 So, let us continue the discussion with regard to the dynamic range problem in FTNMR. We said last time that when you have a huge solvent signal and a small sample signal because of the limited ADC resolution, the solvent signal overrides the ADC and then you will not be able to detect this small signal at all. The small signal will not get representation in the ADC and therefore no amount of signal averaging will help in recovering the sample signals. So, therefore what one needs to do is to suppress the solvent peak and that is what we are going to discuss today. You look here this is the spectrum of a sample which is in water and you have a huge water signal and this is a spectrum without water suppression and you will see only the signal of the water, you do not see anything of your samples at all. So, one way to suppress the solvent signal is to do what is called as a kind of a double resonance experiment where you use the second RF and irradiate the solvent resonance and the second RF frequency matches exactly with the water resonance therefore you put that RF here with a high power and well reasonably high power so that it only affects the energy levels corresponding to the water and all the other signals which are present here are not affected to a reasonable approximation. Therefore the signals from here will get saturated, the water sample water signal will get saturated all the other signal will not be perturbed. This is achieved in the following manner like this. So you have a pre-saturation, this is called a pre-saturation period where you put a second RF on the water resonance following that you apply 90 degree pulse and you collect your FID as usual. So here at this point here the water resonance is reduced to a very significant extent and once you apply the 90 degree pulse along with the other signals you detect the FID and this will not overload the receiver or it will not cause dynamic range problem and this is indicated here, this is the same spectrum what you show in the previous slide and this is with water suppression by pre-saturation here, you see the water signal is substantially reduced and you are able to see the other signals of the sample, this is a spectrum of a protein and you are able to see the other signals of the protein here pretty with good intensities. So this is one technique to suppress the water, there are other techniques to suppress the water signal and one of the popular ones there is the inversion recovery sequence. What you do here is you apply a 180 degree pulse to the entire magnetization which includes the water as well as the other sample signals and you wait for a time now following that you apply 90 degree pulse along the X axis and then you detect your FID as usual. What is to be done here, here the trick is to optimize the value of tau so that in the end you do not get enough water magnetization here and your other magnetizations are not perturbed as much. So that you get your solvent suppression effectively and you get your sample signals in a reasonable good intensity. How does this work? So let us look at that with the following vector diagram. Here the magnetization is represented by these arrows here initially at equilibrium all the magnetization is along the Z axis the pink one represents that of the water and the other ones are of their sample signals although this is very huge for schematic purposes we have drawn them in the same height although this actually the pink arrow has to be very very large compared to the other ones. Now what happens if you apply 180 degree pulse this entire magnetization rotates from the Z axis to the minus Z axis here now this is a non-equilibrium state after the 180 degree pulse you allow the system to relax back what it will do it will relax back to the Z axis and this will be determined by the spin lattice relaxation times of the individual spins different ones will relax differently depending upon the their relaxation times which are characteristics of the individual spins for a small tau if this is extremely small and the entire magnetization is still down here and if you apply a 90 degree pulse to this situation then you have the magnetization rotated into the transverse plane the Y axis and this is the negative Y axis and then we get signal here the negative magnetization which is the three signals these are two these two are the sample signals and the pink one is your water signal for a very small value of tau. So suppose you are adjust your tau in such a way that the solvent signal on the during the recovery has come to 0 here and the blue lawn has gone on the positive axis and the other line is still on the negative axis but it is quite substantially reduced because it has recovered along the Z axis. So now if you apply a 90 degree pulse to this situation your magnetization is rotated in the transverse plane and this fellow will be here this blue one is here the water is at 0 and your other signal is still on the lower side. So if you do a Fourier transformation you will get this signal which has recovered you will have this sign because this is opposite sign compared to this sign here. So this will be your positive side this will be on the negative side if this you apply for a give a very long tau then of course all of them will have recovered back to equilibrium here and if you measure then applying a 90 degree pulse then you will see all of them positive in this axis. So therefore the trick for water suppression is to choose your tau such that the water has come down to 0 here and your other signals have recovered completely or partially or they are along the positive Z axis or the negative Z axis but the trick is that your water signal is at 0 therefore you when you detect it after the FID the after applying the 90 degree pulse the FID does get a representation of these signals. So this is technique called as inversion recovery technique which is also used for other purposes which we will soon see. The other technique which has been commonly used is called the jump and return sequence. The idea here is the following here you have the strong water signal let us say it is at 4.7 ppm if you are recording the spectra and water typically the water appears around 4.7 to 4.8 ppm and you have many other signals