CM Chemical Spectroscopy and Applications. Final Examination Solution Manual AY2013/2014
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1 NANYANG TECHNOLOGICAL UNIVERSITY DIVISION OF CHEMISTRY AND BIOLOGICAL CHEMISTRY SCHOOL OF PHYSICAL & MATHEMATICAL SCIENCES CM Chemical Spectroscopy and Applications Final Examination Solution Manual AY2013/2014 Prepared by EUGENE ONG November 2017
2 Question 1 (a) The assignment of each proton is shown in Figure 1. Figure 1. Assignment of olefinic protons of methyl acrylate. Coupling constants in the olefin: Trans protons: Large. Cis protons: Medium. Geminal protons: Small. (b) The answers are summarized in Table 1. Compound 2 Three types of protons (H a, H b, H c /H c' ). H c and H c' are magnetically equivalent. H a couples with H b and H c /H c'. dt expected. Compound 3 Two types of protons (H a /H a', H b /H b' ). H a and H a' are magnetically inequivalent. H b and H b' are magnetically inequivalent. Compound 4 Four types of protons (H a, H b, H c, H d ). H a couples with H b, H c, and H d. ddd expected. Table 1. Analysis of Compounds 2 to 4. (c) Resonance structures of Compound 6 are shown in Figure 2. Figure 2. Resonance structures of Compound C NMR of asterisked carbon: Higher electron density in 6 than in 5 is expected from the resonance structure A. Thus lower chemical shift of 6 is expected. IR absorption of carbonyl group: Resonance structure C indicates that C=O bond of 6 has more single bond character than that of 5. Thus C=O stretching wavenumber will be smaller for 6. 1
3 UV absorption: NOT EXAMINABLE (d) Rule of 13: 97/13 = 7 + 6/13. Base formula = C7H13. Nitrogen rule. At least one N atom. C7H13 + N CH2 = C6H11N. Does not match with 1 H NMR integration. IR absorption at 1704 cm 1 and 13 C NMR peak at indicate the presence of C=O group(s) (acid, ester, amide). C6H11N + O CH4 = C5H7NO. Still does not match with 1 H NMR integration. Further modification of formula. C5H7NO + O CH4 = C4H3NO2. Match with 1 H NMR integration. U = 4. Cannot be a benzene derivative (only 4 C atoms) but contains multiple unsaturated bonds. Two types of carbons ( 13 C NMR). One is likely from symmetric olefin (XHC=CHX; 1 H NMR 6.73 ppm, 13 C NMR ppm), another from C=O (171.4 ppm). Because there are total 4 carbons, this should contain two equivalent C=O groups. Figure 3. Partial structures leading to Compound 7. Question 2 (a) U = 6. 1 H NMR: Tetrasubstituted benzene with three OCH3 groups. Two of the OCH3 groups should be chemically equivalent (symmetric structure expected) IR: C N bond (~2200 cm 1 ) indicated. Then candidates should be A and B. Figure 4. Partial structures leading to Compound 8 and regioisomer B. 1 H chemical shift is the final key. 1 H chemical shift of the aromatic region of 8 is ca. 6.1, and is much smaller than that of its regioisomer (6.88). Aromatic 1 H chemical shift of A should be smaller than that of B due to "triple" resonance effects of A (compared to "double" resonance effect of B). 2
4 (b) 1 H NMR: -CH2-CH2- unit (triplet-triplet in 2-4 ppm). Monosubstituted benzene (multiplets in 7-8 ppm). Carboxylic acid COOH (singlet at > 12 ppm). And one isolated proton (singlet at ~8.5 ppm). 13 C NMR: Two different C=O groups of acid, ester, or amide (~165, ~175 ppm). Four peaks in ppm are from monosubstituted benzene. Two peaks in ppm are from X-CH2-CH2- Y unit. IR: Two different C=O (~1700, ~1620 cm 1 ). The former should be acid (see 1 H NMR interpretation). This is consistent with broad COOH band (3300~2600). The latter is likely amide (because of small wavenumber). Sharp band at ~3300 indicates secondary amide (RNHC=O). In addition to the above partial structures, the information that "hydrolysis of 9 gives aromatic carboxylic acid" is the final clue to determine its structure. For example, structure like 9' can be excluded. Figure 5. Partial structures leading to Compound 9. (c) 1 H NMR: Two isolated CH3 groups (singlets in ~2-2.5 ppm), which should not be directly bonded to O. Para-disubstituted benzene with different substituents (two app. d signals in ~ ppm). One isolated proton, which may be from NH or OH (but not COOH) (singlet at ~8 ppm). 