CHAPTER 3 SOLUTIONS TO EXERCISES

Transcription

1 CHAPTER 3 SOLUTIONS TO EXERCISES Solution MHz 1 H spectrum (CDCl3): (tt, J = 7.3, 1.1 Hz, 1H). The chemical shift is taken from the central line: Hz / MHz = ppm. The large triplet coupling (ortho coupling to Hm) can be measured in two places: = 7.34 Hz; = 7.35 Hz. The small triplet coupling (meta coupling to Ho) can be measured in six different places: = 1.08 Hz; = 1.08 Hz; = 1.10 Hz; etc. Solution 3.2. Solution

2 Solution

3 Solution 3.5. a. For the para isomer there are only two chemical shifts: the two protons (HA) ortho to OCH3 and meta to NO2; and the two protons (HB) meta to OCH3 and ortho to NO2. The chemical shifts can be predicted from the table in Chapter 3, Section 2.5: = 7.12 ppm for HA; = 8.17 ppm for HB. Since each one has only one neighbor, there will be two doublets (8.0 Hz), each with an integral value of 2H. 3-3

4 b. For the meta isomer there are four chemical shifts: (HA) ortho to both substituents; (HB) ortho to OCH3 and para to NO2; (HC) para to OCH3 and meta to NO2; and (HD) meta to OCH3 and ortho to NO2. The chemical shifts and splitting patterns can be predicted as follows: HA: = 7.79 ppm; singlet (no neighbors) HB: = 7.29 ppm; doublet (one neighbor) HC: = 7.50 ppm; triplet (two neighbors) HD: = 7.83 ppm; doublet (one neighbor) 3-4

5 c. For the ortho isomer there are also four chemical shifts: (HA) ortho to OCH3 and meta to NO2; (HB) meta to OCH3 and para to NO2; (HC) para to OCH3 and meta to NO2; and (HD) meta to OCH3 and ortho to NO2. The chemical shifts and splitting patterns can be predicted as follows: HA: = 7.12 ppm; doublet (one neighbor) HB: = 7.63 ppm; triplet (two neighbors) HC: = 7.16 ppm; triplet (two neighbors) HD: = 8.17 ppm; doublet (one neighbor) 3-5

6 Note how HA and HC are very close in chemical shift. This often happens with ortho disubstituted benzenes because the only difference between them is the difference between the ortho and para effects of the OCH3 substituent. Likewise, HB and HD differ only in the para vs. ortho effects of the NO2 substituent. This means that overlap can be common in ortho disubstituted benzenes, because for some substituents the ortho and para effects are similar. Solution 3.7. There is one very broad peak near 6.0 ppm, and four sharp aromatic proton peaks ( ppm). Three of the aromatic peaks are downfield-shifted from the generic region (7-8 ppm) for aromatic protons, so the aromatic ring is electron-poor. Analysis of the splitting patterns of the aromatic peaks gives the following (peaks are labeled a-f from right to left): c: (ddd, J = 7.9, 4.8, 0.9 Hz, 1H) d: (ddd, J = 7.9, 2.3, 1.7 Hz, 1H) e: (dd, J = 4.8, 1.7 Hz, 1H) 3-6

7 f: (dd, J = 2.3, 0.9 Hz, 1H) The 4.8 Hz coupling stands out because it is not within the range of any benzene ring coupling (Figure 3.28: ortho Hz, meta 1-2 Hz, para Hz). This, coupled with the Hf shift more than 1 ppm downfield of the 7-8 ppm range, leads us to suspect a pyridine ring (Figure 3.62, 2 = 8.524, Figure 3.63, J23 = 4.86 Hz). The molecular mass (122) is not an odd number, so a pyridine ring implies a second nitrogen (odd / even rule, Chapter 1, Section 2.2). With four aromatic protons, the pyridine ring must have a single substituent. Subtracting 92 mass units (C5H4N + N) from the molecular mass of 122 leaves 30, and subtracting the two protons in the broad peak near 6.0 leaves 28 mass units. This could be either two more nitrogens (14 +14) or one carbon and one oxygen ( ). The latter is more likely because a carbonyl group could be part of an amide (CONH2), which would have two separate chemical shifts (broad peaks near 6.0 ppm). The two most downfield peaks (He and Hf) are likely to be protons on C1 and C6 of the pyridine ring, and the small coupling of Hf (2.3, 0.9 Hz) implies that it has no vicinal coupling: the single substituent must be next to Hf. The downfield shift of Hf, relative to H2 of pyridine (8.524 ppm) supports the idea of a CONH2 (ortho effect +0.65, Table 3.1) next to Hf. The coupling constants allow us to construct the spin system of the aromatic ring: Note that in heteroaromatic rings the N-CH=CH coupling is always much lower than the typical ortho coupling ( Hz). Putting it together into a pyridine ring gives the structure: 3-7

