1. CHEMICAL SHIFT AND COUPLING CONSTANTS

2. Unpaired nuclear spins are of importance in NMR. The actual spectral data acquired by NMR is the free- induction decay, or FID. FT Time, sec Frequency, Hz FID An applied magnetic field B0, the strength of which is measured in tesla (T), and the frequency n of radiation used for resonance, measured in hertz (Hz), or megahertz (MHz)—(1 MHz = 10 6 Hz).

3. NMR Spectrometer Schematic diagram of a nuclear magnetic resonance spectrometer.

4. 4 An NMR spectrum is a plot of the intensity of a peak against its chemical shift, measured in parts per million (ppm). 1 H NMR—The Spectrum Protons in different environments absorb at slightly different frequencies, so they are distinguishable by NMR.

5.  High frequency:  High frequency: The shift of an NMR signal to the left on the chart paper.  Low frequency:  Low frequency: The shift of an NMR signal to the right on the chart paper. 1 H-NMR spectrum of methyl acetate. NMR absorptions generally appear as sharp peaks. Increasing chemical shift is plotted from left to right. Most protons absorb between 0-10 ppm. The terms “upfield” and “downfield” describe the relative location of peaks. Upfield means to the right. Downfield means to the left. NMR absorptions are measured relative to the position of a reference peak at 0 ppm on the d scale due to tetramethylsilane (TMS). TMS is a volatile inert compound that gives a single peak upfield from typical NMR absorptions.

6. 1D 1 H NMR Spectra of Skeletal Muscle Tissue

7. Chemical Shift When an atom is placed in a magnetic field, its electrons circulate about the direction of the applied magnetic field. This circulation causes a small magnetic field at the nucleus which opposes the externally applied field. The magnetic field at the nucleus (the effective field) is therefore generally less than the applied field by a fraction , B = B o (1-  ) The electron density around each nucleus in a molecule varies according to the types of nuclei and bonds in the molecule. The opposing field and therefore the effective field at each nucleus will vary. This is called the chemical shift phenomenon.  = (  /2  )B local = (   /2  )(1-  )

8. 8 In the vicinity of the nucleus, the magnetic field generated by the circulating electron decreases the external magnetic field that the proton “feels”. Since the electron experiences a lower magnetic field strength, it needs a lower frequency to achieve resonance. Lower frequency is to the right in an NMR spectrum, toward a lower chemical shift, so shielding shifts the absorption upfield. 1 H NMR—Position of Signals

9. 9 Nuclear Magnetic Resonance Spectroscopy 1 H NMR—Position of Signals

10. 10 Nuclear Magnetic Resonance Spectroscopy 1 H NMR—Position of Signals

11. 11 The less shielded the nucleus becomes, the more of the applied magnetic field (B 0 ) it feels. This deshielded nucleus experiences a higher magnetic field strength, to it needs a higher frequency to achieve resonance. Higher frequency is to the left in an NMR spectrum, toward higher chemical shift—so deshielding shifts an absorption downfield. Protons near electronegative atoms are deshielded, so they absorb downfield. 1 H NMR—Position of Signals

12. Chemical Shift is field dependent The chemical shift of a nucleus is the difference between the resonance frequency of the nucleus and a standard, relative to the standard. This quantity is reported in ppm and given by the symbol delta, . Choice of Solvents: D 2 O, H 2 O, CDCl 3, DMSO Reference Compounds: Aqueous solution: DSS (4,4-dimethyl-4- silapentane-1-sulfonic acid) (TSP) Trimethylsilyl propionate (CH 3 ) 4 Si, usually referred to as TMS used in CDCl 3, DMSO The magnitude of the screening depends on the atom. For example, carbon- 13 chemical shifts are much greater than hydrogen-1 chemical shifts.

13. 13 Protons in different environments absorb at slightly different frequencies, so they are distinguishable by NMR. The frequency at which a particular proton absorbs is determined by its electronic environment. The size of the magnetic field generated by the electrons around a proton determines where it absorbs. Modern NMR spectrometers use a constant magnetic field strength B 0, and then a narrow range of frequencies is applied to achieve the resonance of all protons. Only nuclei that contain odd mass numbers (such as 1 H, 13 C, 19 F and 31 P) or odd atomic numbers (such as 2 H and 14 N) give rise to NMR signals.

