The Physical Basis of the NMR Experiment

Transcription

1 The Physical Basis of the NMR Experiment 1

2 Interaction of Materials with Magnetic Fields F F S N S N Paramagnetism Diamagnetism 2

3 Microscopic View: Single Spins an electron has mass and charge in addition it possesses spin the result of the electron spin is that is makes the electron behave like a very small bar magnet therefore it should interact with a (inhomogenous) magnetic field very similar concepts apply to the atom nuclei the quantitization of the spin was first measured with the famous Stern-Gerlach experiment 3

4 The origin of the magnetic moment circulation of electric currents magnetic moments of electrons and nuclei orbital magnetic moment + - magnetic field generated by orbital motion proton + nuclear magnetic moment electron magnetic moment orbital motion electron -

5 µ µ=γj γ= gyromagnetic ratio

6

7 The Stern-Gerlach Experiment Ofen z Magnet y x 7

8 The Quantum-Mechanical Treatment µ = γi For spins with magnetic quantum number m 2s+1 possible states exist for the z-component: -mħ, -mħ+ħ, -mħ+2ħ,...,mħ Iz = mħ µ = γ mħ s z =+ ћ 2 z Epot = µzbz = γizbz z s=1/2 s=1 s z =+ћ s z =- ћ 2 s z =0 s z =-ћ

9 Classical Spinning Particle and the Spin L z z L z ћ L y L x y x y x 9

10 The principle of the NMR experiment B

11 L τ τ Nord W N ϴ S E

12 H z Ε J z = + h 4π m= +1/2 β + 1/2 γ h/2π B E=hν J z = - h 4π m= -1/2 α - 1/2 γ h/2π B Bo= 0 Bo 0 12

13 Macroscopic View: Coherent and non-coherent states 13

14 Phase Coherence (a) y (b) y μ xy,i ϕ i x x M x 14

15 The signal: Magnetic moments induce a current in the receiver coil (a) z (b) (c) V y x t M V N S t 15

16 Thermal Equilibrium Boltzman distribution: N β /N α = exp(-δe/kt) at 600 MHz ( 1 H) N β /N α = 10000:

17 Deriving Spin Precession from Classical Physics: The Bloch Equations

18 N!!!" =! φ! =!!!!!" =!! =!!!!!!" =!!!!!!! 0 0!! =!!!!!!!!! 0

19 A B = A B sin( A, B )=(Ay B z A z B y )ˆx+(A z B x A x z y )ŷ+(a x B y A y B x )ẑ so that the individual components of the new vector C are: A B = C C x =(A y B z A z B y ) C y =(A z B x A x z y ) C z =(A x B y A y B x ) An example important in the context of NMR is the derivation of Bloch sequations

20 I x ( t) = [ I x (0) cos t I y (0)sin t]e ( t / T 2 ) I y ( t ) = [ I x (0)sin t + I y (0) cos t]e ( t / T 2 ) I z ( t) = I eq + ( I z (0) I eq )e ( t / T 1 ) in these equations terms that account for relaxation have been introduced phenomenologically.

21 Trajectories of Magnetization M x z y M y x M z Time 21

22 Magnetization rotates in the transverse plane z z z x y t x y 2 t x y 22

23 y ω y(t) = A sin(ωt) A x Δt 2Δt 3Δt

24 imag. φ z(x,y) real z = a + i * b real component:= r * cos φ = r cos ωt imag. component = r * sin φ = r sin ωt z = a + ib z = r (cos φ + i sin φ) = r e i φ The quantity e i φ is called the phase in physics. For a rotational movement with constant angular speed this phase is equal to e i φ = e i ωt

25 Absolute and Relative Frequencies Ω Ω C Ω B 0 Ω A C B A ω ω 0 ω ref 0 25

26 90 pase-shifter Splitter HF (MHz) sin ωt cos ωt Mixer SFO1 ( MHz) The detector + Amplifier Audio ( 0-KHz) Analog-Digital Converter ADC FT

27 The rotating frame of reference z z LAB y x y' x' 27

28 The effect of RF pulses

29 z B 0 ω 0 ω RF ω 1 B 1

30 The transformation into the rotating frame B o ω B o ω B o ω x y ω 0 B 1 ω ω 0 ω 0

31 B 0 - ω/γ B eff B B eff B 1 B B eff B eff B 1 B 1 ω o ω o ω o

32 Residual fields in z in the rotating frame z z z B 0 ' y' M C x' B 0 ' y' M B x' Ω A y' M A x' Ω C Ω B B 0 ' Ω Ω C Ω B 0 Ω A 32

33 z z z z B' 0 B eff M C M A y B 1 x B' 0 B 1 M B x M y B 1 x y B 1 x B' 0 B eff On-resonant Ω C Ω 0 Ω B Ω A

34 Time -ω RF +ω RF B RF z ϕ RF t B RF I

35 1 0.5 M x, M y [khz]