8 Note that there is no meta coupling ( 4 J) between He and Hf, across the nitrogen, consistent with the very small literature J coupling for pyridine itself (-0.13 Hz, Figure 3.63). The broad doubletlike peak near 6.0 ppm is really two peaks for the distinct N-H protons in the amide. These are broadened by the slow rotation about the C-N bond, which averages the two chemical shifts in a chemical exchange process. The two environments (cis to oxygen and trans to oxygen) lead to two different chemical shifts, but with rotation each proton jumps into the other environment, exchanging the chemical shifts. The rotation process blurs the chemical shifts, making the peaks broad. We call this slow exchange because there are still two peaks in the 1 H spectrum. If the rotation were much faster, the two chemical shifts would be averaged and there would be only one peak (fast exchange). The molecular formula is C6H6N2O, with an unsaturation number of 5 (equivalent to C8H8), consistent with the aromatic ring (4 unsaturations) and the carbonyl group (1 unsaturation). Because there are an even number of nitrogens (2), the molecular mass is an even number. Solution The odd molecular mass (161) suggests the presence of nitrogen. Subtracting 14 for a single N, we have 147 mass units remaining. If the rest is C and H, division by 12 gives 12.25, corresponding to C12 (144) with three hydrogens. The 1 H spectrum has 7 hydrogens (total of integral areas), so there must be more than just C, H and one N. If there is one oxygen, subtracting NO (30) from 161 leaves 131: C10H11. Two oxygens (NO2) leaves 115: C9H7, matching the 1 H spectrum integral total. The molecular formula is C9H7NO2 (u = 6) and the calculated exact mass of the protonated molecular ion ([MH] + ) is (for C9H8NO2). This matches the experimental value of with an error of (1.2 ppm). The residual proton (d5-dmso) peak is seen at 2.50 ppm and used as a chemical shift reference. Often the d6-dmso solvent has water in it, or the sample is contaminated with water. The H2O peak appears around 3.3 ppm and is usually broad. The aromatic region (7-8 ppm) has five oneproton peaks, with general splitting patterns of (left to right): D, D, T, S, T. This cannot be part of a single benzene ring, since five protons leaves a single substituent in the ring, giving it an AA BB C pattern. Instead we have an ortho-disubstituted benzene pattern (D-T-T-D) and an isolated proton (S) with long-range couplings to the D-T-T-D spin system. The presence of nitrogen in the molecular formula and the singlet at 11.7 ppm suggest an indole ring system. A single substituent in the pyrrole portion (position 2 or 3) would leave an isolated proton (no 3 J couplings to aromatic protons) with long-range couplings to the benzene ring portion of indole. The S peak at ppm is a dd with J couplings of 2.19 and 0.93 Hz. The 2.19 Hz coupling could be to the NH proton, leaving a long-range coupling of 0.93 Hz to the benzene portion of the indole. Indole itself (Figure 3.68) has a 0.70 Hz coupling from H3 to H7, a 2.0 Hz coupling from H3 to the NH, no coupling between H2 and the benzene portion, so this peak must be H2 and the substituent must be at position 3. Subtracting the molecular formula of indole, we get: 3-8

