3 Chemical exchange and the McConnell Equations

Transcription

1 3 Chemical exchange and the McConnell Equations NMR is a technique which is well suited to study dynamic processes, such as the rates of chemical reactions. The time window which can be investigated in NMR is in the range of nanoseconds to seconds. It is sampled using different experimental observables, such as relaxation rates, lineshape analysis, etc. In this chapter, we will describe how we can use NMR to study chemical exchange processes of the type:

2 k A1 A k local field: B B Resonance frequency: ω 01 Ω 01 0 ω 0 1 Ω (lab frame) (rot. frame) Examples include internal rotations about a bond, ring puckering, proton exchange, and bond rearrangements. To describe chemical exchange, we can continue with our semi-classical description and use a modified Bloch equations (also known as McConnell equations, after H.M. McConnell who derived them in 1958). 3.1 McConnell Equations Let s take a spin which can be in state A 1 or A (which is equivalent to two different environments) as shown above. If there is no exchange, we can

3 write Bloch equations for both states, namely: dm 1 dt dm dt = γ M 1 B 1 R 1 ( M 1 M 10 ) = γ M B R ( M M 0 ) (3.1) Beware: R here is NOT 1 T. Similarly for R 1. R 1 and R represent the full relaxation matrix for our spin in state A 1 and A, respectively. If we assume now that chemical exchange occurs instantaneously, i.e. that we do not give the system the time to gradually change from A 1 to A and vice versa as this would affect the parameters in equation 3.1 (think of relaxation, for instance). In this case, then we can add first order kinetics terms to the Bloch equations above, such that the overall change in magnetization over time is given by dm 1 dt dm dt = γ M 1 B 1 R 1 ( M 1 M 10 ) + k( M M 1 ) = γ M B R ( M M 0 ) + k( M 1 M ) (3.)

4 These are the McConnell equations. 3. Lineshape Analysis Keeping on with the example above, let us now apply a β= (π/) x pulse, so that the initial conditions for the transverse components in the rotating frame are given by M r 1y(0) = M r y(0) = M 0 / M r 1x(0) = M r x(0) = 0 M r 1y(0) + M r y(0) = M 0 (3.3) Using equation. from the previous chapter, we can write the modified Bloch equations in the rotating frame as dm r 1x dt dm r 1y dt = Ω 1 M r 1y M r 1x T (1) = Ω 1 M r 1x M r 1y T (1) + k(m r x M r 1x) + k(m r y M r 1y) (3.4) and similarly for the M r x and M r y components. To make life simpler (!), we can use the definition for the signal which we had in equation.8 to write the

5 overall magnetization of 1 and as M 1 = M r 1y im r 1x M = M r y im r x (3.5) such that equation 3.4 can be rewritten as dm 1 dt dm dt = iω 1 M 1 M 1 T (1) = iω M M T () + k(m M 1 ) + k(m 1 M ) (3.6) Often we can assume that relaxation is negligible so that equation 3.6 can be written in matrix form as d M 1 = iω 1 k k M 1 dt M k iω k M (3.7) The solution to this equation (you may be familiar with it already if you have done kinetics) is: M 1 (t) = c 11 e Λ 1t + c 1 e Λ t M (t) = c 1 e Λ 1t + c e Λ t where Λ 1 and Λ are the two eigenvalues of the kinetic matrix and can be obtained by solving the

6 equation (iω 1 k Λ 1 )(iω k Λ ) k = 0 (3.8) so that Λ 1, = i (Ω 1 + Ω ) k ± k 1 4 (Ω 1 Ω ) (3.9) with the subscript 1 corresponding to the sum and the subscript corresponding to the difference. Keeping in mind that we are applying a β= (π/) x pulse, we can apply our initial conditions and solve for the coefficients c ij (i = 1, ; j = 1, ) so that with c 11 = 1 M 0 [ i (Ω 1 Ω ) + k + W ] 4 W c 1 = 1 M 0 [ i (Ω 1 Ω ) k + W ] 4 W c 1 = 1 M 0 [ i (Ω 1 Ω ) + k + W ] 4 W c = 1 M 0 [ i (Ω 1 Ω ) k + W ] 4 W W = (3.10) k 1 4 (Ω 1 Ω ) (3.11)