14.  The difference in resonance frequencies among the various hydrogen nuclei within a molecule due to shielding/deshielding is generally very small.  The difference in resonance frequencies for hydrogens in CH 3 Cl compared to CH 3 F under an applied field of 7.05T is only 360 Hz, which is 1.2 parts per million (ppm) compared with the irradiating frequency.

15. ChemicalShifts 1 H-NMR

16. Chemical shift depends on the (1) electronegativity of nearby atoms, (2) hybridization of adjacent atoms, and (3) diamagnetic effects from adjacent pi bonds. Electronegativity

17. Chemical Shift Hybridization of adjacent atoms.

18. Chemical Shift Diamagnetic effects of pi bonds  A carbon-carbon triple bond shields an acetylenic hydrogen and shifts its signal to lower frequency (to the right) to a smaller  value.  A carbon-carbon double bond deshields vinylic hydrogens and shifts their signal to higher frequency (to the left) to a larger  value.

19. 19 Nuclear Magnetic Resonance Spectroscopy The chemical shift of a C—H bond increases with increasing alkyl substitution. 1 H NMR—Chemical Shift Values

20. 20 Nuclear Magnetic Resonance Spectroscopy Protons in a given environment absorb in a predictable region in an NMR spectrum. 1 H NMR—Chemical Shift Values

21. Chemical Shift  Magnetic induction in the  bonds of a carbon-carbon triple bond shields an acetylenic hydrogen and shifts its signal lower frequency.

22. Chemical Shift  Magnetic induction in the  bond of a carbon-carbon double bond deshields vinylic hydrogens and shifts their signal higher frequency.

23. 23 In some cases, such as the benzene molecule, the circulation of the electrons in the aromatic orbitals creates a magnetic field at the hydrogen nuclei which enhances the B o field. This phenomenon is called deshielding. In a magnetic field, the six  electrons in benzene circulate around the ring creating a ring current. The magnetic field induced by these moving electrons reinforces the applied magnetic field in the vicinity of the protons. The protons thus feel a stronger magnetic field and a higher frequency is needed for resonance. Thus they are deshielded and absorb downfield.

24. 24 Four different features of a 1 H NMR spectrum provide information about a compound’s structure: a.Number of signals b.Position of signals c.Intensity of signals. d.Spin-spin splitting of signals.

25. 25 The number of NMR signals equals the number of different types of protons in a compound. Protons in different environments give different NMR signals. Equivalent protons give the same NMR signal. 1 H NMR—Number of Signals To determine equivalent protons in cycloalkanes and alkenes, always draw all bonds to hydrogen.

26. 26 Nuclear Magnetic Resonance Spectroscopy 1 H NMR—Number of Signals

27. 27 Nuclear Magnetic Resonance Spectroscopy In comparing two H atoms on a ring or double bond, two protons are equivalent only if they are cis (or trans) to the same groups. 1 H NMR—Number of Signals

28. 28 Nuclear Magnetic Resonance Spectroscopy Proton equivalency in cycloalkanes can be determined similarly. 1 H NMR—Number of Signals

29. 29 1 H NMR—Position of Signals Other Factors Affecting Chemical Shift 1.Solvent 2.Presence of electronegative atoms. 3.pH. 4.Temperature 5.Hydrogen bond 6.Conformational Changes 7.Presence of ligand

30. 30 Nuclear Magnetic Resonance Spectroscopy The chemical shift of a C—H can be calculated with a high degree of precision if a chemical shift additivity table is used. The additivity tables starts with a base chemical shift value depending on the structural type of hydrogen under consideration: Calculating 1 H NMR—Chemical Shift Values

31. 31 Nuclear Magnetic Resonance Spectroscopy 1 H NMR—Chemical Shift Values

32. 32 Nuclear Magnetic Resonance Spectroscopy Consider the spectrum below: 1 H NMR—Spin-Spin Splitting

33. Ethyl Bromide

34. Spin-Spin Coupling

35. Spin-Spin Coupling (Splitting) Observation: A nucleus with a magnetic moment may interact with other nuclear spins resulting in mutual splitting of the NMR signal from each nucleus into multiplets.

