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Nuclear Magnetic Resonance

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Presentation on theme: "Nuclear Magnetic Resonance"— Presentation transcript:

1 Nuclear Magnetic Resonance
Yale Chemistry 800 MHz Supercooled Magnet

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

3 The Precessing Nucleus
Ho  spin  spin Rf resonance no resonance

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

5 Observable Nuclei Odd At. Wt.; s = ±1/2 Nuclei H C N F P15 …. Abundance (%) Odd At. No.; s = ±1 Nuclei 2H N7 … Unobserved Nuclei 12C O S16…

6 E = h h = Planck’s constant:1.58x10-37kcal-sec E = h/2(Bo) = Gyromagnetic ratio: sensitivity of nucleus to the magnetic field. 1H = 2.67x104 rad sec-1 gauss-1 Thus: = /2(Bo) For a proton, if Bo = 14,092 gauss (1.41 tesla, 1.41 T), = 60x106 cycles/sec = 60 MHz and EN = Nh= cal/mole

7 Rf Field vs. Magnetic Field for a Proton
Bo E = h/2(Bo) 500 MHz 11.74 T 60 MHz 1.41 T 2.35 T 100 MHz

8 Rf Field /Magnetic Field for Some Nuclei
/2 (MHz/T) Bo (T) Rf (MHz) Nuclei 1H 13C 19F 31P 2H 500.00 11.74 42.58 125.74 11.74 10.71 76.78 11.74 6.54 470.54 11.74 40.08 202.51 11.74 17.25

9 Fortunately, all protons are not created equal!
at 1.41 T 60 MHz 60,000,000 Hz 60,000,600 Hz Me4Si upfield shielded downfield deshielded } 60 Hz 240 Hz 1 2 3 5 7 8 9 6 4 10  scale (ppm) } 500 Hz at 500 MHz  = (obs - TMS)/inst(MHz) = ( )/60 = 4.00 chemical shift

10 Where Nuclei Resonate at 11.74T
76.78 12 ppm 2H 12 ppm 1H (TMS) 500.00 RCO2H CH4 202.51 ~380 ppm 31P (H3PO4) 470.54 ~220 ppm 19F (CFCl3) 220 ppm 13C 125.74 -SO2F -CF2-CF2- CPH2 PX3 RCH3 R2C=O 500 MHz 400 300 200 100

11 Chemical Shifts of Protons

12 The Effect of Electronegativity on Proton Chemical Shifts
1 2 3 5 7 8 9 6 4 10 TMS CHCl3 =7.26 CH2Cl2 =5.30 CH3Cl =3.05 CH4 =-0.20 1 2 3 5 7 8 9 6 4 10 TMS CHBr3 =6.83 CH2Br2 =4.95 CH3Br =2.68

13 Chemical Shifts and Integrals
 2.2 1 2 3 5 7 8 9 6 4 10 TMS area = 2 area = 3  4.0 homotopic protons: chemically and magnetically equivalent enantiotopic protons: chemically and magnetically equivalent

14 Spin-Spin Splitting  = 1.2 area = 9 “anticipated” spectrum 1 2 3 5 7
1 2 3 5 7 8 9 6 4 10 TMS  = 6.4 area = 1  = 4.4 area = 1

15 Spin-Spin Splitting 1 2 3 5 7 8 9 6 4 10  TMS  = 1.2 area = 9
1 2 3 5 7 8 9 6 4 10 TMS  = 1.2 area = 9  = 4.4 area = 1  = 6.4 area = 1 observed spectrum H H H H Jcoupling constant = 1.5 Hz J is a constant and independent of field

16 Spin-Spin Splitting  = 6.4 J = 1.5 Hz = 4.4 J = 1.5 Hz 3 4 7 6 5 8 
local applied observed applied observed H local This spectrum is not recorded at ~6 MHz! 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 1 singlet (s) 1/2 2 1:1 doublet (d) 2/2 = 1 3 1:2:1 triplet (t) 3/2 4 1:3:3:1 quartet (q) 4/2 = 1 5 1:4:6:4:1 quintet (qt)

