## Advanced Nmr Theory Noesy And Dqfcosy

The purpose of this book is to provide a deep and satisfying understanding of how NMR experiments work, while maintaining the practical perspective of an NMR user (organic chemist or structural biologist) rather than a theoretical approach, which would be more appropriate for an analytical chemist, physical chemist, or physicist. The simple vector model for net magnetization, along with the energy diagrams with open and filled circles to represent population differences, has provided a strong conceptual basis for understanding the NOE (nuclear Overhauser enhancement) and many of the complex tricks of evolution of single-quantum coherence (spin echoes, APT, BIRD, TANGO, gradients, etc.). In order to describe coherence transfer (INEPT), we needed the additional theoretical tools of the product operators (Ix, 2IySz, etc.): a simple mathematical approach that is firmly tied to the visual representation of the vector model. We have touched upon the idea of multiple quantum coherences (ZQC and DQC) and defined them in terms of product operators (2Ix Sx, 2Iy Sx, etc.), but many things had to be taken on faith. Why we multiply the operators together (2IySz, 2IySx, etc.) in a particular way to represent certain coherences (DQC, antiphase SQC, etc.) has been until now just a set of rules.

In this chapter, we will introduce a new level of theoretical tools—the density matrix— and show by a bit of matrix algebra what the product operators actually represent. The qualitative picture of population changes in the NOE will be made more exact, the precise basis of cross-relaxation will be revealed, and a new phenomenon of cross-relaxation— chemical exchange—will be introduced. With these expanded tools, it will be possible to understand the 2D NOESY (nuclear Overhauser and exchange spectroscopy) and DQF-COSY experiments in detail.

The tricks of selecting desired coherences and rejecting unwanted (artifact) peaks by phase cycling or gradients will be formalized by introducing the spherical product operators and defining the coherence order precisely. This gives us a very simple way of describing an

NMR Spectroscopy Explained: Simplified Theory, Applications and Examples for Organic Chemistry and Structural Biology, by Neil E Jacobsen Copyright © 2007 John Wiley & Sons, Inc.

NMR pulse sequence without getting tied up in the details of pulse phases and a mountain of sine and cosine terms: only the essential elements of the sample net magnetization will be described at each point. Finally, the formal Hamiltonian description of solution-state NMR will be described and applied to explain two related phenomena: strong coupling ("leaning" of multiplets) and TOCSY mixing (the "isotropic" mixing sequence).

10.1 SPIN KINETICS: DERIVATION OF THE RATE EQUATION FOR CROSS-RELAXATION

In Chapter 5, we demonstrated qualitatively how DQ relaxation alone (ftft ^ aa) leads to a negative NOE (saturation or inversion of Ha leads to enhancement of Hb's z magnetization), and ZQ relaxation alone (aft ^ fta) leads to a positive NOE (reduction of Hb's z magnetization). We also showed how the distribution of tumbling rates changes as molecular size is increased, leading to a change from relaxation dominated by DQ transitions (small molecules) to relaxation dominated by ZQ transitions (large molecules). We will now consider more quantitatively how this happens by looking at the "kinetics" of molecules (or proton pairs Ha and Hb) moving among the four homonuclear spin states aa, aft, fta, and ftft.

For simplicity, we assume that Ha and Hb are close to each other in a molecule (rab < 5 A) but have no J coupling. We will use Pftft, Pap, Ppa, and Paa to represent the populations of the four spin states ftft (Ha = ft, Hb = ft), aft (Ha = a, Hb = ft), fta (Ha = ft, Hb = a), and aa (Ha = a, Hb = a), respectively. At equilibrium, Pftft = N/4 — 28, Paft = N/4, Pfta = N/4, and Paa = N/4 + 28, where N is the total number of molecules and 8 is a very small fraction of N determined by the Boltzmann condition (Fig. 10.1).

Because z magnetization is the result of population differences between spin states, we can equate z magnetization with population difference (actually it is proportional, but for simplicity the proportionality constant is omitted):

Ml = Paft — Pftft = Paa — Pfta (Ha(2) and Ha(1) transitions)

m\ = Paa — Paft = Pfta — Pftft (Hb(1) and Hb(2) transitions)

mi + Mb = (Paft — Pftft) + (Paa — Paft) = Paa — Pftft (DQ transition)

M\ - Mb = (Paß - Pßß) - (Pßa - Pßß) = Paß - Pßa (ZQ transiton)

Mo = 28 (equilibrium)

Note that in each case we subtract the population of the higher energy (less populated at equilibrium) state from the population of the lower energy (more populated at equilibrium) state. We can also define the amount of "disequilibrium" as the difference between the actual z magnetization and the equilibrium z magnetization.

ami = mi - Mo = Paß - Pßß - 28 = Paa - Pßa - 28

AMbz = Mb - Mo = Paa - Paß - 28 = Pßa - Pßß - 28

Note that AMz? and AM^ both tend toward zero (AP = 28) as the nuclei relax.

