1. Nuclear Magnetic Resonance Yale Chemistry 800 MHz Supercooled Magnet

2. Atomic nuclei in The absence of a magnetic field BoBo Atomic nuclei in the presence of a magnetic field  spin - with the field  spin - opposed to the field

3. The Precessing Nucleus HoHo  spin  spin RfRf resonance no resonanceno resonanceno resonanceno resonance

4. The Continuous Wave Spectrometer The Precessing Nucleus Again The Fourier Transform Spectrometer

5. Observable Nuclei Odd At. Wt.; s = ±1/2 Nuclei 1 H 1 13 C 6 15 N 7 19 F 9 31 P 15 …. Abundance (%) 99.98 1.1 0.385 100 100 Odd At. No.; s = ±1 Nuclei 2 H 1 14 N 7 … Unobserved Nuclei 12 C 6 16 O 8 32 S 16 …

6.  E = h h = Planck’s constant:1.58x10 -37 kcal-sec  E =  h/2  (B o )  = Gyromagnetic ratio: sensitivity of nucleus to the magnetic field. 1 H = 2.67x10 4 rad sec -1 gauss -1 Thus:  =  /2  (B o ) For a proton, if B o = 14,092 gauss (1.41 tesla, 1.41 T),  = 60x10 6 cycles/sec = 60 MHz and  E N = Nh  = 0.006 cal/mole

7. E BoBo 500 MHz 11.74 T 2.35 T 100 MHz 60 MHz 1.41 T  E =  h/2  (B o ) R f Field vs. Magnetic Field for a Proton

8. R f Field /Magnetic Field for Some Nuclei  /2  (MHz/T) B o (T)R f (MHz)Nuclei 1H1H 13 C 19 F 31 P 2H2H 500.0011.7442.58 125.7411.7410.71 76.7811.746.54 470.5411.7440.08 202.5111.7417.25

9. Fortunately, all protons are not created equally! at 1.41 T 60 MHz 60,000,000 Hz upfield shielded downfield deshielded 60,000,600 Hz Me 4 Si } 60 Hz } 500 Hz at 500 MHz 012357896410  scale (ppm)  = ( obs - TMS )/ inst(MHz) = (240.00 - 0)/60 = 4.00 240 Hz chemical shift

10. Where Nuclei Resonate at 11.74T 470.54 ~220 ppm 19 F (CFCl 3 ) 12 ppm 1 H (TMS) 500.0076.78 12 ppm 2 H 220 ppm 13 C 125.74 500 MHz400300200100 RCO 2 H CH 4 -SO 2 F-CF 2 -CF 2 - 202.51 ~380 ppm 31 P (H 3 PO 4 ) CPH 2 PX 3 RCH 3 R 2 C=O

11. Chemical Shifts of Protons

12. The Effect of Electronegativity on Proton Chemical Shifts 012357896410  TMS CH 4  =-0.20 CH 3 Cl  =3.05 CH 2 Cl 2  =5.30 CHCl 3  =7.26 CH 2 Br 2  =4.95 CH 3 Br  =2.68 CHBr 3  =6.83 012357896410  TMS

13. Chemical Shifts and Integrals 012357896410  TMS homotopic protons: chemically and magnetically equivalent enantiotopic protons: chemically and magnetically equivalent  2.2  4.0 area = 2 area = 3

14. Spin-Spin Splitting 012357896410  TMS  = 1.2 area = 9  = 4.4 area = 1  = 6.4 area = 1 “anticipated” spectrum

15. Spin-Spin Splitting observed spectrum 012357896410  TMS  = 1.2 area = 9  = 4.4 area = 1  = 6.4 area = 1 HH HH HH HH J coupling constant = 1.5 Hz J is a constant and independent of field

16. Spin-Spin Splitting  = 4.4 J = 1.5 Hz  = 6.4 J = 1.5 Hz 347658  H   local  applied  observed  applied  observed H  local This spectrum is not recorded at ~6 MHz! HH HH  and J are not to scale.