which are little far away from there somewhere around 10 ppm. Now let us assume that we are interested in these resonances and not any others which are close by here and things like that. So therefore we want to maximize the signal intensity of these ones minimize the water intensity we do not care so much about the ones which are in the close vicinity of the water signal here. So therefore we want to get maximum intensity of these signals because we are interested in these ones. Assume that this distance from here to here in terms of frequency is nu max if this is 10 ppm 4.7 ppm this difference is 5.3 ppm. So we use that trick so that we get maximum intensity for these resonances for others we may get more reduced intensity but that is not a concern to us because we are interested in these particular peaks. So how does this experiment work? The pulse sequence for this experiment is like this you have a first a 90 degree pulse which is applied along the x axis then you wait for a time tau then you apply as another 90 degree pulse but now with a opposite phase this is with minus x axis if it is applied along x this is along minus x and following that you deduct your FID and Fourier transform it to get your spectrum. So let us look at it from the vector picture how does the magnetization rotate? So here are two arrows drawn one is for the water and the other one is for your interested signal. And you put the carrier on water which means the carrier is at zero frequency all your other frequencies are away from the water and they will have precessional frequencies. The water will not have precessional frequency because you put the offset on water which means that it is zero frequency with respect to the carrier. Now when you apply 90 degree pulse along the x so let us say the magnetization is rotated onto the minus y axis so both the magnetizations are here. Now what happens during this period tau after that the individual components will start precessing with their characteristic frequencies. Now after a time tau they would have precess to different extents but the water would have remained there because water is at zero frequency which means it has no precessional frequency it stays there and your sample signal which is of interest to you would have moved. Now you wait for such a time that this magnetization which is this orange magnetization which is of interest to you has rotated by 90 degrees. So for maximization you allow the magnetization to rotate by 90 degrees so that it has come here and that is dependent on your tau period. So adjust your tau period so that the desired magnetization is rotated through 90 degrees. So what is the angle of rotation here how do we calculate that? Now if nu max is your frequency of interest then the angle of rotation of this transverse magnetization is 2 pi tau nu max times tau, tau is the period for which the magnetization is precessing. Now if you want it to be rotated by pi by 2 with 90 degrees then you said theta is equal to pi by 2 if you put this equation in this equation here then you get tau is equal to 1 by 4 times nu max. So if you adjust your tau to this value then your desired magnetization has rotated by 90 degrees and is along the x axis. Now you apply a 90 minus x, 90 minus x meaning you apply it along the other axis here, here this magnetization is rotated here along the minus x axis if you apply. So this magnetization the orange magnetization does not rotate because it is along the minus x axis itself it stays there only. But the water magnetization which was here now rotates back to the z axis it comes here. So the water magnetization is come back to the z axis your desired magnetization has remained in the transverse plane now if you detect the signal you will get the signals from the desired magnetization only. Therefore this is one other way of suppressing the solvent signal and detecting the signals of interest. This is called as jump and return. Notice one thing here in this case we had kept the carrier on the water. Now there are signals which are on the left side of the water and on the right side of the water as well. So here we have put the signals which the water is here and we had we are interested in the signals which are at this PPM that is 5.3 PPM away. So if you go 5.3 PPM on this side there will be signals on this side as well. So with respect to this if these frequencies are positive frequencies these will be negative frequencies. So whatever applies to this 5.3 PPM here will also apply to the 5.3 PPM on this side and those also will get excited how will that show up in your spectrum. That is I have not shown it here but one can visualize this how it could have happened what would happen to those. In this magnetization the orange has moved in this direction during this period tau the other ones which are a negative frequency they would have moved in this direction. They will come like this right and at 5.3 PPM they would have aligned along this axis here the opposite axis here the ones which are shown with orange they are here the other ones would have been here. Therefore these two have opposite signs. So if you Fourier transform this 5D which you collect from here the signals which are here would show up with one sign and the signals which are here will show up with the opposite sign. Therefore in your spectrum finally you will get on one side of the water signal you will have peaks like this and the other side of the water signal you will have peaks like this. In the center you have the water there are peaks on this side peaks on this side as a result of the water suppression and maximizing at 5.3 PPM peaks on this side of the water will go like this and the peaks on this side will go like this and that is of course not a very big advantage you get maximum signal nonetheless and one can select only one portion of the spectrum for your analysis depending upon which one is interested on. So that is there are other techniques for water suppression we will come back to it later but before we actually do that we have to consider one other important concept in multiple experiments or in the Fourier transform NMR or various other experiments as well and that is called as the spin echo. We will now consider another important invention namely