13 C NMR: Two different C=O groups of ester, acid, or amide (two peaks at ~170 ppm). But acid is impossible (see 1 H NMR and IR). Four peaks in ppm are from benzene ring. Two peaks in ppm are from two different CH3 groups. IR: Two different C=O (~1750, ~1690 cm 1 ). The former may be ester (definitely not acid because of 1 H NMR and absence of broad COO-H band). The latter is likely amide. Sharp band at ~3400 indicates secondary amide (RNHC=O). You are almost complete if you get the structure 10 or 10'. The key to distinguish these two is the C=O wavenumber of amide. 10' should have even smaller C=O wavenumber because of double resonance effects (phenyl and amino groups bonded to C=O). 3
5 Figure 6. Partial structures leading to Compound 10. (d) α-cleavage at the amide moiety of Compound 9 gives an acylium ion of m/z = 105, as shown in Figure 7. Figure 7. Mechanism for the fragmentation of Compound 9. Therefore, Compound 9 = Spectrum A α-cleavage at either the amide or ester moiety of Compound 10 gives an acylium ion of m/z = 43, as shown in Figure 8. Figure 8. Mechanism for the fragmentation of Compound 10. Therefore, Compound 10 = Spectrum C The remaining Spectrum B belongs to Compound 11. 4
6 Question 3 (a) The chemical shifts, coordination shifts and coupling constants of the given spectrum are determined, as shown below: 1. Chemical shifts: δp a (Cis-isomer) = [( )/2] = ppm δp a (Trans-isomer) = [( )/2] = ppm δp b (Trans-isomer) = [( )/2] = ppm δp b (Cis-isomer) = [( )/2] = ppm 2. Coupling constants: 2 JPP(Cis) = {[( ) + ( )]/2} x 161 MHz = Hz 2 JPP(Trans) = {[( ) + ( )]/2} x 161 MHz = Hz 3. Coordination shifts: Given: δligand 12 = ppm; δligand 13 = ppm. ΔP a (Cis-isomer) = (- 8.70) = ppm ΔP a (Trans-isomer) = (- 8.70) = ppm ΔP b (Trans-isomer) = (- 5.30) = ppm ΔP b (Cis-isomer) = (- 5.30) = ppm (b) Two isomers of Complex 11 are observed in solution and their structures are shown in Figure 9. Figure 9. Two isomers of Complex 11. (c) Assignment of signals: 1. Trans-isomer: δp a = ppm (doublet) δp b = ppm (doublet) 2. Cis-isomer: δp a = ppm (doublet) δp b = ppm (doublet) (d) There are two isomers in solution, the trans-isomer is the major isomer. From the given spectrum, the trans-isomer signals have larger peak area and intensity compared to the cisisomer signals. Integration of the peaks would give the isomeric ratio. 5
7 (e) Due to chelate effect, the bidentate Ligand 12 is not as labile as the monodentate triphenylphosphine ligand. Since Ligand 12 is not prone to dissociation, its P Pd bond is much more stable than the triphenylphosphine P Pd bond. (f) The expected 31 P { 1 H} NMR signals of Complex 11 at an operating frequency of 100 MHz is calculated below and the resulting spectrum is shown in Figure 10. We know that 2 JPP(Cis) = Hz and 2 JPP(Trans) = Hz. At an operating frequency of 100 MHz, the distance between the doublet peaks are ppm for the cis-isomer and ppm for the trans-isomer. We know that the chemical shifts do not change, so we can determine the position of the new signals based on the information obtained from (a) and the calculated peak distances. δp a (Cis-isomer) = ppm δp a (Trans-isomer) = ppm δp b (Trans-isomer) = ppm δp b (Cis-isomer) = ppm Let s start from the downfield region first: (0.2576/2) = ppm (0.2576/2) = ppm (4.2504/2) = ppm (4.2504/2) = ppm (4.2504/2) = ppm (4.2504/2) = ppm (0.2576/2) = ppm (0.2576/2) = ppm Figure P { 1 H} NMR spectrum of Complex 11 at an operating frequency of 100 MHz. xxx End of Solution Manual xxx 6
8 Supporting Information Disclaimer: The following spectra are identical to the given spectra in the final examination. The spectra are available through the Spectral Databases for Organic Compounds: Compound 7 7
9 8
10 9
11 Regioisomer of Compound 8 10
12 Compound 8 11
13 12
14 13 m/z Intensity (%)
15 Compound 9 14
16 15
17 16 m/z Intensity (%)
18 Compound 10 17
19 18
20 19 m/z Intensity (%)
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