 Hello, everyone. It's a pleasure to welcome you all once again to MSB lecture series on Interpretive Spectroscopy. In my last couple of lectures, I started discussion on 30 NMR spectroscopy. So let me continue from where I had stopped. And let's look into molecules having more than one NMR active nuclei. For example, if you look into phosphorus compounds, we can have 19F fluorine. We can have H. And of course, we can have 13C. Let's try to look into more interesting molecules and try to interpret the data obtained from different NMR nuclei. So one such example I'm going to show you here. Look into this phosphorus compound here. We have on phosphorus one fluorine is there and one H is there and one OH group is there. And also I have displayed one H NMR spectrum, 19F NMR spectrum and also 30 NMR spectrum. So let us see how one can split the lines to see multiplets are OK or not. For example, if you just look into one H NMR, there are two types of hydrogen atoms are there. One is OH and other one is directly attached to phosphorus H. And we know that phosphorus to fluorine and phosphorus to hydrogen with one bond separation show huge or large amount of coupling constant and they come in the order of 800 to 900 and sometime at up to 1200 hertz. And in this example, if you just look into the OH signal, it comes around 14. It is pretty deshielded. And then when you look into hydrogen, now hydrogen can split by two ways. One is with one bond separated phosphorus that splits into a doublet. And then each line in the doublet will be further split into doublets because of two bond coupling with fluorine. So let us ignore this one time being. Let us focus our attention on signal of this directly bound phosphorus bound hydrogen. First it shows a doublet and this we call it as 1 J pH. And of course, magnitude is also given here. It is 780 hertz. Then it will be coupled with fluorine. So this would be a doublet and this is 115 hertz. I think 115 hertz. So this is again 115 hertz. This is 2 J pH coupling. So now the spectrum should look like. So you can see here this is a spectrum. So this is how you can interpret the splitting. And now let us look into 19 F here. 19 F when we consider again this is first coupled to phosphorus and then it will be coupled to hydrogen. And it is very similar to what we saw in case of 1 H NMR signal for this one. This distance is called 1 J P F. 1 J P F is 1030 hertz here. And then each line is further split because of 2 J a coupling here. This is 115 hertz. So this is also a doublet of doublet. So phosphorus and NMR we can focus our attention now. Again first it will be coupled to P F because here P F coupling is much larger. So it first splits into doublet. And here this is the coupling this is 1030 hertz. And then each line in the similar fashion it will be split by hydrogen and this is 1 J pH coupling. This is 780 hertz. So all of them look identical except if you omit this portion all of them look identical doublet of doublets because they all have two couplings with one bond and two bond couplings. So this is how you can interpret data very easily. And we must have understood now how it is easy to interpret data obtained from multinuclear NMR spectra. Let us look into more such examples here. Of course here I have shown you. So it is doublet of doublet here. First it splits by fluorine 1 J P F and then each line will be split into further two lines because of pH coupling. This is 1030 and then this is 780 hertz. So it appears like this. Now I have to spread all of them. You can clearly see here. Now let us move on to another example where instead of 1 F we have 2 Fs are here and then instead of OH we have a H here. And again we can interpret data in a similar fashion. First let us take up 1 H NMR. 1 H NMR if you take first it will be split into a doublet. This is 1 J pH and 1 J pH is about 8080 hertz here given. So now each line will be split into triplet because we have two equivalent chemical and magnetically equivalent fluorine atoms are there. They are two bond apart from hydrogen. So they split this one to triplet. That means basically it happens something like this. And this spacing is 2 J pH coupling. And this is in the order of 115 hertz. And the spectrum should look like. So this is 1 is to 2 is to 1 ratio you should remember from Pascal. Or if you have forgotten I can simply show you. So we have two of them are there. One is like this. One will be like this. And so this ratio is 1 is to 2 is to 1. So what you get is this is what I have shown here. So it is very easy to interpret. This is about 1 H NMR spectrum. Now let us look into 19 F NMR spectrum here. 19 F if you see these two fluorines are equivalent. So they will show one chemical shift. First they will be coupled with phosphorus to show a doublet. And each line in the doublet will be further split into a doublet because of 2 J pH coupling. And as a result what we get is a doublet of doublet. And this 1 J pH value is given. 1 J pH is 1110 hertz. And then here this is F H coupling this is about 115 hertz. So this is again a doublet of doublet. So what is left is now 31 P NMR. So 31 P NMR if you see here we saw from the previous 19 F NMR that PF coupling is larger compared to pH coupling. As a result first P signal is split into triplet. And then each line will be split by same 1 J pH coupling the magnitude of that one is little less. So it comes second. First we will take something like this. So this spacing is same as this spacing. This is our 1 J PF and this is 1110 hertz. And now each line will be split into a doublet here. So now all are identical. These spacings are identical. This is 1 J pH. This is 8080 hertz. But the spectrum should have looked like something like this. But if you just see there is little bit of overlapping it appears for this one and these two are one. And for this one little bit overlapping is there because what happens if you see the difference is much less here. As a result what happens it little bit it comes here and it comes here little bit and same thing it comes little bit here and it little bit here. As a result what happens the spacing doesn't look like uniform. So now it appears like this. So but still we should be able to interpret the data to understand the splitting pattern. So you can see that one that is pH coupling. So now let us look into two examples given here. Two isomers of a squabbener complex is shown here. We have on platinum two PME3 groups are there and one bromine and one chlorine is there. And you know that squabbener complexes of the type MA2 B2 or MA2 BC. So they can show isomerization. And when you look into isomerization two type of isomers are possible. One is cis, one is trans here. And then how to distinguish them? For example I have made these compounds in a particular reaction in that one I do not know which isomer I got or I must have got both the isomers. Then how to interpret data? Just if you just look into this trans compound here trans compound we can do C2 axis rotation. As a result what happens these two are identical indistinguishable whereas here you cannot do that one. So if you do the C2 axis rotation we are goes here and CL goes here. So it does not have C2 axis of rotation and either in this direction or in this direction. So here two phosphorus are chemically