Lecture #6 Chemical Exchange

Transcription

1 Lecture #6 Chemical Exchange Topics Introduction Effects on longitudinal magnetization Effects on transverse magnetization Examples Handouts and Reading assignments Kowalewski, Chapter 13 Levitt, sections 15.5 and 15.6 van de Ven, sections 2.4, 2.5, and

2 Chemical Exchange Cross relaxation can lead to exchange of magnetization between coupled spins I and S. However, uncoupled spins can manifest themselves as an apparent coupled spin systems, if the spins are engaged in chemical exchange. Consider spins and B on two molecules undergoing chemical exchange with rate constants k and k B respectively. k k B B 1 τ ex = k ex = k kb Exchange time Exchange rate We ll assume any transition from to B is instantaneous, but happens at an average rate of 1/τ ex. 2

3 Chemical Exchange Examples Intramolecular: inorganic phosphate ( 31 P MRS) H 2 PO 4 HPO H + Different chemical shifts Intramolecular: water and hydration layers Free water Hydration layer Protein Intermolecular: CEST Different correlation times Different chemical shifts 3

4 Chemical exchange Chemical exchange and τ c stochastic modulations relaxation Exchange rates (μs to ms time scales) << molecular tumbling - Too slow to effect anisotropic interactions such as CS or dipole coupling - Can effect isotropic interactions such as chemical shift or J coupling Example: Let spins I and S be J coupled with the bond between them regularly broken by an exchange process. J-coupling is modulated and 1 = 2πJ T 1,sc 2 The name for this particular effect is scalar relaxation of the 1 st kind. If 1 τ e >> J ( ) 2 then τ e 1+ ( ω I ω S )τ 2 e In a few lectures, we ll see why J(ω) is probed at ω I -ω S G( τ ) = J( t)j( t +τ ) = J 2 e τ τ e Hence, the exchange time can look just like a rotational correlation time! 4

5 Longitudinal Magnetization Chemical exchange can lead to the flow of longitudinal magnetization between sites. Bloch-McConnell equations dm z dt = M z M z,0 T 1 M z τ + M B z τ B and B dm z dt = M B B z M z,0 T 1B M B z τ B + M z τ More compactly d dt! # # " M z M z B Cross relaxation $! & & = α 1 τ # B 1 τ % "# α B Direct relaxation $! &# %& # " M z M z,0 M B B z M z,0 Note, same form as the Solomon equations for dipolar coupling. $ & & % with α = T 1 τ α B = T 1B τ B 5

6 Longitudinal Magnetization Some interesting limiting cases Case 1: slow exchange τ 1 +τ B 1 << T 1 1 +T 1 1B α T 1 1, α B T 1 1B Case 2: T 1 >> τ 1 1B,τ 1 B >> T 1 1 α τ 1, α B T 1 1B lmost all relaxation at site B Case 3: site B has very rapid relaxation and very small population, e.g. free water rapidly exchanging with a small pool of bound water. α = 1 T 1 + p B p T 1B + p B τ fractional pool sizes Important case for water in tissue as well as contrast agents 6

7 Transverse Magnetization Chemical reactions can have profound effects on NMR linewidths, with the effects strongly dependent on the exchange rate. T 2 T 2 + chem. exchange E 7

8 Lineshape Calculations k k B B Modified Bloch equations with coupling: dm x dt dm y dt = 1 T M x + cos( Ω t)m y k M B x + k B M x 2 = 1 T M y sin( Ω t)m x k M B y + k B M y 2 (analogous equations for the B spin) Similar to the Bloch- McConnell equations, but now for transverse magnetization. Using, compact notation for both and B spins: M xy = M x + im y dm xy dt B dm xy dt = ( iω +1 T 2 )M xy k M B xy + k B M xy = ( B iω B +1 T 2 )M B xy k B M B xy + k M xy 8

9 Lineshape Calculations Rewriting d M! xy = L M! xy dt where L = Ω + k and # Ω = iω +1 T 2 0 & % B ( $ 0 iω B +1 T 2 ' # k = k % $ k k B & ( k B ' Solution:!! M xy (t) = e Lt M xy (0) 9

10 Slow Intermediate Exchange Before giving the general solution, let s look at two special cases. Spin (similar for spin B) Slow exchange: Ω Ω B >> k,k B (a) (b) slow intermediate ΔΩ k M xy for species involved in slow exchange: (a) t=0, (b) t=t slow Linebroadening due to chemical exchange. s k increases lines get broader. dm xy dt = ( iω +1 T 2app )M xy 1 T 2app =1 T 2 + k 10