the spin echo. This was made by Erwin Hahn in the 1950s and it turned out to be such an important technique and that was a period when many many important techniques were important concepts were being developed. So this was almost worth a Nobel Prize but however Erwin Hahn missed that one but we all realize how important this invention was. It now appears in all kinds of multiple experiments it is an integral part of many multiple experiments and therefore it is very crucial to understand the principles of this spin echo and see how it can be used in various experiments. It is a very simple experimented technique that it now consists of two pulses a 90 degree expulse and a 180 degree pulse it could be 90x, 180y or 180x it does not matter. You apply a 90x pulse to a spin system wait for a period tau then you apply a 180x pulse to the spin system and wait for a time tau. It turns out that the signal which is actually decaying here because immediately after the 90 degree pulse we have seen that there is there will be an FID there will be a free precession and the magnetization decays in the following manner as this indicated here. It turns out that application of this 180x pulse reverses this decay of the magnetization and starts building it up. So as a result at the end of this period tau which is exactly equal to this tau there will be a refocusing of the entire magnetization and this is called as the echo and the magnetization here is called as the echo amplitude and following that of course it starts decaying once more which is the normal FID and the data is actually collected from here onwards. We do not actually collect the data during this period we collect the data from here onwards. The important point to note here is this echo amplitude is essentially a recovery of the magnetization which had decayed during the first tau period and the 180 pulses reverses this decay and you get this echo of the main magnetization. How does this work? So let us look at that and try to understand how it works. First let us consider two uncoupled spins and these two have two different frequencies there is one frequency here there is another frequency here and initially the magnetization of the two spins is along the z axis when I apply 90x pulse to this magnetization this is rotated into the xy plane comes along the minus y axis if it is a 90x pulse. Now during the period tau the two spins will start processing with the respective frequencies the one which is here is a higher frequency which is here is a lower frequency therefore this orange one has moved more than this cyan one so therefore these are going in this direction processional direction and the B has moved further than A in the transverse plane. Now we apply 180 degree pulse to this which is along the x axis now the magnetization rotates out of the plane and comes back here to this position. So the A which is here comes here and the B which is here comes here. The sense of rotation remains the same so they will continue to move in the same directions therefore B will cover a wider angle than A and eventually after the time tau here they will come back to the same y axis so this is called as the refocusing. So no matter how much was the difference between the A and B in this motion they will come back to the same position in phase at the end of this period tau and that is called as refocusing. So the chemical shifts we say are refocused because this is regardless of what the frequency difference between A and B is essentially this is the chemical shift difference. So therefore regardless of what these differences are they are coming back to the same point at the end of this period tau. So 180 pulse actually causes a time reversal and we get a refocusing of the magnetization at the end of the period 2 tau. Here we consider these 2 spins are 2 independent spins or 2 separate spins however it can also be that these are different frequencies of the same spin located in different portions of the sample if there is in homogeneity in your magnetic field. The magnetic field is inhomogeneous one portion of the sample sees one field other portion of the sample sees another field and that may result in 2 different frequencies. And for example water water at one portion of the sample sees one frequency another portion of the sample sees another frequency and they may be 2 different frequencies because of the inhomogeneity in the magnetic field. But then as a result of the echo they come back to the same position that is what we say that the echo amplitude is unaffected by field inhomogeneities. This is an important idea important concept and we will see how this will be used later for the measurement of transverse relaxation times. Now let us take 2 coupled spins we saw in the case of one coupled spins the magnetic the chemical shifts were refocused. To look at the coupled spins we will have to look at this energy level diagram and understand spin system little bit better. So let us draw the energy level diagram for 2 spins see if there are 2 spins there are 4 energy states. So these are alpha a alpha x we are considering 2 spins a and x alpha a and alpha x alpha a beta x beta a alpha x and beta a beta x. Now if you look at the transition from here to here which is which spin is flipping it is the x spin which is flipping alpha x goes to beta x alpha a remains the same. So therefore this we call it as a x 1 transition. Here also it is the same thing alpha of x spin flips to the beta and therefore this is the x transition we call this as x 2. If you look at the transition between these 2 levels then here alpha of a spin flips to beta and therefore this we call it as a transition and label it as a 1 and this is again the a transition and we label this as a 2. If there is no coupling between the spins then these 2 energy differences are identical and therefore the a 1 and the a 2 frequencies will be identical. Similarly the x 1 and the x 2 frequencies will also be identical. So if you look at the NMR spectrum of this then there will be one line at the chemical shift on u a which will have the 2 transitions a 1 and a 2 likewise for the x spin also there will be 2 transitions x 1 and x 2 will be degenerate and they will be at the same position which we call it as the chemical