9 C9 H7 N O2 - C8 H7 N = C O2 Inserting CO2 into the C2-H2 bond, we get indole-2-carboxylic acid. The assignment of peaks and detailed analysis of coupling constants helps to confirm this structure: Ha: (ddd, J = 8.00, 6.92, 1.00 Hz) Hb: (dd, J = 2.19, 0.93 Hz) Hb: (ddd, J = 8.19, 6.93, 1.16 Hz) Hd: (dq, J = 8.29, 0.94 Hz) He: (dq, J = 8.03, 0.97 Hz) The coupling constants are kept at the ±0.01 Hz accuracy until the analysis is complete. Note that peaks a and c (7.050 and ppm) are slightly mismatched T patterns; the mismatch means that the two pairs of lines in the center overlap perfectly in the center to give a 1:2:1 pattern. In peaks d and e (7.441 and ppm), measurements are made using the taller lines in the center of each quartet; this minimizes errors due to the poorly defined shoulder lines at the sides. The spin system in the benzene portion is easily assigned as Hd-Hc-Ha-He with the smaller Jac (6.9 Hz) in the center and the slightly different Jcd and Jae on the outside (8.2 and 8.0, respectively). Comparison with the indole J values (Figure 3.68) gives us a slight preference to put Hd in position 7 and He in position 4. The order of chemical shifts in indole (H5 upfield of H6, H7 upfield of H4) supports this arrangement (Ha upfield of Hc, Hd upfield of He). Peaks d and e look like double-quartets, but there is probably some mismatching of the three small couplings, and there is a kind of leveling effect in the barely resolved pattern. For Ha, only one of the three long-range couplings can be accurately measured (Jce = 1.16 Hz), and the other two are probably around 1.0 Hz to give the appearance of a 1.0-Hz quartet pattern. For Hd, two small 3-9

10 couplings (0.93 Hz and 1.00 Hz) can be measured accurately, and the third (Jde, the para coupling) is probably around 0.9 Hz. The better definition of the quartet pattern for Hd is probably due to a closer match of the three small couplings. The differences between the indole-2-carboxylic acid chemical shifts and the indole chemical shifts give us a measure of the effect of the CO2H group as a substituent. Position: : This can be compared to the values for CO2H as a substituent in the benzene ring (Table 3.1): ortho, +0.84; meta, +0.17; para, ppm. With CO2H in the indole position 2, the H3 shift is strongly affected, with smaller effects at H4 and H6 and the smallest at H5 and H7. Solution a. terephthalaldehyde, C8H6O2. There are only two peaks with an integration ratio of 1:2. Peak b is an aldehyde proton, and peak a is a downfield-shifted aromatic proton. With one aldehyde (CHO) and one aromatic ring (C6H2) we have a mass of 103 ( ), which is 31 mass units short. A second aldehyde and two more aromatic protons (CHO + CHO + C6H4) gives us the molecular mass of 134 ( ). In order to have a singlet for peak a, the substitution on the benzene has to be para. Any structure that includes olefins (3 C=C) instead of an aromatic ring would not give a singlet for peak a and would lack the extra unsaturation of the ring. c. 2-nitrophenol, C6H5NO3. Peaks a-d form a classic D T T D spin system, indicating an ortho-disubstituted benzene ring. Peak e could be an aldehyde proton. The odd molecular weight suggests the presence of one nitrogen atom. Two of the aromatic protons (peaks a and b) are upfield-shifted, ortho and para to an electron-donating substituent, and two more (peaks c and d) are downfield-shifted, ortho and para to an electron-withdrawing substituent. One possibility is an aldehyde (CHO) and an amine (NH2). But the mass (C6H4 + CHO + NH2 = 121) is 18 units short, and there is no NH2 peak in the 1 H spectrum. Reversing the roles of the two substituents, the nitrogen substituent could be NO2 (ewg) and the electron-donating substituent could be OH. The OH could form an intramolecular hydrogen bond with the oxygen of the nitro group, blocking exchange and giving a sharp peak (peak e) for the phenolic OH. From Table 3.1 we have predicted shifts of 7.08, 7.61, 7.17 and 8.15 for peaks b, c a and d, respectively ( D T T D ). The intramolecular hydrogen bond holds the OH proton in a position to have a zigzag coupling to Hc (0.45 Hz). 3-10

