1. © 2016 Cengage Learning. All Rights Reserved. John E. McMurry www.cengage.com/chemistry/mcmurry Chapter 13 Structure Determination: Nuclear Magnetic Resonance Spectroscopy

2. © 2016 Cengage Learning. All Rights Reserved. Learning Objectives (13.1)  Nuclear magnetic resonance spectroscopy (13.2)  The nature of NMR absorptions (13.3)  The chemical shift (13.4)  Chemical shifts in 1 H NMR spectroscopy

3. © 2016 Cengage Learning. All Rights Reserved. Learning Objectives (13.5)  Integration of 1 H NMR absorptions: Proton counting (13.6)  Spin–spin splitting in 1 H NMR spectra (13.7)  1 H NMR spectroscopy and proton equivalence (13.8)  More complex spin–spin splitting patterns

4. © 2016 Cengage Learning. All Rights Reserved. Learning Objectives (13.9)  Uses of 1 H NMR spectroscopy (13.10)  13 C NMR spectroscopy: Signal averaging and FT–NMR (13.11)  Characteristics of 13 C NMR spectroscopy (13.12)  DEPT 13 C NMR spectroscopy

5. © 2016 Cengage Learning. All Rights Reserved. Learning Objectives (13.13)  Uses of 13 C NMR spectroscopy

6. © 2016 Cengage Learning. All Rights Reserved. Nuclear Magnetic Resonance Spectroscopy  Nuclei are positively charged and interact with an external magnetic field denoted by B 0  Magnetic rotation of nuclei is random in the absence of a magnetic field  In the presence of a strong magnet, nuclei adopt specific orientations

7. © 2016 Cengage Learning. All Rights Reserved. Nuclear Magnetic Resonance Spectroscopy  When exposed to a certain frequency of electromagnetic radiation, oriented nuclei absorb energy and causes a spinflip from a state of lower energy to higher energy  Nuclear magnetic resonance - Nuclei are in resonance with applied radiation  Frequency that causes resonance depends on:  Strength of external magnetic field  Identity of the nucleus  Electronic environment of the nucleus

8. © 2016 Cengage Learning. All Rights Reserved. Nuclear Magnetic Resonance Spectroscopy  Larmor equation  Relation between  Resonance frequency of a nucleus  Magnetic field and the magnetogyric ratio of the nucleus

9. © 2016 Cengage Learning. All Rights Reserved. Worked Example  Calculate the amount of energy required to spin- flip a proton in a spectrometer operating at 300 MHz  Analyze if the increase of spectrometer frequency from 200 MHz to 300 MHz increases or decreases the amount of energy necessary for resonance

10. © 2016 Cengage Learning. All Rights Reserved. Worked Example  Solution:  Increasing the spectrometer frequency from 200 MHz to 300 MHz increases the amount of energy needed for resonance

11. © 2016 Cengage Learning. All Rights Reserved. The Nature of NMR Absorptions  Absorption frequencies differ across 1 H and 13 C molecules  Shielding: Opposing magnetic field produced by electrons surrounding nuclei to counteract the effects of an external magnetic field  Effect on the nucleus is lesser than the applied magnetic field  B effective = B applied – B local  Individual variances in the electronic environment of nuclei leads to different shielding intensities

12. © 2016 Cengage Learning. All Rights Reserved. Figure 13.3 - NMR Spectrum of 1 H and 13 C

13. © 2016 Cengage Learning. All Rights Reserved. Working of an NMR Spectrometer  Organic sample dissolved in a suitable solvent is placed in a thin glass tube between the poles of a magnet  1 H and 13 C nuclei respond to the magnetic field by aligning themselves to one of the two possible orientations followed by rf irradiation  Constant and varied strength of the applied field causes each nucleus to resonate at a slightly varied field strength  Absorption of rf energy is monitored by a sensitive detector that displays signals as a peak

14. © 2016 Cengage Learning. All Rights Reserved. Figure 13.4 - Operation of a Basic NMR Spectrometer

15. © 2016 Cengage Learning. All Rights Reserved. NMR Spectrometer  Time taken by IR spectroscopy is about 10 –13 s  Time taken by NMR spectroscopy is about 10 –3 s  Provides a blurring effect that is used in the measurement of rates and activation energies of vary fast processes