36. 36 The Origin of 1 H NMR—Spin-Spin Splitting The frequency difference, measured in Hz, between two peaks of the doublet is called the coupling constant, J.

37. 37 Spin-spin splitting occurs only between nonequivalent protons on the same carbon or adjacent carbons. The Origin of 1 H NMR—Spin-Spin Splitting Let us consider how the doublet due to the CH 2 group on BrCH 2 CHBr 2 occurs: When placed in an applied field, (B 0 ), the adjacent proton (CHBr 2 ) can be aligned with (  ) or against (  ) B 0. The likelihood of either case is about 50% (i.e., 1,000,006  vs 1,000,000  ). Thus, the absorbing CH 2 protons feel two slightly different magnetic fields—one slightly larger than B 0, and one slightly smaller than B 0. Since the absorbing protons feel two different magnetic fields, they absorb at two different frequencies in the NMR spectrum, thus splitting a single absorption into a doublet, where the two peaks of the doublet have equal intensity.

38. 38 The Origin of 1 H NMR—Spin-Spin Splitting Let us now consider how a triplet arises: When placed in an applied magnetic field (B 0 ), the adjacent protons H a and H b can each be aligned with (  ) or against (  ) B 0. Thus, the absorbing proton feels three slightly different magnetic fields—one slightly larger than B 0 (  a  b ). one slightly smaller than B 0 (  a  b ) and one the same strength as B 0 (  a  b ).

39. When determining the spin-spin coupling, look at the number of protons on the adjacent carbon. For the methyl group, look at the methylene group. There are 2 protons, so using the N+1 rule tells us that the peak should be a triplet in a 1:2:1 ratio. CH2 b

40. 40 The Origin of 1 H NMR—Spin-Spin Splitting

41. For the methylene group, look at the methyl group. There are 3 protons, so using the N+1 rule tells us that the peak should be a quartet in a 1:3:3:1 ratio. CH3 a

42. 4. The Coupling Constant, J - the separation between peaks in a multiplet measured in units of Hz. CH3CH2Br CH2 b CH3 a Jba Jab Jba Jab a b Jba = Jab = 7.3 Hz

43. The energy of the interactions between two spins A and B can be found by the relationship: E = J AB * I A * I B I A and I B are the nuclear spin vectors, and are proportional to  A and  B, the magnetic moments of the two nuclei. J AB is the scalar coupling constant. So we see a very important feature of couplings. It does not matter if we have a 60, a 400, or an 800 MHz magnet, the coupling constants are always the same!!!

44. Lets do a more detailed analysis in term of the energies. Lets think a two energy level system, and the transitions for nuclei A. When we have no coupling (J = 0), the energy involved in either transition (A 1 or A 2 ) is equal (no spin-spin interaction). A X A1A1 A2A2 A1A1 A2A2 BoBo E J = 0 J > 0 E4E3E2E1E4E3E2E1 When J > 0, the energy levels of the spin system will be either stabilized or destabilized. Depending on the relative orientations of the nuclear moments, the energies for the A 1 and A 2 transition will change giving two different frequencies (two peaks for A).

45. A1A1 A2A2 A 1 = A 2 A2A2 A1A1 A 2 A 1 A1A1 A2A2 A 1 A 2 J = 0 J > 0J < 0 As mentioned before, the choice of positive or negative J is a definition. However, we see that we won’t be able to tell if we have a positive or negative J, because the lines in the spectrum corresponding to the different transitions basically change places. Unless we are interested in studying the energies, this is not important for structure elucidation…

46. Now look at some simple examples. Examine the size of the peaks in the splitting. H b is augmenting external field causing a larger energy gap. H b decrementing external field causing a smaller energy gap. H a is being excited. H b is causing spin- spin splitting by slightly increasing or decreasing the magnetic field experienced by H a. Spin-Spin Splitting

47. Figure 13.15b, p.512 Two neighboring atoms assist external field. More energy needed to excite. Peak is “downfield”. One neighbor assists, one hinders. No effect. Both neighbors oppose. Less energy needed to excite, “upfield”. Again H a is flipping, resonating. The two H b are causing spin- spin splitting by slightly changing the magnetic field experienced by H a. Recall that for the two H b atoms the two states (helping and hindering the external field) are almost equally likely. This give us the 1 : 2 : 1 ratio.

