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Nuclear Magnetic Resonance

Nuclear Magnetic Resonance. Proton NMR. Theory. An atomic nucleus can be viewed as a spinning, positively charged sphere. If the nucleus has an odd mass number or atomic number, it generates a magnetic field just like a spinning bar magnet would. Theory.

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Nuclear Magnetic Resonance

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  1. Nuclear Magnetic Resonance Proton NMR

  2. Theory • An atomic nucleus can be viewed as a spinning, positively charged sphere. • If the nucleus has an odd mass number or atomic number, it generates a magnetic field just like a spinning bar magnet would.

  3. Theory • When placed in the magnetic field generated by a larger bar magnet, a small bar magnet can either align with the field (lower energy) or against the field (higher energy).

  4. Theory • In the absence of a magnetic field, the spins of a group of nuclei are randomly oriented. • When a magnetic field is applied, the nuclei align either: • with the field • a spin state • lower energy • against the field • b spin state • higher energy

  5. Theory • The difference in energy (DE) between the spin states depends on both the strength of the magnetic field and the identity of the nuclei. • Absorption of a photon with the correct energy causes the nuclei to flip between the a and b spin states

  6. Theory • When the nucleus is subjected to the right combination of both magnetic field and the right frequency of electromagnetic radiation, the spin of the nucleus changes. • a state  b state • “in resonance” • NMR spectrometer detects the absorption (or emission) of energy at a specific magnetic field.

  7. Simple NMR Spectrometer RF is held constant, most common u = 60, 100 or 300 MHz Strength of magnetic field is varied.

  8. Theory • The nuclei present in an organic compound are not “naked”; they are surrounded by electrons. • The electrons generate a small induced magnetic field that opposes the external field and partially “shields” the nuclei from the applied magnetic field. • Partially “shielded” • The magnetic field experienced by the nucleus is weaker than the applied magnetic field.

  9. Theory • In order for “resonance” to occur (i.e. in order for the nucleus to flip from one spin state to the other), the strength of the applied magnetic field must be increased in order for the shielded proton to “flip” at the same frequency.

  10. Theory • Protons in an organic molecule are shielded by different amounts due to differences in their chemical environment. More shielded Absorbs at higher field Less shielded due to high electronegativity of O; Absorbs at lower field

  11. Theory • Proton NMR Spectrum of Methanol UPFIELD More Shielded Lower chemical shift DOWNFIELD Less shielded Higher chemical shift

  12. Information from NMR Spectrum • Number of signals • Number of different kinds of protons • Location of signals • Chemical environment of protons • Degree of shielding or deshielding • Intensity of signal • Number of protons of a particular type • Splitting • The number of protons on adjacent carbon atoms

  13. Information from NMR Spectrum • Number of signals = number of different types of protons • Chemically equivalent protons • same chemical environment • exact same chemical shift • How many groups of chemically equivalent protons are present in each of the following?

  14. Information from NMR Spectrum • The location of each signal is described by its chemical shift • the ratio of how far a signal is shifted downfield relative to a standard (TMS) to the total spectrometer frequency • d = shift downfield from TMS (Hz) instrument frequency in MHz • Since MHz = 1,000,000 Hz, the chemical shift is reported as ppm

  15. Chemical Shifts • “Substituents” with electronegative atoms (O, halogens, N) or p systems (like aromatic rings and vinyl groups) result in deshielding of protons. • Downfield shift • Downfield shift increases as the electronegativity of a substituent increases.

  16. Chemical Shifts • The amount of deshielding decreases as the distance between the electronegative atom increases. • Shielding and deshielding effects are additive.

  17. Information from NMR Spectrum

  18. Alkane C – H Chemical Shifts CH3 ~0.9 ppm CH2 ~ 1.2 ppm CH ~ 1.5 ppm http://riodb01.ibase.aist.go.jp/sdbs/ (National Institute of Advanced Industrial Science and Technology, 10/16/09)

  19. http://riodb01.ibase.aist.go.jp/sdbs/ (National Institute of Advanced Industrial Science and Technology, 10/17/09) Carbonyl Compounds • CHn with an a Carbonyl c Typical range: 1.9 – 3.3 ppm with 2.0-2.5 being quite common a b

  20. http://riodb01.ibase.aist.go.jp/sdbs/ (National Institute of Advanced Industrial Science and Technology, 10/17/09) Carbonyl Compounds • CHn with an a Carbonyl Typical range: 1.9 – 3.3 ppm with 2.0-2.5 being quite common

  21. http://riodb01.ibase.aist.go.jp/sdbs/ (National Institute of Advanced Industrial Science and Technology, 10/17/09) Aromatic Compounds a Benzylic CHn: 2.2 – 3.0 ppm b Aromatic CH: 6.5 -8.5 ppm

  22. http://riodb01.ibase.aist.go.jp/sdbs/ (National Institute of Advanced Industrial Science and Technology, 10/17/09) Alkyl Halides CHn with a halogen: 2.1 – 4.5 ppm (3 – 4 ppm most common) c a Downfield shift decreases as distance from halogen increases. b Downfield shift increases as e.n. of halogen increases.