7 Finally the signal we measure experimentally is the sum of the signal we get from our spin in state A 1 and in state A, or mathematically s(t) = M 1 (t) + M (t) (3.1) with the two time-dependent magnetization components given by equation 3.8, i.e. depending on c ij and Λ 1,. To try to make sense of these equations and to understand the physical implications, let us now consider two cases: Case 1: slow exchange In the case of slow exchange, the following condition applies: k << Ω 1 Ω (3.13) i.e. the NMR spectrum will consist of two lines, one corresponding to state A 1 at Ω 1 and one for state A at position Ω. In this case, Λ 1 = iω 1 k Λ = iω k

8 since W Therefore c 11 = c = M 0 / c 1 = c 1 = 0 (3.14) 1 4 (Ω 1 Ω ) = i (Ω 1 Ω ) (3.15) s(t) = 1 M 0[e (iω 1 k)t + e (iω k)t ] (3.16) Fourier transforming this gives the frequency domain signal: S(ω) = 1 M 1 0[ k i(ω 1 ω) + 1 k i(ω ω) ] (3.17) with a real part Re{S(ω)} = 1 M k 0[ k + (Ω 1 ω) + k k + (Ω ω) ] (3.18) In other words, the absorptive peaks measured in the NMR spectrum will consist of two Lorentzian lines with a half width at half height of k.

9 k k Ω 1 Ω Case : fast exchange In the fast exchange limit, k >> Ω 1 Ω (3.19) i.e. only an average of the two species/lines will be seen, at an average position of (Ω 1 + Ω )/. In this case, Λ 1, = i (Ω 1 + Ω ) k ± k 1 4 (Ω 1 Ω ) (3.0) can be simplified to Λ 1, i (Ω 1 + Ω ) k ± k[1 1 8k (Ω 1 Ω ) ] (3.1)

10 and c 11 = c 1 = M 0 / c 1 = c = 0 (3.) As a result, the signal is given by s(t) = M 0 exp{[i Ω 1 + Ω (Ω 1 Ω ) 8k which when Fourier transformed gives ]t} (3.3) with Re{S(ω)} = M 0 [ 1 ( Ω ω) + ( 1 ) ] (3.4) or graphically 1 Ω = Ω 1 + Ω = (Ω 1 Ω ) 8k (3.5)

11 1/ Ω 1 + Ω One can obtain a general equation for the real part of the frequency domain signal from Re{S(ω)} = M 0 k(ω 1 Ω ) [(ω Ω 1 ) (ω Ω ) + 4k (ω Ω 1+Ω ) ] (3.6) Therefore with increasing rate constants (or increasing temperature), the lineshapes vary in the following manner: In the slow exchange limit, we have two lines with the full width in half given by π F W HH = 1 = k or k = π F W HH where F W HH is in Hz and k is in s 1 (BEWARE OF THE UNITS and the factors

12 of π!!!). As the temperature increases the lines broaden, until we reach the coalescence temperature - at this temperature, the two lines merge into one broad flat-topped line. The rate at which coalescence occurs is given by k c = π Ω 1 Ω, where k c is in s 1 and Ω 1 Ω is in Hz. Past this temperature, we have one line, with a linewidth which is inversely proportional to the rate of exchange, i.e. k= π(ω 1 Ω ) F W HH where again k is in s 1 and F W HH and Ω 1 Ω are in Hz. This is equivalent to the second equation in 3.5. For example:

13 (What is k c?).

14 Given experimentally determined exchange rate constants as a function of temperature, we can determine the activation energy E a or the activation enthalpy H and entropy S from the Arrhenius equation k(t ) = Aexp( E a /RT ) (3.7) or the Eyring equation respectively. k(t ) = k BT h e S R 3.3 Exchange from D NMR e H R, (3.8) The chemical exchange problem is a nice way to introduce the concept of higher dimensional NMR, so in this section, we will show how D experiments can be created and applied to the chemical exchange problem. Let us start again from our exchange process: A1 local field: B B Resonance frequency: ω 01 Ω k k A 01 0 ω 0 1 Ω (lab frame) (rot. frame)

15 which gives rise to two lines when the exchange is slow. We will now perturb our system in such a way as to invert the magnetization for the A 1 state selectively (see box 1), so that the initial condition at time 0 is M 1z (0) = M 0 M z (0) = M 0 (3.9) In this case, the exchange equation can be written as: d M 1z(t) = k k M 1z(t) (3.30) dt M z (t) k k M z (t) or d[m 1z M z ] dt d[m 1z + M z ] dt = k(m 1z M z ) = 0. (3.31) The latter equation describes the conservation of the total magnetization of the system over time.

