 So, we have been discussing two-dimensional separation experiments in NMR and in the last class we talked about 2D heteronuclear separation experiments where the chemical shifts and the coupling constants were separated along the two-dimensions of the 2D experiment. So, we are going to extend this discussion to homonuclear experiments, homonuclear 2D j-resolved experiments. So, this is generally represented as j-res NMR or 2D j-resolved NMR. So, the difference here will be because of the homonuclear coupling constants that will be present. In the heteronuclear experiments which we discussed earlier, C13 was the heteronucleus and proton was the other nucleus. We are talking about the coupling constants between carbon and proton and we did not talk about the homonuclear carbon-carbon couplings because we are working at natural abundance and therefore the chances of carbon-carbon coupling occurring is very, very small. But in homonuclear proton experiments, there is always be proton-proton couplings. So, therefore, homonuclear 2D j-resolved experiments with protons actually leads to some additional complications and we are going to discuss that today. So, the pulse sequence is very similar to the previous ones except that of course you only have one channel here. So, we have the proton channel. So, the experimental sequence is like the spin echo sequence as before. It starts with a 90 degree expulse followed by the T1 by 2 period and then a 180 expulse followed by another T1 by 2 period. These two together constitute the evolution period. The 180 pulse is put in the middle of the evolution period and then the detection happens during the T2 period here. So, this is the detection period. Now, of course, I have put here the numbers 1, 2, 3, 4 to indicate the time points and these will be used when we actually calculate the product operators at the different time points. Notice here during the T1 period chemical shifts are refocused. This we have seen before when we discussed the spin echo. This chemical shifts are refocused during the spin echo. This is the spin echo happens at this point. So, chemical shift evolution happens only during T2. On the other hand, coupling evolution occurs all through T1 and T2. So, this will be reflected in the 2D spectrum. So, consider a two-spin system for understanding the experiment. A weakly coupled two-spin system K and L. So, each one will produce a doublet. So, if this is the chemical shift of K spin, it has a doublet with two lines separated by coupling constant J. Similarly, the L spin also is a doublet with the lines separated by the coupling constant J. Let us try and understand this experiment using product operator calculations. At time point 1, the density operator is ikz plus ilz. The magnetization is along the z axis and this is the density operator for the magnetization at time point 1. And when you apply 90 expels, now you apply to both the spins and this for this will result in y magnetization of the both the spins. So, minus iky plus ily. Now, this row 2 evolves under coupling for the period T1 plus T2. You remember, we can calculate the evolutions of the spins under the influence of the chemical shift of the coupling constant independently. It does not matter which one we calculate first and which one we calculate later. So, in this case, we will first calculate the coupling evolution and this happens for the whole period T1 plus T2. Let us consider for one of the spins, the same applies to the second spin as well. So, we consider the calculation for the spin K. So, therefore, this minus iky evolving under the coupling Hamiltonian gives you minus iky cosine pi jkl T1 plus T2, where jkl is the coupling constant between the spins K and L and plus 2 ikx ilz sin pi jkl T1 plus T2. Among these, you also recall our previous discussion that this anti-phase term is not observable because of the trace of this with whether you take with ikx or ily, it goes to 0 and therefore, this is not an observable term. Whereas, this is it represents in-phase magnetization of K transitions and therefore, this will be an observable term and therefore, we have only this term to consider for further calculations. Now, what we do? We take this term and calculate chemical shift evolution of the observable term during T2. So, this gives me cosine pi jkl T1 plus T2 iky evolves under the chemical shift as this you keep it out cosine pi jkl T1 plus T2 and inside the bracket you evolve chemical shift iky cosine omega kT2 minus ikx sin omega kT2. Now, let us assume that we detect Y magnetization which is without loss of any generality. This we can do this to understand the experiment and the principles behind it. Therefore, the signal here will be minus cosine pi jkl T1 plus T2 cosine omega kT2. Of course, notice here this sign appears because we applied a X pulse but it could be considered without the this minus sign as well if you applied a 90 minus X in the beginning then you would have only the iky term. So, then the force this minus sign is of not much consequence for the understanding of the experiment just now. So, therefore, the signal that we will detect will be just this cosine omega kT2 cosine pi jkl T1 plus T2. Notice here that the coupling constant evolution modulates the detected signal and it has both the T1 and the T2 dependence. Therefore, this is a very interesting situation what is the consequence of this we will see. So, now let us expand this cosine pi jkl T1 plus T2 cosine omega kT2 and this is minus cosine pi jkl T1 cosine pi jkl T2 minus sin pi jkl T1 sin pi jkl T2 that comes