36 Relaxation

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38 Longitudinal (T1) Relaxation z z z z z y y y y y x x x x x #(α) #(β) = e (E α E β )/ kt ->enthalpic process M (t) = M (t ) e t /T1 z z 0

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40 The origin of broad lines: Efficient transverse (T2) relaxation Bo M x,y (t) = M x,y (t 0 ) e t /T 2 Δν 1 / 2 = 1 πt 2

41 M x x M x y M z

42 The chemical shift ω = γ B 0 ω = γ B eff σ dia (local) + σ para (local) + σ m + σ rc + σ ef + σ solv σdia the diamagnetic contribution σpara the paramagnetic contribution σm the neighbor anisotropy effect σef the electric field effect σsolv the solvent effect

43 The chemical shift tensor ind

44 B eff =!B 0 + B 0!!"#!!!!!!"!!!!!!!!!!!"!!!!!"!!!!!!!

45 x y =

46 B 0 nuclear spin flow of electrons induced field

47 Bo Lamb Shift σ iso = µ oe 2 3m e o rρ(r)dr

48 The paramagnetic contribution Bo LUMO HOMO r radius of 2p orbital ΔE energy difference (LUMO-HOMO) Q electron densities and bond orders σ i para = µ o µ B 2 2πΔE r 3 ' Q + ) i ( i j Q j *, +

49 Chemical Shift Anisotropy σ iso = 1 3 (σ xx +σ yy + σ zz )

50 Chemical Shifts of paramagnetic compounds Fermi contact shift: Electrons generate an additional magnetic field at the nucleus through the electron magnetic moment located at the nucleus itself. Electron density at the nucleus is important (s-character of orbitals!) (g is the electronic g factor, and α N is the electron nuclear spin coupling constant) limited to close vicinity of the paramagnetic center! Large shifts (up to 50 ppm) Δσ = 2πα N gµ B S(S +1) 3γ I KT

51 Chemical Shifts of paramagnetic compounds (II) Pseudo contact shift: is the dipolar interaction between the nucleus and the electron Δσ = µ B 2 S(S +1)(g 2! g 2 ) (1 3cos2 θ) 9KTr 3 r is the distance of the paramagnetic center to the nucleus and θ is the angle between r and the symmetry axis of the g factor. The pseudocontact shift can be an intermolecular effect. Since its interaction depends on the r 3, it can reach much further than the direct contact term (up to 20 Å), but the shifts are usually much smaller. Pseudocontact shifts are used in Bio-NMR to derive structural restraints

52 The Use of Pseudocontact Shifts (PCS) >> gives distance and spatial information! signals from close residues are bleached, those further away are shifted in a distance and orientationdependent manner Isosurfaces depicting the pseudo-contact shifts(pcss) induced by Dy3+ plotted on a ribbon representation of the crystal structure of ε Rigid body docking of the proteins ε186 and θ by superposition of the χ tensors of Dy3+ and Er3+ determined with respect to the individual proteins. Otting, Acc. Chem. Res. 40 (2007), 210

53 Scalar couplings (line splittings) The splittings of signals are due to interactions with other spins. The Magnitude of these couplings depends on the number of intervening bonds

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55 J (2 µ o g e µ B 3 ) 2 γ A γ B ψ A (0) 2 ψ B (0) 2 c A 2 c B 2 1 Δ T ΔT

56 Reduced scalar couplings κ ik = J ik h % 2π (% ' * 2π ( ' * & γ i )& γ k )

57 2J couplings X Hund s rule Fermi-contact 57

58 3J couplings electron negative substituents decrease 3 J increasing HCC bond angles decrease 3 J increasing C--C bond lengths decrease 3 J

59 The dependance of the 3J vicinal coupling is due to differences in orbital overlap φ

60 The Karplus relationship 3J(X,Y) X θ Y θ

61 The 3 J HN,Hα coupling reveals stereochemical information Sekundärstruktur θ 3 JHN-αH α Helix Hz Helix Hz antiparalleles β- Blatt Hz paralleles β-blatt Hz

62 Long-range couplings require polarization of pi-electrons (or zigzag arrangements) C H H C C H H C C C H H C C C H

63 Energy-level diagram for a two-spin system I S S I

64 Multiple-Quantum Transitions ββ ββ ββ β β βα αβ βα αβ βα αβ βα αβ αα αα αα αα single -quantum transitions zero -quantum transition double -quantum transition ω=ω I ω=ω I -ω S ω=ω I +ω S

65 Quadrupolar coupling

66 Quadrupolar nuclei may result in very broad lines H 3 N H 3 N NH 3 Co NH 3 NH 3 NH khz Co [ppm] Co [ppm]

67 Dipolar coupling B 0 B 0 r I,S μ S S I θ μ I!!!!!!!!!!!!!!!!!!!!