6 The noe effect c Gerd Gemmecker, 1999 Nuclear OVERHAUSER Effect(NOE): resonance line intensity changes caused by dipolare cross- relaxation from neighbouring spins with perturbed energy level populations To understand the nature of the NOE, we have to look at a two-spin system I' and 12. Since NOE does not involves coherences, but merely polarization, i.e. population differences between the a and p states, we can use the energy level diagram here W W1() W1(I) W1() The possible transitions for this two-spin system can be classified into three groups W, transitions involving a spin flip of only one of the two spins(either I or 1), corresponding merely to T relaxation of the spin a Wotransition involving a simultaneous spin flip a-B for one spin and a>p for the other one (i.e, in summa a zero-quantum transition) a W2 transition involving a simultaneous spin flip of both spins in the same direction corresponding to a net double-quantum transition We are just contemplating spin state transitions here caused by relaxation, which do not involve a coherent process(like, e.g., Ix/Iv coherences, which require a phase coherent transition between two states that is generated by an r f. pulse)!
92 6 The NOE Effect © Gerd Gemmecker, 1999 Nuclear OVERHAUSER Effect (NOE): resonance line intensity changes caused by dipolare crossrelaxation from neighbouring spins with perturbed energy level populations. To understand the nature of the NOE, we have to look at a two-spin system I1 and I2 . Since NOE does not involves coherences, but merely polarization, i.e. population differences between the a and b states, we can use the energy level diagram here: The possible transitions for this two-spin system can be classified into three groups: - W1 transitions involving a spin flip of only one of the two spins (either I1 or I2 ), corresponding merely to T1 relaxation of the spin. - a W0 transition involving a simultaneous spin flip a®b for one spin and a®b for the other one (i.e., in summa a zero-quantum transition). - a W2 transition involving a simultaneous spin flip of both spins in the same direction, corresponding to a net double-quantum transition. We are just contemplating spin state transitions here caused by relaxation, which do not involve a coherent process (like, e.g., Ix / Iy coherences, which require a phase coherent transition between two states that is generated by an r.f. pulse)!
If we perturb one spin, e.g., I, i.e., change its populations of the a and B state(e. g, by saturating the resonance= creating equal population of both states ), then relaxation will force I' back to the equilibrium BOLTZMANN distribution. With the W, mechanism, spin I' will just relax without effecting spin 12. However, the other two mechanism will effect F2 B一个个个一个二阝 E B一个个个一个B W a4个 个个个个a a个个 个个个个个α With the I polarization going back from saturation to the BOLTZMANN equilibrium, the Wo mechanism will cause the neighbouring(so far unperturbed) spin to deviate from its BOLtZmann equilibrium towards a decrease in a, B population difference. After a 90 pulse, this will result in a decrease in signal intensity for 12-a"negative NOE effect" On the other hand, the W2 mechanism will cause the population difference of the undisturbed spin 12 to increase, corresponding to an increase in signal intensity a"positive NOE effect These effects can be directly observed in a very simple experiment, the id difference Noe sequence AQ CW One spin is selectively saturated by a long, low-power CW(continuous wave)irradiation. As soon as the spin deviates from its BOLTZMANN population distribution, it starts with TI relaxation. Via the Wo or w2 mechanisms it causes changes in the population distribution of neighbouring spins. After a 90o pulse, these show up as an increase or decrease in signal intensity
93 If we perturb one spin, e.g., I1 , i.e., change its populations of the a and b state (e.g., by saturating the resonance = creating equal population of both states), then relaxation will force I1 back to the equilibrium BOLTZMANN distribution. With the W1 mechanism, spin I1 will just relax without effecting spin I2 . However, the other two mechanism will effect I2 : With the I1 polarization going back from saturation to the BOLTZMANN equilibrium, the W0 mechanism will cause the neighbouring (so far unperturbed) spin to deviate from its BOLTZMANN equilibrium towards a decrease in a,b population difference. After a 90° pulse, this will result in a decrease in signal intensity for I2 — a "negative NOE effect". On the other hand, the W2 mechanism will cause the population difference of the undisturbed spin I2 to increase, corresponding to an increase in signal intensity: a "positive NOE effect". These effects can be directly observed in a very simple experiment, the 1D difference NOE sequence: One spin is selectively saturated by a long, low-power CW (continuous wave) irradiation. As soon as the spin deviates from its BOLTZMANN population distribution, it starts with T1 relaxation. Via the W0 or W2 mechanisms it causes changes in the population distribution of neighbouring spins. After a 90° pulse, these show up as an increase or decrease in signal intensity
