This preview shows page 1. Sign up to view the full content.
Unformatted text preview: Chemistry 30B Sample NMR Problems Key
1) Label the indicated pairs of atoms or groups as Homotopic, Enantiotopic or Diastereotopic. A O Me Me B
Enantioto pic O En anti otop ic
OH OH D OH OH HH E
Me OH HH En an tiotop ic
Me OH Me Diastereoto pic Di astereotopi c En an tiotop ic 2) Draw the 1H NMR and 13C NMR for the following molecules. Note 1: The following numerical values are calculated so that you can see what the real values should be. However, you do not need this level of accuracy. A O 1H NMR: 1.1 ppm (6H, d) 2.7 ppm (1H, septet) NMR: 15 ppm (Me) 28 ppm (CH2) 1.8 ppm (1H, s (or t due to 42 ppm (CH) coupling thru alkyne t o CH2)) 68 ppm (Alkyne CH) 3.2 ppm (2H, s (or d due to coupling thru alkyne t o CH)) 80 ppm (Alkyne CH) 210 ppm (C=O ) 13C 1H B O NH 2 13C N MR : 1 8 p pm (Me ) NMR: 1. 0 ppm (6H, d) 1.5 ppm (1H, mult iplet ) 29 ppm (CH) 2.0 ppm (2H, s, NH2) 2.8 ppm (2H, t , CH2N) 3.3 ppm (2H, d) 3.6 ppm (2H, t) 43 ppm (CH2N) 75 ppm (CH2O ) 79 ppm (CH2O ) O 1H C H Cl O NMR: 4.5 ppm (1H, dd) 4. 7 ppm (1H, dd) 6. 6 ppm (1H, dd) 6. 9 ppm (1H, d) 7. 5 ppm (1H, d) 7. 7 ppm (1H, s) 9. 9 ppm (1H, s) 13C NMR: 95 ppm (CH2) 118 ppm (CH) 123 ppm (CCl) 128 ppm (CH) 131 ppm (CH) 132 ppm (C) 149 ppm (OCH=) 163 ppm (CO) 190 ppm (C=O) May see long range coupling (4J) between the two meta aromatic protons. D
Cl Br O O 1H NMR: 3. 7 ppm (3H, s) 13C NMR: 49 ppm (CH2Cl) 50 ppm (CH3O) 51 ppm (CHBr) 172 ppm (CO) 4. 0 ppm (1H, dd (large geminal coupling)) 4. 1 ppm (1H, dd(large geminal coupling)) 4.6 ppm (1H, dd, CHBr) 3) Draw the 1H NMR specta for concentrated and dilute solutions of isopropanol. Concentrated solution: 1 ppm (6H, d), 2 ppm (1H, broad), 3.5 ppm (1H, heptet) Dilute solution: 1 ppm (6H, d), 2 ppm (1H, d), 3.5 ppm (1H, doublet of heptets) (Only if two coupling constants happened to be equal, then would see octet) Note 2: For clarity the answers are tabulated rather than drawn. d = doublet t = triplet dd = doublet of doublets
2 3 J (geminal coupling) is usually large - 15 Hz. 4 J is typically about 7 Hz. J is typically only about 0-2 Hz. So in problem D the geminal coupling is about 15 Hz. The geminal coupling on the terminal alkene in C is unusual at about 2 Hz. There may be long range coupling in A (Alkyne CH to CH2) and C (meta aromatic protons). The signal for the CH at 1.5 ppm in problem B is likely a multiplet since it would formally be a triplet of septets. Answers to Spectral Problems
Spectral Problem 1: The compound is 1-amino-2-bromobenzene.
NH2 Br The NMR signal at 4 ppm is the NH2. The coupling patterns of the four aromatic protons clearly prove the 1,2 substitution pattern. 7.4 ppm dd (regular and small long range.) 7.1 ppm ddd (regular, regular and small long range.) 6.8 ppm dd (regular and small long range.) 6.6 ppm ddd (regular, regular and small long range.) Spectral Problem 2: The compound is 1,2-dibromo-1-trimethylsilylethane. Me3Si-CHBr-CH2Br The NMR signal at 0.2 ppm is the SiMe3. The two protons of the CH2 unit are diastereotopic so an NMR signal for each and they couple to each other. 4.0 ppm dd (large and small.) (geminal + vicinal coupling) 3.7 ppm dd (large and large.) (geminal + vicinal coupling) 3.4 ppm dd (large and small.) (vicinal + vicinal coupling) The two vicinal couplings are so different due to the conformation of the molecule, but that is somewhat unusual. Spectral Problem 3:
Me Me Si O Me Hb
13 Ha O Hd He O Me Hc C NMR: 171.7 ppm C=O 65.7 ppm O-CH2 of ester 60.1 ppm O-CH 44.8 ppm CH2 alpha to ester 25.6 ppm Methyls of tert-butyl group 23.8 ppm Methyl of ester 17.8 ppm Methyl 14.0 ppm C of tert-butyl group -4.6 ppm Diastereotopic Me on Si -5.1 ppm Diastereotopic Me on Si 1 H NMR: 0.03 ppm s Diastereotopic Me on Si 0.06 ppm s Diastereotopic Me on Si 0.66 ppm s Methyls of tert-butyl group 1.20 ppm d Methyl on left end 1.25 ppm t Methyl of the ester 2.35 ppm dd (large geminal and regular vicinal coupling) CH2 unit diastereotopic proton 2.45 ppm dd (large geminal and regular vicinal coupling) CH2 unit diastereotopic proton 4.10 ppm two overlapping dq for CH2 unit diastereotopic protons of ethyl ester group 4.25 ppm ddq CH of stereogenic center. ...
View Full Document
This note was uploaded on 08/11/2010 for the course CHEM 30B Chem 30B U taught by Professor Miguelgarciagaribay during the Spring '10 term at UCLA.
- Spring '10