11 e. 3-bromochlorobenzene, C6H4ClBr. The isotope ratios, with large M+2 and M+4 isotope peaks, requires two halogens (Br2, BrCl or Cl2) in the molecule (Chapter 1, Section 2.3). The ratio for Br2 is 1:2:1, and for Cl2 it is 9:3:1. For one Br and one Cl, using the approximate ratios of 1:1 for Br and 3:1 for Cl: M: 79 Br 35 Cl (1 x 3) M+2: 79 Br 37 Cl (1 x 1) + 81 Br 35 Cl (1 x 3) M+4: 81 Br 37 Cl (1 x 1) For a ratio of 3:4:1 for M : M+2 : M+4. Subtracting 79 and 35 from the M + ion mass, we have a remainder of = 76, or C6H4. Peaks a-d form a S and D T - D spin system, indicating a meta di-substituted benzene ring. Peak b is assigned ortho to Cl and para to Br because it is upfield of peak c (ortho to Br and para to Cl), based on predicted shifts of 7.16 and 7.27 ppm, respectively. g. ibuprofen, C13H18O2. Peak a (d, 6H) corresponds to the two equivalent methyls of an isopropyl group, with peak c being the CH proton split by the six methyl protons. But the weak outer lines of peak c means that Hc is next to a CH2 group, giving it a 9-line pattern (1:8:28:56:70:56:28:8:1 ratio). This additional CH2 group is probably peak d, a doublet. The entire spin system is: CH2- CH(CH3)2, an isobutyl group. The chemical shift of peak d is consistent with attachment to a benzene ring, truncating the spin system. The remaining spin system, peaks b and e, corresponds to the fragment CH3-CH. Peak e is in the correct region to be singly-oxygenated. Peaks f and g 3-11

12 indicate a para-disubstituted benzene ring (AA BB spin system) with mildly electron-donating substituents, such as alkyl groups. Adding up the masses of these fragments, we have C6H4 (76), CH2-CH(CH3)2 (57) and CH3-CH (28) for a total of 161, 45 mass units short of the molecular mass of 206. The broad peak at 11 ppm (peak h) suggests a carboxylic acid or phenol OH, and CO2H corresponds to 45 mass units. The CH3-CH fragment has two unused valences, so it could be attached to CO2H and to the benzene ring. These two influences (carbonyl and aromatic ring) combine to bring peak e downfield into the singly-oxygenated region. The AA BB system was analyzed by first locating the centers of the Hf/Hf triplets (split by Hd) and the Hg/Hg doublets (split by He) and then applying the spreadsheet analysis for a para disubstituted benzene. The values of Jff and Jgg shown are the average of the two J values. i. 2-bromotoluene, C7H7Br. The M+2 ion peak with intensity equal to the M + ion means that there is one bromine atom (Chapter 1, Section 2.3). Peak a is a methyl group attached to an alkene, aromatic ring or carbonyl group; it has complex long-range couplings. The left side of the T pattern of peak c (7.19 ppm) is overlapped with peak d (7.23 ppm). Peaks b-e form a classic D T T D spin system, indicating an ortho-disubstituted benzene ring. Adding up the fragments (CH3 + Br + C6H4) gives for a total of 170. J coupling values with asterisk are couplings to CH3. Peak b is ddq, peak c is a dtq, peak d is a dd quintet, and peak e is a dd quintet. Peak e can be assigned by the ortho effect of Br (+0.16), and the remaining assignments can be made by matching J values. k. 3-bromobenzaldehyde, C7H5BrO. The M+2 ion with intensity equal to the M+ ion indicates a single bromine atom. Peaks a-d have a classic S and D T D spin system, corresponding to a meta-disubstituted benzene ring. Peak e is probably an aldehyde proton, confirmed by adding up the fragments: CHO + Br + C6H4 = = 184. Peak c is assigned ortho to CHO using Table 3.1 (pred. 7.80, para to CHO pred. 7.77). 3-12

13 m. ethyl 4-aminobenzoate, C9H11NO2. The odd molecular mass means that there is probably one nitrogen in the molecule. Peaks a and c give us the fragment CH3CH2, with the CH2 attached to oxygen. The downfield shift of peak c (4.31 ppm vs. 3-4 ppm generic) indicates that the oxygen is part of an ester, or attached to an aromatic ring. The broad peak b corresponds to two protons in OH or NH groups. Peaks d and e indicate a para-disubstituted benzene ring with one electrondonating substituent and one electron-withdrawing substituent. Adding the fragments C6H4 (76), CH3CH2O (45) and NH2 (16) gives 137 mass units, leaving 28 unaccounted for. This could be a carbonyl group (CO = 28) between the aromatic ring and the CH3CH2O fragment. This makes the two substituents electron-donating (NH2) and electron-withdrawing (CO2CH2CH3). Aromatic J values were obtained by spreadsheet analysis of the AA BB system. J values with asterisk are an average of the two meta couplings. o. 3-aminophenol, C6H7NO. Peaks c-f form a classic S and D T D spin system, indicating a meta-disubstituted benzene ring with electron-donating substituents. The odd molecular mass indicates a single nitrogen in the molecule. If peak a corresponds to an NH2 group, subtraction of C6H4 (76) and NH2 (16) from the molecular mass leaves 17, an OH group. Peak b must be the phenolic OH peak. Peak d and e assignments can be reversed, since both are predicted to be 6.29 ppm. 3-13

