CHEM 2423. Chapter 13. Nuclear Magnetic Spectroscopy (Homework) W Short Answer 1. For a nucleus to exhibit the nuclear magnetic resonance phenomenon, it must be magnetic. Magnetic nuclei include: a. all nuclei with even numbers of neutrons and protons b. all nuclei with odd numbers of protons c. all nuclei with odd numbers of neutrons d. both b and c 2. Nuclear magnetic resonance spectroscopy provides information about a molecule's: a. conjugated pi electron system. b. size and formula. c. carbon-hydrogen framework. d. functional groups. 3. Explain why all protons in a molecule do not absorb rf energy at the same frequency. Exhibit 13-1 The following question(s) pertain to the charting of NMR spectra. MATCH a term to each description below. Place the letter of the term in the blank to the left of the description. a. TMS e. low-field or downfield side b. high-field or upfield side f. chemical shift c. MHz g. specific absorption d. delta ( ) 4. When looking at an NMR chart the right-hand part of the chart is the. 5. The exact place on the chart at which a nucleus absorbs is called its. 6. The calibration standard for 1 H and 13 C NMR is. 7. The NMR charts are calibrated using an arbitrary scale that is divided into units. Exhibit 13-2 For each of the compounds below tell how many signals you would expect the molecule to have in its normal, broadband decoupled 13 C NMR spectra. 8. 1 CHEM 2423. Chapter 13. Nuclear Magnetic Spectroscopy (Homework) W
9. 10. 11. 12. 13. 14. 15. Exhibit 13-3 Identify the indicated sets of protons as unrelated, homotopic, enantiotopic, or diastereotopic. 16. 2 CHEM 2423. Chapter 13. Nuclear Magnetic Spectroscopy (Homework) W
17. 18. 19. Exhibit 13-4 For each compound below tell how many types of nonequivalent protons there are. 20. 21. 22. 23. 24. 3 CHEM 2423. Chapter 13. Nuclear Magnetic Spectroscopy (Homework) W
25. 26. 27. Exhibit 13-5 Predict the splitting patterns you would expect for each proton in the molecules below. 28. 29. 30. 31. The 1 H NMR spectrum of styrene oxide shows that protons 1, 2, and 3 all have different chemical shift values. Proton 1 is coupled to both proton 2 (J = 5.8 Hz) and proton 3 (J = 2.5 Hz). Draw a tree diagram for the proton 1 signal. 4 CHEM 2423. Chapter 13. Nuclear Magnetic Spectroscopy (Homework) W
Exhibit 13-6 Refer to the structure of 3-methyl-2-butanone below to answer the following question(s). 32. Refer to Exhibit 13-6. What is the splitting pattern for the hydrogens in 3-methyl-2-butanone labeled a.? a. septet b. quartet c. doublet d. singlet 33. Refer to Exhibit 13-6. What is the splitting pattern for the hydrogens in 3-methyl-2-butanone labeled b.? a. septet b. quartet c. doublet d. singlet 34. Refer to Exhibit 13-6. The carbonyl-carbon resonance of 3-methyl-2-butanone occurs at 208.7 ppm downfield from TMS. How many hertz downfield from TMS would this carbonyl-carbon absorb if the spectrometer used to measure this absorption were operating at 200 MHz? 35. Treatment of tert-butyl alcohol with hydrogen chloride yields a mixture of tert-butyl chloride (S N 1 product) and 2-methylpropene (E1 product). After chromatographic separation, how would you use 1 H NMR to help you decide which was which? Exhibit 13-8 Below are three isomeric chlorobutanes and their 13 C NMR spectral data. MATCH the spectral data to the correct structures by placing the letter of the spectrum in the blank to the left of the corresponding structure. a =, 55.4, 36.2, 19.3 b =, 56.2, 35.3 c =, 56.0, 36.0, 24.2, 8.2 5 CHEM 2423. Chapter 13. Nuclear Magnetic Spectroscopy (Homework) W
