Transient Noe Using Dpfgse

We are finally able to apply our fancy selective excitation building block, DPFGSE, to a real experiment that can give us structural information. Remember that a 90o nonselective (hard) pulse followed by the DPFGSE (containing 180o shaped pulses) is just a 90o excitation pulse for the resonance we are selecting. For all other peaks in the spectrum, it totally destroys all net magnetization. How can we use this to make an NOE experiment? We will add a 90o hard pulse at the end of the DPFGSE to flip the selected resonance magnetization down from the x'-y' plane to the — z axis. This is the largest perturbation of populations (z magnetization) possible, and if we now wait a while (the mixing time tm), we will see that this perturbation from equilibrium propagates to nearby (<5 A away) protons in the molecule. As all the other protons in the molecule have no net magnetization at all at the end of the DPFGSE, they are unaffected by the 90o hard pulse and have no z magnetization at all at the start of the mixing time. Any z magnetization that is transferred from the selected spin via NOE during the mixing time will then show up as positive z magnetization (—I^ ^ Ib) and we can "read" it out with a final 90o hard pulse at the end of the mixing time. The full sequence is shown in Figure 8.27 (omitting the repeat of the PFGSE) with the magnetization vectors shown at various levels within the NMR tube for the selected and nonselected spins. The spectrum will show a very large upside-down peak for the inverted peak, which is selected (Ha), and very small in-phase positive peaks for the resonances which receive z magnetization transfer from the selected peak. We have done an NOE experiment in a single scan, with no subtraction of spectra!

This technique differs from the old NOE difference experiment we looked at in Chapter 5, where we selectively saturate one resonance (make Mz = 0) over a long period of time (the mixing time) whereas other spins are perturbed and reach a steady-state level of

z magnetization, which is slightly enhanced over the equilibrium value Mo. The PFGSE method introduces a sudden perturbation (inversion of the selected resonance) and then waits for this perturbation to propagate to nearby protons. During the mixing time there is no RF; we just are waiting for the NOE to develop. Furthermore, because the PFGSE kills all magnetization on the nonselected spins, any z magnetization that we "read" with the final 90° pulse has to be an NOE, transferred from the selected proton.

8.9.1 Transient NOE

The rapid perturbation method is called the "transient NOE." Let's look at the process in detail. At thermal equilibrium in a strong magnetic field, there is a slight excess of population of nuclei in the lower energy (aligned with the magnetic field) state and a slight depletion of nuclei in the higher energy state (opposed to the magnetic field). If this equilibrium is perturbed for one group of nuclei (corresponding to a peak in the XH spectrum), this perturbation is propagated to nearby nuclei in the molecule due to the NOE. Because the intensity of a peak in an NMR spectrum is directly proportional to this population difference, the perturbation can be measured by simply recording a spectrum.

The traditional 1D NOE experiment (Section 5.12) involves irradiating with low-power radio frequency at the resonant frequency of one peak in the 1H spectrum in order to equalize the populations of the two states ("saturation"). This saturated state is maintained by continued irradiation until the perturbation of populations of nearby nuclei in the molecule reaches a steady state and does not change any further. Then a 90° pulse is applied and an FID is recorded to measure the amount of perturbation on the nearby nuclei. As the enhancement of signals is quite small (a few percent), it is necessary to record a control spectrum with irradiation away from any peaks in the spectrum, and then subtract the control spectrum from the NOE spectrum. There are a number of disadvantages to this approach:

1. In any difference spectrum, the conditions (temperature, RF power, sensitivity, magnetic field, and vibration) must be identical in the two experiments in order to get perfect subtraction of the signals that are not affected. This subtraction is always imperfect as the two spectra are recorded at different times, so there are always big subtraction artifacts in the difference spectrum.

2. The magnitude of the NOE is proportional to the inverse sixth power of the distance between two nuclei only for very short times between the perturbation and the measurement of the effect on other nuclei. The magnitude of the steady-state NOE is dependent on many other competing relaxation processes, so it cannot be used as an accurate measure of distance. To accurately measure distances, you need to measure the transient NOE with a number of different times between perturbation and measurement ("mixing times") and measure the initial slope of the curve as the effect increases with time.

3. The selectivity of continuous-wave (CW) irradiation is limited, and in crowded regions of the spectrum nearby peaks are also affected. This sometimes makes the results ambiguous.

With shaped (selective) pulses, we specifically invert (overall 180° pulse) a single peak in the spectrum. This is the most dramatic perturbation you can create, as the excess population in the lower energy level is now in the higher energy level and the depleted population is now in the lower energy level. If we then wait a short time for this perturbation to propagate to nearby nuclei, a 90o pulse will "read" the effect on the other nuclei in the form of a spectrum with enhanced peak areas. We can avoid having to subtract two spectra, as the gradients in the PFGSE kill all magnetization on the other nuclei at the same time that we invert the desired peak. Thus, the only thing that will be detected with the 90o "read" pulse is the perturbation due to the NOE (i.e., the transferred magnetization).