11 Fast Exchange Fast exchange: Ω Ω B << k,k B In many ways, the opposite of slow exchange. Spins hop back and forth so fast that we observe a single resonance at the weighted average chemical shift: Ω = f Ω + f B Ω B where f and f B are the molar fractions of and B Linebroadening due to chemical exchange. s k and k B increases line gets sharper. 1 T 2app = 1 T 2 + Δν with Δν ΔΩ2 k + k B ( ) ΔΩ k 11

12 Detailed Calculations Starting with! M xy (t) = e Lt! M xy (0) Taking the Fourier Transform and assuming that k >>1 T 2 and k B >>1 T 2 B (i.e. chemical exchange is much faster than T 2 relaxation) yields (after considerable algebra): Real part of spectrum where Ω = f Ω + f B Ω B, molar fractions of and B S(ω) = f f B (Ω Ω B ) 2 1 τ ex (Ω ω) 2 (Ω B ω) 2 + (Ω ω) 2 τ M 2 0 ex f = k B f + f B =1,, τ ex = f B k molar fractions related to reaction rates: k k B B 1 k + k B, measure of interconversion rate between and B 1 = τ ex τ τ B τ =1 k, τ B =1 k B : lifetimes of and B 12

13 Detailed Calculations S(ω) = f f B (Ω Ω B ) 2 1 τ ex (Ω ω) 2 (Ω B ω) 2 + (Ω ω) 2 τ M 2 0 ex Three values of ω which correspond to spectral peaks S(ω Ω ) = f τ (Ω ω) 2 + τ 2 M 0 S(ω Ω B ): analysis same as for. Lorentzian at Ω with width 1/τ =k. If τ is very short, peak very broad. peak visible under slow exchange S(ω Ω) = f f B (Ω Ω B ) 2 τ ex f 2 f B2 τ 2 ex (Ω Ω B ) 4 + (Ω ω) M 2 0 Lorentzian at Ω = f Ω + f B Ω B Linewidth: πδν = f a f b (Ω a Ω b ) 2 τ ex increases with τ ex peak visible under fast exchange 13

14 2-Spin System with Chemical Exchange S(ω) = f f B (Ω Ω B ) 2 1 τ ex (Ω ω) 2 (Ω B ω) 2 + (Ω ω) 2 τ M 2 0 ex Ω Ω B f Ω + f B Ω B 14

15 Example 1: Fast Exchange S(ω) = f f B (Ω Ω B ) 2 τ 1 (Ω ω) 2 (Ω B ω) 2 + (Ω ω) 2 τ 2 M 0 derived under assumptions that k >>1 T 2 and k B >>1 T 2 B (i.e. chemical exchange much faster than T 2 ) Not necessarily true for contrast agents. Some parameters to consider. Chemical shift difference between water when free and when coordinated with the agent/metal. B T 2 of water bound to the agent (typically dominated by the unpaired electron spin). Lifetime, τ B, of the water in the coordination sphere of the contrast agent. To be discussed in detail later 15

16 Example 2: Intermediate Exchange Chemical exchange saturation transfer (CEST) 16

17 Example 3: Fast Exchange H HPO H + 2 PO 4 Ω =3.2 ppm Ω B =5.7 ppm Under fast exchange, the 31 P Pi peak will be at Ω = f Ω + f B Ω B Henderson-Hasselbach relationship # f ph = pk + log & % ( $ 1 f ' Combining the above and expressing things in terms of chemical shift yields % ph = pk + log ω Ω ( ' * & Ω B ω ) (inorganic phosphate) 17

18 31 P Exercise Study What acid is forming to drive down the ph? ph PCr Pi TP 18

19 Example 4: Fast Exchange Temperature mapping via water chemical shift The resonance frequency of the in vivo water 1 H peak is known to shift with temperature at a rate of ~0.01ppm/ºC. This affect can be explained via a two-site exchange process. 19

20 Temperature mapping with H 2 O The earliest reference I found was Hindman JC, Proton Resonance Shift of Water in the Gas and Liquid States, J. Chemical Physics, 44, 4583 (1966). Hydrogen bonds decrease the electron density at the involved proton site and hence lead to a positive frequency shift. Liquid water can be modeled as a mixture of two components: a hydrogen-bonded ice-like fraction and a non-hydrogen-bonded monomeric fraction. The chemical shifts for these two components, which are in fast exchange, are shielding constant for monomeric water, σ w -0.4x10-6 shielding constant hydrogen-bonded water, σ p σ w - 5.5x

21 Temperature mapping with H 2 O Under fast exchange, water chemical shift is Ω = f Ω + f B Ω B Combining with the data provided below, Table VI. Calculation of the fraction of zero-bonded water from thermal, dielectric, and chemical-shielding data. Temp (ºC) Thermal Shielding Dielectric 0 (0.155) (0.155) yields a water proton frequency shift of Δ ppm/ºc. What would we expect to see if the hydrogen-bonded and zero-bonded water were NOT in fast exchange? 21

22 Next Lecture: In vivo water 22