Figure 5.21

Figure 5.21

Pa and decreasing Pp—but as the Hb spins drop down they are actually "pumping up" the Hb population difference beyond the equilibrium difference! In Fig. 5.21 (center) we start the relaxation process ("mixing time") with AP(Ha) = 0 and AP(Hb) = 8. If four of the Ha spins drop down from p to a we get back to the equilibrium population difference of AP(Ha) = 8, but we also drag four of the Hb spins down from the p state to the a state, giving a population difference of AP(Hb) = 16, or twice the equilibrium population difference! This means that Mz for the Hb resonance has increased from Mo (AP = 8) to 2Mo (AP = 16) as a result of this cross-relaxation process. If we rotate the net magnetization into the x-y plane with a 90o pulse and record a spectrum at this point, we would see the Hb peak twice as large as it would be if we had not saturated Ha first. This is an exaggerated version of the simple NOE experiment for small molecules (^o tc < 1): preirradiation of one resonance leads to enhancement of the 1H peak for another resonance representing a proton that is nearby (<5 A) in space.

While we are looking at this figure, consider another extreme where the only route for relaxation is the zero-quantum pathway, whereby Ha drops down from the p state to the a state while "dragging up" an Hb spin from the a state to the p state. This transition is stimulated by molecules tumbling at the zero-quantum frequency, which is just the difference in frequency between the two spins: vo(Ha) — vo(Hb). This frequency is very low in the audio frequency range for two protons, and for large molecules there are significant numbers of molecules tumbling at this rate (Fig. 5.13). Again we start (Fig. 5.22, center) with AP(Ha) = 0 (saturation) and AP(Hb) = 8 (equilibrium) and allow Ha spins to drop from the p state to the a state, each time pulling up an Hb spin from the a state to the p state. By the time Ha has reached equilibrium (Fig. 5.22, right: AP = 8), we have destroyed the population difference for Hb (AP = 0). The effect of irradiating Ha has been to reduce the intensity of the Hb peak in the final spectrum. This is the case for large molecules (&otc > 1). This might seem like a stupid thing to do, but we will see that it is the difference in peak intensity

Figure 5.23

compared to a normal spectrum that defines the NOE, and this difference can be either an increase or a decrease. Either way it can be measured and quantified and interpreted in terms of a close approach (<5 A) of two protons.

Although this is a useful exercise, it is not really accurate to draw two energy diagrams— one for Ha and one for Hb. Because the two spins are interacting in a single molecule, we use an energy level diagram with four spin states for the two nuclei Ha and Hb: one with both spins in the a state ("aa"), one with both spins in the ¡ state ("¡¡"), one with Ha in the a state and Hb in the ¡ state ("a¡") and one with Ha in the ¡ state and Hb in the a state ("¡a"). In an energy diagram the state is highest in energy, the aa state is lowest, and the a¡ and ¡a states are in the middle at the same energy level (Fig. 5.23). The transitions can be identified as four "single-quantum" transitions, where only one of the protons changes state while the other remains the same (a¡ ^ aa, ¡a ^ aa, ^ a¡, ^ ¡a), a "double-quantum" transition, where both spins drop down (or move up) simultaneously (¡¡ ^ aa), and a "zero-quantum" transition where one spins drops down and the other spin moves up (a¡ ^ ¡a). The single-quantum transitions are observable, so they can be associated with the peaks in the spectrum. The lower right and upper left SQ transitions give rise to the Ha peak in the spectrum because only Ha undergoes a transition and Hb does not change. The lower left and upper right SQ transitions give rise to the Hb peak in the spectrum, with Hb changing its spin state and Ha doing nothing. Note that the energy difference for any SQ transition is just the energy difference for one proton to change state: AE = hvo = hyBo/2n. The energy difference for a DQ transition is twice this amount, and the energy difference for a ZQ transition is zero (actually it is h [va — vb] due to the slight difference in Larmor frequencies of Ha and Hb). We will be using this energy diagram throughout the book whenever we discuss populations and NOE interactions in a two-spin homonuclear (two proton) system.