14 q. 2-bromoanisole, C7H7OBr. The M+2 ion with equal intensity to the M + peak indicates a single bromine atom in the molecule. Peak a is a CH3O group with a long-range coupling (0.3 Hz). Peaks b-e form a classic D T T D spin system, indicating an ortho-disubstituted benzene ring with one electron-withdrawing substitutent (peaks d and e) and one electron-donating substituent (peaks b and c). Adding up the fragments, we have CH3O + Br + C6H4 ( / ) = 186/188. Peak c is a dd quintet (J = 8.3, 1.4, 0.3 Hz) with long-range coupling to CH3O and He, and peak e is a ddd (J = 7.8, 1.7, 0.2 Hz). s. 4-chlorotoluene, C7H7Cl. Peak a is a methyl group with long-range couplings (tt, J = 0.73, 0.35 Hz), attached to a carbonyl, alkene or aromatic ring. Peaks b and c indicate a paradisubstituted benzene with electron-donating substituents. The M+2 ion with intensity one-third of the M + ion indicates a single chlorine atom in the molecule. Some of the groups of lines in peaks b and c can be analyzed as quartets, split by the CH3 group on the aromatic ring. The fragments (CH3 + Cl + C6H4 = / = 126/128) account for all of the molecular mass. The AA BB system can be analyzed using peak c, averaging to find the center position of each quartet and plugging these line frequencies into the spreadsheet. J values shown with asterisks are averages of the two meta coupling values. 3-14

15 u. 3-nitrobenzyl bromide, C7H6NO2Br. The odd molecular weight suggests a single nitrogen atom, and the M+2 peak with equal intensity to the M + ion indicates a single bromine atom in the molecule. Peak a (2H) is a CH2 group with long-range couplings into the aromatic ring. Peaks c- e form a classic S and D T D spin system, indicating a meta-disubstituted benzene ring with electron-withdrawing substituents. Peak e (8.266 ppm) is especially far downfield, suggesting an NO2 substituent. Adding up these fragments (CH2 + Br + NO2 + C6H4 = / = 215/217) accounts for all of the molecular mass. The peak a chemical shift (4.542 ppm) would suggest a singly-oxygenated carbon, but in fact the Br and the aromatic ring combine to give this downfield shift. Peak a is a quintet due to 4 similar couplings to the aromatic protons: Hc (0.44 Hz), He (0.46), Hb (0.40), and Hd (0.30). The para coupling combines with the coupling to the CH2 group to give quartet fine structure for peaks b (ddq) and e (ddq). Peak c is a dddt (J = 7.7, 1.8, 1.1, 0.4 Hz). 3-15

16 w. 2-n-dodecylphenol, C18H30O. Peaks g-j form form a classic D T T D spin system, indicating an ortho-disubstituted benzene ring with two electron-donating substitutents. Peak a is a distorted triplet at the generic position (0.85 ppm), indicating attachment to a long, straight saturated hydrocarbon chain. The 18H peak at ppm can be divided into a symmetrical broad singlet at ppm (10H, peak b) and a complex multiplet ( ppm, 8H, peak c). Virtual coupling effects distort the peak a triplet and the quintet-like peak d. Peaks d and e form an AA BB system (see Figure 6.56 in Chapter 6) due to a preferred conformation with the aromatic ring and the remainder of the aliphatic chain (peaks a-c) in an anti relationship. Together peaks a-e form a straight-chain CH3-(CH2)11- group, attached to a carbonyl, alkene or aromatic ring. Peak f could be a phenolic OH peak. Together the fragments add up (C6H4 + C12H25 + OH = = 262) to account for all of the molecular mass. Peak g (6.752 ppm) is assigned ortho to the OH group. All of the aromatic proton peaks are broadened, possibly by unresolved coupling to the peak e CH2 group. y. 3-nitrobenzaldehyde, C7H5NO3. The odd molecular mass indicates a single nitrogen in the molecular formula. Peaks a-d form a classic S and D T D spin system, indicating a meta-disubstituted benzene ring with electron-withdrawing substituents. Peak e could be either an aldhyde or a phenolic OH peak. Because we are looking for electron-withdrawing substituents, CHO and NO2 are preferable to OH and NH2 to explain peak e and the odd molecular mass, respectively. The molecular mass is accounted for using CHO, C6H4 and NO2 fragments ( = 151). Peak c (8.50 ppm) is assigned ortho to NO2, with peak b (7.245 ppm) ortho to the less strongly electron-withdrawing carbonyl group. The center part of peak a is the superposition of two triplets, offset by the mismatch of the two ortho couplings. Note the zigzag 5-bond coupling from the aldehyde proton to Ha. 3-16