16. © 2016 Cengage Learning. All Rights Reserved. Worked Example  Explain why 2-chloropropene shows signals for three kinds of protons in its 1 H NMR spectrum  Solution:  2-Chloropropene has three kinds of protons  Protons b and c differ because one is cis to the chlorine and the other is trans

17. © 2016 Cengage Learning. All Rights Reserved. The Chemical Shift  The left segment of the chart is the downfield  Nuclei absorbing on the downfield have less shielding as they require a lower field for resistance  The right segment is the upfield  Nuclei absorbing on the upfield have more shielding as they require a higher field strength for resistance

18. © 2016 Cengage Learning. All Rights Reserved. Figure 13.5 - The NMR Chart

19. © 2016 Cengage Learning. All Rights Reserved. The Chemical Shift  Chemical shift is the position on the chart at which a nucleus absorbs  The delta (δ) scale is used in calibration of the NMR chart  1 δ = 1 part-per-million of the spectrometer operating frequency  The delta scale is used as the units of measurement can be used to compare values across other instruments

20. © 2016 Cengage Learning. All Rights Reserved. Worked Example  The 1 H NMR peak of CHCl 3 was recorded on a spectrometer operating at 200 MHz providing the value of 1454 Hz  Convert 1454 Hz into δ units  Solution:

21. © 2016 Cengage Learning. All Rights Reserved. Chemical Shifts in 1 H NMR Spectroscopy  Chemical shifts are due to the varied electromagnetic fields produced by electrons surrounding nuclei  Protons bonded to saturated, sp 3 -hybridized carbons absorb at higher fields  Protons bonded to sp 2 -hybridized carbons absorb at lower fields  Protons bonded to electronegative atoms absorb at lower fields

22. © 2016 Cengage Learning. All Rights Reserved. Table 13.2 - Regions of the 1 H NMR Spectrum

23. © 2016 Cengage Learning. All Rights Reserved. Worked Example  CH 2 Cl 2 has a single 1 H NMR peak  Determine the location of absorption  Solution:  For CH 2 Cl 2, δ = 5.30  The location of absorption are the protons adjacent to the two halogens

24. © 2016 Cengage Learning. All Rights Reserved. Integration of 1 H NMR Absorptions: Proton Counting  In the figure, the peak caused by (CH 3 ) 3 C– protons is larger than the peak caused by –OCH protons  Integration of the area under the peak can be used to quantify the different kinds of protons in a molecule

25. © 2016 Cengage Learning. All Rights Reserved. Worked Example  Mention the number of peaks in the 1H NMR spectrum of 1,4-dimethyl-benzene (para-xylene or p-xylene)  Mention the ratio of peak areas possible on integration of the spectrum

26. © 2016 Cengage Learning. All Rights Reserved. Worked Example  Solution:  There are two absorptions in the 1H NMR spectrum of p-xylene  The four ring protons absorb at 7.05 δ and the six methyl-groups absorb at 2.23 δ  The peak ratio of methyl protons:ring protons is 3:2

27. © 2016 Cengage Learning. All Rights Reserved. Worked Example

28. © 2016 Cengage Learning. All Rights Reserved. Spin-Spin Splitting in 1 H NMR Spectra  Multiplet: Absorption of a proton that splits into multiple peaks  The phenomenon is called spin-spin splitting  Caused by coupling of neighboring spins

29. © 2016 Cengage Learning. All Rights Reserved. Spin-Spin Splitting in 1 H NMR Spectra  Alignment of –CH 2 Br proton spins with the applied field can result in:  Slightly larger total effective field and slight reduction in the applied field to achieve resonance  There is no effect if one of the –CH 2 Br proton spins aligns with the applied field and the other aligns against it  Alignment of –CH 2 Br proton spins against the applied field results in:  Smaller effective field and an increased applied field to achieve resonance

30. © 2016 Cengage Learning. All Rights Reserved. Figure 13.8 - The Origin of Spin-Spin Splitting in Bromoethane