If all four energy states had the same population, they would each have N/4 molecules (or N/4 "spin pairs" Ha-Hb). But at thermal equilibrium the upper spin state (¡¡) will lose a few molecules to the middle states, leaving a slightly depleted population of N/4 — 28 (represented by two open circles in Fig. 5.24). The two middle levels (Ha = a, Hb = ¡ and Ha = ¡, Hb = a) will initially gain 28 molecules from but will lose these to the lower state aa, leaving them with populations of N/4 (represented by the absence of any circles). Finally, the lower spin state (aa) will gain 28 molecules from the middle states to get a slightly augmented population N/4 + 28 (represented by two filled circles). The

Figure 5.24

populations can be derived from the Boltzmann distribution using the energies: zero for the aft and fta states, +AE for the ftft state and-AE for the aa state, where AE is the energy gap of a single proton transition. But it should make sense that the two middle states (aft and fta) have the same population because they are equal in energy, whereas the upper ftft state is depleted a bit and the lower aa state's population is enhanced by the same amount: 28 (Fig. 5.24). Note that the population difference across each of the observable transitions is exactly 28, with the lower energy level of each transition having the greater population. The Ha transitions are those in which Ha changes from a to ft or from ft to a without any change in Hb, and the Hb transitions are labeled where Hb changes and Ha remains the same. This is the population diagram corresponding to a sample that has been placed in the magnetic field and been given time to come to thermal equilibrium. Because z magnetization (Mz) is proportional to the population difference and is equal to Mo at equilibrium, we can associate this single-quantum population difference 28 with a z magnetization of +Mo. Thus if we perturb the system at this moment with a 90o pulse and record an FID, the Ha net magnetization vector will be rotated into the x-y plane where it will precess according to its Larmor frequency, va, and induce an oscillating and decaying voltage in the FID. The same thing will happen to the Hb net magnetization, which will induce a decaying voltage corresponding to a slightly different frequency, vb, in the FID. Fourier transformation of this FID will give a spectrum with two peaks, one at frequency va and one at frequency Vb. The height of each of these two peaks will be taken as our "control" experiment: 100% because we started with an equilibrium distribution of populations (Fig. 5.24, right). By executing a 90o pulse, recording the FID and doing the Fourier transform we effectively "read out" the population differences across the single-quantum transitions. Any deviation from the equilibrium population difference (28) across any of the SQ transitions will change the z magnetization just before the pulse, which will change the length of the vector in the x-y plane after the pulse and change the height of the peak in the resulting spectrum. For example, if the populations are equal across the Ha transitions (AP = 0), the Ha peak will disappear from the resulting spectrum. If, on the contrary, the populations are inverted across the Hb transitions (AP = -28), the Hb peak in the spectrum will be upside-down (-100% peak height).

Now let's us have some fun by perturbing this equilibrium population distribution! Selective saturation of the Ha transitions (aa ^ fta and aft ^ ftft) by low-power

Figure 5.25

continuous irradiation at frequency va (Fig. 5.25) will promote exactly 8 Ha spins in each transition, decreasing the aa state population from N/4 + 28 to N/4 + 8 (one filled circle) and increasing the fta state population from N/4 to N/4 + 8 (one filled circle). Likewise for the other Ha transition (upper left) the aft state population is decreased from N/4 to N/4 — 8 (one open circle) and the ftft state population is increased from N/4 — 28 to N/4 — 8 (one open circle) as exactly 8 spins are promoted to the higher level. At this point, the population differences across the Hb transitions (aa ^ aft and fta ^ ftft) are still all exactly equal to 28, but the Ha transitions have no population difference. If we were to execute a 90o pulse and collect the FID at this point we would see a normal peak for Hb (AP = 28 for both of the Hb transitions before the pulse, so Mb = Mo) and no peak at all for Ha (AP = 0 for both of the Ha transitions before the pulse, so M^ = 0). This makes sense because we have saturated the Ha spins selectively without affecting the Hb spins.