36. 37. 38. Exhibit 13-8 Propose structures for compounds that fit the following 1 H NMR data. 39. C 8 H 9 Br 3H doublet at 2.0, J = 7 Hz 1H quartet at 5.0, J = 7 Hz 5H singlet at 7.3 40. C 7 H 14 O 6H triplet at 0.9, J = 7 Hz 4H sextet at 1.6, J = 7 Hz 4H triplet at 2.4, J = 7 Hz 41. C 3 H 6 Br 2 2H quintet at 2.4, J = 6 Hz 4H triplet at 3.5, J = 6 Hz 42. C 10 H 14 6H doublet at 1.2, J = 7 Hz 3H singlet at 2.3 1H septet at 2.9, J = 7 Hz 4H singlet at 7.0 43. C 6 H 14 12H doublet at 0.8 2H septet at 1.4 Exhibit 13-9 To answer the following question(s), consider the data and 1 H NMR spectrum below: The mass spectrum of this compound shows a molecular ion at m/z = 113, the IR spectrum has characteristic absorptions at 2270 and 1735 cm 1, and the 13 C NMR spectrum has five signals. 6 CHEM 2423. Chapter 13. Nuclear Magnetic Spectroscopy (Homework) W
Spectrum obtained from: SDBSWeb: http://www.aist.go.jp/riodb/sdbs/ 44. Refer to Exhibit 13-9. Based on the mass spectral data and the IR data, what functional groups are present in this compound? 45. Refer to Exhibit 13-9. How many types of nonequivalent protons are there in this molecule? 46. Refer to Exhibit 13-9. Describe the signal at 3.5 in terms of its integration, splitting pattern and chemical shift. 47. Refer to Exhibit 13-9. Describe the signals at 4.35 and 1.3 in terms of their integration, splitting and chemical shift. 48. Refer to Exhibit 13-9. What is the significance of the 13 C NMR data? 49. Refer to Exhibit 13-9. Propose a structure for this compound. 50. How would you use 1 H and 13 C NMR to help you distinguish between these two isomeric structures? 7 CHEM 2423. Chapter 13. Nuclear Magnetic Spectroscopy (Homework) W
Exhibit 13-10 Answer the question(s) for the compound whose 1 H NMR spectra is shown below. C 4 H 8 O Spectrum obtained from: SDBSWeb: http://www.aist.go.jp/riodb/sdbs/ 51. Refer to Exhibit 13-10. Describe each signal in terms of its integration, splitting and chemical shift. 52. Refer to Exhibit 13-10. Propose a structure for this compound. Exhibit 13-11 Answer the question(s) for the compound whose 1 H NMR spectra is shown below. C 5 H 12 O 8 CHEM 2423. Chapter 13. Nuclear Magnetic Spectroscopy (Homework) W
Spectrum obtained from: SDBSWeb: http://www.aist.go.jp/riodb/sdbs/ 53. Refer to Exhibit 13-11. Calculate the degree of unsaturation for this compound. 54. Refer to Exhibit 13-11. Describe each signal in the 1 H NMR in terms of its integration, splitting and chemical shift. 55. Refer to Exhibit 13-11. Propose a structure for this compound. Exhibit 13-12 Answer the question(s) for the compound whose 1 H NMR spectra is shown below. C 8 H 7 ClO 9 CHEM 2423. Chapter 13. Nuclear Magnetic Spectroscopy (Homework) W
Spectrum obtained from: SDBSWeb: http://www.aist.go.jp/riodb/sdbs/ 56. Refer to Exhibit 13-12. Calculate the degree of unsaturation in this compound. 57. Refer to Exhibit 13-12. Describe the signals that occur between 7 and 8 in terms of integration, splitting and chemical shift. 58. Refer to Exhibit 13-12. Describe the signal at 2.6 in terms of integration, splitting and chemical shift. 59. Refer to Exhibit 13-12. Propose a structure for this compound. 60. Propose a structure for a compound, C 6 H 14 O, with the following 13 C NMR spectral data: Broadband decoupled 13 C NMR: 23.0, 68.4 DEPT-90: 68.4 DEPT-135: positive peaks at 23.0, 68.4 ; no negative peaks 61. Propose a structure for Compound X, which has M + = 120 and (M + 2) = 122 of approximately equal intensity in its mass spectrum and has the following 13 C NMR spectral data: Broadband decoupled 13 C NMR: 32.6, 118.8, 134.3 DEPT-90: 134.3 DEPT-135: positive peaks at 134.3 ; negative peaks at 32.6, 118.8 10 CHEM 2423. Chapter 13. Nuclear Magnetic Spectroscopy (Homework) W