Now consider the "kinetics" of the flow of spins between spin states. For each pair of spin states, we calculate the difference in population and compare it to the equilibrium difference. If these two are not the same, there will be a flow of spins from the "overpopulated" state to the "underpopulated" state at a rate that is proportional to the "rate constant" for that transition and to the amount by which the transition is out of equilibrium. The rate constants, or relaxation rates, for each transition are determined by

1. the distance r between protons Ha and Hb in the molecule (1/r6 effect);

2. the number of molecules in solution that are tumbling at a rate corresponding to the frequency (v) of that transition: va and vb for the SQ transitions, va + vb for the DQ transition, and va - vb for the ZQ transition.

The four single-quantum transitions relax at a rate Wi, or more specifically W* for the Ha transitions and Wb for the Hb transitions. The double-quantum transition (aa ^ fifi) relaxes with rate W2 and the zero-quantum transition (afi ^ ¡5a) relaxes at a rate Wo (Fig. 10.2). For small organic molecules in nonviscous solvents (e.g., cholesterol in CDCl3), the molecule tumbles rapidly compared to the Larmor frequency and the ratio of W 1:Wo:W2 is about

3:2:12, meaning that double-quantum relaxation is the fastest pathway. For a protein with a molecular weight of 13,690 Da (Ribonuclease A) in H2O, the ratio W 1:Wo:W2 is 1:28:1 on a 500-MHz spectrometer. Thus, for large molecules, the zero-quantum pathway is fastest.

Consider first the single-quantum transition between the aft and ftft states (Ha(2) transition, Fig. 10.1). This is an Ha transition with relaxation rate Wa. The equilibrium difference in population for this transition is Paft — Pftft = 28. If this equality does not hold, then the "overpopulation" of the ftft state is given by Pftft — Paft + 28, and the rate of spins dropping down from the ftft state to the aft state is Wf(Ppp — Paft + 28). If this were the only transition available (i.e., if there were no double-quantum or zero-quantum pathways), we could write down the rate of change of population as dPaft/dt = —dPftft/dt = W1a(Pftft — Pap + 28) We can calculate from this the rate equation for the relaxation of the Ha spins:

dAM?/dt = d(Mza — Mo)/dt = dM\/dt = d(Paft — Pftft )/dt

= dPaft/dt — dPftft/dt = 2W1a(Pftft — Paft + 28) = —2W1a AM\

The overall equation is a simple first-order decay with rate constant 2Wa. This corresponds to the longitudinal relaxation rate for Ha in the absence of cross-relaxation: R\ = 1/Ta. Looking at the other Ha transition, aa state to fta state, gives the same result, and the Hb transitions yield the analogous result Rb = 1/rb = 2 Wb. These self-relaxation times, T* and rb, will decrease when we introduce the cross-relaxation pathways (DQ and ZQ relaxation).

Now let's look at the more interesting situation where the cross-relaxation pathways (single quantum and double quantum) are available. Spins in the ftft state can relax by any of three pathways: they can drop down to the aft state (rate Wa), drop down to the fta state (rate W1b), or follow the double-quantum pathway down to the aa state (rate W2). All of these pathways will contribute to the change in population of the ftft state as a function of time (Fig. 10.3). Considering all three pathways leading away from the ftft state, we can

write the three terms contributing to the rate of loss of spins from this state:

Wf(Ppp - Pap + 25) via the Ha SQ (single quantum) transition

W2(Ppp - Paa + 45) via the DQ (double quantum) transition

Note that the equilibrium difference across the DQ transition (Paa - Ppp) is 45 because the energy separation is twice that of an SQ transition. Combining all three terms, dPpp/dt = - Wf(Ppp - Pap + 25) - Wb(Ppp - Ppa + 25) - W2(Ppp - Paa + 45)

The minus signs reflect the fact that all three pathways remove spins from the pp state when the spins flow "downhill" to the more stable states ap, pa, and aa. Likewise for the other three spin states dPap/dt = Wf(Ppp - Pap + 25) - Wb(Pap - Paa + 25) - W0(Pap - Ppa) dPpa/dt = Wb(Ppp - Ppa + 25) - Wl(Ppa - Paa + 25) + W0(Pap - Ppa) dPaa/dt = Wb(Pap - Paa + 25) + Wf(Ppa - Paa + 25) + W2(Ppp - Paa + 45)

Note that the equilibrium population difference for the zero-quantum transition is zero because the two states have (essentially) the same energy. Now we can substitute the (indirectly) measurable quantities and M^ for the population differences. For the "disequilibrium" of Ml we have dAM^/dt = d(Mza - Mo)/dt = dMza/dt = d(Pap - Ppp)/dt = dPap/dt - dPpp/dt