48. Figure 13.15c, p.512 All H b augment Two augment, one decrement. One augment, two decrement. All decrement. H a being excited. Three equivalent H b causing spin spin splitting. Three neighboring H b ’s causing splitting when H a is excited.

49. The Origin of 1 H NMR—Spin-Spin Splitting

50. Spin-Spin Splitting in 1 H NMR Spectra Peaks are often split into multiple peaks due to magnetic interactions between nonequivalent protons on adjacent carbons, The process is called spin-spin splitting The splitting is into one more peak than the number of H’s on the adjacent carbon(s), This is the “n+1 rule” The relative intensities are in proportion of a binomial distribution given by Pascal’s Triangle The set of peaks is a multiplet (2 = doublet, 3 = triplet, 4 = quartet, 5=pentet, 6=hextet, 7=heptet…..)

51. p. 491 Attempt to anticipate the splitting patterns in each molecule.

52. Magnitude of Coupling Constant, J The magnitude of the coupling constant, J, can vary from 0 to about 20 Hz. This represents an energy gap (E = h ) due to the interaction of the nuclei within the molecule. It does not depend on the strength of the external field. J is related to the dihedral angle between bonds. J largest for 0 (eclipsed) or 180 (anti), smallest for 90, intermediate for gauche.

53. Table 13.4, p.511 anti gauche vinyl systems

54. Figure 13.17, p.513 Naturally if there are two non-equivalent nuclei they split each other.

55. Figure 13.19, p.513 Three nonequivalent nuclei. H a and H b split each other. Also H b and H c split each other. Technique: use a tree diagram and consider splittings sequentially.

56. Figure 13.20, p.514 More complicated system

57. Figure 13.21, p.514 Not equivalent (R 1 is not same as R 2 ) because there is no rotation about the C=C bond. Return to Vinyl Systems

58. J AB = 11-18 Hz, BIG J AC = 0 – 5 Hz, SMALL J BC = 5 – 10 Hz, MIDDLE Each of these patterns is different from the others. Analysis Now examine the left most signal….

59. Fast Exchange ethanol  Expect coupling between these hydrogens. Three bond separation.  There is no coupling observed especially in acid or base.  Reason: exchange of weakly acidic hydrogen with solvent.  The spectrometer sees an “averaged hydrogen”. No coupling and broad peak.

60. J is independent of magnetic field strength Mutually coupled hydrogen nuclei have the same J value. The magnitude of J gives information about structure. J = 7 Hz J = 15 Hz J = 10 Hz J = 2 Hz

61. Assignment 10Hz Karplus equation for determining dihedral angle Coupling consts. J, Hz H 1’ -H 2’ 5.9 H 2’ -H 3’ 5.5 H 3’ -H 4’ 3.0 H 4’ -H 5’ 4.1 H 4’ -H 5” 3.5 H 5’ -H 5” 12.3 H 2’ -C 2’ -OH6.6 H 3’ -C 3’ -OH4.7 H 5’ -C 5’ -OH7.2 H 5” -C 5’ -OH4.4 decoupled

62. 13 C NMR 13 C has spin states similar to H. Natural occurrence is 1.1% making 13 C- 13 C spin spin splitting very rare. H atoms can spin-spin split a 13 C peak. ( 13 CH 4 would yield a quintet). This would yield complicated spectra. H splitting eliminated by irradiating with an additional frequency chosen to rapidly flip (decouple) the H’s averaging their magnetic field to zero. A decoupled spectrum consists of a single peak for each kind of carbon present. The magnitude of the peak is not important.