Physikalische Chemie IV (Magnetische Resonanz) HS Solution Set 2. Hand out: Hand in:

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1 Solution Set Hand out:.. Hand in:.. Repetition. The magnetization moves adiabatically during the application of an r.f. pulse if it is always aligned along the effective field axis. This behaviour is observed when the orientation of the effective field changes slowly compared to its strength, i.e. dθ /dt << ω eff. A resonance line is inverted adiabatically by an r.f. pulse if the irradiation frequency is continuously changed from far below the resonance frequency to far above the resonance frequency.. The H T relaxation rate constant can be estimated from a one-dimensional inversion recovery experiment (Figure.8 PC IV script) in which the 9 pulse and the receiver are phase cycled in 9 increments. The recycle delay should be greater than 6T to ensure complete relaxation between transients. The magnitude of the transverse magnetization following the 9 pulse varies with the delay τ as: The value of T is given by M z (τ ) = [ exp( τ /T )] () T = τ /ln () Where τ represents the delay when the signal is zero. More generally, a monoexponential fit of the build-up of the magnetization yields T (see Figure ). can be determined using a spin echo experiment (Figure.6 PC IV script) by measuring the intensity of the FID for a series of values for τ.

2 inversionrecovery experiment including best fit 3 intensity / a.u log ( / s) Figure. 3. Possible reasons for the decay of the x-y magnetization are: - relaxation; - B inhomogeneities; - Differences in chemical shift; The Hahn echo pulse sequence (Figure.6 PC IV script) refocuses the last two mechanisms. Transverse magnetization created by the first pulse decays during the τ period due to the loss of in phase coherence of the individual spin vectors. The 8 pulse which is applied at the end of the τ period flips the relative orientations of the spin vectors. effects are not refocused by the spin echo. 4. It is possible to measure chemical-exchange processes in the dynamic equilibrium regime using NMR because the difference in energy between two multiple states matches the exchange rates we are interested in. Furthermore, the system is at equilibrium throughout the measurement. It is conceivable that electron paramagnetic resonance can also probe chemical exchange in dynamic equilibrium as long as the electronic structure is not affected (i.e. molecular rotations and translations).

3 Problem : Continuous-wave (CW) Spectroscopy. With an effective field of r B eff = (, B, ΔB ) this leads to the following three equations: = γm r y ΔB γb M r z M r x = γm r x ΔB M r y ( ) = γm r x B M r z T (3) which have the following solutions: M x r = γb / (+ (γb )/ + (γδb M y r = γb γδb (+ (γb )/ + (γδb M z r = / + (γδb (+ (γb )/ + (γδb (4). If the applied B field is weak in the sense that (γb <<, the solutions can be simplified: M x r = γb / / + (γδb M y r = γb γδb / + (γδb M z r = / + (γδb / + (γδb = (5) 3. In the weak r.f. limit, M x r and M y r are given by the absorptive and dispersive parts of a Lorentzian line: 3

4 Figure. M x r and M y r components as a function of ΔB assuming that γ is negative. 4. The line shape is given by M x r = γb / ((γb / )+ (γδb (6) and the derivatives are r dm x = γ 3 ΔB dδb B (γb / [( ) + ( γδb ] (7) 4

5 dm x r [( + ( γδb ] γb M /T = γm db (γb / / [((γb / ) + ( γδb ] d ((γb / ) db ((γb [( + ( γδb ] / ) + ( γδb T ( γb T /T = γm (γb / Setting the right-hand side of equation (7) equal to zero yields: ΔB = for B and. This shows M x r becomes extremal for perfect on-resonance pulses. By calculating the second derivative, one can show that the absorption line reaches a maximum for ΔB =. Analogously, equation (8) gives the extremum with respect to B : (8) B = γ T + T γδb ( (9) 5. The maximum for the absorption line with respect to ΔB and B is M x r = T () with ΔB =, B = γ T insert in equation (6). Problem : Composite pulses. The z-magnetization after a (π) y pulse is: M '= R y (θ) M cosθ sinθ = sinθ cosθ The z-component after the pulse is: M z ' = cos(θ) (I) where θ=-γb τ (plot given in Figure 3). 5