8.9.2 Populations After a Selective Inversion Pulse

We saw the effect of cross-relaxation (DQ relaxation for small molecules) after a selective saturation of one resonance in Chapter 5 (Figs. 5.24-5.26) by analyzing the four-state population diagrams. A selective inversion of one resonance (Ha) is twice the perturbation of saturation, reducing Mz from Mo (equilibrium) to -Mo (inverted) rather than to zero (saturated). The resulting population diagram is shown in Figure 8.28, with all of the N/4 + 28 spins originally in the aa state (Fig. 5.24) now in the /a state (Ha = /, Hb = a) and all of the N/4 spins originally in the /a state now in the aa state (Fig. 8.28, lower right). Remember that inversion (a 180o pulse) affects every single spin in the ensemble (in this case all of the Ha spins), switching spins in the a state to / and spins in the / state to a. The other Ha transition is affected in the same way, moving the N/2 - 28 spins in the // state to the a/ state and the N/4 spins in the a/ state to the // state (Fig. 8.28, upper left). After the selective 180o pulse, the population difference across the Hb transitions is unaffected (AP = 28) and the population difference across the Ha transition is inverted (AP = -28). We can say that the z-magnetization of Hb is Mo (equilibrium) and the z-magnetization of Ha is —Mo (inverted). If we acquire a spectrum at this point (90o pulse and FID), we will see a normal peak for Hb and an upside-down peak for Ha, both with 100% of normal peak height (Fig. 8.28, right).

Instead of acquiring an FID, we wait for a period of time tm (the mixing time) and allow relaxation to occur, dominated (for small molecules) by the DQ relaxation pathway: 3/ ^ aa. What is the equilibrium population difference between these two states? From Figure 5.24, we see that AP = (N/2 + 28) — (N/2 — 28) = 48, or counting the circles

Figure 8.29

we have 2 — (-2) = 4. This makes sense because the energy difference is twice that of a single-quantum transition, so the population difference should be twice as large according to the Boltzmann distribution. After the selective inversion pulse (Fig. 8.28), there is no population difference between // and aa, so the molecules want to drop down across that transition. If we let 8 molecules drop down, we will have one open circle (N/2 — 8) in the // state and one closed circle (N/2 + 8) in the aa state, for a population difference of 28 (two circles). To reach the equilibrium population difference, we let another 8 molecules drop down, leaving two open circles in // and two filled circles in aa (Fig. 8.29). This is the equilibrium population difference, so no more molecules will drop down. In the real world, you would not allow enough time to reach the equilibrium population difference, and you would also have competition from other relaxation pathways: ZQ (a/ ^ //a) and SQ (Ha and Hb transitions). For simplicity, we are allowing relaxation to proceed to completion via the DQ pathway, and we are blocking any relaxation by any other pathway—this will give a greatly exaggerated NOE effect but will simplify the explanation. At this point, we can take stock of the population differences (Fig. 8.29): for the Ha transitions we have AP = 0, corresponding to Mz = 0, and for the Hb transitions we have AP = 48 (2 — (—2) = 4 circles), corresponding to Mz = 2Mo. We have enhanced the z magnetization on the Hb spins by a factor of 2 (100% NOE), increasing it from the equilibrium value of Mo to 2Mo. At this point, a 90° nonselective pulse will lead to no peak at all for Ha and a peak of twice the normal height for Hb (Fig. 8.29, right). Overall, the effect of reducing Ha's z magnetization by 2Mo (from Mo to —Mo) has increased Hb's magnetization by Mo (from Mo to 2Mo). Using the product operator notation, we can describe the experiment as follows:

By writing the result, 2lb, as Ib + Ib we can see what happened during the mixing period: —Iaz was converted into Ib, a transfer of magnetization! The cross-relaxation that occurs in an NOE experiment can be described as transfer of z magnetization from one spin (Ha) to another (Hb). Note that the sign changes when magnetization transfers: this is characteristic of small molecules, a direct result of the dominant pathway being DQ relaxation. Although the effect on Hb is enhancement of its z magnetization, the NOE can be described as negative because the effect on Hb (increase in Mz) is opposite to the original perturbation of Ha (decrease of Mz). The "negative" NOE is clearly seen in the product operator representation of cross-relaxation: Iaz ^ — Ibb.

Exercise: Go through the same thought experiment for large molecules, allowing only ZQ relaxation (aft ^ fta) during the mixing time. What is the equilibrium population difference between these two states? Allow complete relaxation to this difference during tm. What is the effect on the final spectrum? Describe the experiment using product operators and show the net effect of cross-relaxation (transfer of z magnetization) in terms of IJ and Ibb. Would you call this a positive or negative NOE? Now try the experiment for large molecules and for small molecules with relaxation during the mixing time (ZQ or DQ, respectively) proceeding only half of the way to the equilibrium population difference. How does this affect the percent change in z magnetization at the end? This is a slightly more realistic thought experiment.