Substituting the expressions above for the "flow" toward and away from the ap and pp states dAMza/dt = WKPpp - Pap + 25) - Wb(Pap - Paa + 25) - wq(Pap - Ppa)

+ WKPpp - Pap + 25) + Wb(Ppp - Ppa + 25) + W2(Ppp - Paa + 45) = 2Wf(Ppp - Pap + 25) - Wb(Pap - Paa - Ppp + Ppa)

+ W2(Ppp - Paa + 45) - Wo(Pap - Ppa) = 2WI(-M1 + Mo) - Wb(-Mzb + Mb)

+W2(-Mza - Mb + 2mq) - wq(mi - Mb) = -2WaAMa - W2(AMza + AMzb) - Wq(AM^ - AMhz) = -AMaz(2W^ + W2 + wq) - AMzb(W2 - wq) (10.1)

Likewise for AM?, dAM?/dt = d(Mb - Mo)/dt = dM?/dt = d(Pßa - Pßß)/dt = dPßa/dt - dPßß/dt

Substituting the expressions above for the "flow" toward and away from the ßa and ßß states dAMb/dt = Wb(Pßß - Pßa + 25) - Wf(Pßa - Paa + 25) + W0(Pctß - Pßa)

+ Wa(Pßß - Paß + 25) + Wb(Pßß - Pßa + 25) + W2(Pßß - Paa + 45) = 2Wb(Pßß - Pßa + 25) - Wj1 (Pßa - Paa - Pßß + Paß)

+ W2(Pßß - Paa + 45) + Wq(Paß - Pßa) = 2Wb(-Mb + Mo) - WK-Ml + Ma)

+ W2(-Mj - Mb + 2Mq) + Wq(Mj - Mb) = -2WbAMb - W2(AMj + AM?) + Wq(AMj - AMb) = -AM?(2Wb + W2 + Wq) - AMa(W2 - Wq) (10.2)

The two results can be written together as a system of two linked (coupled) first-order differential equations. This means that the return of My to equilibrium depends on how far M? is from equilibrium, and vice versa.

where the self-relaxation rates Raa and Rbb are the longitudinal relaxation rates for Ha and Hb, respectively, and the cross-relaxation rate Rab depends on the competition of the DQ and ZQ pathways:

Rab = W2 - Wo cross-relaxation

For a small organic molecule, W2 is about six times as fast as Wq, so Rab is positive. This means that if we start at equilibrium (AM? = 0) and saturate Ha (AMzi = - Mq) the z magnetization of Hb is enhanced:

dAM-b/dt = -RabAM_a - RbbAM_b = -Rab(-Mo) (initial)

a positive

This is because the cross-relaxation term (-RabAMp becomes positive and makes M° grow initially and become greater than Mo. This leads to enhancement of the Hb signal in a 1D NOE experiment, and the initial rate of growth as a function of time ("mixing time") is a direct measure of Rab, which is proportional to 1/r6. This is the "classical" NOE familiar to the organic chemist. Note that inversion of the Ha spins (AMa = -2Mo) has the same effect but twice as strong. We call this a negative NOE because decreasing Ma has the effect of increasing MZb; for this reason the crosspeaks in a 2D NOESY spectrum of a small organic molecule are negative with respect to the positive diagonal.

For a large molecule such as a protein, Wo is much faster than W2, so Rab is negative. This means that if we saturate Ha (AMa = -Mo) when the Hb spins are at equilibrium (AMb = 0), the z magnetization on Hb will begin to decrease:

n negative

This is because the cross-relaxation term (-RabAMZ) becomes negative (-negative x negative) and makes M^ decrease initially below Mo. This leads to reduction of the Hb signal in a 1D NOE experiment, and an NOE "buildup" study using a series of different mixing times can be used to measure Rab as the initial rate. We call this a positive NOE because decreasing AMI has the effect of decreasing AMZb; for this reason the crosspeaks in a 2D NOESY spectrum of a large molecule are positive with respect to the positive diagonal.

There is a molecular size in between these extremes for which W2 and Wo are the same. In this case Rab = 0 and there is no NOE. For an ideal spherical protein in water at 27 °C on a 500-MHz instrument, this occurs at a molecular weight of 2370 Da, or a typical 20-residue peptide. Fortunately, there is another 2D experiment called ROESY (rotating frame nuclear overhauser effect spectroscopy) that carries out the NOE transfer in a weak B1 field (typically yB1/2n ~ 3000 Hz) rather than in the Bo field (e.g., yBo/2n = 500 MHz). Under these conditions, the SQ and ZQ transition frequencies are so low (3000 and 6000 Hz) that even large molecules such as proteins have significant populations tumbling at these rates. This means that all molecules, regardless of size, have W2 > Wo and a positive Rab; that is, all molecules behave like small molecules. For small and large molecules alike the NOE effect ("ROE") is an enhancement of Mz, and the crosspeaks in the 2D ROESY experiment are negative with respect to the positive diagonal.

## Post a comment