Usually, the experiment is repeated without saturation, giving the normal ID spectrum. This is then subtracted from the irradiated spectrum, so that the small intensity changes from the noe effects can be easier distinguished: spins with a positive NOE (i.e, higher intensity in the NOE spectrum than in the reference ID) show a small positive residual signal, spins with a negative NOE yield a negative signal, spins without an NOE cancel completely What determines if we see a positive or a negative NOE? First of all, an noe requires that there is a significant interaction between the magnetic dipoles of the two spins. Dipolar interactions drop of ery fast with distance, so a H, H Noe can only occur with a distance of <5-6 A (500-600 pm) Due to the very small energy differences between NMR spin states, a spontaneous transition from a higher to a lower energy state is very improbable(with average lifetimes of several years for the spin states!). All transitions have to be induced by electromagnetic fields which are in resonance, i.e., have the right frequency corresponding exactly to the energy difference of the two spin states involved in the transition For the wI transitions, this frequency is identical with the larmor frequency of the spin that is undergoing the spin flip, i. e, the resonance frequencies o(I )and o(). For the Wo transition, the energy difference is identical to the difference o(I)-o(12), and for W2 it is the sum of the resonance frequencies for the two spins, o(I)+o(12) If we consider, e.g., H in a 500 MHz spectrometer, then the W, mechanism would require frequencies of 500 MHz to induce transitions, the w, mechanism(corresponding to a simulatneous flip of two H spins)requires fields of 1000 MHz, and the Wo transitions frequencies equal to the resonance frequency difference between the two protons, i.e., a few ppm(some 100 or 1000 Hz) Where do we get these electromagnetic fields? If we imagine two small magnets(e.g, compass gether at a fixed distance(like tv field(e.g, the Earth's magnetic field), then the interaction between the two depends on the orientation of the "molecule"in the external field
94 Usually, the experiment is repeated without saturation, giving the normal 1D spectrum. This is then subtracted from the irradiated spectrum, so that the small intensity changes from the NOE effects can be easier distinguished: spins with a positive NOE (i.e., higher intensity in the NOE spectrum than in the reference 1D) show a small positive residual signal, spins with a negative NOE yield a negative signal, spins without an NOE cancel completely. What determines if we see a positive or a negative NOE? First of all, an NOE requires that there is a significant interaction between the magnetic dipoles of the two spins. Dipolar interactions drop of very fast with distance, so a 1H, 1H NOE can only occur with a distance of <5-6 Å (500-600 pm). Due to the very small energy differences between NMR spin states, a spontaneous transition from a higher to a lower energy state is very improbable (with average lifetimes of several years for the spin states!). All transitions have to be induced by electromagnetic fields which are in resonance, i.e., have the right frequency corresponding exactly to the energy difference of the two spin states involved in the transition. For the W1 transitions, this frequency is identical with the Larmor frequency of the spin that is undergoing the spin flip, i.e., the resonance frequencies w(I1 ) and w(I2 ) . For the W0 transition, the energy difference is identical to the difference w(I1 ) – w(I2 ) , and for W2 it is the sum of the resonance frequencies for the two spins, w(I1 ) + w(I2 ) . If we consider, e.g., 1H in a 500 MHz spectrometer, then the W1 mechanism would require frequencies of 500 MHz to induce transitions, the W2 mechanism (corresponding to a simulatneous flip of two 1H spins) requires fields of 1000 MHz, and the W0 transitions frequencies equal to the resonance frequency difference between the two protons, i.e., a few ppm (some 100 or 1000 Hz). Where do we get these electromagnetic fields? If we imagine two small magnets (e.g., compass needles) close together at a fixed distance (like two 1H spins in a molecule) in an external magnetic field (e.g., the Earth's magnetic field), then the interaction between the two depends on the orientation of the "molecule" in the external field:
In position a), there will be a repulsive interaction between the two, in position c)it will be attractive, and somewhere in between(position b))the interaction will be zero. Now our molecule is not static, but pushed around all the time in our sample solution. A rotation of the molecule will change its orientation in the static magnetic field and cause a modulation of the field strenght a spin actually through the effects of the neighbouring spins. Thus, the rotation of the other spins generates an electromagnetic field -just like in an electric generator, but not with a well-defined frequency, but with a wide range of frequency components, due to the random nature of the molecular rotation The easiest approximation to describe the rotation of a molecule in solution is the assumption of the molecule as a rigid sphere. If we calculate how"much"of a certain frequency is generated in the random rotation, we find the following expression for the"spectral density"at a frequency o J()= It means that we have a maximum at @=0 and the spectral density then drops off for higher frequencies. How fast is controlled by the parameter tc, the molecular rotaional correlation time(or just" correlation time"). A long correlation time means a rather sluggish rotation, and a fast drop off of J(very small high-frequency contributions). A short t, corresponds to a fast random rotation causing a much wider frequency distribution/more high-frequency contributions
95 In position a), there will be a repulsive interaction between the two, in position c) it will be attractive, and somewhere in between (position b)) the interaction will be zero. Now our molecule is not static, but pushed around all the time in our sample solution. A rotation of the molecule will change its orientation in the static magnetic field and cause a modulation of the field strenght a spin actually "experiences" – through the effects of the neighbouring spins. Thus, the rotation of the other spins generates an electromagnetic field – just like in an electric generator, but not with a well-defined frequency, but with a wide range of frequency components, due to the random nature of the molecular rotation. The easiest approximation to describe the rotation of a molecule in solution is the assumption of the molecule as a rigid sphere. If we calculate how "much" of a certain frequency is generated in the random rotation, we find the following expression for the "spectral density" at a frequency w: J c c (w) t w t = + 2 1 2 2 It means that we have a maximum at w=0 and the spectral density then drops off for higher frequencies. How fast is controlled by the parameter t c , the molecular rotaional correlation time (or just "correlation time"). A long correlation time means a rather sluggish rotation, and a fast drop off of J (=very small high-frequency contributions). A short t c corresponds to a fast random rotation, causing a much wider frequency distribution / more high-frequency contributions
The correlation time is determined by different factors: mainly, the molecular weight. The larger a molecule. the slower its re-orientation i.. the longer its t. as a rule-of-thumb. the correlation time of a molecul can be estimated from the molecular weight MW: tc [ps]=0.5 MW [Da](. g, a MW of 1000 Da corresponds to ca. 0.5 ns). But tc is also influenced by other parameters: temperature(the higher the temperature, the shorter tc), solvent viscosity(the more viscous, the longer tc) aggregation in solution et Back to the NOE the observable Noe is determined by the cross-relaxation rate (W2-Wo) and with the dependence on the spectral densities at the"resonance frequencies"for Wo(o 0)and 6J(200)-J(0)} For 6J(200)>J(O), we get positive NOEs. This requires a relatively high spectral density at 0=2 i.e., a short tc. In the opposite case(long tc ) Wo will dominate: negative NOEs. The crossing point is at ootc =31, where the effects from Wo and W2 cancel each other and there is no