H 3, has three kinds of inequivalent toms, six H1/H3, one H2, and one OH. We expect to see three resonances at 3.6, 1.2, and anywhere between 1 and 6, in 1:6:1 ratio. One tends not to see coupling between 1H s attached to heteroatoms and atoms, so we expect to see a doublet for the H1/H3 resonance (one magnetically active, inequivalent neighbor), a septet for the H2 resonance (six neighbors, not counting the OH), and a singlet (possibly broad) for the OH resonance. Other problems to work: 3-hloropentane (diastereotopic H s), 1- chloropentane. We need to discuss one more subject. Suppose we have a compound like trans-cinnamaldehyde, PhH=HHO. H1 gives us a doublet at 9.67 with J= 7 Hz. H3 gives us a doublet at 7.42 with J= 15 Hz. (It is shifted downfield from the normal vinyl region because it is next to the aryl ring. The coupling constant for trans toms on an alkene is always particularly large.) What kind of pattern can we expect to see for H2? It has two neighbors, but they are inequivalent. The first neighbor, H3, splits the resonance for H2 into a doublet, with J= 15 Hz. The second neighbor, H1, splits each of the doublet peaks into another doublet, with J= 7 Hz. The pattern that is seen is called a doublet of doublets. If the H1-H2 and H2-H3 coupling constants had been the same, we would have seen an apparent triplet. Often many signals overlap each other and clean spectra aren t observed. This is especially true when you have two sets of 1H s with similar chemical shifts and that are coupled to one another. For example, toluene. We expect to see a singlet for the H 3 group, and we do. In the aryl region we expect to see three resonances in a 1:2:2 integrated ratio for the three kinds of aryl H s, but what we see instead is either a singlet (low frequency spectrometer) or an uninterpretable mess (high frequency spectrometer) that integrates to 5 with respect to the Me resonance s 3. The reason is that the aryl H s all have similar chemical shifts, and they are all coupled to one another, so the splitting patterns are very complex (so-called first-order coupling). Higher frequency spectrometers tend to simplify coupling patterns, because the J values in ppm are made smaller with respect to the chemical shift differences between peaks. 15.7. Interpreting 13 NMR Spectra. If you stop to think about it, the fact that we can measure 13 NMR spectra at all is quite remarkable. Only 11 out of 1000 atoms is 13. In contrast, almost 100% of toms are 1 H. Thus, if we have 10 mmol of, say, Hl 3 in solution, we have 10 mmol of 1 toms but only 100 µmol of 13 atoms. When we do 13 NMR spectroscopy, we try to make our solutions as 15

concentrated as possible, and we also take many spectra and add them up to increase our signal to noise ratio. When we take a 13 NMR spectrum, we obtain a certain number of resonances corresponding to the number of inequivalent atoms in the compound. The chemical shifts of each of the resonances gives us information about the nature of each of the atoms (especially the hybridization) and the groups to which they are attached. Finally, spin-spin coupling gives us information as to the number of 1toms attached to each 13 atom. 13 nuclei resonate at a far wider range (220 ppm) of frequencies than do protons (ca. 10 ppm), but, like 1 H s, 13 nuclei with certain hybridizations tend to resonate in certain regions. (sp 3 ) atoms resonate in the region between 0 and 80 ppm, (sp) between 65 and 120 ppm, aryl and alkenyl (sp2) between 100 and 160 ppm, and =O between 165 and 220 ppm. There is much more overlap of ranges in 13 NMR than there is in 1 H NMR, and it s harder to predict exactly where particular resonances will be found in each range. In the alkyl range, (sp3) O tends to resonate further downfield than alkyl (sp3), and alkyl (sp3) tends to resonate further upfield when it is 1 and further downfield when it is 3 or 4. In the carbonyl region, ketones and aldehydes resonate further downfield than esters. Problem to work in class. Methyl acetate shows peaks at δ 170, 50, and 20 ppm. Identify the atom corresponding to each peak. Integration of 13 NMR spectra is rarely done. The area under 13 resonances is not proportional to the number of atoms contributing to that resonance unless special experimental techniques are used. However, it is true that 4 atoms tend to give signals of much weaker intensity than 1, 2, and 3 atoms. One more feature of 13 NMR spectra needs to be discussed. If we look at a compound like chloroform Hl 3, we notice that the atom is attached to a magnetic atom, the 1 H. This atom should affect the magnetic environment of the 13 nucleus, splitting it into a doublet. Actually, in normal 13 NMR operating mode, called spin-decoupled mode, we blast the 1 H s with radio waves at the same time that we measure the 13 resonances. (The radio waves that excite 1 H nuclei are of different energy from those that excite 13 nuclei, so the two pulses don t interfere with one another.) The purpose of blasting the 1 H s is to cause them to flip their spins so quickly that the 13 nuclei can t tell whether the 1 H s are aligned with or against the magnetic field, and so no splitting is observed. 16