17. Multiplicity of Spin-Spin Splitting for s = ±1/2 multiplicity (m) = 2  s + 1 # equiv. neighbors spin (1/2)multiplicity pattern (a + b) n symbol 0011 singlet (s) 11/221:1doublet (d) 22/2 = 131:2:1triplet (t) 33/241:3:3:1quartet (q) 44/2 = 151:4:6:4:1quintet (qt)

18. 1 H NMR of Ethyl Bromide (90 MHz) CH 3 CH 2 Br  = 3.43 area = 2  = 1.68 area = 3 for H spins     for H spins    90 Hz/46 pixels = 3J/12 pixels : J= 7.83 Hz

19. 1 H NMR of Isopropanol (90 MHz)  4.25  3.80 1H septuplet; no coupling to OH (4.25-3.80)x90/6 = 7.5 Hz (CH 3 ) 2 CHOH 6H, d, J = 7.5 Hz 1H, s

20. Proton Exchange of Isopropanol (90 MHz) (CH 3 ) 2 CHOH+ D 2 O(CH 3 ) 2 CHOD

21. Diastereotopic Protons: 2-Bromobutane homotopic sets of protons Diastereotopic protons R pro-R pro-S

22. Diastereotopic Protons: 2-Bromobutane at 90MHz  1.03 (3H, t)  1.70 (3H, d)  1.82 (1H, m)  1.84 (1H, m)  4.09 (1H, m (sextet?))

23. 1 H NMR of Propionaldehyde: 300 MHz  9.79 (1H, t, J = 1.4 Hz)  1.13 (3H, t, J = 7.3 Hz) Is this pattern at  2.46 a pentuplet given that there are two different values for J? No!

24. 1 H NMR of Propionaldehyde: 300 MHz  9.79 (1H, t, J = 1.4 Hz)  1.13 (3H, t, J = 7.3 Hz) 1.4 Hz 7.3 Hz  2.46 quartet of doublets 7.3 Hz

25. 1 H NMR of Ethyl Vinyl Ether: 300 MHz 3H, t, J = 7.0 Hz2H, q, J = 7.0 Hz J = 6.9 Hz J = 1.9 Hz J = 14.4 Hz  3.96 (1H, dd, J = 6.9, 1.9 Hz)  4.17 (1H, dd, J = 14.4, 1.9 Hz)  6.46 (1H, dd, J = 14.4, 6.9 Hz)

26.  3.96 (1H, dd, J = 6.9, 1.9 Hz)  6.46 (1H, dd, J = 14.4, 6.9 Hz)  4.17 (1H, dd, J = 14.4, 1.9 Hz)  6.46 14.4 6.9  4.17 14.4 1.9  3.96 1.9 6.9 ABX Coupling in Ethyl Vinyl Ether Another example

27. Dependence of J on the Dihedral Angle The Karplus Equation

28. 1 H NMR (400 MHz): cis-4-t-Butylcyclohexanol H e  4.03, J(H a ) = 3.0 Hz J(H e ) = 2.7 Hz H e  4.03, J(H a ) = 3.0 Hz J(H e ) = 2.7 Hz 400 Hz HeHe 11.4 Hz triplet of triplets 2.7 3.0

29. 1 H NMR (400 MHz): trans-4-t-Butylcyclohexanol H a  3.51, J(H a ) = 11.1 Hz J(H e ) = 4.3 Hz HaHa 30.8 Hz 11.1 4.3 triplet of triplets

30. Peak Shape as a Function of  vs. J 012357896410   = 0 012357896410    >> J J 012357896410   ~= J

31. BoBo Downfield shift for aromatic protons (  6.0-8.5) Magnetic Anisotropy H H HH HH

32. BoBo Upfield shift for alkyne protons (  1.7-3.1) Magnetic Anisotropy C C R H

33. 13 C Nuclear Magnetic Resonance 13 C Chemical Shifts

34. Where’s Waldo?

35. One carbon in 3 molecules of squalene is 13 C What are the odds that two 13 C are bonded to one another? ~10,000 to 1

36. 13 C NMR Spectrum of Ethyl Bromide at 62.8 MHz 01020 30 ppm (  ) TMS J CH = 118 Hz J CH = 151 Hz C1C1 J CH = 5 Hz 26.6 J CH = 3 Hz 18.3 J CH = 126 Hz C2C2

37. 13 C NMR Spectrum of Ethyl Bromide at 62.8 MHz 01020 30 ppm (  ) C1C1 26.618.3 J CH = 5 Hz J CH = 3 Hz C2C2 Off resonance decoupling of the 1 H region removes small C-H coupling. Broadband decoupling removes all C-H coupling. TMS J CH = 118 Hz J CH = 151 Hz J CH = 126 Hz

38. 13 C Spectrum of Methyl Salicylate (Broadband Decoupled)

39. 13 C Spectrum of Methyl p-Hydroxybenzoate (Broadband Decoupled)

40. The 13 C Spectrum of Camphor

41. The End F. E. Ziegler, 2004