Now suppose that double-quantum relaxation (ftft ^ aa) is the only mechanism of relaxation. Of course, simple one-nucleus relaxation is also going on but because it does not lead to an NOE we will ignore it. At this point, the population difference between the aa and the ftft states is just 28 (N/4 + 8 vs. N/4 — 8). Because the equilibrium population difference between the ftft and aa states is 48 (Fig. 5.24), we will see 8 molecules drop down from ftft to aa to restore the equilibrium. The result is shown in Figure 5.26: the ftft state population has been reduced to N/4 - 28 and the aa state population has been increased to N/4 + 28. Note that the Ha transitions are now halfway back to their equilibrium distribution (population difference of 8 versus equilibrium difference of 28). This corresponds to a net z magnetization of Mza = Mo/2. More importantly, the Hb transitions now have a population difference even greater than equilibrium (38 versus an equilibrium population difference of 28). The net z magnetization of the Hb spins is therefore Mzb = 3Mo/2, or 50% enhanced from its equilibrium value. Thus the process of saturation of Ha followed by allowing time for Ha to relax, with double-quantum relaxation predominating, leads to a 50% NOE enhancement of Hb. Note that as our model becomes more accurate, the theoretical maximum NOE enhancement drops!

Exercise: Go through the same thought experiment with relaxation occurring to completion only by the zero-quantum pathway (aft ^ fta, the dominant pathway for a large molecule).

Figure 5.26

What is the equilibrium population difference between aft and fta? After cross-relaxation, what is the percentage change in the z magnetization of Hb, and is it increased (enhanced) or decreased?

Because z magnetization is not observable, we need to convert it to observable (x-y plane) magnetization in order to measure the NOE enhancement. At this point a 90o hard pulse on all protons will yield a spectrum in which the Ha peak is reduced to 50% of its normal intensity and the Hb peak is enhanced by 50% of its normal intensity (Fig. 5.26, right). Note that the amount of z magnetization gained by the Hb nuclei is exactly equal to that lost by the Ha nuclei. We can say that the NOE experiment has transferred z magnetization from the Ha to the Hb nuclei. This concept of magnetization transfer from one nucleus to another is the key to understanding all 2D NMR experiments. Actual NOEs are quite a bit less than 50% because cross relaxation by this pathway (ftft ^ aa) is not the only relaxation pathway available. Also, the 1D NOE difference experiment does not instantly saturate the Ha resonance and then stop irradiation and wait for cross relaxation to happen. Instead, cross relaxation and normal relaxation are happening as the Ha transitions are being irradiated, so that eventually a steady state is reached when all of these processes are going on simultaneously and the population levels are constant. It is this steady-state population distribution that is then sampled by the 90o "read" pulse. The enhancement in the Hb signal observed is referred to as the "steady state NOE," and it is typically in the range of a fraction of a percent to 10%.

Our crude picture of the distribution of tumbling rates explains why zero-quantum relaxation dominates for large molecules because they have negligible populations tumbling at the SQ (single-quantum or T1 relaxation) frequency (vo) or the DQ frequency (2vo = va + Vb) and large numbers tumbling at the ZQ frequency (va — vb). Effectively the SQ and DQ frequencies are above the cutoff tumbling frequency for these large molecules (Fig. 5.13). But for small molecules it looks like ZQ, DQ, and SQ would all have the same rates because all three frequencies lie in the flat region of the tumbling rate distribution. But there are inherent differences in the efficiency of ZQ and DQ relaxation relative to SQ relaxation, and a more detailed mathematical analysis gives the following relative rates for ZQ, SQ, and DQ relaxation:

Each of these numbers is multiplied by 1/r6, reflecting the distance dependence of the dipole-dipole interaction. Now we see that double-quantum relaxation does in fact dominate the dipole-dipole relaxation of small molecules, and our cartoon model of relaxation exclusively by the DQ pathway during the mixing time is not that far off. Likewise, the assumption that only ZQ relaxation occurs for large molecules (see exercise above) is also qualitatively correct.