17 Solution a. phenanthrene, C14H10. There are five aromatic protons, with overall splitting patterns D - T - T - D (an ortho-disubstituted benzene pattern) and S. The C6H4 fragment accounts for 76 mass units, less than half of the total mass. This suggests that there might be symmetry in the molecule. The aromatic system appears to be electron-poor if it is a benzene ring. There is either no nitrogen or an even number of nitrogens (odd-even rule). If the molecule contains only carbon and hydrogen, the formula would be C14H10 (u = 10), accounting for the 5 proton peaks if there is a mirror plane in the molecule (C13H22, u = 3, is not reasonable for an aromatic compound). If there were two oxygens (e.g., carbonyls) the formula could be C12H2O2, u = 12, or C11H14O2, u = 5, neither of which is consistent with the number of 1 H peaks. Returning to the formula C14H10, we could have two ortho-disubstituted benzene rings (C12H8) with a ring in the center adding C2H2. For 10 unsaturations, we have 4 each for the benzene rings, one for the additional ring, and one more for a double bond in the C2H2 fragment. The structure is phenanthrene (see below). The two central olefinic protons are equivalent by symmetry, so they give a singlet (2H). All of the integral values in the spectrum are actually doubled by symmetry. The downfield shift of the aromatic protons (7.56 to 8.70 ppm) relative to benzene is due to the larger ring current in the aromatic (14 electrons = 4n + 2) system, not to electron-withdrawing substituents. The most downfield peak (peak e) is located in the bay region of the polycyclic aromatic hydrocarbon (PAH), experiencing the through-space (anisotropic) effect of the opposite benzene ring as well as its own ring. c. 2-chlorotoluene, C7H7Cl. The proton peaks are very messy and overlapped. There is a methyl peak (singlet-like) at ppm, a triplet-like peak at ppm (leaning heavily towards the left, with the left side of the peak overlapped), another triplet-like peak at ppm (leaning heavily towards the right, with the right side of the peak overlapped) and two doublet-like peaks (

18 and ppm). Thus the pattern ( D T T D ) indicates an ortho-disubstituted benzene. The methyl group and benzene ring together (CH3 + C6H4) account for 91 mass units, leaving 35 / 37 (3:1 ratio), or Cl. Peak b is a triplet (7.4 Hz) of dq patterns (J = 2.1, 0.55 Hz). Peak c shows quartet coupling (0.3 Hz) as well, but strong coupling to peak b makes it non-first order. Peaks d and e are very messy, with virtual coupling effects due to the strong b/c coupling, and a para coupling between them. Long-range coupling values (*) are tentative due to these nonfirst order effects. e. acenaphthene, C12H10. There are only four proton peaks, in the integral ratio 2:1:1:1. The aromatic peaks can be classified as D T D, suggesting a 1,2,3-trisubstituted benzene ring. The mass of 154 can accommodate as many as 12 carbons (154 / 12 = ), so there must be symmetry in order to have such a simple spectrum. Two C6H3 units would account for 150 mass units, leaving no room for the two CH2 groups (peak a). A naphthalene unit with two substituents, C10H6, would leave room for 28 mass units, or CH2-CH2. These form a bridge between C1 and C8 of naphthalene and account for peak a. The two CH2 groups, a and a, are magnetically non-equivalent due to their different couplings to Hb, and unlike the CH=CH bridge in phenanthrene (Exercise 3.13a) the CH2-CH2 system is not isolated from the rest of the molecule. g. salicylic acid, C7H6O3. The spectrum shows a classic pattern for an ortho-disubstituted benzene ring, with one electron-donating substituent (with Ha and Hb ortho and para to it) and one electronwithdrawing substituent (with Hc and Hd ortho and para to it). The broad peak at could be the OH of a carboxylic acid or phenol. Subtracting C6H4 (76) from the molecular mass of 138 leaves 62, and taking away a further 17 (OH) leaves 45, which corresponds to CO2H. Matching the ortho J couplings gives the spin system order: b c a d. Only one OH is observed, probably the other is so broad as to be lost in the noise. Peak d has a minor component shifted slightly upfield; this could be due to major and minor conformations of the CO2H group. 3-18

