 So, in the previous classes we have discussed in detail the product operator formalism for describing density operator evolution through an NMR experiment and we have seen how it becomes easier to use this formalism for analyzing the NMR experiments. Similarly we considered the spin echo in the previous classes and I am going to extend this further describing to you one more experiment as an example and that is the inept. Inept stands for insensitive nuclear enhancement by polarization transfer. This experiment has actually been described to you in the previous classes by Professor Ashutosh Kumar and this is the polarization transfer experiment where the magnetization is transferred from proton to an insensitive nucleus like carbon-13 or nitrogen-15. So here I will consider proton and carbon-13 here to see how the polarization or magnetization gets transferred from the proton to the carbon and we detect the carbon. So, the pulse sequence for this is shown here. So, on the proton channel you apply these pulses you have a 90 X followed by tau and then you have a 180 degree pulse which is applied simultaneously on both proton and the carbon channels. Then you have another time period tau and which is followed by a 90 degree pulse on proton channel and a 90 degree pulse on the carbon channel and this is the FID. Notice here you will have to be careful with regard to the phases of the pulses. Here it is a 90 X pulse this is the 180 this can be X or Y does not matter but this has to be a Y if this is X this has to be a Y pulse. This can be X or Y does not matter this carbon pulse 90 X pulse can be 90 X or 90 Y does not matter but it is very crucial here that if this is 90 X this has to be a 90 Y there has to be a 90 degree phase shift between these two 90 degree pulses and the various time points here are represented like this 0, 1, 2, 3, 4, 5 and this is the time point 6 which is in your detection period. You also recall that this is very similar to the spin echo sequence. The spin echo also had this 90, 180, 90 and separated by the time tau and except that we are now using it as a hetero experiment for transferring polarization from proton to carbon. How does it happen? Let us see this. We will analyze that using the product operator formalism. So, at time point 0 we consider the density operator which consists of the proton magnetization along the Z axis and so this we call I write here now as HZ. So, do not confuse this with the Hamiltonian. This is the operator for the proton Z magnetization and therefore I write it as HZ. When I apply 90 X pulse on the proton channel, notice there is no proton no pulse applied on the carbon channel. So, therefore the carbon magnetization which is along the Z axis is unaffected. So, therefore we do not even write the carbon magnetization here. We will only write the proton magnetization here HZ and the pulse is applied only on the proton channel and therefore we get a rho 1 which is minus HY. And during the period 1 to 4 what happens? This is the spin echo. We have seen that this is the spin echo. The spin echo sequence is exactly in that manner. So, two things happen here. First, the proton chemical shift is refocused completely. So, the 90 X tau 180 X tau at the end of that there is a spin echo. So, at the point 4 time point 4 this is the spin echo and the proton chemical shift is therefore refocused. However, the proton magnetization evolves under the proton carbon coupling. So, this Hamiltonian is the operative. This is not refocused. So, the coupling is not refocused in the spin echo experiment when the 180 pulse is applied on both of the channels both of the nuclear. Therefore, we will directly write rho 4 because we do not consider the chemical shift evolution at all because it is refocused. Therefore, we from rho 1 to rho 4 we will only consider only the evolution under the coupling Hamiltonian between proton and carbon. So, how does this H y evolve? This is through the product operator formalism. The rho 4 turns out to be like this. H y evolves as H y in cosine pi j H c 2 tau because of the time period of the whole thing is 2 tau minus 2 H x C z sin pi j H c 2 tau. Notice now these are the operators. This represents the antiphase magnetization of proton with respect to the carbon and this is the in-phase magnetization of proton of along the y axis. So, these are the operators H y and 2 H x C z are the operators and tau is the time and j H c is the coupling constant between proton and carbon. Now what we do is we set tau equal to 1 by 4 times j H c. So, here if I put tau is equal to 1 by 4 j H c what happens? So, this H c H c will cancel. So, this will become cosine pi by 2 and similarly here this will become sin pi by 2. Therefore, this goes to 0 and only this term remains. Therefore, rho 4 becomes 2 H x C z and notice this minus sin and this minus sin becomes plus and therefore, here I have 2 H x C z and this is now the proton magnetization antiphase to carbon. This is we learnt it from the product operator descriptions. Now, when I apply 90 pulse on proton and carbon, we get rho 5 and rho 5. So, I am applying a 90 y pulse on proton. So, therefore, this H x goes to H z and C z goes to C y because I am applying this the carbon channel applying along the on the x axis. Therefore, this goes to C y. Now, this is carbon magnetization