5.12.1 NOE Difference

The NOE interaction between two protons in a molecule can conveniently be measured by applying a low power, continuous-wave RF irradiation at the exact resonance frequency of one nucleus (i.e., one peak in the spectrum: Ha) for a period of seconds (usually during the entire relaxation delay). During the irradiation period, the NOE builds up at another nucleus that is close (<5 A) in space (Hb), meaning that its Mz value increases by a few percent beyond Mo (Fig. 5.27). Eventually a steady state is reached and the Mz values of nearby nuclei do not increase any more, whereas the Mz of the irradiated nucleus remains near zero. At the end of the irradiation period (also called the "mixing time"), a 90o high power pulse is applied to excite all of the nuclei in the sample, rotating all z magnetization down to the x-j plane where it precesses and generates signals in the FID. The nucleus being irradiated (Ha) gives essentially no signal because its Mz was near zero at the start of the 90o pulse, and the nuclei that are close to it in space (e.g., Hb) give signals that are enhanced by a few percent over their normal peak intensities. This difference in peak intensity is often difficult

Time Figure 5.28

to detect directly, so normally a reference spectrum is collected in which the irradiation occurs in a region of noise rather than on any peak in the spectrum. This reference spectrum (which should be identical to a normal 1D spectrum) is then mathematically subtracted from the NOE spectrum at every point to yield a difference spectrum. In the difference spectrum, the peak that was irradiated will be upside down (zero minus a normal peak), peaks that show no NOE enhancement will be missing (normal minus normal), and peaks that show an NOE will appear as weak positive peaks (enhanced minus normal). This method is called the NOE difference experiment.

An analogy to heat flow helps to explain how the experiment works. Consider a container of water divided into two compartments by a glass partition (Fig. 5.28, top). The left-hand side represents proton Ha and the right-hand side represents a nearby proton Hb. At the start of the experiment (equilibrium), both compartments are at 25 0C, the temperature of the surrounding water bath (the "lattice" temperature). Then we rapidly heat the left-hand compartment (Hb) to 50 0C and hold it constant at that temperature for several minutes. During this time, heat flows from compartment A through the glass sides of the container to the environment (Ha's T1 relaxation), but we keep adding heat to maintain the temperature at 50 0C (saturation: continuous input of RF energy). Some heat also flows through the partition to compartment B (cross-relaxation) and raises the temperature of compartment B (NOE enhancement of Hb). As the temperature of compartment B rises, heat begins to flow through the sides to the environment (Hb's T1 relaxation), limiting the rise in temperature of compartment B to a small amount. Eventually we reach a steady state where the temperature in compartment B is no longer rising and the heat flow from compartment A (cross relaxation) exactly equals the heat loss to the environment (Hb's T1 relaxation). The temperature in compartment B is steady at 30 0C (Mz of Hb is steady at 1.10 Mo). At this point we measure the temperature in compartment B (900 pulse to "read" the Hb z magnetization: the Hb peak is enhanced by 10%).

Figure 5.29

The time course of these changes is shown in at the bottom of Fig. 5.28 for the heat flow analogy. The temperature of compartment A rises very quickly from 25 to 50 °C, and then the temperature of compartment B rises slowly to reach its steady-state value of 30 °C (the "mixing time"). From the point of view of z magnetization, the CW irradiation quickly reduces Mz of Ha from Mo to zero (Fig. 5.29). During the mixing time, the Mz of Hb "builds up" from the equilibrium value of Mo to the steady-state enhanced value of 1.10 Mo. When the steady state is reached, we sample the z magnetization with a 90° pulse, recording an FID. The spectrum will show no peak for Ha (M£ = 0 before the pulse), and a 10% taller than normal peak for Hb (M^ = 1.10Mo before the pulse). Note that for small molecules (i.e., molecules for which «orc « 1) the effect of "heating up" Ha is actually to "cool down" Hb, moving spins from the j state down to the a state and increasing the population difference above the equilibrium value of 25. For this reason the small molecule NOE is often referred to as a "negative" NOE. For large molecules, the effect is opposite: heating up Ha leads to a loss of z magnetization (heating up) for Hb. We call this a "positive" NOE. For "medium-sized" molecules (motc ~ 1, MW ~ 1000 Da) the sign of the NOE crosses zero and we see no NOE at all! We will see later on that there are other experimental techniques available to get around this problem.