19 i. 3-nitrotoluene, C7H7NO2. Peak a is a methyl group (singlet-like). Peak b ( T ) and peak c D ) are aromatic protons, and peaks d and e are the overlap of an aromatic D peak and an aromatic S peak (left side). Together, this indicates a meta-disubstituted benzene ring with one of the substituents being CH3. Subtracting C6H4-CH3 (91) from the molecular mass of 137 leaves 46. The odd mass number (137) leads us to suspect one nitrogen, and subtracting 14 from 46 leaves 32, or O2. The other substituent is NO2, accounting for the downfield position of peaks d and e. Coupling between the methyl protons and each of the aromatic protons makes the peaks complex, and the very close approach of peak d and peak e leads to strong coupling effects. Peak a (CH3) is a dq (J = 0.3, 0.7 Hz), with 0.7 Hz couplings to the ortho and para protons and a 0.3 Hz coupling to the meta proton. The meta couplings can be analyzed as follows: peak c has a quartet coupling (0.68 Hz) and from the sum of couplings in one-half of the peak ( = 4.93 Hz) we have: Jce + Jcd = *0.68 = 2.89 Hz. Peak d has a quartet coupling (0.59 Hz) and from the sum of couplings in the right half of the peak ( = 5.26 Hz) we have: Jcd + Jde = *0.59 = 3.49 Hz. Unfortunately this is two equations in three unknowns, so we cheat a little by comparing to Exercise 3.12u (3-nitrobenzyl bromide), which has a 2.3 Hz meta coupling across the NO2 substituent. Plugging this value in for Jde we have Jcd = 1.2 Hz and Jce = 1.7 Hz. Peak b has a dd fine structure visible on the right and left peaklets of the triplet-like peak, with J = 0.74, 0.35 Hz. The smaller coupling matches the doublet coupling in peak a and the larger must be the para coupling to peak e. k. 3-methoxyacetophenone, C9H10O2. Peak a is a CH3 ( S, attached to C=C or C=O) and peak b is a CH3O singlet. Peaks c, d, e and f are aromatic D, T, S and D peaks, respectively, indicating a meta-disubstituted benzene ring. Peak c is upfield-shifted, peak d is slightly downfield-shifted, and peaks e and f are downfield-shifted from the benzene (7.28 ppm) chemical shift. The pieces identified so far (C6H4 + CH3 + CH3O) add up to 122 mass units, leaving 28 unaccounted for. This can be a carbonyl group (C=O) between the CH3 and the aromatic ring, consistent with peak a s chemical shift. The other arrangement would be a methyl ester (CO2- CH3) and a methyl group directly attached to the ring, which is also consistent with the chemical 3-19

20 shifts of peak a and b. The important piece of evidence against a methyl group on the ring is the lack of long-range (quartet) couplings in any of the aromatic proton peaks. Peaks e and f are downfield-shifted by being ortho to the acetyl group, and peak c is upfield-shifted by its position ortho to the methoxy group. Peaks c-f are all ddd splittings, with the central line overlapped in peak d. m. 2-chloroanisole, C7H7OCl. Peak a is a CH3O group with a 0.32 Hz coupling to an ortho ring proton. Peaks b-e form a classic D T T D spin system, so we have an orthodisubstituted benzene ring. The isotope ratio (M : M+2 = 3:1) in the mass spectrum implies a single chlorine atom. Adding up the pieces (C6H4 + OCH3 + Cl) gives us /37 = 142/144. Peaks b and c are upfield-shifted by their para and ortho relationships to the methoxy group (-0.36 and substituent effects, respectively). The ortho relationship of OCH3 and Cl interferes with additivity, as shown by the slight downfield substituent effect on peak e (-0.05 in Table 3.1). Peak c is a dd quintet (J = 8.3, 1.4, 0.3 Hz). o. o-anisaldehyde, C8H8O2. Peak a is a CH3O group with a 0.3 Hz coupling to an ortho ring proton. Peak b ( D ) is slightly overlapped with peak c ( T ). Peaks b-e form a classic D T T D spin system indicative of an ortho-disubstituted benzene ring. Peak f is an aldehyde proton with a 0.8 Hz coupling to a ring proton. Adding up the pieces (CH3O + C6H4 + CHO) gives = 136, accounting for all of the molecular mass. The central peaklet of the tripletlike peak consists of two triplets, slightly offset due to the mismatch of ortho couplings (7.36 and 7.70 Hz). The coupling from Hc to the aldehyde proton (0.80 Hz) and the meta coupling to Hb (0.97 Hz) combine to give an apparent triplet (0.87 Hz). Peak b is a dd quintet (J = 8.4, 1.0, 0.35 Hz). Peaks b and c are shifted upfield by the methoxy group, and d and e are shifted downfield 3-20