from the expansion of this term and cosine omega kT2. So, we see here that along the T2 axis there are both chemical shifts and coupling constants. You see the chemical shift is here and the coupling constant is here. Whereas, along the T1 axis there is only coupling and similar calculations are valid for the L spin also. So, we considered here for the K spin and the same kind of terms will appear for the L spin except that here you will have omega L instead of omega k. Now, let us look at this in somewhat greater detail what sort of a spectrum it will produce. Now, to understand this let us try and expand the term which I wrote in more explicit detail. Let us now calculate explicitly the appearance of the 2Dj resolved spectrum by expanding this the density operator what we got in explicit terms. We can assume without loss of generality that the chemical shift of the case spin is 0. So, in which case the density operator what I have is cosine by the signal what I have is cosine pi j T1 plus T2 and I am not writing the Jkl here because it is implicit that it is between the 2 spins k and L. So, this if I were to expand this gives me cosine pi j T1 into cosine pi j T2 minus sin pi j T1 into sin pi j T2. Now, so this is equal to 1 by 4 e to the i pi j T1 plus e to the minus i pi j T2 into T1 here sorry this is T1 into e to the i pi j T2 plus e to the minus i pi j T2 and minus this terms here and this will give me 1 by 4 here e to the i pi j T1 minus e to the i pi j T1 and the fourth this term will give me e to the i pi j T2 minus e to the minus i pi j T2. So, therefore this expansion will have total of 8 terms let us write this 8 terms explicitly and y e to the i pi j T2 into e to the i pi j T1. Then the second term is e to the i pi j T2 into e to the minus i pi j T1 and the third term is e to the minus i pi j T1 T2 into e to the minus e to the plus i pi j T1 and the fourth term is e to the minus i pi j T2 into e to the minus i pi j T1 and the fifth 6 terms we will write here and the next one will be plus e to the i pi j T1 into e to the i pi j T2 then minus e to the i pi j T1 into e to the minus i pi j T2 and then minus e to the minus i pi j T1 into e to the i pi j T2 and finally plus e to the minus i pi j T1 into e to the minus i pi j T2. So, these are the 8 terms we have here and let us see what sort of a spectrum this will produce. So, we have assumed that the chemical shift is 0. So, assume the chemical 0 frequency is here then along the f2 axis this is the f2 axis this is the f1 axis I have the j information here and let us assume that the peaks the doublet the peaks will be shifted by j by 2 this will be j by 2 and this will be j by 2 again this is on the plus axis this is the plus side and this is the minus side. So, the 0 frequency if it is here 0 and on this axis I will have a plus j and on this side I will have the minus j. So, because they will be shifted from the center one side we take it as a positive frequency other side will be the negative frequency we take this side as plus and let us say we take this side as minus here then I have the 0 line at this point and now let us see what each term gives me. The first term gives me along the T2 axis plus j by 2 that is at this point and it will produce a peak at plus j by 2 on f1 axis. So, it will produce me a peak here the second term will produce me at the same point but on the minus j by 2 here and that will produce a peak here. The third term will produce at minus j by 2 and at plus j by 2 on the T1 axis and that will be peak here and the fourth term will be at minus j by 2 along f2 minus j by 2 along f1 and that will be peak here. Now, the fifth term is plus j by 2 along the T1 and plus j by 2 along this. So, this will be at the same place here. So, I will add here to that. So, there is a fifth term produces an addition at this point and the sixth term will be e to the minus i pi j T1 into e to the minus i pi j T2 this will be minus j by 2 along this axis and plus j by 2 along this. So, that will be here. So, this will be a minus here because this will be subtraction here and the seventh term this will be minus j by 2 along the f1 and that is here and plus j by 2 along the f2 and that will be here and this is again a minus sign. So, that will be this will subtract here and the last term is minus j by 2 along the f2 and minus j by 2 along the f1 and that will be here this will add. So, therefore, this will be plus. So, now what happens therefore these 2 will cancel and these 2 will co-add therefore in the end you will see that in the spectrum these peaks are not present and you will only have peaks here. Now these peaks are if you want to look at it in this manner they are actually tilted with respect to this by an angle which is 45 degrees because this is J by 2 and this is also J by 2. So, therefore this is an isosceles triangle so this angle is 45 degrees. So, this is the way this spectrum will appear and see in the explicitly in the schematically I have shown this in the next slide. So, this is how the schematically the spectrum will appear and we have actually calculated why this comes in this manner. We have shown it for the k spin assuming the k chemical shift is here whether it does not matter that is only for convenience no matter what the chemical shift is the same thing will be valid and therefore you will have here this tilted pattern here and we will have for both the spins we have the 2 lines which are now shifted along the F1 axis at this particular angle. Now, so the detector signal has both cosine and sine modulations as we saw before along