31. © 2016 Cengage Learning. All Rights Reserved. Spin-Spin Splitting in 1 H NMR Spectra  n + 1 rule: Protons that exhibit n + 1 peaks in the NMR spectrum possess  n = number of equivalent neighboring protons  Coupling constant is the distance between peaks in a multiplet

32. © 2016 Cengage Learning. All Rights Reserved. Spin-Spin Splitting in 1 H NMR Spectra  It is possible to identify multiplets in a complex NMR that are related  Multiplets that have the same coupling constant can be related  Multiplet-causing protons are situated adjacent to each other in the molecule

33. © 2016 Cengage Learning. All Rights Reserved. Rules of Spin-Spin Splitting  Chemically equivalent protons do not show spin- spin splitting  The signal of a proton with n equivalent neighboring protons is split into a multiplet of n + 1 peaks with a coupling constant  Two groups of photons coupled together have the same coupling constant, J

34. © 2016 Cengage Learning. All Rights Reserved. Worked Example  The integrated 1 H NMR spectrum of a compound of formula C 4 H 10 O is shown below  Propose a structure

35. © 2016 Cengage Learning. All Rights Reserved. Worked Example  Solution:  The molecular formula (C 4 H 10 O) indicates that the compound has no multiple bonds or rings  The 1 H NMR spectrum shows two signals, corresponding to two types of hydrogens in the ratio 1.50:1.00, or 3:2  Since the unknown contains 10 hydrogens, four protons are of one type and six are of the other type  The upfield signal at 1.22 δ is due to saturated primary protons

36. © 2016 Cengage Learning. All Rights Reserved. Worked Example  The downfield signal at 3.49 δ is due to protons on carbon adjacent to an electronegative atom - in this case, oxygen  The signal at 1.23 δ is a triplet, indicating two neighboring protons  The signal at 3.49 δ is a quartet, indicating three neighboring protons  This splitting pattern is characteristic of an ethyl group  The compound is diethyl ether, CH 3 CH 2 OCH 2 CH 3

37. © 2016 Cengage Learning. All Rights Reserved. 1 H NMR Spectroscopy and Proton Equivalence  Proton NMR is much more sensitive than 13 C and the active nucleus ( 1 H) is nearly 100 % of the natural abundance  Shows how many kinds of nonequivalent hydrogens are in a compound  Theoretical equivalence can be predicted by comparing structures formed by replacing each H with X gives the same or different outcome  Equivalent H’s have the same signal while nonequivalent are different  There are degrees of nonequivalence

38. © 2016 Cengage Learning. All Rights Reserved. 1 H NMR Spectroscopy and Proton Equivalence  One use of 1 H NMR is to ascertain the number of electronically non-equivalent hydrogens present in a molecule  In relatively small molecules, a brief look at the structure can help determine the kinds of protons present and the number of possible NMR absorptions  Equivalence or nonequivalence of two protons can be determined by comparison of structures formed if each hydrogen were replaced by an X group

39. © 2016 Cengage Learning. All Rights Reserved. 1 H NMR Spectroscopy and Proton Equivalence  Possibilities  If the protons are chemically unrelated and non- equivalent, the products formed by substitution would be different constitutional isomers

40. © 2016 Cengage Learning. All Rights Reserved. 1 H NMR Spectroscopy and Proton Equivalence  If the protons are chemically identical, the same product would be formed despite the substitution

41. © 2016 Cengage Learning. All Rights Reserved. 1 H NMR Spectroscopy and Proton Equivalence  If the hydrogens are homotopic but not identical, substitution will form a new chirality center  Hydrogens that lead to formation of enantiomers upon substitution with X are called enantiotopic

42. © 2016 Cengage Learning. All Rights Reserved. 1 H NMR Spectroscopy and Proton Equivalence  If the hydrogens are neither homotopic nor enantiotopic, substitution of a hydrogen at C3 would form a second chirality center

43. © 2016 Cengage Learning. All Rights Reserved. Worked Example  How many absorptions will (S)-malate, an intermediate in carbohydrate metabolism have in its 1 H NMR spectrum? Explain

44. © 2016 Cengage Learning. All Rights Reserved. Worked Example  Solution:  Because (S)-malate already has a chirality center(starred), the two protons next to it are diastereotopic and absorb at different values  The 1 H NMR spectrum of (S)-malate has four absorptions