6 . This problem can be solved immediately just by thinking about the rotations. Irrespective of how far the first pulse around the x axis rotates the magnetization the perfect (π) y pulse will invert the z and x components of the magnetization, so that the final pulse will align matrices: M with the z-axis (Fig..7 PC IV script). We can also show this with the rotation M '= R x (φ)r y (π)r x (φ) M = cosφ sinφ cosφ sinφ sinφ cosφ sinφ cosφ = cosφ sinφ sinφ = sinφ cosφ + sinφ cosφ sinφ cosφ cosφ sin φ cos = φ (II) using cos φ + sin φ = for all φ. 3. We calculate M z analogously to. Using the rotation matrix R y (θ), with θ=φ. M '= R x (φ)r y (θ)r x (φ) M cosθ sinθ = cosφ sinφ cosφ sinφ sinφ cosφ sinθ cosθ sinφ cosφ cosθ sinθ = sinφ sinθ cosφ sinφ cosθ sinφ cosφ sinθ sinφ cosφ cosθ cosφ We obtain the M z ' component using φ = θ : θ M ' z (θ) = sin θ + cos cosθ (III) The functions (I), (II) and (III) are plotted in Figure 3. 6

7 Figure 3. Effect of composite pulses. Graphical representation of M z = f (θ). Problem 3: Chemical Exchange in acetyl-lysine (D). The exchange rate can be calculated using = k diff c(oh ) + pk Lys pk Water and c(oh ) mol /l ph 4 = The difference of the chemical shifts in angular frequencies is given by: Ω Lys Ω Water = πν δ Lys δ Water = 8859 rad / s To decide whether we are in the limit of fast or slow exchange, we have to evaluate r: r = Ω Lys Ω Water If r <<, which means << Ω Lys Ω Water we are in the limit of slow exchange and we have two separate lines and we can calculate the position δ and the line width Δν as follows: δ = δ Water = 4.78 ppm; δ = δ Lys = 7.6 ppm; Δν = Δν = π If r >>, which means >> Ω Lys Ω Water we are in the limit of fast exchange and we have only one line and we can calculate the position δ and the line width Δν as follows: 7

8 δ = δ Lys +δ Water ( Δν = Ω Lys Ω Water π 4 The results for the different ph values are given in the table below. ph / s - r n width / Hz position /ppm and x and x x Coalescence appears when: r = = Ω Lys Ω Water.35 Combining this equation with the ph-dependence of the exchange rate, we obtain: c(oh ) mol /l = Ω Lys Ω Water (+ pk Lys pkwater ) k diff = This corresponds to a ph of In an aqueous solution it is quite unlikely that there is symmetric exchange between the water and the amine protons. Symmetric exchange means that = k' ex, which implies under equilibrium conditions that the populations of both sites are the same. However, in an aqueous solution there are more water molecules than acetyl-lysine molecules. At ph 9. the exchange is in the fast limit and so only one line is observed. Its position can be used to get an estimation of the relative populations of acetyl-lysine protons and water protons. p Lys = 4.9 ppm δ Water δ Lys δ Water.5; p Water = 4.9 ppm δ Lys δ Water δ Lys.95 8

9 This means that there are about times as many water protons in the solution then amine protons. Therefore, should be twenty times larger than k' ex, so that there is no net transfer from the water to the acetyl-lysine or in the other direction. At coalescence we can write ' + = +.5 = Ω Lys Ω Water = 665 s.5 = 5966 s Using the relationship between the ph and the exchange rate, we get for the c(oh - ) c(oh ) mol /l = k diff (+ pk Lys pk Water ) = which corresponds to a ph of