Any change in the experimental parameters, such as temperature, Bo field (lock feedback loop variation), RF power level or phase coming from the amplifiers, or vibration, will degrade the subtraction process because we are looking for very small differences in peak intensities. For this reason, the NOE spectrum and the control spectrum are usually collected in an interleaved manner: for example, eight scans with the decoupler frequency (Varian dof, Bruker o2) set to the resonance of interest, eight scans with the decoupler set to a region of noise, and then repeating this sequence as many times as required for the desired signal-to-noise ratio. Each set of eight scans is added into the appropriate "NOE" or "control" FID. In this way, the two spectra are acquired essentially simultaneously and subtraction artifacts are minimized. The process can be expanded to include each peak in the 1H spectrum and a single control frequency, giving a map of all NOE interactions in the

Figure 5.30

molecule, which identifies all of the close approaches of one proton to another (distances over 5 A are generally too weak to be detected as an NOE). This NOE correlation map can also be obtained by the two-dimensional NOESY experiment (Chapter 10). Artifacts are a big problem with the NOE difference experiment, as with any subtraction experiment, and currently the much cleaner transient NOE experiment is used with selective (shaped) pulses and pulsed field gradients (PFGs). This technique will be discussed in Chapter 8.

Figure 5.30 shows the steady-state NOE difference spectrum of sucrose in D2O, selecting the H-f1 singlet at 3.62 ppm for saturation. We can see that the crowded region around H-f1 is also affected to some extent, giving small negative peaks for H-g3 (triplet just downfield of H-f1), for H-g6 and H-f6 (a bit further downfield) and for H-g2 (just upfield of H-f1). Because these protons are partially saturated, they will also give NOE enhancement to the peaks representing protons near them in space. This makes the data harder to interpret, so it is best to select peaks that are clearly resolved and far from any other peaks in chemical shift. In this case, the small negative peaks are so weak compared to the negative H-f1 peak that these unwanted NOE interactions are probably not even measurable. We see very strong NOE enhancements to H-f3 (doublet at 4.16 ppm, 5% peak area relative to -100% for H-f1) and to H-g1 (7% peak area). The NOE to H-f3 is due to the cis-1,3 relationship within the five-membered fructose ring, and the NOE to H-g1 is across the glycosidic linkage from the fructose ring to the glucose ring. This type of NOE across a glycosidic

linkage is extremely useful in establishing the linkage (connectivity of monosaccharides) within an oligosaccharide. Before NMR was available, this kind of information had to be obtained by a series of chemical protection and degradation steps to find which position on each side was contributing to the glycosidic linkage. This NOE also indicates that our structure diagram (Fig. 5.30) is probably not accurate as drawn: the fructose ring must rotate around the glycosidic linkage to put the CH2 group of position f1 close to H-g1 in the glucose unit. This illustrates how NOE measurements can be used to determine the conformation of biomolecules.

What if we try the NOE experiment the other way around, irradiating H-g1 on the other side of the glycosidic linkage? Figure 5.31 shows the result of this NOE difference experiment. Note that H-g1 is cleanly selected, with no saturation of any other resonances in the spectrum. Anomeric protons in sugars are useful "handles" for this kind of experiment because they are rare (no more than one per monosaccharide) and shifted to another region of the spectrum (4.5-6 ppm) away from the simple -CH-O protons on singly oxygenated carbons. We see a strong and clean NOE enhancement of the H-f1 singlet at 3.62 ppm (3.5% NOE—we divide by two because there are two protons in the H-f1 resonance), proving that the NOE interaction is a mutual effect: if saturating Ha affects Hb, then saturating Hb should affect Ha to the same extent. We see an even stronger NOE for the cis 1,2-related H-g2 resonance (14% enhancement). There are also some artifacts due to imperfect subtraction: the H-f3 doublet at 4.16 ppm shows a "dispersive" line shape due to imperfect horizontal alignment of the identical doublets in the two raw spectra (NOE and control). Subtraction then gives the "dispersive" appearance. The integral area is zero, so we know it is not a true NOE. The H-f4 triplet at 3.99 ppm looks somewhat dispersive but is more positive than negative. The peak area is positive, representing a 2% NOE enhancement. Thus H-g1 "talks" to both H-f1 and H-f4 across the glycosidic linkage, suggesting that the fructose ring can adopt two different conformations relative to the glucose ring.

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