It is possible, though, to run the 13 spectra in spin-coupled mode. Under these conditions, each atom experiences the magnetic field splitting caused by the magnetic nuclei directly attached to it. In practice, one only needs to worry about splitting from 1 H, 2 H, 19 F, and 3`P. (Since 13 has only 1.1% natural abundance, the probability of finding a 13 nucleus adjacent to another 13 nucleus is very small, so 13 13 splitting is not normally observed.) Just as we saw in 1 H spectra, a atom attached to n toms has a multiplicity of n+1, with intensities determined by the coefficients of the polynomial expansion. In 13 NMR spectra, coupling patterns are usually quite simple, because one needs only worry about 1 H s attached directly to the 13 atom. In other words, one-bond coupling only is seen. The resonances of 3 atoms are split into doublets, those of 2 atoms split into triplets, those of 1 atoms split into quartets, and those of 4 atoms are not split at all. H coupling constants in a particular compound are independent of the spectrometer frequency when they are expressed in Hz. H coupling constants are usually not measured or reported unless there is a particular reason to do so. To summarize: The spin-decoupled (normal) 13 NMR spectrum detects a single resonance for every non-equivalent atom in the compound. The chemical shift of the resonance is usually expressed in ppm, which is independent of spectrometer frequency. The chemical environment of the atoms in a compound can be deduced by examining the chemical shifts of the resonances. In the spin-coupled 13 NMR spectrum, each resonance may be split into two to four peaks, depending on the number of toms attached to the in question. The splitting distance, called the coupling constant or J, is usually expressed in Hz, which is independent of spectrometer frequency. I want you to be able to predict the 1 nd 13 NMR spectra of a given compound to the extent demonstrated above. I also want you to be able to deduce the structure of a compound if you are given a 1 H NMR spectrum and perhaps other data such as IR, MS, and 13 NMR spectra. 15.8. NMR Spectroscopy in Biology. Magnetic resonance imaging, or MRI, is a new medical diagnostic technique based on the principles of NMR spectroscopy. The only reason MRI isn t called NMR is that the public is afraid of the word "nuclear". In MRI, a person is placed inside a giant magnet, and an NMR spectrum of the H2O in his or her body is obtained. The basis of MRI is that different body tissues have different chemical environments, so the water in, say, a liver tumor has a different resonant frequency from the water in the rest of the liver. This allows the doctor to "look" into a person s body in a non-invasive manner and look for abnormal biological structures. NMR spectroscopy is unique among diagnostic techniques in that it provides information about the environment of specific atoms in a molecule in solution. This is important when trying to elucidate the interactions of proteins with one another and with the small molecules to which they bind. 17

Example Problems and Solutions (1) Identify the number of 1 nd 13 resonances each of these compounds will display. (a) cis-1,3-dimethylcyclohexane. There is a plane of symmetry containing 2 and 5 that relates the two H3 s to one another. The plane relates 1 to 3 and 4 to 6. There is no plane that relates the H s on top of the ring to those on the bottom of the ring. Five 13 and eight 1 H resonances will be seen. (b) Benzene. Benzene is a flat molecule. Its atoms are sp2-hybridized, so all the atoms in benzene are in a single plane. Benzene has a six-fold axis of symmetry through the center of the ring. Rotating benzene about this axis maps every and H onto the next and H. (The double bonds don t lower the symmetry in benzene, because two equally good resonance structures for benzene can be drawn where the double bonds are drawn in different places. As a result all the bonds in benzene are equivalent to one another. Often a "toilet bowl" resonance structure is drawn for benzene for this reason.) Benzene also has seven planes of symmetry. Every is equivalent to the next, and likewise for the H s. One 13 and one 1 H resonance will be seen. (c) Toluene (methylbenzene). Toluene no longer has the six-fold axis of symmetry of benzene, but it has a plane of symmetry containing the Me group and two s of the ring. Five 13 and four 1 H resonances will be seen. 18

