1. Chapter 3 Nuclear Magnetic Resonance Spectroscopy Many atomic nuclei have the property of nuclear spin. When placed between the poles of a magnet, the axis of rotation of the spinning nuclei precess around the axis of the field. This precession can be detected by irradiation with energy in the radiofrequency region of the spectrum. When the precession and irradiation frequencies are equal the system is said to be in resonance.

2. Nuclear Magnetic Resonance (NMR)- Absorption of radiowave EM energy by nuclei in a magnetic field. Background and Theory –Electron spin observed - 1922 –By 1926, became apparent that nuclear spin also existed. –1939, Rabi observes absorption of radio frequency (RF) energy by nuclei of H 2 gas; gets Nobel prize in physics in 1944.

3. Nuclei are charged and if they have spin, they are magnetic No Field Applied Magnetic Field = B Energy of transition = energy of radiowaves Higher energy state: magnetic field opposes applied field Lower energy state: magnetic field aligned with applied field 3.1 How NMR Works

4. Nuclear Spin Do all nuclei have spin movement? Nuclear spin depends on its spin angular momentum number I If the protons and neutrons comprising these nuclei are not paired(one is odd number, the other even number), the overall spin of the charged nucleus generates a magnetic dipole along the spin axis. I=1/2, 3/2, 5/2 … Example: 1 H, 13 C, 15 N, 19 F, 31 P

5. Nuclear Spin (cont.) If both the protons and neutrons are even numbers, I=0, there is no nuclear spin movement. Example: 12 C, 16 O, 32 S If both the protons and neutrons are odd number, I=1, 2, 3…, the nucleus will spin. Example: 2 H, 14 N

6. NMR Background & Theory (cont.) Observed in bulk samples- 1946 –Bloche (Stanford) for H in H 2 O –Purcell (Harvard) for H in C n H 2n+2 (paraffin) –Bloche & Purcell got Nobel prize in 1952 in Physics. Basic NMR relation

7. NMR Theory (cont.) In 1949 & 50, it was observed that ethanol gave 3 signals. Not due to experimental error. 3 signals from H’s on 3 different groups or chemical environments. Since signal is proportional to # of H’s, the peaks can be assigned. 20,000,000 Hz 20,000,050 Hz 20,000,085 Hz

8. These variations in frequency (  ) are due to different magnetic environments. Explanation: The electrons surrounding the H nucleus provide shielding of applied magnetic field so that:- NMR Theory (cont.)

9. Shift in , due to shielding by electrons, is called the 60 MHz NMR spectrum of CH 3 CH 2 OH 60,000,000 Hz 60,000,220 Hz 60,000,320 Hz 60,000,072 Hz (parts per million)

10. The variations in frequency (  ) is depended on the externally applied magnetic field of strength Bo CH 3 Br 60MHz(Bo=14100)  =162 Hz 100MHz(Bo=23500)  =270 Hz To correct the effect of Bo, Chemical shift of sample was compared with a standard TMS (tetramethyl silane)

11. Chemical Shift δ on the left side of TMS + δ on the right side of TMS -

12. TMS Only one peak on NMR spectrum High electronic density of H in TMS. Almost all the H peaks of organic compounds appear on the left of the TMS peak TMS is volatile compound(bp 26.5 o C), very easy to remove. It dissolves in most organic solvents δ of TMS is set as 0

13. Calculation of chemical shift 1,1,2-trichloropropane

14. 3.3 Factors affecting chemical shift Depends on adjacent group For protons on carbon attached to an electronegative atom or group X, the chemical shift increases with the electronegativity of X. This is due to the inductive effect on the shielding of the protons and is apparent in the methyl halides.

15. Depends on atom attached Fig 1 NMR Chemical Shifts and Splitting Patterns Compound CH 3 XCH 3 FCH 3 OHCH 3 ClCH 3 BrCH 3 ICH 3 C- 3 CH 4 (CH 3 ) 4 Si Element XFOClBrICHSi Electronegativity of X 4.03.53.12.82.72.52.11.8 Chemical shift, ppm 4.263.403.052.682.160.90.230.0 ppm

16. Depends on adjacent group (cont.) The inductive (deshielding) effect of a substituent on a proton decreases as the separation between the proton and substituent is increased. CH 3 - CH 2 - CH 2 - OH 0.92 1.57 3.58

17. Depends on carbon group attached:    

18. Depends in hybridization:

19. Depends on anisotropy( 各向异性） Protons on an aromatic ring appears at very low field (7.27), due to the aromatic ring current. In addition, substituents with significant(diamagnetic) anisotropy e.g. NO 2 and C=O deshield the ortho protons even more

23. Chemical Shifts of aliphatic protons

24. H bond δ increases while H bond was formed among molecules. That is due to the decrease of electronic density around the nucleus Example: δ of OH in n-butene-2-ol 1% 1 100% 5

25. Peak areas The relative peak areas are given by the changes in height of the integration curve. These may be obtained by counting lines if the spectrum is recorded on ruled paper The signal strength, or peak area, as measured by electronic integration, is directly proportional to the number of identical protons in the sample which produce the signal.

27. 3.4 Spin-spin coupling

28. The signal of a proton may split into several peaks in high resolution spectrum, as shown before. This is caused by spin coupling. The distance among these peaks (Hz) is Coupling Constants. Spin coupling is due to the spin of H on the adjacent C

29. Actual H = H 0 + H’

30. Actual H = H 0 - H’

32. N+1 rule Proton will split into n+1 peaks by n protons on the adjacent carbon. The intensity of each peak can be expressed with (a + b) n

34. Coupling Constants The spacing of the lines (splitting) in the multiple peaks is called coupling constants J (Hz). They are independent on B and ν Spinning coupling can be attribute to ortho, meta and same carbon etc. Coupling constants can help us to conclude the structure of compound Δ ν/J >10 weak coupling (primary spectra) Δ ν/J <10 strong coupling (advanced spectra)

35. Primary spectra The number of split peaks depend on H of adjacent carbon. n+1 Area ratio of split peaks follow (x+1) n Advanced spectra are very complicated.

36. Interpretation of NMR Elemental analysis, UV, IR results before interpreting NMR 1Check the signal of TMS (whether it’s sharp, symmetric and at original point) 2Some peaks may from solvent (CDCl3 may have peak at δ7.25) 3Determine the H number from area

37. Interpretation of NMR 4Analyze signal peak (CH 3 O, CH3N, CH 3 - ph, CH 3 -C=O, CH 3 -C=C, CH 3 CR 3 etc) first. Then, analyze the methyl (methylene) peak with spin coupling 5 Determine the signal of COOH, CHO, OH at low field region 6The active H atom in OH, NH, COOH can be verified with addition of D 2 O

38. Interpretation of NMR 7The signal of benzene ring in aromatic compounds is complicated

39. C 4 H 10 O, 6:1:2:1

40. C 8 H 14 O 4 6:4:4

41. C 4 H 7 O 2 Cl

42. C 3 H 6 O 3, δ7.8 peak can exchange with D 2 O

43. C 11 H 13 O 2 N, 5:2:6H