noe to be observed In these equations we are looking at o j, the cross-relaxation rate between two spins i and j However, in the id difference noe experiment described above, we cannot really measure the rate of Noe build-up, since we need a relatively long (in the order of seconds), low-power selective radiation to saturate one spin. During this time, the noe will gradually build-up, as we start saturating I more and more, with no defined starting point. What we get after sufficiently long irradiation is the so-called"steady state NOE", i.e. the steady state population distribution of the neighbouring spins that derives from two counter-acting effects: 1. the noe build-up from cross- relaxation with the irradiated spin I!, and 2. the T, relaxation of the spins that have deviated from their Boltzmann equilibrium due to the noe, trying to bring them back to the equilibrium state
96 The correlation time is determined by different factors: mainly, the molecular weight. The larger a molecule, the slower its re-orientation, i.e., the longer its tc . As a rule-of-thumb, the correlation time of a molecul can be estimated from the molecular weight MW: tc [ps] » 0.5 MW [Da] (e.g., a MW of 1000 Da corresponds to ca. 0.5 ns). But tc is also influenced by other parameters: temperature (the higher the temperature, the shorter tc), solvent viscosity (the more viscous, the longer tc), aggregation in solution etc.. Back to the NOE: the observable NOE is determined by the cross-relaxation rate s = (W2 - W0 ) and with the dependence on the spectral densities at the "resonance frequencies" for W0 (w » 0) and W2 (w = 2 w0 ) we get: sij w ij r µ J - J 1 6 2 0 6 0 { ( ) ( )} For 6J(2w0 ) > J(0) , we get positive NOEs. This requires a relatively high spectral density at w = 2 w0 , i.e., a short tc . In the opposite case (long tc ), W0 will dominate: negative NOEs. The crossing point is at w0 tc = 5 2 » 1 , where the effects from W0 and W2 cancel each other and there is no NOE to be observed. In these equations we are looking at sij , the cross-relaxation rate between two spins i and j . However, in the 1D difference NOE experiment described above, we cannot really measure the rate of NOE build-up, since we need a relatively long (in the order of seconds), low-power selective irradiation to saturate one spin. During this time, the NOE will gradually build-up, as we start saturating I1 more and more, with no defined starting point. What we get after sufficiently long irradiation is the so-called "steady state NOE", i.e. the steady state population distribution of the neighbouring spins that derives from two counter-acting effects: 1. the NOE build-up from crossrelaxation with the irradiated spin I1 , and 2. the T1 relaxation of the spins that have deviated from their BOLTZMANN equilibrium due to the NOE, trying to bring them back to the equilibrium state
97 The steady state NOE resulting from this balance now does not relly depend on the distance between the two spins, only so far that the observation of any effect proves the distance to be shorter than ca 5-6 A. The intensity of the observed NOE can vary widely depending on the neighbourhood of other protons which act as a relaxation source. a few examples will show this(from: Sanders/Hunter, Modern NMr Spectroscopy, Oxford University press, Oxford 1987) two-spin system I-12 saturated 50% for small molecules(W2 dominating) saturated saturated -100% for large molecules(Wo -100% aturated There is no distance dependence for these steady-state NOes! linear three-spin system 1-12-1(r23=r12)* saturated 25% -11.5% 49.2% saturated 49.2% -11.5% only shown for positive NOEs, with W2>> Wo Generally, the NOEs are weaker, because they now have to compete with relaxation from two neighbouring protons. This can be seen esp for the first case, where 12 gets an NOE contribution from the saturated spin I, but now has two equally close neighbours causing T, relaxation(I' and 13) Since the NOE on 12 leads to an increased population difference for this spin, it causes itself(being