(d) Diethyl ether (ethoxyethane). In its highest-symmetry conformation, diethyl ether has two planes of symmetry. (It also has a two-fold axis of symmetry through O in the plane of the paper.) One relates the left half of the compound to the right half; the other, containing the s and O, relate the H s on each to each other. Two 13 and two 1 H resonances will be seen. (e) 2-Methylpentane. This compound can be drawn in a conformation in which it has one plane of symmetry, the plane of the paper. This relates the two Me groups on the one to each other, and it relates the two H s in each H 2 to one another. Five 13 and five 1 H resonances will be seen. (f) 3-Ethylpentane. This compound can be thought of as three ethyl groups hanging off of one. It has a three-fold axis of symmetry containing the central and its H. It also has several planes of symmetry; each contains the central nd one of the Et groups, and it relates the two H s of the contained Et group to one another and the two uncontained Et groups to one another. Three 13 and three 1 H resonances will be seen. (2) Predict the approximate chemical shift, integration, and multiplicity of each resonance in the 1 H NMR spectrum of each of the compounds in 19

problem (1). Assume that each set of H s resonates at a different field strength, so no overlap between signals is observed. (a) cis-1,3-dimethylcyclohexane. In reality, many of these resonances will overlap with one another and give very complex splitting patterns. (b) Benzene. (c) Toluene (methylbenzene). (d) Diethyl ether (ethoxyethane). (e) 2-Methylpentane. Again, many of these resonances will overlap with one another. 20

(f) 3-Ethylpentane. (3) Predict the approximate chemical shift and multiplicity of each resonance in the 13 NMR spectrum of each of the compounds above. (a) cis-1,3-dimethylcyclohexane. (b) Benzene. (c) Toluene (methylbenzene). 21

(d) Diethyl ether (ethoxyethane). (e) 2-Methylpentane. (f) 3-Ethylpentane. More Problems For each problem, draw a structure or structures consistent with all of the information provided up to that point. In other words, your answer to (a) should be consistent with the information given in (a), your answer to (b) should be consistent with the information given in (a) and (b), your answer to (c) should be consistent with the information given in (a), (b), and (c), etc. Assume that the unknowns may consist of, H, N, O, l, or Br. For best results, try to think of all the different possibilities at each stage of the question before moving on to the next piece of information. Problem 1 (a) The compound has an M+ ion at 132 amu. There is a peak at 133 amu 22

whose intensity is 6.6% of the M+ peak and one at 134 amu whose intensity is 33% of the M+ peak. a) We know from the M+2 peak that this compound has a l atom in it. (If it were a Br atom, the M+2 peak would be the same size as the M+ peak.) We know from the M+1 peak that there are six atoms in this compound. That makes 6l so far. 132-6*12-35 = 25, so we have 25 amu left to consider. We can't possibly have a compound with the formula 6 H 25 l (negative degrees of unsaturation). However, we could have a compound with the formula 6 H 9 lo. Such a compound would have two degrees of unsaturation (of course, there are other possibilities). (b) The only IR absorbances in the interpretable region (1500-4000 cm-1) are at 2900 and 1715 cm-1. These absorbances are characteristic of (sp3)-nd either an unstrained ketone or an ester next to a double bond. We don't have any (sp2)-h bonds (no peak at 3000-3100 cm-1), according to the IR, nor do we have any double bonds next to carbonyls (no peak at 1600-1650 cm-1), so we are most likely going to have an unstrained ketone. A ketone uses up one degree of unsaturation, so we must have another double bond or a ring. onsidering the lack of (sp2)-h stretches, the latter is more likely. Either the ring has to be six-membered or the ketone carbon can't be part of the ring, because cyclopentanones and smaller cycloalkanones stretch at higher frequency. Possibilities include (there are others): 2-chlorocyclohexanone, 3- chlorocyclohexanone, and 4-chlorocyclohexanone (c) The 13 NMR spectrum shows six resonances at δ 200 (s), 55 (d), 30 (t), 26 (t), 24 (t), and 20 (t). The resonance at δ 200 confirms the ketone. We have five other resonances, showing that the compound isn't symmetrical. None of the resonances are quartets, showing there are no Me groups. Four of the resonances are H2 groups, while one is a H group. Also, the H group is shifted considerably far downfield, suggesting it is bound to l. With four H2 groups, one Hl group, and one =O group, the most likely possibilities are: 2-chlorocyclohexanone, or 3-chlorocyclohexanone. Note that 4- chlorocyclohexanone is not a possibility, because it would show only four resonances in the 13 NMR spectrum. It has a plane of symmetry that makes 2 equivalent to 6, and 3 equivalent to 4. (d) In the 1 H NMR spectrum, there is a resonance at δ 3.8 ppm that integrates to 1 nd whose multiplicity is dddd. The resonance at 3.8 ppm is due to the Hl group. With a multiplicity of dddd, it must have four different inequivalent neighbors that are two or three bonds away, i.e. on the same or on the adjacent 's. In 2- chlorocyclohexanone, the H in the Hl group has just two such neighbors, but in 3-chlorocyclohexanone it has four such neighbors. The compound is 3- chlorocyclohexanone. 23

