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Chapter 14 Nuclear Magnetic Resonance Spectroscopy

Chapter 14 Nuclear Magnetic Resonance Spectroscopy. “study of interaction of light with molecules ”. 14.1 Proton Magnetic Resonance Spectroscopy. Two or more E states are generated when external magnetic field is applied radio wave absorption between these states NMR spectroscopy.

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Chapter 14 Nuclear Magnetic Resonance Spectroscopy

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  1. Chapter 14 Nuclear Magnetic Resonance Spectroscopy “study of interaction of light with molecules”

  2. 14.1Proton Magnetic Resonance Spectroscopy • Two or more E states are generated when external magnetic field is applied • radio wave absorption between these states • NMR spectroscopy Hydrogen nucleus (1H) and the isotope of carbon (13C) 1H NMR and 13C NMR Three types of information for H NMR spectrum Chemical shift: the carbon to which proton is attached  Multiplicity( # of the peaks in each group): neighboring hydrogens  Integral: # of hydrogens

  3. Two spin states are generated when external magnetic field is applied ( 546 Figure 14.3) • E (E difference between the two states) = hBo/2 h = plank constant,  = magnetogyric ratio (characteristic value for each nucleus) Bo = external magnetic field 14.2 Theory of 1H-NMR

  4. Typical values of E of proton (reference TMS) • External magnetic field 14,000 gauss (or 1.4-tesla) then E =10-6 kcal/mol 60 x 106 s-1 or 60 MHz 46,700 gauss (4.67-tesla) magnet  200 MHz (in this book) 70,000 gauss  300 MHz 9.33-tesla magnetic  400 MHz (in this book) Magnetic field , resolution of NMR   to get high magnetic field superconducting maget is needed (superconducting magnet is operated using liquid helium) Magnetic field , $ 

  5. 13C, 19F, 2H and 31P have nuclear spins  NMR technique can be applied 12C and 16O do not have nuclear spins  no NMR absorption Why? Ask god!

  6.  the field required for the hydrogen to resonance varies slightly with the chemical environment of the hydrogen • The required field difference between the hydrogen of the sample and that of TMS (tetramethylsilane) 14.3The Chemical Shift Chemical shift () = ex) 200MHz ( 46,700 gauss) 1H-NMR : aceton peak at 436 Hz (436 x 106/200x106 =2.16  or 2.16 ppm in 300 MHz NMR, this peak will be detected at 654 Hz  548 Figure 14.4

  7. Factors on Chemical Shift () downfield: left on an NMR spectrum  deshielded  increased external field  higher  upfield: right on an NMR spectrum  shielded  decreased external field  lower  Inductive effect

  8. Pi Electron Effect deshielding effect  higher  • Exchangeable protons: O-H/N-H (H-bonding) • O-H: 2~5 (ppm), N-H: 1~3, CO2H: ~12 (10~15) • peak position varies

  9. Protons with chemically identical environment Chemical equivalence Three different sets of chemically equivalent protons  Three NMR absorptions

  10. Enantiomer: identical chemical shift Diastereomer: different chemical shift Two different sets of chemically equivalent protons  two NMR absorptions Three different sets of chemically equivalent protons  Three NMR absorptions

  11. one NMR absorption peak two NMR absorption peaks five NMR absorption peak

  12. A useful summary of the chemical shifts

  13. Estimation of the chemical shift approximate chemical shift of hydrogens

  14. Empirical Correlation for predicting Chemical Shifts  Shoolery’s Rule (old text book) Ex) To the value from Table 14.1, add 0.3 ppm if CH2 0.7 ppm if CH 2-5 ppm according to concentration (actual value= 2.3 ppm) O-CH3 = 3.3 ppm, 3.3 + 0.3 (due to CH2) = 3.6 ppm (actual value= 3.5 ppm) C-CH3 = 0.9 ppm, 0.9 + 0.3 ( CH2) + 0.3 (CH3) = 1.5 ppm (actual value= 1.5 ppm) 0.9 ppm (C-CH3 )

  15. Practice

  16. Geminal coupling = Two-Bond Coupling If they are diastereotopic, then have differentchemical shifts 14.4Spin Coupling (H,H Coupling) Vicinal coupling = Three-Bond Coupling Very thoroughly studied, both experimentally and theoretically Long-range Coupling: very small (not easily detected)

  17. spin coupling: change in the local magnetic field by nearbyH atoms having different d multiplicity: Bo BH;  559 & Figure 14.6~7 splitting pattern: Jax in 0~20 Hz (typical 6);s (singlet), d (doublet), t (triplet), q (quartet), m (multiplet), br (broad) Differing spin relationships between HA and HB

  18. The origin of spin-spin spliting in proton A’s NMR spectrum NMRspectrum of this compound

  19. Spin-spin splitting in ethyl iodide (CH3CH2I) Splitting pattern of methyl hydrogens Splitting pattern of ethyl hydrogens

  20. Spin-spin splitting rule: n+1 rule Two neighboring give triplet (three peaks) n+1=3 One neighboring give douplet two peaks n+1=2 Pascal’s triangle  Peak number and area (height)

  21. The Coupling Constant (J) In 1H-NMR, peak area indicates the number of hydrogen for the corresponding absorption See  559 Fig. 14.6~7

  22. Proton is coupled to nonequivalent protons

  23. 14.5Complex Coupling Leaning  The downfield peak of the triplet is slightly larger than the upfield peak.  The quartet leans toward the triplet As the difference in chemical shift becomes smaller, the leaning effect becomes larger. 562 Figure 14.8 vs  569 Figure 14.10

  24. NMR spectrum at RT  one peak NMR spectrum at low temp  two peaks 14.6Chemical Exchange No coupling in NMR spectrum of methanol (no splitting) unless the methanol is extremely pure In general, H in oxygen or nitrogen are subject to the rapid exchange and do not couple to nearby H’s  SLOW CAMERA EFFECT

  25. Solvent for NMR  no signal not to interfere the spectrum of the sample CCl4 can be used  poor solubility • Deuterated solvents (CDCl3, (CD3)2SO, C6D6) are used in NMR • Deuterium is invisible in NMR 1H-NMR measures E + small chemical shift • E (E difference between the two states) = hBo/2 • (magnetogyric ratio) of the nucleus of deuterium is different from that of hydrogen  it does not appear in 1H-NMR carbocations by NMR:  565 Focus On 14.7Deuterium

  26. Step 1. Examine the general position of the peak. Step 2. Examine the integral for the ratios of the different kinds of hydrogen. Step 3. Examine the coupling pattern. Step 4. Construct a tentative structure Step 5. Determine whether all the information is consistent with this structure 14.8Interpretation of 1H-NMR Spectra

  27. 14.913C-NMR Spectrum • Chemical shift range: 0~240 ppm • Broader than that of 1N-NMR • resolution is much higher There is no, if any, 13C -13Ccoupling natural abundance: 13C 1.1% & 12C 98.9% There is 13C –1Hcoupling, while this is removed by broad band decoupling technique Then in 13C-NMR, all peaks are singlet

  28. Other 13C-NMR Techniques Off-resonance decoupling which allows hydrogens and carbons that are directly bonded to couple CH3:quartet, CH2:triplet, CH:doublet, C:singlet  Old technique DEPT-NMR: Three spectra are obtained: New technique Broad band decoupled spectrum DEPT 90o spectrum: only CH’s appear DEPT 135o: CH’s and CH3’s appear as positive absorptions and CH2’s appears as negative absorptions see next page

  29. Chemical equivalence in 13C-NMR

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