equivalent but they are not magnetically equivalent. As a result and of course even chemically they are not equivalent because one is cis to chlorine, one is trans to bromine. Otherwise this one is cis to bromine and trans to chlorine. As a result both are different so both of them show separate signals. And then we know that platinum 195 platinum. We have two isotopes 196 platinum and this is roughly 34 percent NMR active and I with I equals half and then this is 66 percent is NMR inactive with I equals 0. So that means basically if I take any of these isomers 100 molecules if I take out of 100 molecules 66 molecules do not show any interaction with platinum because of 196 which is having I equals 0. And then if you consider the other 34 percent so they will be interacting with 195 platinum with I equals half that can split the phosphorus signal. So first let us consider this one we get a singlet like this for 66 percent of that one no coupling with platinum whereas this one will split into a doublet here when it splitting into doublet something like this they will be coming and if this is 66 percent this will be 17 percent intensity this will be 17 percent. So something like this and then this separation we call it as 1 J VTP. So this is how a trans compound will look like but on the other hand when we look into cis one we will get two signals here two signals and also because of difference in their chemical and magnetic behavior they split each one into a doublet. So that means basically you get a doublet here and you get a doublet here one is for this one one is for this one. This platinum also splits each into further a doublet here something like this and this one might come here and something like this will be there very similar to here and this spacing is same as this spacing and then here this spacing is same as this spacing. So here you can if you assign A and B this is delta P A and this is delta P B and then if you take here this is 1 J P T P A and then if I take this one is here if it has come. So for example the center of the middle line if I take this is 1 J P T P B. So it is basically having two sets of this that is it because they have again coupling between them. So each line will be split into doublet. So this is how we can interpret and the moment we look into spectrum we should be able to tell whether we have obtained cis isomer or trans isomer or both of them are present in certain stoichiometric ratio by looking into the intensity we should be able to tell whether the cis is 75 percent trans is 25 percent or vice versa. So this gives some idea about the type of isomers we have obtained and also how the spectrum look like. So it is very easy to distinguish. On the other hand we carry out this reaction we do not know whether we got cis isomer or trans isomer then if the spectra is provided then we should be able to tell which isomer is present or there is a possibility of the presence of both isomers in different stoichiometric ratio. I have another interesting molecule here 31 P N M R spectrum of tetrachystrimethyl phosphine methyl rhodium compound shows some sort of flexionality at room temperature it shows a doublet at minus 80 degree centigrade it shows two signals one is doublet of doublets and another one is quartet of doublets. So how that happens first we should look into the molecule and how many ligands are there and of course by looking into the molecule we should be able to tell that it is rhodium one. Let me write the possible structures first let me write square pyramidal geometries all possible. So this is one possibility where all phosphines are in the plane and methyl in the axial position so this is another possibility I do not think I should be able to write any more isomers for square pyramidal geometry here only two are possible. Now another possibility is trigonal bipyramidal so here let me put two phosphines the axial position and two in the equatorial plane and then one methyl group here or other possibilities one methyl group is here one pph 3 is here these. So now these are the possibilities I do not think any other isomeric forms are possible for this one. So now let us look into room temperature spectrum in the at room temperature what we are getting is a simple doublet. Why we are getting a simple doublet here we should remember one or three rhodium is NMR active it is 100 percent abundant and I equals half and then phosphorus to rhodium coupling constant would be anywhere between 160 to 300 hertz. As a result in this case if I write it couples so all these phosphorus atoms are equivalent and they couple with rhodium to give a doublet here. So that means if the correct structure is for this one this is the doublet. So at room temperature it shows a doublet means the structure is something like this where 4 trimethyl phosphine groups are in the plane and one methyl group in the axial position. So now let us look into doublet of doublets and quadrate of doublets here if you just look into this structure here in this structure one is axial and this is you know making cis relationship with remaining three. So first let us say this will couple with this one to give a doublet this one will couple with this one to doublet and each line will be a quadrate same thing is to increase of this also if you see here. So this is coupled with rhodium to give a doublet and each line in the doublet will be split into 1, 2, 3 so quadrate here. So that means two possibilities are there whether it has this structure or whether it has this structure that we have to analyze now. First let us look into these three in the plane when you look into these three in the plane what actually happens is so this one is trans to this one and these two are trans to each other as a result what happens this may not be a right structure for interpretation of this data provided here quadrate of doublets and probably here these three are identical. So these three would couple equally to rhodium and then they couple with this one to show a doublet. So the doublet is here these three in the equatorial plane would couple with rhodium to give a doublet and then this will be split again into doublet here. So this is 1 j rhodium first first coupling and then this is p p coupling 2 j p p coupling p p coupling. So yes this is a doublet of doublet is there this is fine these are taken care now and now this will be split into this doublet first and then it will first this is split by rhodium to give a doublet and each line in the doublet will be a quadrate here and then this is 2 j p p and then this is 1 j rhodium first process. So that means the axial one this is for axial one and then this is for equatorial ones p e. Now we can say at room temperature this molecule exists in square permeable geometry but at minus 80 degree centigrade this turns into trigonal bipyramid geometry having three phosphines in the equatorial plane and one being axial. So this explains that way we can also understand in the frictionality and the type of geometrical changes that happens with the temperature. So here 31 PNMR comes very handy in understanding the both the geometry and also the position of ligands in the coordination sphere. With this let me stop here let me continue with more interesting examples in my next lecture. Thank you.