antiphase to proton and that is the interesting thing. Therefore, rho 4 to rho 5 represents a coherence transfer from proton to carbon which all the polarization which is present on the proton has now become carbon magnetization. What is the meaning of this? This means there is a significant sensitivity enhancement of the carbon magnetization. Sensitivity enhancement by factor gamma H by gamma C because proton magnetization is determined by the gyromagnetic ratio of proton and the carbon magnetization is normally determined by the gyromagnetic ratio of the carbon. But here the proton magnetization is appearing as carbon magnetization. Therefore, there is a sensitivity enhancement of the carbon signal by a factor gamma H by gamma C and this is a factor of 4. So, this is a factor of 4. If this were an hydrogen for example, then this will be a factor of 10 and that implies a substantial saving in time and the great improvement in the signal to noise ratio in your experimental time because experimental time goes as a square of this factor. Now, after this rho 5 evolves during the detection period under the influence of chemical shift and coupling Hamiltonians of carbon. So, these 2 HZCY during the detection period evolves under the chemical shift as well as the coupling. Let us first consider the coupling evolution because you remember I had told you that it does not matter which evolution you will consider first and which evolution you will consider later. So, we will consider the coupling evolution first. There is a purpose in it. So, I will illustrate this to you very quickly. So, 2 HZCY now this is the product operator and you are considering its evolution under the influence of the coupling Hamiltonian between proton and carbon. So, this gives you 2 HZCY cosine pi j HCT. Now, it is a function of time this is during the FID during the FID this is what you are getting 2 HZCY cosine pi j HCT minus CX sine pi j HCT. So, the same thing is written once more here and now you notice that this particular term is not observable this is anti phase magnetization is not an observable term. So, therefore, if we did not have this coupling evolution we would not have any magnetization which is observable this is not observable term because you recall our discussion that to be observable it is a trace with CX or CY has to be non-zero. Now, if you take the trace of this with either CX or CY this is 0 and we had actually described this earlier also that the anti phase magnetization is not an observable operator whereas this one is an observable term. Therefore, the coupling evolution leads to an observable magnetization here observable term. So, this is the observable term and this is not observable. Now, therefore, after this we need not consider the evolution of this at all for the chemical shift because this is anyway not observable. So, this will go away. Now, we will only consider this term therefore we will now consider minus CX sine pi j HCT will evolve this under the chemical shift. So, keep the minus sine pi j HCT outside and evolve this CX operator under the chemical shift of carbon. So, this gives you CX cosine omega CT plus CY sine omega CT this is the normal chemical shift evolution under the influence of the carbon chemical shift the magnetonion. Now, so I have a CX term and a CY term right. So, if I observe the X magnetization I get this term if I have to observe the Y magnetization then I get this term. So, therefore, rho 6 the density operator if I were to do X detection then I get this minus CX sine pi j HCT cosine omega CT and if I were to do a Y detection I will get minus CY sine pi j HCT sine omega CT. What is our signal? To see what is the signal I have to take the trace of this with CX or CY. So, when I take the trace of this is CXC operators this CX terms will vanish I will only be left with this coefficients which actually are functions of time and that is my observed signal. So, therefore, the signal what we get after taking the trace with the CX or CY. So, this will be for X detection sine pi j HCT cosine omega CT. Now, let us try to expand this and this sine pi j HCT cosine omega CT will be expanded as the sum of two sine terms. So, when we expand this as sum of two we get this here I have to put this there is a factor of 1 by 2 here I have to put it in. So, I will get two frequencies here there are two sine terms here. So, these are two frequencies the frequencies are omega C plus pi j HC and the other one is omega C minus pi j HC of course, these are in radiance. So, if I take out the if I want to express them in hertz in terms of numbers then I take out this 2 pi here I write this as sine 2 pi nu C nu C is actually in terms of the chemical shift plus j HC by 2 T and this is minus sine 2 pi nu C minus j HC by 2 T. So, I have two sine terms here with the frequencies nu C plus j HC by 2 and nu C minus 2 HC by 2. So, if I would do Fourier transformation of this signal then I will get two frequencies the frequency spectrum. But this frequency spectrum will have dispersive line shape because these ones are sine terms this is what we have seen in the very early discussions in the course. So, this sine term is giving me two dispersive signals and now because of this minus sign these two also have opposite phases. So, this goes like this positive and negative with and