  23. Alcohols OH: 0.5 – 5.0 ppm a b Carbinol CHn: 3.2 – 5.2 ppm with 3 – 4 ppm most common c Downfield shift for protons decreases as distance from hydroxyl group increases. http://riodb01.ibase.aist.go.jp/sdbs/ (National Institute of Advanced Industrial Science and Technology, 10/17/09)

  24. Ethers a CHnOR: Similar to alcohol’s carbinol CH 3.2 – 5.2 ppm (3 -4 ppm most common) b & d c

  25. Esters a -CHnCO2R 1.9 – 3.3 ppm (2-2.5 most common) c O-CHn 3.5 – 5 ppm b http://riodb01.ibase.aist.go.jp/sdbs/ (National Institute of Advanced Industrial Science and Technology, 10/17/09)

  26. Amines N – H : Variable position 1.5 – 4.0 ppm CHnN: ~ 2 – 4 ppm http://riodb01.ibase.aist.go.jp/sdbs/ (National Institute of Advanced Industrial Science and Technology, 10/17/09)

  27. Alkenes Allylic protons: 1.6 – 2.5 ppm a d b c Vinyl protons: 4.5 – 7.3 ppm http://riodb01.ibase.aist.go.jp/sdbs/ (National Institute of Advanced Industrial Science and Technology, 10/17/09)

  28. Aldehydes a Aldehyde proton: 9.5 – 10.5 ppm c b http://riodb01.ibase.aist.go.jp/sdbs/ (National Institute of Advanced Industrial Science and Technology, 10/17/09)

  29. Carboxylic Acids a R-COOH 9.7 - 12.5 ppm b http://riodb01.ibase.aist.go.jp/sdbs/ (National Institute of Advanced Industrial Science and Technology, 10/17/09)

  30. Integration • The area under each NMR signal is proportional to the number of protons that give rise to that signal. • The relative area is measured by an integrator and is depicted as an integration curve above the spectrum. • The height of the integration curve is proportional to the area

  31. Integration • The height of the integration curve is proportional to the relative number of protons.

  32. Splitting • The magnetic fields generated by non-equivalent protons on adjacent carbon atoms will either align with or against the external magnetic field. • Magnetic coupling causes the proton to absorb at a slightly different chemical shift: • Slightly downfield when external field is reinforced • Slightly upfield when external field is opposed

  33. Splitting • The protons (Hb) on carbon 2 of 1,1,2-tribromoethane appear as a doublet. • If Ha has spin aligned with b, then b is reinforced and peak shifts downfield. • If Ha has spin opposed to b, then the peak shifts upfield.

  34. Splitting • Since all possibilities occur, the signal is split into multiple peaks when non-equivalent protons are present on an adjacent carbon atom. Notice that the hydroxyl proton is not split by the protons on the adjacent methylene and vice versa!

  35. Splitting • A signal that is split by N equivalent protons will be split into N + 1 peaks.

  36. Splitting • Equivalent protons do not split each other. • Protons attached to the same carbon will not split each other unless they are non-equivalent. • Protons separated by four or more bonds do not couple (split each other). • Protons attached to O or N will not split the protons attached to an adjacent atom (and vice versa).

  37. “Classic” Ethyl Group Pattern

  38. “Classic” Isopropyl Pattern

  39. Example • Predict the splitting pattern and approximate chemical shifts for the following compounds:

  40. 2,5-hexanedione http://riodb01.ibase.aist.go.jp/sdbs/ (National Institute of Advanced Industrial Science and Technology, 10/18/09)

  41. Isopropyl alcohol http://riodb01.ibase.aist.go.jp/sdbs/ (National Institute of Advanced Industrial Science and Technology, 10/18/09)

  42. Propiophenone http://riodb01.ibase.aist.go.jp/sdbs/ (National Institute of Advanced Industrial Science and Technology, 10/18/09)

  43. Example • Assign each peak in the NMR spectrum to the appropriate protons. http://riodb01.ibase.aist.go.jp/sdbs/ (National Institute of Advanced Industrial Science and Technology, 10/19/09)

  44. Example • Draw the structure for the following compound if its molecular formula is C4H10O. http://riodb01.ibase.aist.go.jp/sdbs/ (National Institute of Advanced Industrial Science and Technology, 10/18/09)

  45. Example • Draw the structure of the following compound. http://riodb01.ibase.aist.go.jp/sdbs/ (National Institute of Advanced Industrial Science and Technology, 10/18/09)

  46. Example A compound with the formula C6H10O3 has strong peaks at 1746 and 1719 cm-1 in its IR spectrum. Identify its structure using the following proton NMR. 3 3 2 2 http://riodb01.ibase.aist.go.jp/sdbs/ (National Institute of Advanced Industrial Science and Technology, 10/18/09)

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