8.9.3 The Heat Flow Analogy

As we did for the steady-state NOE (Chapter 5), we can look at the transient NOE using the heat flow analogy (Fig. 8.30). As before, we have two beakers filled with water, immersed in a tub of water at 250 C. The beaker on the left (A) represents the selected proton, and the beaker on the right (B) is a nearby proton. A shared glass wall between them allows heat flow (NOE transfer of z magnetization) between the two beakers. The transient perturbation (inversion of Ha) is represented by dropping a hot stone into beaker A. The temperature immediately rises from 25 to 500C. Heat begins to flow out of beaker A into the surrounding tub of water at 250C (self-relaxation of Ha through T1 relaxation), bringing the temperature back

Time Figure 8.30
Figure 8.31

toward the equilibrium value (250C). But some heat flows through the partition to beaker B, and its temperature begins to rise slightly, in a linear fashion at first because the heat flow is constant. But as beaker A begins to cool down, the rate of heat flow to beaker B slows and the rise in temperature begins to occur at a slower rate. In addition, as beaker B is now above the temperature of the surroundings, it begins to lose heat to the tub of water (self-relaxation of Hb through T1 relaxation). At some point, the temperature in beaker B reaches a maximum (300 C) and begins to fall as the heat flow out to the environment exceeds the heat flow in from beaker A. After a long time, both beakers return to the temperature of the surroundings (250C).

From the standpoint of NMR z magnetization, the experiment is diagramed in Figure 8.31. For simplicity, we use a 1800 shaped pulse alone, rather than a PFGSE. At equilibrium Mz = Mo for both Ha and Hb. Immediately after a selective inversion of Ha, Mz = —Mo for H a and Mo for Hb. The z magnetization of Ha recovers in an exponential fashion with a time constant slower than T1 as it is donating some of its z magnetization to Hb, at the same time it is gaining it by T1 relaxation. The z magnetization of Hb is increasing above Mo because magnetization transfer by NOE for small molecules is in the opposite sense of the heat analogy: \az ^ — . "Heating up" Ha leads to a "cooling down" of Hb. This happens in a linear fashion at first, and the slope of this line is the rate of z-magnetization transfer from Ha, which is proportional to 1/r6, where r is the distance between the two protons. This rate of increase falls off until Mz reaches a maximum for Hb and begins to fall. The optimal mixing time depends on what you are trying to measure: for accurate distance measurements you will need to repeat the experiment a number of times with various short values of tm in order to measure the initial rate of increase of Mz for Hb. This "NOE buildup study" is the only way to get accurate distances from an NOE experiment. More commonly, however, we just want to know if there is a NOE or not, and the mixing time is set to correspond to the maximum of the NOE buildup curve.

1D TRANSIENT NOE USING DPFGSE 327 8.9.4 Simulation of the Transient NOE Experiment

Both the heat flow analogy and the NMR experiment can be represented by a pair of linked differential equations:

dAMb/dt = — Rab AMI — RbbAMhz where AM^ = M^ — Mo is the "disequilibrium" or perturbation of the z magnetization of Ha from equilibrium and AMZ = M^ — Mo is the "disequilibrium" of Hb. Raa is the self-relaxation rate of Ha and Rbb is the self-relaxation rate of Hb, whereas Rab is the cross-relaxation rate or the rate at which Ha "disequilibrium" is propagated to Hb and vice versa. Rab is actually proportional to 1/r6, the "Holy Grail" of NOE experiments. So all this says is that the rate of change of Hb's "disequilibrium" depends on its own "disequilibrium" (the self-relaxation or T1 process) and on the "disequilibrium" of Ha (the NOE process). It will pick up z magnetization from Ha to the extent that Ha is out of equilibrium, and it will tend to recover from any disequilibrium of its own and move back to Mo.

Figure 8.32 shows a simulation of these equations starting right after the selective 180° pulse that inverts Ha: at this moment = —Mo and Mb = Mo. The T1 value is set to 0.7 s and the rates are Raa = Rbb = 0.95 s—1 and Rab = —0.45 s—1 (negative because it is a small molecule). The behavior is very much like the cartoon of Figure 8.31; M^ rises at an initial linear rate of 0.7 Mo/s, which is just Rab times the initial perturbation (—2Mo) of Ha: —0.45 s—1 x —2Mo. This eventually slows down and MZ reaches a maximum after 1.125 s with an NOE enhancement of 36.7%. M£ passes through zero after 0.8 s, slower than predicted by its T1 value alone (t1/2 = ln 2T1 = 0.693T1 = 0.485 s). Note that the sum of M^ and M\ recovers from 0 (—Mo + Mo) to 2Mo (Mo + Mo) in the same way as it would if there were no cross-relaxation: a simple exponential curve based on a T1 value of 0.7 s (2Mo e—t/T 1:D symbols in Fig. 8.32), passing through the halfway point (Mo) at

exactly 0.485 s. Any z magnetization transferred from Ha to Hb is invisible in the sum M^ + Mb; the slower recovery of Mb. Transfer of magnetization c; Ha loses is exactly gained by Hb.

Ml + Mb; the slower recovery of M^ is exactly compensated by the growth of the NOE for Mb. Transfer of magnetization can be accounted for precisely: whatever z-magnetization

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