both the T1 and T2 axis the cosine modulation results in the absorb to a line shape and the sine modulation results in dispersive line shape after Fourier transformation. We actually looked at it explicitly taking the as e to the i pi j T1 and things like that. You notice from the previous discussion that when you have such a kind of a phase modulation e to the i pi j T1 it already implies that we are going to get mixed phases. Therefore, here along both the axis F1 and F2 axis we will have mixed phases mixed line shapes and therefore we cannot do a phase sensitive experiment in this situation and we have to do a magnitude mode calculation that means the square of the absorb to signal and the square of the dispersive signal take the sum and then take the square root and that is called as the magnitude mode calculation of the spectra. So, when we did that of course, you get a spectrum which is like this. However, we can do a further manipulation of the spectrum by what is called as a shearing transformation. We do not want this to be oriented like this. If we were to take a projection of the spectrum like this it will produce me the same one dimensional spectrum as before. But we want to remove this coupling information along this axis keep it only along the F1 axis so that you have complete separation of the chemical shift and the coupling constants along the two orthogonal axis. So, what we do we can do a manipulation of the spectrum after after data collection after Fourier transformation this is called as a shearing transformation. So, we move this peak to this point and move this peak to this point. Similarly, we move this peak to this point move this peak to this point this is artificially done. So, when you do that then you get peaks like this this separation is Jkl and we have removed the coupling information along the F1 axis completely. So, therefore if we were to take a projection now so we have only one line here you also have only one line and you have the coupling constant present along the effort F1 axis. So, this is a clear separation of the chemical shifts and the coupling constants along the two orthogonal axis. You can extend this calculation to more complex spin systems we did it for two spins now but you can have more you can have two spins three spins coupled and then you will see you will have a multiplets appearing like this. This will be the doublet as we saw before and here we will have a doublet of a doublet and here we will have a triplet and so depending upon the nature of the spin system we will have different fine structures here and that is what will be helpful to measure the coupling constants in a accurate manner. So, what are the advantages of this? So, you can measure the coupling constants in the chemical shifts they are clearly separated the homonuclear broadband decoupled spectra along the F2 axis this is the unique feature is extremely difficult to obtain homonuclear decoupling homonuclear broadband decoupling. We have discussed earlier how to selectively decouple one spin from the other but here where decoupling every spin from every other spin. So, therefore along the F2 axis you have homonuclear broadband decoupled spectra and this improves the resolution in the spectrum because along one axis you do not have the coupling information at all and then coupling constants can be measured with high accuracy and notice also that because we have used the spin echo the inhomogenities are refocused along the F1 axis and this is an extremely useful feature and you will have better line shapes and accuracy of the measurements can be high although of course we lose some of this because of the mixed phases because of the magnitude mode calculations some resolution will be lost. However, along the T1 axis now you see the spectral width is extremely small it is only determined by the J coupling information and the J values are not too large the homonuclear couplings are of the order of 8 to 10 hertz. So, even if you go like doublet of a doublet or doublet of a doublet of a doublet some surface then it will go to not more than 40-50 hertz. Therefore, your increment along the T1 axis will be extremely large compared to what you have when you have the full chemical shift information along the F1 axis. Therefore, within a small number of experiments you can have very high acquisition time along the T1 domain and that will produce very high resolution in the J resolved spectrum. Now, let us look at some experimental data here. So, here is an experimental spectrum of some particular molecule it does not matter what it is you can see here very nicely resolved peaks and here you have a doublet of a doublet a doublet of a doublet and this is more complex spin system here. Now, you do 45 degree shearing that means you move all of these peaks to this line. So, move all of these to this line. Notice of course of the extent to which you move will depend upon the peak where it is appearing how much it is tilted away from here. So, you do a shearing transformation of all of these all of these will move along this axis these will move here along this axis. So, you get now all of them along one line here one line here and one line here. So, if you take a projection you will have a fully decoupled spectrum and if you take cross sections here then you get the fine structure. So, the individual spins and that will allow you to determine what is the coupling constants and what sort of a nucleus it is to how many protons it is coupled and so on. So, we stop here and continue with other methods in the next classes.