shift of x. So this is nu x and this is nu a. Now what happens if the spins are coupled? If the spins are coupled these energy levels get changed. For example this energy level which was here will now go up in energy by an amount j by 4. This energy level which was here will come down by an amount j by 4. j is the coupling constant between the 2 spins. This energy level which was here comes down again by an amount j by 4 and likewise this energy level which was here will now go up by an amount j by 4. Now what is the consequence of this? Now we look at the a 1 transition. The a 1 transition has reduced in energy by an amount j by 2, j by 4 here and j by 4 here therefore the energy is reduced by j by 2. So we say nu a 1 is reduced to the nu a minus j by 2. Likewise the a 2 transition which was from here to here has now gone up by the same amount j by 2. So the nu a 2 is now nu a plus j by 2. So a 2 is always taken to be a higher frequency compared to the a 1. That is a convention. You could have labeled them the other way around as well. And therefore the separation between these 2 transitions is j, nu a 2 minus nu a 1 is j and that this separation is j. Similarly the x 2 transition has the frequency nu x plus j by 2 and the x 1 transition has the frequency nu x minus j by 2 and nu x 2 minus nu x 1 is equal to j. Now notice here one more thing that is the higher frequency a 2 transition has the x spin in the beta orientation in the beta polarization state. And the lower frequency which is a 1 which has the x spin in the alpha orientation. Sometimes we also call them as faster spin and slower spin because if the energy is increasing in this manner we call this as a faster spin, faster moving spin and slower moving spin. If you sit in the rotating frame of the nu a here for example then this one is to the left, this is to the right, this is a higher frequency then and this is a lower frequency. Sometimes you also refer to this as a faster moving spin and this one as a slower moving spin. Same applies to this situation as well. For the x 2 transition the a spin is in the beta polarization and for the x 1 transition the a spin is in the alpha polarization. What happens next? In the case of the spin echo experiment we had this 90 pulse and then followed by 180 pulse after the certain evolution period and what is the consequence of that? Let us look at that once more here. Here you remember from the previous slide the a 2 transition has the beta polarization of the x spin and the a 1 transition has the alpha orientation of the x spin. Now an a spin with a frequency a 2 moves faster than an a spin with a frequency a 1 in the transverse plane. After you applied a 90 degree pulse the magnetizations corresponding to these two transitions are in the transverse plane and as they start processing in the transverse plane this one moves faster compared to this one, that is what we are trying to say. In the rotating frame of the a spin there is a chemical shift nu a that is in the middle here this is a nu a position. The two spins move in opposite directions in the transverse plane because these in the transverse plane once you put the magnetization in the transverse plane the magnetizations corresponding to these two spins move in the opposite directions. We treat the faster moving spin as moving in the anticlockwise direction and the slower moving one in the clockwise direction. Now what happens when I apply a 180 pulse on the x spin? X spin when I do I flip the x spin right x spin alpha goes to the beta and beta goes to the alpha therefore the one which was moving faster has to become slower. The one which was going slower now becomes faster. So the x spin flip converts a faster spin into a slower spin and vice versa that means the transitions change labels in the transverse plane. Let us look at that explicitly here we have the same diagram as before in the case of uncoupled spins except that we are looking at the a spins in the rotating frame of the spin a. So this is initial rotation here and we are looking at the a1 and the a2 transitions until at this point then the two transitions have moved in opposite direction. We said the a2 goes in anticlockwise direction and a1 goes in clockwise direction therefore this goes in this manner this goes in this direction. If I apply 180 y pulse then this one gets rotated onto this this one gets rotated here. So a2 goes here a1 comes here and they continue to move in those directions a2 goes here that will continue to go here and a1 which was going like this is moved here and that will move in the opposite direction a2 goes here a1 comes here a2 goes there and a1 comes here but the sense of rotation remains the same after the 180 degree y pulse. If I apply 180 x pulse now then what happens we said the labels will get changed the faster one becomes slower and the slower one becomes faster. So this comes back here the a2 comes back the a1 comes back to a2 and this a2 goes back to a1 and this will continue to go in this manner this will continue to go in this manner as before because this is a faster moving transition and then you will see in the during the next tau period these ones have spanned out more than before instead of refocusing they have actually moved apart away from each other as a result of this evolution during the tau period. Therefore what we say is that the four J couple spins evolution due to J coupling is not refocused angle between the two lines a1 and a2 at the time of echo is given by 4 pi J times tau this is easily possible to calculate and because of the separation between the two transitions is J and during the time J the cover the angle J times tau and multiply by J times 2 tau and multiply by 2 pi to calculate that in terms of radians. Now so the spin echo has become an extremely important technique and it is also used for water separation which we discussed in the previous class and that experiment is called as water gate to use spin echo for water separation. The pulse sequence for that is given here you have a 90 degree X pulse which is a hard pulse which is applied on all the spins and then you give a certain time tau during which you apply a field gradient pulses field gradient pulse this is applied for small period of let us approximately 1 millisecond