18 1H NMR of Ethyl Bromide (90 MHz)
CH3CH2Br  = 1.68 area = 3 for H spins      = 3.43 area = 2 90 Hz/46 pixels = 3J/12 pixels : J= 7.83 Hz for H spins   

19 1H NMR of Isopropanol (90 MHz)
(CH3)2CHOH 6H, d, J = 7.5 Hz 1H septuplet; no coupling to OH 4.25 3.80 ( )x90/6 = 7.5 Hz 1H, s

20 Proton Exchange of Isopropanol (90 MHz)
(CH3)2CHOH + D2O (CH3)2CHOD

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

22 Diastereotopic Protons: 2-Bromobutane at 90MHz
d1.70 (3H, d) d1.03 (3H, t) d1.82 (1H, m) d1.84 (1H, m) d4.09 (1H, m (sextet?))

23 1H NMR of Propionaldehyde: 300 MHz
d9.79 (1H, t, J = 1.4 Hz) d1.13 (3H, t, J = 7.3 Hz) Is this pattern at d2.46 a pentuplet given that there are two different values for J? No!

24 1H NMR of Propionaldehyde: 300 MHz
quartet of doublets 7.3 Hz d9.79 (1H, t, J = 1.4 Hz) d1.13 (3H, t, J = 7.3 Hz)

25 1H NMR of Ethyl Vinyl Ether: 300 MHz
J = 1.9 Hz 3H, t, J = 7.0 Hz J = 14.4 Hz J = 6.9 Hz d3.96 (1H, dd, J = 6.9, 1.9 Hz) 2H, q, J = 7.0 Hz d4.17 (1H, dd, J = 14.4, 1.9 Hz) d6.46 (1H, dd, J = 14.4, 6.9 Hz)

26 ABX Coupling in Ethyl Vinyl Ether
d6.46 14.4 6.9 ABX Coupling in Ethyl Vinyl Ether d4.17 14.4 1.9 d3.96 (1H, dd, J = 6.9, 1.9 Hz) d4.17 (1H, dd, J = 14.4, 1.9 Hz) d3.96 1.9 6.9 d6.46 (1H, dd, J = 14.4, 6.9 Hz) Another example

27 Dependence of J on the Dihedral Angle
The Karplus Equation

28 1H NMR (400 MHz): cis-4-t-Butylcyclohexanol
triplet of triplets He 3.0 He  4.03, J(Ha) = 3.0 Hz J(He) = 2.7 Hz He 4.03, J(Ha) = 3.0 Hz J(He) = 2.7 Hz 2.7 400 Hz

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

30 Peak Shape as a Function of  vs. J
1 2 3 5 7 8 9 6 4 10   >> J J 1 2 3 5 7 8 9 6 4 10  ~= J 1 2 3 5 7 8 9 6 4 10  = 0

31 Magnetic Anisotropy H Bo
Downfield shift for aromatic protons ( )

32 Magnetic Anisotropy R C H Bo
Upfield shift for alkyne protons ( ) Bo

33 13C Nuclear Magnetic Resonance
13C Chemical Shifts

34 Where’s Waldo?

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

36 13C NMR Spectrum of Ethyl Bromide at 62.8 MHz
JCH = 151 Hz C1 JCH = 5 Hz 26.6 JCH = 3 Hz 18.3 JCH = 126 Hz C2 TMS JCH = 118 Hz 10 20 30 ppm ()

37 13C NMR Spectrum of Ethyl Bromide at 62.8 MHz
TMS JCH = 118 Hz JCH = 151 Hz JCH = 126 Hz Off resonance decoupling of the 1H region removes small C-H coupling. JCH = 5 Hz JCH = 3 Hz C1 Broadband decoupling removes all C-H coupling. C2 30 20 10 26.6 18.3 ppm ()

38 13C Spectrum of Methyl Salicylate (Broadband Decoupled)

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

40 The 13C Spectrum of Camphor

41 The End F. E. Ziegler, 2004


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