then thus this will be opposite in sign compared to this and therefore this goes in this manner and we have two dispersive signals which are anti-phase in nature when it is called an anti-phase doublet with dispersive line shapes. Now, during the detection during the detection period these two signals will be present as they are if you do not decouple you cannot decouple this if you want to decouple this then of course these two will collapse fall on top of each other and then this will be zero signal. So, therefore in this situation you cannot decouple the during the detection period and this the center of this is the chemical shift and the separation between these two is the coupling constant JCH that is from here notice the center is here the center is not here or here this is center is here this center is here. So, from here to here it is JCH. Now, if I wait to do a Y detection then my term what I have is sin pi J H C T sin omega C T once again the operator part has gone off because after I take the trace with the C Y that the trace of C Y square is 1. So, therefore that will not appear and I will have only have the time dependent function here which actually is part of the FID. So, once again here I will have a factor of 1 half. So, I have now once again the same two frequencies omega C plus pi J H C and omega C minus pi J H C these are two frequencies but now the difference is these ones are cosine functions. So, representing them in this in terms of the hertz. So, again I write it here as cosine 2 pi nu C plus J H C by 2 and this is nu C minus J H C by 2 both are cosine terms and once again these two have opposite signs. So, this one is if I wait to take it as positive and this will because of the minus sign this will be negative. Now, it will represent absorptive signals because these are cosine terms. So, the cosine terms will produce me an absorptive signal here at this particular frequency and this will produce a negative signal at this particular frequency which is this. So, once again because of this antiphase signals C H cannot be decoupled during detection. What happens if you decouple because this they will overlap they come on top of each other this will come here and this will come here therefore they will cancel. So, therefore you cannot decouple if the signals are antiphase in nature the antiphase doublet will appear like this. So, here is an experimental example this was the very first data which was recorded this came from Morrison Freeman and this is published in 1979 in Journal of American Chemical Society and you can see here there are three antiphase doublets. So, one here this is negative positive negative positive negative positive and of course you are seeing some fine structure here and of some molecule which is there and because this is because of the carbon-carbon couplings which also evolve during the detection period. So, we have not done nothing about the carbon-carbon couplings. So, wherever there is a carbon-carbon coupling it will lead to a splitting and therefore the whole set here will contain this carbon-carbon couplings here. So, there are three carbons C2, C4, C3 and they are coupled to each other and that results in this kind of a fine structure here and you have antiphase doublets for each of the carbons. Now, but this antiphase nature is often a disturbance and they can if there are two antiphase which are close by then they may lead to cancellations and therefore we do not want to have that thing to happen. So, therefore we need to do something further to improve upon this. So, therefore what we do we do what we called as the refocused inept. So, here the pulse sequence is extended see the inept was ending here and you are having the FID collected from here. Now, for the refocused inept what you do is you extend this pulse sequence by introducing another spin echo sequence here tau 180 tau and the 180 pulse is applied on both the sequences and both of the channels both proton and carbon. So, you have tau 180 tau applied and the data is collected as a function of time on the carbon channel and you can also decouple here on the proton channel how it happens we will I will show you. So, until rho phi the product operator description is the same. So, we do not want to do that once more here. Now, rho phi is 2 HZ CY this is the carbon magnetization anti phase to proton. So, this is what we had now evolution under coupling. So, now where is rho 6 here see the rho 6 is at this point this is the extended point we want to calculate was the density operator at this point. So, rho phi to rho 6 what happens there is no chemical shift evolution because of the 180 pulse the magnetization is on carbon magnetization is here and this gets refocused chemical shift is refocused because of the 180 pulse and then because of the 180 pulse is on both carbon and proton channels this evolution under the coupling continues and that is not refocused. Therefore, at this point rho 6 we have to consider evolution under the coupling only. So, therefore if I do evolution under coupling that gives me the rho 6 and what is that 2 HZ CY evolution under a coupling Hamiltonian leads me to 2 HZ CY cosine phi H is AC 2 tau once again because of the 2 tau period of the spin echo sequence minus CX sine pi JHC 2 tau. Now, once again I adjust my tau because this is under