or so and following that you have a soft pulse on water soft pulse this indicates this is a soft pulse on water this is a 90 minus X on water 180 X pulse this is a hard pulse again this is applied on all the spins and followed by again a soft pulse on water again 90 minus X and the carrier is placed on water therefore water is at zero frequency. Now what is the consequence here of this pulse let us look at this pulse more explicitly this is very important pulse it essentially tells you that on water there is effectively no pulse zero degree pulse on water 90 minus X 90 minus X is 180 minus X and this is 180 X together it means there is essentially zero degree pulse on water whereas a hard 180 pulse is applied on your sample signals. How does it help in water suppression this we will see in this slide the field gradients as I mentioned this is the field gradients are applied along the Z axis. So if I have the main field H0 like this B0 or H0 we call it and we apply a field gradient along the same axis called as Z. So if I have my sample here various spins at this point they now see different fields at different positions for example the water here will see one field water here will see another field water here will see another field likewise the sample spins as well. So they will see different kinds of fields which will manifest itself in their precessional frequencies. So let us first of all look at how this affects the precession of the sample signals. So to begin with you have three different positions and all of them have the same frequency the sample frequency is the same the magnetization is in the transverse plane here oriented in this direction. Now during the field gradient pulse which is for an approximately 1 millisecond this is something which one needs to optimize with reference to this we keep this as constant these two would have moved to different extents because the fields have changed because the field here and the field there are different. Therefore they move at different extents let us assume that this moves here with 180 degree rotation and this has moved in this direction by 120 degrees and this is can be simply calculated by the field strength how much they have moved. Now if I apply 180 degree pulse to this what happens this spin moves over to here and it comes here this spin moves over to here and it comes here. Now I apply another field gradient pulse the same GZ which is identical to this what it will do it will continue to introduce a phase angle to this they will continue to move it because of the field change and this will move to the same extent as it moved here from here to here it will come back to this position. Similarly this one which was moving in this direction it was moving in this direction here it will continue to move like this and it will come back to the same position this is our reference this has remained the same with respect to this now all of them have come back to the same place. So the sample signals have refocused because you remember we are applying a 180 pulse on the sample signals the other 2 pulses were soft pulses they were not applied on the sample signals they were applied only on to the water. So therefore the sample signals have refocused here at the end of the water gate pulse sequence. So they are all in phase and therefore you produce a good signal from the sample. What happens to the water? Until this point it is the same these are water positions here now it is the water signal it is not the sample signal this is the water signal here I have applied a 0 degree pulse 90 minus x 180 90 minus x means I have applied a 0 degree pulse essentially no pulse. So at the end of this therefore this will stay here only and this will stay here only. So when I apply the next gradient pulse GZ so it will go further defacing of this magnetization. So this one which has moved here it will continue to move like this and it will come here to the exactly it will make the same amount of rotation it will come here and this will make the same amount of rotation and it will come here. So therefore these are not refocused now and we have only taken few particular positions here and illustrations. So typically what happens? There are spins all over the place and none of them will actually refocus therefore you will get at the end of this if you consider as water signals from all over the place then you will have in the transverse plane I have water signals here, here, here, here, here, here, here, here, everywhere there is a water signal and therefore they all cancel each other and that is why we say the water signal will be completely cancelled out at the result of this pulse sequence. Initially it causes a defacing here and we do not apply any pulse to the water and apply another gradient pulse here it causes further defacing therefore when you consider an ensemble of the water signals from all over the place there will be complete cancellation of the transverse magnetization due to the water. So this is what we wanted to achieve and that is what has happened here and we see the result in this experimental spectrum. So this is a spectrum without water suppression see this is a huge signal and nothing well the other things cannot be seen at all and here is water suppression with pre-saturation this is what we discussed in the very beginning and you do pre-saturation saturate the water here and then you reduce it to a certain extent alright and you start seeing some signals which are present here. Now you see with the water gate you will see so many more signals and you are getting all the sample signals with good intensities in this experimental example. This is of a blood serum and you have so many signals here some with higher intensity some with small intensity and all of them can be clearly seen. And another important point to note here is that in the case of water pre-saturation the signals which are under the water will also get saturated and you will lose them. In this case that is unlikely to happen and it is only the water that is defaced and the other signals are not defaced and therefore you will get all the will recover all the signals. With that we stop here and we will continue with the things discussion on multiple experiments in the next classes.