62. Propose a structure for Compound Z, which has the following spectroscopic properties: MS: M + = 88 IR: 3380 cm 1 1 H NMR: 0.85 (6H doublet); 1.40 (3H multiplet); 2.68 (1H singlet); 3.55 (2H triplet) Broadband decoupled 13 C NMR: 22.7, 25.0, 41.8, 60.5 DEPT-90: 25.0 DEPT-135: positive peaks at 22.7, 25.0 ; negative peaks at 41.8, 60.5 63. Describe how you could differentiate the following compounds using a DEPT NMR experiment. 64. Which structure of molecular formula C 4 H 8 Cl 2 fits both the 1 H NMR and 13 C NMR spectra shown below? a. b. c. d. 11 CHEM 2423. Chapter 13. Nuclear Magnetic Spectroscopy (Homework) W
Spectra obtained from: SDBSWeb: http://www.aist.go.jp/riodb/sdbs/ 12 CHEM 2423. Chapter 13. Nuclear Magnetic Spectroscopy (Homework) W
CHEM 2423. Chapter 13. Nuclear Magnetic Spectroscopy (Homework) W Answer Section SHORT ANSWER 1. ANS: d 2. ANS: c 3. ANS: All nuclei in molecules are surrounded by electron clouds. When a uniform external magnetic field is applied to a molecule, the circulating electron clouds set up tiny local magnetic fields of their own. These local magnetic fields act in opposition to the applied field, so that the effective field actually felt by a nucleus is a bit smaller than the applied field. B effective = B applied B local This effect is termed shielding. Each nucleus is shielded to a slightly different extent, so each unique kind of proton in a molecule resonates at a slightly different frequency and gives rise to a unique NMR signal. 4. ANS: b 5. ANS: f 6. ANS: a 7. ANS: d 8. ANS: 13 CHEM 2423. Chapter 13. Nuclear Magnetic Spectroscopy (Homework) W
three 9. ANS: two 10. ANS: five 11. ANS: six 12. ANS: five 13. ANS: 14 CHEM 2423. Chapter 13. Nuclear Magnetic Spectroscopy (Homework) W
three 14. ANS: five 15. ANS: three 16. ANS: homotopic 17. ANS: enantiotopic 18. ANS: diastereotopic 19. ANS: unrelated 20. ANS: 15 CHEM 2423. Chapter 13. Nuclear Magnetic Spectroscopy (Homework) W
four 21. ANS: one 22. ANS: four 23. ANS: four 24. ANS: 16 CHEM 2423. Chapter 13. Nuclear Magnetic Spectroscopy (Homework) W
four 25. ANS: two 26. ANS: five 27. ANS: four 28. ANS: No. of adjacent Proton Protons Splitting 1 1 doublet 2 6 septet 29. ANS: 17 CHEM 2423. Chapter 13. Nuclear Magnetic Spectroscopy (Homework) W
No. of adjacent Proton Protons Splitting 1 0 singlet 2 3 quartet 3 2 triplet 30. ANS: No. of adjacent Proton Protons Splitting 1 0 singlet 2 1 doublet 3 1 doublet 31. ANS: 32. ANS: d 33. ANS: c 34. ANS: 18 CHEM 2423. Chapter 13. Nuclear Magnetic Spectroscopy (Homework) W
35. ANS: 2-Methylpropene has two kinds of hydrogens. It will have a vinylic absorption (4.5 6.5 ) representing two hydrogens and an unsplit signal (1.0 1.5 ) due to the six equivalent methyl hydrogens. tert-butyl chloride has only one kind of hydrogen, which results in one unsplit signal. 36. ANS: b 37. ANS: c 38. ANS: a 39. ANS: 40. ANS: 41. ANS: BrCH 2 CH 2 CH 2 Br 42. ANS: 19 CHEM 2423. Chapter 13. Nuclear Magnetic Spectroscopy (Homework) W
43. ANS: 44. ANS: A molecular ion of mass 113 indicates an odd number of nitrogen atoms in the compound. Coupled with the IR absorption at 2270 cm 1, a nitrile functional group is indicated. The IR absorption at 1735 cm 1 indicates a carbonyl group, possibly an ester. 45. ANS: three 46. ANS: The integration of the signal at 3.5 indicates that it is due to two equivalent hydrogens, probably a CH 2 group. Since the signal is a singlet (not split) there are no nonequivalent hydrogens attached to the atoms adjacent to the carbon to which these two hydrogens are bonded. The chemical shift to 3.5 indicates that this CH 2 group has at least one electronegative atom or group bonded to it, possibly a carbonyl group. 47. ANS: The signal at 4.35 is owing to two equivalent hydrogens split by three adjacent hydrogens or a CH 2 next to a CH 3. It is shifted by attachment to an electronegative atom like oxygen. The signal at 1.3 is three equivalent hydrogens split by two adjacent hydrogens of a CH 3 next to a CH 2. It is shifted slightly downfield by the presence of an electronegative atom bonded to the adjacent CH 2. 48. ANS: The 13 C NMR has five signals, which means that there are five different kinds of carbon in this compound. 49. ANS: 20 CHEM 2423. Chapter 13. Nuclear Magnetic Spectroscopy (Homework) W