(e) an this compound be optically active? Yes, it's chiral. The compound has no plane of symmetry, and it has one stereocenter, at 3. Problem 2 (a) The compound has an M+ ion at 158 amu. There is no M+2 peak. The compound can have at most 13 's and no l's or Br's. If it has 13 's, its formula is 13H2, and it has 13 degrees of unsaturation. Not likely. Subtract one, add 12 H's to get 12H14, with six degrees of unsaturation. This is more likely, especially if the compound has a phenyl ring, which uses up four degrees of unsaturation. Subtract one, add 12 H's to get 11H26, an unreasonable structure because it has too many H's, but from this subtract H4, add O to get 10H22O (no degrees of unsaturation), or subtract H4, add O again to get 9H18O2 (one degree of unsaturation). Or take 10H22O, subtract O, add N2 to get 9H22N2 (no degrees of unsaturation). (b) The intensity of the M+1 peak is 9.9% of the M+ peak. The compound has nine 's. This narrows the possibilities down to 9H18O2 and 9H22N2. (c) The only IR absorbances in the interpretable region are at 2850 and 1740 cm-1. The compound has no OH, no NH2, and no (sp2)-h. It has a carbonyl group of some sort. This rules out 9H22N2. The carbonyl group might be an ester, or it might be a cyclopentanone. However, a cyclopentanone would have to have at least two degrees of unsaturation, and 9H18O2 has only one. So we have a nine-carbon ester with no rings. Possibilities: (d) The spin-coupled 13 NMR spectrum shows seven resonances at δ 168 (s), 70 (t), 35 (d), 30 (d), 25 (t), 10 (q), and 8 (q) ppm. [How can this information be consistent with the information given in (b)?] There are nine atoms, but only seven of them show in the 13 NMR spectrum. This suggests either that there are two pairs of equivalent 's, or that three 's are equivalent. The resonance at 168 ppm confirms the ester. The triplet at 70 ppm suggests (=O)-O-H 2. In addition to this there are two kinds of Me groups, two H groups, and one more H 2 group. If there were three equivalent H 3 groups, they would all have to be attached to the same, but since no singlet is seen, it is much more likely that there are two pairs of H 3 groups. Each set of H 3 's would be attached to a H group, so we have two inequivalent i-pr groups. The only question is the location of the last H 2 group. There are two possible structures: 2-methylpropyl 3-24

methylbutanoate, and 3-methylbutyl 2-methylpropanoate. (e) The structure of the compound has been narrowed down to two possibilities. Now predict the 1 H NMR spectrum of the compound you drew for (d). Label each tom Ha, Hb, etc., giving equivalent H's identical letters. Then predict the approximate chemical shift, multiplicity, and integration of each resonance that you would see in the 1 H NMR spectrum. If you can see the two possibilities, how might you distinguish them by 1H NMR? H b H d H d O O H c H c H e 2-methylpropyl 3-methylbutanoate H c Hc H e O O H d H d H b 3-methylbutyl 2-methylpropanoate 25

The biggest distinguishing factor in these two spectra is the multiplicity of the resonance at d 4.1. In the first compound it is a doublet, while in the second it is a triplet. (f) an the compound be optically active? Nope! It's achiral. This is most easily seen by noting its lack of stereocenters. 26