16 Box 1. Selective Pulses As we saw in the last chapter, the magnitude of B 1 has an effect on the frequency domain profile of the rf pulse applied. To make a pulse selective for one region of the spectrum, we can therefore use a low power pulse (small B 1 ). Alternatively, we can use rf pulses which have a special pulse profile (in the frequency domain) - called shaped pulses. Examples include Gaussian pulses: and others (e.g. BURP, EBURP, REBURP) which have a range of inversion/excitation profiles.

17 ' $ Solving these equations gives [M1z Mz ](τm ) = [M1z Mz ](0)e kτm (3.3) [M1z + Mz ](τm ) = [M1z + Mz ](0) (3.33) or M1z (τm ) & = + 1 [M1z + Mz ](0) 1 [M1z Mz ](0)e kτm %

18 M z (τ m ) = 1 [M 1z + M z ](0) 1 [M 1z M z ](0)e kτ m Therefore measuring the magnetization using the sequence: (3.34) τ m π/ as a function of τ m, one can determine k, as long as relaxation effects can be neglected. This experiment can also be applied to more complex systems (more than two state exchange) but requires that a series of experiments, each with the appropriate selective pulse, be run. In the example above, we have used a selective pulse to distinguish the spins in the A 1 state from those in the A state. Is there another way to make this distinction? We know that if we apply a single

19 (π/) y pulse and let the system evolve, we can label our magnetization in terms of Ω (recall equation.6). So if we perform the following experiment, ( π/) y ( π/) y ( π/) y τ m t 1 A B C D E then for the spins in the A 1 state, we have F M 1 A = 1 M 0 e z M r,b 1 = 1 M 0 e x M r,c 1 = 1 M 0[ e x cos(ω 1 t 1 ) + e y sin(ω 1 t 1 )] M 1z D = 1 M 0cos(Ω 1 t 1 ) (3.35) and similarly for the spins in the A state. Therefore at point D we have a similar initial condition as we had in equation 3.9, but now it is

20 modulated (or labelled ) by a cos(ωt 1 ) term, so that for the exchange process, the initial condition is M 1z (t 1, 0) = 1 M 0cos(Ω 1 t 1 ) M z (t 1, 0) = 1 M 0cos(Ω t 1 ) (3.36) The components of the magnetization after exchange M 1z (t 1, τ m ) and M z (t 1, τ m ) will be measured in the transverse using the last (π/) y pulse such that M r,f 1 = M 1z (t 1, τ m ) e x M r,f = M z (t 1, τ m ) e x (3.37) The observed x-component of the magnetization is thus M r x(t 1, τ m, t ) = M 1z (t 1, τ m )cos(ω 1 t ) + M z (t 1, τ m )cos(ω t ) (3.38)

21 ω 1 ω where the right hand side can be rewritten as 1 4 M 0 [ (1 + e kτ m )cos(ω 1 t 1 )cos(ω 1 t ) + (1 + e kτ m )cos(ω t 1 )cos(ω t ) + (1 e kτ m )cos(ω 1 t 1 )cos(ω t ) + (1 e kτ m )cos(ω t 1 )cos(ω 1 t )] (3.39) In other words, the first two terms oscillate with the same frequency in both t 1 and t and the two last terms represent a change in frequency and therefore represent the chemical exchange phenomenon. If one Fourier transforms the data in t and t 1, then the following spectrum is obtained

22 The diagonal peaks, which have a relative amplitude of 1 (1 + e kτ m ), represent the non-exchanged magnetization, whereas the cross-peaks, with relative intensity of 1 (1 e kτ m ), represent the exchanging magnetization from the chemical exchange process. Plotting the diagonal and cross-peak intensity as a function of the mixing time (τ m ), we get 1 (0.5*(1+exp(-*1*x))) (0.5*(1-exp(-*1*x))) diagonal cross peak τ m If one takes into account relaxation effects, then this is changed to

23 1 (0.5*(exp(-x/10))*(1+exp(-*1*x))) (0.5*(exp(-x/10))*(1-exp(-*1*x))) diagonal cross peak 0 τ m where the diagonal peak intensities are 1 (1 + e kτ m )e t/t 1 and the cross-peak intensities are 1 (1 e kτ m )e t/t 1 Questions: Why is T 1 only included here? Does T contribute? In what way?