45. © 2016 Cengage Learning. All Rights Reserved. More Complex Spin-Spin Splitting Patterns  Some hydrogens in a molecule possess accidentally overlapping signals  In the spectrum of toluene (methylbenzene), the five aromatic ring protons produce a complex, overlapping pattern though they are not equivalent

46. © 2016 Cengage Learning. All Rights Reserved. More Complex Spin-Spin Splitting Patterns  Splitting of a signal by two or more nonequivalent kinds of protons causes a complication in 1 H NMR spectroscopy

47. © 2016 Cengage Learning. All Rights Reserved. Figure 13.14 - Tree Diagram for the C2 proton of trans-cinnamaldehyde

48. © 2016 Cengage Learning. All Rights Reserved. Worked Example  3-Bromo-1-phenyl-1-propene shows a complex NMR spectrum in which the vinylic proton at C2 is couples with both the C1 vinylic proton (J = 16 Hz) and the C3 methylene protons (J = 8 Hz)  Draw a tree diageam for the C2 proton signal and account for the fact that a live-line multiplet is observed

49. © 2016 Cengage Learning. All Rights Reserved. Worked Example  Solution:  C2 proton couples with vinylic proton (J = 16) Hz  C2 proton’s signal is split into a doublet  C2 proton also couples with the two C3 protons (J = 8 Hz)  Each leg of the C2 proton doublet is split into a triplet to produce a total of six lines

50. © 2016 Cengage Learning. All Rights Reserved. Worked Example

51. © 2016 Cengage Learning. All Rights Reserved. Uses of 1H NMR Spectroscopy  The technique is used to identify likely products in the laboratory quickly and easily  NMR can help prove that hydroboration-oxidation of alkenes occurs with non-Markovnikov regiochemistry to yield the less highly substituted alcohol

52. © 2016 Cengage Learning. All Rights Reserved. Figure 13.15 - 1 H NMR Spectra of Cyclohexylmethanol

53. © 2016 Cengage Learning. All Rights Reserved. Worked Example  Mention how 1 H NMR is used to determine the regiochemistry of electrophilic addition to alkenes  Determine whether addition of HCl to 1- methylcyclohexene yields 1-chloro-1- nethylcyclohexane or 1-chloro-2- methylcyclohexane

54. © 2016 Cengage Learning. All Rights Reserved. Worked Example  Solution:  Referring to 1 H NMR methyl group absorption  The unslpit methyl group in the left appears as a doublet in the product on the right  Bonding of a proton to a carbon that is also bonded to an electronegative atom causes a downfield absorption in the 2.5–4.0 region  1 H NMR spectrum of the product would confirm the product to be 1-chloro-1-methylcyclohexane

55. © 2016 Cengage Learning. All Rights Reserved. 13 C NMR Spectroscopy: Signal Averaging and FT–NMR  Carbon-13 is the only naturally occurring carbon isotope that possesses a nuclear spin, but its natural abundance is 1.1%  Signal averaging and Fourier-transform NMR (FT–NMR) help in detecting carbon 13  Due to the excess random electronic background noise present in 13 C NMR, an average is taken from hundreds or thousands of individual NMR spectra

56. © 2016 Cengage Learning. All Rights Reserved. Figure 13.16 - Carbon-13 NMR Spectra of 1-Pentanol

57. © 2016 Cengage Learning. All Rights Reserved. 13 C NMR Spectroscopy: Signal Averaging and FT–NMR  Spin-spin splitting is observed only in 1 H NMR  The low natural abundance of 13 C nucleus is the reason that coupling with adjacent carbons is highly unlikely  Due to the broadband decoupling method used to record 13 C spectra, hydrogen coupling is not seen

58. © 2016 Cengage Learning. All Rights Reserved. Characteristics of 13 C NMR Spectroscopy  13 C NMR provides a count of the different carbon atoms in a molecule  13 C resonances are 0 to 220 ppm downfield from TMS