my control I adjust my tau to be equal to 1 by 4 times JHC for assuming JHC is of 150 hertz and this will be only tau will be approximately 1.66 milliseconds or 1.7 milliseconds. So, this is approximate therefore if it is 140 hertz so 150 hertz and things like that. So, approximately one can adjust this to 1.67 milliseconds. Now, when you put that this term goes to 0 as before because this is cosine pi by 2 and this will become 1. Therefore, 2 HZ C phi will now become just minus CX all the time dependent this terms are gone I only have C rho 6 is therefore simply equal to minus CX. Now, suppose the proton was not decoupled. Suppose the proton was not decoupled now CX will evolve under a chemical shift as well as coupling during the detection period. So, now we are considering what happens during the detection period rho 6 was at the end of the spin echo. Now, this will now evolve under the coupling this is rho 6 the rho 6 was here there is no pulse applied here. So, this will evolve during the phi D it will evolve under the influence of the chemical shift and the coupling and depending upon what you do. So, if I wait considering without proton decoupling then I will consider the evolution under chemical shift as well as the coupling. Let us consider first the evolution under chemical shift. So, this will give me CX cosine omega CT plus CY sine omega CT just I ignored the sign here so it does not matter. So, CX cosine omega CT plus CY sine omega CT assuming X detection we will ignore this term this because we are only demonstrating in principle of course when you are collecting the data and evolving so you will have to consider all the terms which are present but this is to demonstrate how things develop depending upon what term you select. So, if I were to collect X data that is X detection and evolve under the J coupling now. So, shift evolution has produced CX and CY now consider the evolution of the X part under J coupling and this gives me rho 7. So, the rho 7 gives me cosine omega CT CX cosine pi J HCT now it is a function of time plus 2 CY HZ sine pi J HCT once again this term is not observable this is anti phase carbon magnetization with respect to the proton and this is not observable because this trace will be 0 with respect to CX or CY. Therefore, my signal will be cosine omega CT cosine pi J HCT. So, once more I write this signal as this and I will decompose this into 2 terms into the 2 frequency terms put here once more a factor of half. So, what do you get here now? So, this is the same as before we get 2 frequencies omega C plus pi J HC and omega C minus pi J HC because if I take it in terms of the 2 pi out in terms of the radiance we take it out your nu C plus J HC by 2 and nu C minus J HC by 2. But now notice I have a plus sign here earlier this was a minus sign now I have a plus sign which means both these signals have the same phase. So, both are positive. Therefore, this is an in phase doublet of carbon. So, we are starting from the proton magnetization which is the refocused inept during the detection period. So, I have here in phase doublet as a measurement. Now, this separation is J HC J CH. Now, I can decouple this if I decouple these 2 will collapse into a singlet decoupling collapses this doublet into a singlet and this will be at the center here and which is the twice the intensity because both this will contribute to the intensity and intensity of this will be twice. So, this will be a big advantage for the carbon signal. So, now here it is a comparison of an experimental spectra under the various conditions. This is the carbon 13 spectra of methanolic acid there is only 1 carbon and 1 proton and so you have here a coupled spectrum which is a doublet normal coupled spectrum carbon detection direct carbon detection and if you decouple this you have this appearing in the center here the carbon this is a decoupled spectrum. Now, if it is an inept experiment now it is the magnetization is now coming from the proton remember it is not the we are not taking the carbon magnetization as it is done here the magnetization is coming from the proton therefore this has a gamma H by gamma C signal enhancement therefore in the normal inept you have this anti phase signals which are one positive here and one negative here. Now, if I did a refocused inept this will turn out to be positive it will go up like this so this is positive here and this is positive here and now refocused and decoupled now you decouple this so this will do collapse again once more here and you have a huge signal coming here. So, this is the significant advantage in the signal to noise ratio through this refocused decoupled inept experiment and that is how this has become an extremely important tool extremely important technique for heteronuclear experiments for all organic chemistry people this is becoming extremely useful technique for determining how many carbons there are in their molecules how to get signal to know and you can do with the natural abundance you can do carbon 13 spectra at natural abundance you use proton magnetization to enhance the signal to noise ratio and you can identify the number of carbons in your molecule and if you couple record proton coupled spectra also then of course you can see the multiplied structures in all of those things. So, I think I will stop here.