97 The steady state NOE resulting from this balance now does not relly depend on the distance between the two spins, only so far that the observation of any effect proves the distance to be shorter than ca. 5-6 Å. The intensity of the observed NOE can vary widely, depending on the neighbourhood of other protons which act as a relaxation source. A few examples will show this (from: Sanders / Hunter, Modern NMR Spectroscopy, Oxford University press, Oxford 1987): two-spin system I1 – I2 I 1 I 2 saturated 50 % for small molecules (W2 dominating) 50 % saturated saturated -100 % for large molecules (W0 dominating) -100 % saturated There is no distance dependence for these steady-state NOEs! linear three-spin system I1 – I2 – I3 (r23 = r12) * I 1 I 2 I 3 saturated 25 % -11.5 % 49.2 % saturated 49.2 % -11.5 % 25 % saturated *only shown for positive NOEs, with W2 >> W0 ! Generally, the NOEs are weaker, because they now have to compete with relaxation from two neighbouring protons. This can be seen esp. for the first case, where I2 gets an NOE contribution from the saturated spin I1 , but now has two equally close neighbours causing T1 relaxation (I1 and I3 ). Since the NOE on I2 leads to an increased population difference for this spin, it causes itself (being
not in the BOLtZMAnn equilibrium anymore)an NOE at spin 13, of opposite sign than the direct NOE caused by the decreased population difference of I linear three-spin system I-I-I([23=2 12) saturated 49.2% 18.6% 50% saturated 50% -0.4% 0.8% saturated for Wo>>Wo! In the last case, I and 1- are close together and relax each other very efficiently, while the Noe build-up from the more distant i is too slow to lead to a significant stady state NOE. This is often true,e.g,for CH2 groups, where the distance between the two methylene protons(ca. 1.78 A)is much shorter than to any other proton 2D NOESY Any quantitative interpretation of steady state NOEs requires knowledge of the arrangement of all protons relative to each other! Otherwise, instead of staedy state NOEs, the Noe build-up rate has to be determined, which depends on the interproton distance with r-6. It is also called the transient noE or kinetic noe While some variations of the id difference Noe experiment have been used to measure build-ip rates, the easiest way to do this is the 2D NOESY experiment t2 Like, e.g., the 2D COSY or TOCSY experiments, it starts with a H 90 pulse followed by aH indirect evolution time, at the end of which the following terms exist
98 not in the BOLTZMANN equilibrium anymore) an NOE at spin I3 , of opposite sign than the direct NOE caused by the decreased population difference of I1 . linear three-spin system I1 – I2 –––– I3 (r23 = 2 r12) * I 1 I 2 I 3 saturated 49.2 % -18.6 % 50 % saturated 50 % -0.4 % 0.8 % saturated *for W2 >> W0 ! In the last case, I1 and I2 are close together and relax each other very efficiently, while the NOE build-up from the more distant I3 is too slow to lead to a significant staedy state NOE. This is often true,e.g., for CH2 groups, where the distance between the two methylene protons (ca. 1.78 Å) is much shorter than to any other proton. 2D NOESY Any quantitative interpretation of steady state NOEs requires knowledge of the arrangement of all protons relative to each other! Otherwise, instead of staedy state NOEs, the NOE build-up rate has to be determined, which depends on the interproton distance with r –6 . It is also called the transient NOE or kinetic NOE. While some variations of the 1D difference NOE experiment have been used to measure build-ip rates, the easiest way to do this is the 2D NOESY experiment: Like, e.g., the 2D COSY or TOCSY experiments, it starts with a 1H 90° pulse followed by a 1H indirect evolution time, at the end of which the following terms exist:
IIx cos Q2,ti cos TJt+ Ily sin &2 tr cos tJt+ 211y2z cos S2,t sin TJt-211xI2z sin $ tI sin TJtu
99 I 1x cos W1 t1 cos pJt1 + I1y sin W1 t1 cos pJt1 + 2I1yI 2z cos W1 t1 sin pJt1 – 2I1xI 2z sin W1 t1 sin pJt1
After the 90 H pulse at the end of t, we get -Iiz cos &2,tI cos IJt +lly sin Q2,tI cos IJt +21Ivl2x cos &2,tI sin TJt:+ 211,2x sin Q2,tI sin Jt1 polarization DQC+zQC SO The phase cycle is set up so that only coherences with coherence order zero during tm are selected i.e., the first term(polarization)and the zQc part of the third term( I1yl2x-211x2y) The first term is a z magnetization which is modulated with the cosine of the chemical shift(and coupling), i.e., it oscillates between +liz and -1lz. This non-equilibrium polarization will cause ar NOE on neighbouring protons I1z cos a,t cos TJt1->-(1-k12)11z cos Q2,tr cos Tt: -k12 12z cos 22,t cos TTr The amount of NoE transfer is, of course, determined by the cross-relaxation rate 12 and the duration of tm. After the final 'H 90 pulsethese z components are converted into detectable signal - (1-k12IIx cos Q2,tI cos TJt -k12 l2x cos 2,tI cos TTr The first term results in a diagonal peak after 2D FT(at Q2,), the second one in a cross-peak at Q2, in F1(from the tI modulation) and at Q2, in F2 (since it is an 2 coherence now ) The sign of the cross peak depends on the sign of the cross-relaxation rate, i.e., the dominance of either Wo or W2(the diagram shows the maximum NOE cross-peak intensity relative to the diagonal system) 05 NOE with a negative NOE effect, the cross-peaks will 025 be negative with a negative diagonal signal (like in the Id difference NOE experiment!) 00 for a positive NOE effect, one gets positive cross-peaks(with a negative diagonal) T.,0 for ootes l, the NOE effect will disappear, and no NOESY cross-peaks will be visible