By compiling all the information deduced from the data provided, we note that this compound is a) a nitrile and an ester [two carbons accounted for], and b) has two CH 2 groups and one CH 3 group [three more carbons accounted for]. One of the CH 2 groups is attached to the oxygen of the ester and one is attached to the carbonyl of the ester. The singlet CH 2 is also bonded to the nitrile. Based on the chemical shifts, the singlet CH 2 must be bonded to the carbonyl and the quartet CH 2 must be bonded to the oxygen of the ester. So the compound is ethyl cyanoacetate: 50. ANS: # peaks Distinguishing Absorptions 1 H 5 methyl doublet at 1.5 (overlaps other methylene signals), no vinyl protons 13 C 6 one carbonyl carbon, no vinyl carbons 1 H 4 split vinylic peak; rel. area 2 methyl singlet between 3.5 4.0 13 C 4 one vinylic carbon, no carbonyl carbon 51. ANS: The signals at 1.05 and 2.45 are characteristic, in their integration and splitting (a 2H quartet and a 3H triplet), for an ethyl group. The shift of the CH 2 to 2.45 indicates that the ethyl group is attached to a carbonyl group. The signal at 2.1, a 3H singlet, is characteristic for a CH 3 attached to a carbonyl group. 52. ANS: 2-butanone, CH 3 COCH 2 CH 3 53. ANS: The base formula for this compound is C 5 H 12, which corresponds to a saturated compound, so there is no unsaturation in this compound. 54. ANS: 21 CHEM 2423. Chapter 13. Nuclear Magnetic Spectroscopy (Homework) W
In looking at the spectrum, it is important to note that the relative number of hydrogens for each signal adds up to four. The formula of the compound indicates that there are a total of 12 hydrogens. Therefore, the integration of 3:1 corresponds to 9H:3H. The signal at 1.2 is a 9H singlet, which is characteristic for a tert-butyl group. The slight downfield shifts indicates that this group is attached to an electronegative atom, in this case, oxygen. The signal at 3.2 is a 3H singlet which is characteristic of a methyl group bonded to an oxygen. 55. ANS: tert-butyl methyl ether, (CH 3 ) 3 COCH 3 56. ANS: The base formula for this compound is C 8 H 8 ; the saturated formula is C 8 H 18 ; the degree of unsaturation is five [(18 8) 2]. 57. ANS: The chemical shift of the signals between 7 and 8 indicates that these hydrogens are aromatic. The integration of four hydrogens and the splitting pattern of two doublets is characteristic of a para disubstituted aromatic compound. 58. ANS: This signal is a 3H singlet shifted by attachment to a carbonyl group; a methyl ketone, probably. 59. ANS: 4-chloroacetophenone 60. ANS: 61. ANS: 62. ANS: (CH 3 ) 2 CHCH 2 CH 2 OH 22 CHEM 2423. Chapter 13. Nuclear Magnetic Spectroscopy (Homework) W
63. ANS: Both compounds contain two different kinds of carbon, so an ordinary broadband decoupled 13 C NMR spectrum of either compound will show two peaks. The DEPT-90 spectrum, however, will show no peaks for 1,4-dichlorobutane while the DEPT-90 spectrum of 2,3-dichlorobutane will show one peak. The DEPT-135 spectrum for 1,4-dichlorobutane will show two negative peaks, while the DEPT-135 spectrum of 2,3-dichlorobutane will show two positive peaks. 64. ANS: b 23 CHEM 2423. Chapter 13. Nuclear Magnetic Spectroscopy (Homework) W