21 by the aldehyde carbonyl group. The relatively large (0.8 Hz) 5 J coupling from Hc to the aldehyde proton is due to the zigzag arrangement enforced by steric preference for the carbonyl oxygen to be away from the methoxy substituent. q/r. o-tolualdehyde, C8H8O. In spectrum q we have a CH3 group attached to a C=O or aromatic ring (peak a), a D T T D spin system (peaks b-e) and an aldehyde proton (peak f). All peaks show complex splitting patterns due to long-range couplings. Combining the pieces (C6H4 + CH3 + CHO) gives a mass of 120 ( ), accounting for all of the mass. The aldehyde proton is a quartet (0.34 Hz), split by the ortho-related CH3 group ( 5 J coupling). Peak c has double quintet fine structure (J = 1.28, 0.54 Hz) and peak d has dq fine structure (J = 1.55, 0.36 Hz). Homonuclear decoupling applied to peak a (spectrum r) removes the quartet splitting from peak d, but peak c retains a 0.4 Hz splitting ( zigzag coupling to Hf). Peak b has a triplet fine structure (J = 0.4 Hz para coupling to He and zigzag coupling to Hf) and peak e has a doublet fine structure (J = 0.4 Hz para coupling to Hb). t. benzil, C14H10O2. The three peaks in the spectrum form a classic AA BB C system for a monosubstituted aromatic ring. Peak c ( D, ortho, 7.97 ppm) is 0.69 ppm downfield of benzene, matching the C(O)CH3 substituent best in Table 3.1. Peak b ( T, para, 7.65 ppm) is 0.37 ppm downfield of benzene, also consistent with a ketone substituent (substitutent effect for C(O)CH3). The fragment C6H5-C=O (105 mass units) is exactly half of the molecular mass, so we can connect two of them together to get the correct structure. 3-21

22 v. benzophenone, C13H10O. The spectrum is almost identical to spectrum t, with all three peaks a bit upfield of the corresponding peaks in spectrum t. We can build the same fragment: C6H5- C=O (105 mass units). Subtracting from the molecular mass of 182 leaves 77, corresponding to C6H5. So there is only one carbonyl group in this molecule. The J values shown were obtained by simulation, adjusting the J values to match the simulated spectrum to the experimental data. x. 9-fluorenone, C13H8O. The four peaks in the aromatic region ( ppm) form a classic D T T D spin system, indicating an ortho-disubstituted benzene. The downfield shift of peak d ( D ) corresponds to an electron-withdrawing substitutent next to Hd, but the other substituent (next to Hc) is less electron-withdrawing. Subtracting C6H4 (76) from the molecular mass of 180 leaves 104 mass units. This much mass (more than half of the molecular weight) is unlikely to have no protons, so we have to think about symmetry. Subtracting two C6H4 units (2 x 76) from the molecular mass leaves 28, corresponding to a carbonyl group (C=O). The molecular formula would be C13H8O, with 10 unsaturations. The two aromatic rings and the carbonyl bond account for only 9, so there must be another ring. Attaching the two benzene rings to each other (biphenyl) leaves the two ortho substituent positions to bond to the C=O group, forming a fivemembered ring. Peak c shows splitting more complex than the expected ddd (J = 7.4, 1.4, 0.8 Hz), possibly due to an Hc to Hc coupling. 3-22