100 After the 90°y 1H pulse at the end of t1 we get –I1z cos W1 t1 cos pJt1 + I1y sin W1 t1 cos pJt1 + 2I1yI 2x cos W1 t1 sin pJt1 + 2I1zI 2x sin W1 t1 sin pJt1 polarization SQC DQC+ZQC SQC The phase cycle is set up so that only coherences with coherence order zero during tm are selected, i.e., the first term (polarization) and the ZQC part of the third term (= I1yI 2x – 2I1xI 2y). The first term is a z magnetization which is modulated with the cosine of the chemical shift (and coupling), i.e., it oscillates between +I1z and –I1z . This non-equilibrium polarization will cause an NOE on neighbouring protons: –I1z cos W1 t1 cos pJt1 ¾® – (1-k12) I1z cos W1 t1 cos pJt1 – k12 I2z cos W1 t1 cos pJt1 The amount of NOE transfer is, of course, determined by the cross-relaxation rate s12 and the duration of tm . After the final 1H 90° pulsethese z components are converted into detectable signals: ¾® – (1-k12) I1x cos W1 t1 cos pJt1 – k12 I2x cos W1 t1 cos pJt1 The first term results in a diagonal peak after 2D FT (at W1 ) , the second one in a cross-peak at W1 in F1 (from the t1 modulation) and at W2 in F2 (since it is an I2 coherence now). The sign of the crosspeak depends on the sign of the cross-relaxation rate, i.e., the dominance of either W0 or W2 (the diagram shows the maximum NOE cross-peak intensity relative to the diagonal system): - with a negative NOE effect, the cross-peaks will be negative with a negative diagonal signal (like in the 1D difference NOE experiment!) - for a positive NOE effect, one gets positive cross-peaks (with a negative diagonal). - for w0 tc » 1, the NOE effect will disappear, and no NOESY cross-peaks will be visible
Unfortunately, the ZQc term will also survive the phase cycling, and lead to dispersive antiphase signals at the diagonal and at the cross-peak position between two protons with a scalar coupling These signals can feign a NOESY cross-peak even in the absence of an NOE However, unlike the in-phase NOE cross-peaks, they do not have a net integral(antiphase! ) so that they can be distinguished from real NOESY cross-peaks(but only if really the whole peak area is integrated, including the wide dispersive tails of the ZQc peaks! ) Of course, ZQC peaks do not interfere with an noe between two protons that do not show a mutual scalar coupling How does one now measure o, i.e., the Noe build-up rate? Let's consider the behaviour of the NOE (i.e, the integral of the NOESY cross-peak)as a function of t 1.0 0.9 0.8 0.7 0.6 0.5 0. 0.1 0.0 0 0.2 4 8 0.9 1.0 Here, the build-up of two NOEs is shown(solid lines), with the o of the first one twice as large as the second one. In addition, an indirect NOE (i.e, an NoE caused by the perturbation from a direct NOE) is depicted with a broken line
101 Unfortunately, the ZQC term will also survive the phase cycling, and lead to dispersive antiphase signals at the diagonal and at the cross-peak position between two protons with a scalar coupling. These signals can feign a NOESY cross-peak even in the absence of an NOE! However, unlike the in-phase NOE cross-peaks, they do not have a net integral (antiphase!), so that they can be distinguished from real NOESY cross-peaks (but only if really the whole peak area is integrated, including the wide dispersive tails of the ZQC poeaks!). Of course, ZQC peaks do not interfere with an NOE between two protons that do not show a mutual scalar coupling! How does one now measure sij , i.e., the NOE build-up rate? Let's consider the behaviour of the NOE (i.e., the integral of the NOESY cross-peak) as a function of tm : 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 tm [s] NOE (arbitrary units) Here, the build-up of two NOEs is shown (solid lines), with the s of the first one twice as large as the second one. In addition, an indirect NOE (i.e., an NOE caused by the perturbation from a direct NOE) is depicted with a broken line