59. © 2016 Cengage Learning. All Rights Reserved. Characteristics of 13 C NMR Spectroscopy  General factors that determine chemical shifts  The electronegativity of nearby atoms  The diamagnetic anisotropy of pi systems  The absorption of sp 3 -hybridized carbons and sp 2 carbons

60. © 2016 Cengage Learning. All Rights Reserved. Figure 13.18 - Carbon-13 Spectra of 2- butanone and para-bromoacetophenone

61. © 2016 Cengage Learning. All Rights Reserved. Worked Example  Classify the resonances in the 13 C spectrum of methyl propanoate, CH 3 CH 2 CO 2 CH 3

62. © 2016 Cengage Learning. All Rights Reserved. Worked Example  Solution:  Methyl propanoate has four unique carbons that individually absorb in specific regions of the 13 C spectrum

63. © 2016 Cengage Learning. All Rights Reserved. DEPT 13 C NMR Spectroscopy  DEPT-NMR (distortionless enhancement by polarization transfer)  Stages of a DEPT experiment  Run a broadband-decoupled spectrum  Run a DEPT-90  Run a DEPT-135  The DEPT experiment manipulates the nuclear spins of carbon nuclei

64. © 2016 Cengage Learning. All Rights Reserved. Figure 13.20 – DEPT-NMR Spectra for 6-methyl-5-hepten-2-ol

65. © 2016 Cengage Learning. All Rights Reserved. Uses of 13 C NMR Spectroscopy  Helps in determining molecular structures  Provides a count of non-equivalent carbons  Provides information on the electronic environment of each carbon and the number of attached protons  Provides answers on molecule structure that IR spectrometry or mass spectrometry cannot provide

66. © 2016 Cengage Learning. All Rights Reserved. Figure 13.21 - 13 C NMR Spectrum of 1-methylcyclohexane

67. © 2016 Cengage Learning. All Rights Reserved. Worked Example  Propose a structure for an aromatic hydrocarbon, C 11 H 16, that has the following 13 C NMR spectral data:  Broadband decoupled: 29.5, 31.8, 50.2, 125.5, 127.5, 130.3, 139.8 δ  DEPT-90: 125.5, 127.5, 130.3 δ  DEPT-135: positive peaks at 29.5, 125.5, 127.5, 130.3 δ; negative peak at 50.2 δ

68. © 2016 Cengage Learning. All Rights Reserved. Worked Example  Solution:  Calculate the degree of unsaturation of the unknown compound  C 11 H 16 has 4 degrees of unsaturation  Look for elements of symmetry  7 peaks appearing in the 13 C NMR spectrum indicate a plane of symmetry  According to the DEPT-90 spectrum, 3 of the kinds of carbons in the aromatic ring are CH carbons

69. © 2016 Cengage Learning. All Rights Reserved. Worked Example  The unknown structure is a monosubstituted benzene ring with a substituent containing CH 2 and CH 3 carbons

70. © 2016 Cengage Learning. All Rights Reserved. Summary  Nuclear magnetic resonance spectroscopy or NMR is the most important spectroscopic technique used in the determination of molecular structure  Magnetic nuclei such as 1 H and 13 C spin-flip from a lower energy state to a higher energy state when they absorb radiofrequency waves  Each 1 H or 13 C nucleus possesses a unique electromagnetic field that causes it to resonate at different values of the applied field causing peaks whose exact position is termed a chemical shift

71. © 2016 Cengage Learning. All Rights Reserved. Summary  Delta (δ) is the unit of calibration in NMR charts  Tetramethylsilane (TMS) is a reference point on the NMR chart  TMS absorption that occurs at the right-hand (upfield) side of the chart is assigned a value of 0 δ  Fourier-transform NMR (FT–NMR) spectrometers are used to obtain 13 C spectra using broadband decoupling of proton spins

72. © 2016 Cengage Learning. All Rights Reserved. Summary  Electronic integration of the area under each absorption peak in 1 H NMR spectra is used to determine the number of hydrogens that cause each peak  Neighboring nuclear spins can couple to cause the spin-spin splitting of NMR peaks into multiplets  The NMR signal of a hydrogen neighbored by n equivalent adjacent hydrogens splits into n + 1 peaks (the n + 1 rule) with coupling constant J