Yale Chemistry 800 MHz Supercooled Magnet. Nuclear Magnetic Resonance

Transcription

1 Yale Chemistry 800 Mz Supercooled Magnet Nuclear Magnetic Resonance

2 B o Atomic nuclei in The absence of a magnetic field Atomic nuclei in the presence of a magnetic field α spin - with the field β spin - opposed to the field

3 The Precessing Nucleus resonance o α spin resonance β spin no resonance no resonance R f

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

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

6 ΔE = hν ΔE = γh/2π(b o ) h = Planck s constant:1.58x10-37 kcalsec γ = Gyromagnetic ratio: sensitivity of nucleus to the magnetic field. 1 = 2.67x10 4 rad sec -1 gauss -1 Thus: ν = γ/2π(b o ) For a proton, if B o = 14,092 gauss (1.41 tesla, 1.41 T), ν = 60x10 6 cycles/sec = 60 Mz and ΔE N = Nhν = cal/mole

7 R f Field vs. Magnetic Field for a Proton E 60 Mz 100 Mz ΔE = γh/2π(b o ) 500 Mz 1.41 T 2.35 T T B o

8 R f Field /Magnetic Field for Some Nuclei Nuclei R f (Mz) B o (T) γ/2π (Mz/T) 1 13 C 2 19 F 31 P

9 Fortunately, all protons are not created equal! 60 Mz 60,000,000 z at 1.41 T 60,000,600 z downfield deshielded 60 z } upfield shielded 240 z Me 4 Si δ scale (ppm) at 500 Mz } 500 z δ = (ν obs - ν TMS )/ν inst(mz) = ( )/60 = 4.00 chemical shift

10 Where Nuclei Resonate at 11.74T 12 ppm 2 RCO 2 12 ppm 1 (TMS) ~380 ppm 31 P ( 3 PO 4 ) ~220 ppm ppm 19 F (CFCl 3 ) C 4 13 C PX 3 CP 2 -SO 2 F -CF 2 -CF R 2 C=O Mz

11 Chemical Shifts of Protons

12 The Effect of Electronegativity on Proton Chemical Shifts TMS CCl 3 C 2 Cl 2 C 3 Cl C 4 δ=7.26 δ=5.30 δ=3.05 δ= δ TMS CBr 3 δ=6.83 C 2 Br 2 δ=4.95 C 3 Br δ= δ

13 Chemical Shifts and Integrals δ 4.0 δ 2.2 area = 3 TMS area = δ homotopic protons: chemically and magnetically equivalent 3 C Cl C C 2 Cl enantiotopic protons: chemically and magnetically equivalent Cl

14 Spin-Spin Splitting anticipated spectrum δ = 1.2 area = 9 TMS δ = 6.4 area = 1 δ = 4.4 area = δ Br Br C 3 Br C C C C 3 C 3

15 Spin-Spin Splitting observed spectrum δ = 1.2 area = 9 TMS δ = 6.4 area = 1 δ = 4.4 area = 1 β α α β δ J coupling constant = 1.5 z Br Br C Br C C 3 C C 3 C 3 J is a constant and independent of field

16 Spin-Spin Splitting δ = 6.4 J = 1.5 z δ = 4.4 J = 1.5 z β α α ν local β ν local ν observed ν observed δ ν applied ν applied Br Br C 3 This spectrum is not recorded at ~6 Mz! δ and J are not to scale. Br C C C C 3 C 3

17 Multiplicity of Spin-Spin Splitting for s = ±1/2 multiplicity (m) = 2Σs + 1 # equiv. neighbors spin (1/2) multiplicity pattern (a + b) n symbol singlet (s) 1 1/2 2 1:1 doublet (d) 2 2/2 = 1 3 1:2:1 triplet (t) 3 3/2 4 1:3:3:1 quartet (q) 4 4/2 = 1 5 1:4:6:4:1 quintet (qt)

18 1 NMR of Ethyl Bromide (90 Mz) C 3 C 2 Br for spins ααβ αββ αβα βαβ ααα βαα ββα βββ for spins αβ αα βα ββ δ = 3.43 area = 2 90 z/46 pixels = 3J/12 pixels : J= 7.83 z δ = 1.68 area = 3

19 1 NMR of Isopropanol (90 Mz) (C 3 ) 2 CO 1 septuplet; no coupling to O 6, d, J = 7.5 z ( )x90/6 = 7.5 z 1, s δ4.25 δ3.80

20 Proton Exchange of Isopropanol (90 Mz) (C 3 ) 2 CO + D 2 O (C 3 ) 2 COD

21 Diastereotopic Protons: 2-Bromobutane Br homotopic sets of protons pro-r Diastereotopic protons R pro-s

22 Diastereotopic Protons: 2-Bromobutane at 90Mz Br δ1.03 (3, t) δ1.70 (3, d) δ1.82 (1, m) δ1.84 (1, m) δ4.09 (1, m (sextet?))

23 1 NMR of Propionaldehyde: 300 Mz O δ1.13 (3, t, J = 7.3 z) δ9.79 (1, t, J = 1.4 z) Is this pattern at δ2.46 a pentuplet given that there are two different values for J? No!

24 1 NMR of Propionaldehyde: 300 Mz δ2.46 O 7.3 z 7.3 z 7.3 z 1.4 z 7.3 z 7.3 z quartet of doublets δ9.79 (1, t, J = 1.4 z) δ1.13 (3, t, J = 7.3 z)

25 1 NMR of Ethyl Vinyl Ether: 300 Mz J = 14.4 z J 3, t, J = 7.0 z = 1.9 z O J = 6.9 z δ3.96 (1, dd, J = 6.9, 1.9 z) δ4.17 (1, dd, J = 14.4, 1.9 z) 2, q, J = 7.0 z δ6.46 (1, dd, J = 14.4, 6.9 z)

26 ABX Coupling in Ethyl Vinyl Ether δ O δ δ3.96 (1, dd, J = 6.9, 1.9 z) δ4.17 (1, dd, J = 14.4, 1.9 z) δ3.96 δ6.46 (1, dd, J = 14.4, 6.9 z) Another example

27 Dependence of J on the Dihedral Angle The Karplus Equation

28 1 NMR (400 Mz): cis-4-t-butylcyclohexanol O e e a e triplet of triplets a e e δ e δ4.03, J( J( a ) a = ) = z z J( J( e ) e = ) = z z 11.4 z 400 z

29 1 NMR (400 Mz): trans-4-t-butylcyclohexanol a e O a e triplet of triplets a a a δ3.51, J( a ) = 11.1 z J( e ) = 4.3 z z

30 W X Y Z Peak Shape as a Function of Δδ vs. J Δδ >> J Δδ J δ Δδ ~= J δ Δδ = δ

31 Magnetic Anisotropy Downfield shift for aromatic protons (δ ) B o

32 Magnetic Anisotropy R C C Upfield shift for alkyne protons (δ ) B o

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 Mz J C = 126 z J C = 151 z C 1 J C = 3 z C 2 Br J C = 5 z TMS J C = 118 z Si ppm (δ)

37 13 C NMR Spectrum of Ethyl Bromide at 62.8 Mz Off resonance decoupling of the 1 region removes small C- coupling. J C = 126 z J C = 151 z C 1 J C = 3 z Broadband decoupling removes all C- coupling. C 2 Br J C = 5 z TMS J C = 118 z Si ppm (δ)

38 13 C Spectrum of Methyl Salicylate (Broadband Decoupled)

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

40 The 13 C Spectrum of Camphor O

41 F. E. Ziegler, 2004 The End