 Once again, I welcome you all to MSB lecture series on interpretive spectroscopy. And today let me discuss again on 31 PNMR spectroscopy. In my previous lecture, I gave you very good examples of identifying different isomers of square panor complex and also a square pyramidal complex and also how to distinguish if the coordination number is 5 for a D8 system, two possibilities are there in having an appropriate geometry that is square pyramid geometry and trigonal bipyramidal geometry and how this geometrical changes can happen with temperature we saw in case of rhodium complex having tetraquus, trimethyl, phosphine, methyl composition in its coordination sphere. So let me continue with more interesting examples in this lecture. So now I have given another interesting molecule here. Let us try to analyze the spectra given. This spectrum is of 195 platinum NMR. In this spectrum, if you see, we have two triplets of equal intensity. Then if you look into the structure here, we have platinum. Platinum is directly attached to N15. That means here N-richid molecule with N15 and we have another molecule here isocyanate where we have carbon to nitrogen triple bond is there and we have CH2's are there and it seems hydrogen is not interfering in this NMR, 195 platinum NMR. So let us try to analyze and first thing we should look into is this is N-richid with 15 and 15N we know that I equals half if it is interacting with platinum which is one bond apart it should split the platinum signal into a doublet and this should be 1J platinum 15N coupling. This will split into a doublet. Now we have each line is split into triplet of equal intensity and here if triplet of equal intensity means we should think of nuclear spin value something else and of course the enclosed vicinity we have 13C carbon here. Carbon does not split something like this here because 13C has I equals half and also its abundance is only 1 percent if there is a possibility of seeing satellites. But on the other hand here we have 14N and 14N we know that for 14N I equals 1. Now if it is splitting this one we can use 2 Ni plus 1 rule 2 only 14 nitrogen is there and then its spin is 1 plus 1 so it will be 3 peaks. We have 3 peaks and when I equals 1 and triplet is their intensities are 1 is to 1 is to 1. So in this case it should split something like this unlike the splitting with nucleus with I equals half where intensively 1 is to 2 is to 1 here it is 1 is to 1 is to 1 and then if you assume this is 2 J this is 2 J a platinum N coupling then it should appear something like this we can correlate this one this is 1 J P T N 15 and then if you take any of this separation this is 2 J P T 14 N. So this is how although it appears complicated when you understand it becomes very easy. So this is 195 platinum NMR this involves both 1 J platinum 15 N coupling as well as 2 bonds platinum 14 N coupling. Normally we do not see 14 N coupling and in case if we get something like this interpretation would be rather easy. So this is a typical 195 platinum NMR spectrum and again when we look into 195 platinum here it is very similar to carrying out 13 C NMR measurement. Now I have given set of chemical shifts or spectra here and then write the appropriate structures and match the corresponding 31 P NMR spectrum for the following inorganic cages and explain. So that means 4 spectra are given and I would be providing 4 molecules here form molecules. Now we have to interpret and we have to identify which molecule has which spectrum here. Then if you look into P4 it is a tetrahedral molecule with P4 having each one having 3 bonds connected to each other and now this is all the phosphorus atoms are identical they are chemical and magnitude equivalent as a result one should expect a single resonance in its that even PNMR spectrum. So here one of that one it could be this one if I say one here it can be this one or it can be this one here come back to that one and now let us look into this molecule and when we react P4 with excess of oxygen and we can get P4 O 10 in the same way if you react white phosphorus with sulfur excess of sulfur we can get the analogous sulfide compound that is P4 S 10 and here all phosphorus atoms are pentavalent and tetra coordinated having one phosphorus to sulfur double bond and three phosphorus to sulfur single bonds. As a result all of them are chemically and magnetically equivalent and molecule is very symmetric as a result we also expect for this one to show one peak here. So it could be this one or this one and next let us look into this molecule here in this molecule two phosphorus are in pentavalent state and two phosphorus are trivalent state that means we have two different type of phosphorus atoms are there we have here and then these two are identical and these two are identical. So these two phosphorus are equidistant from trivalent phosphorus here trivalent phosphorus that means these two phosphorus can be coupled with trivalent phosphorus to give a triplet and similarly pentavalent phosphorus also can couple with two trivalent phosphorus to give a triplet that means the spectrum should have two triplets and yes this is two triplets and this can be corresponding to this molecule here. Now if you look into fourth molecule here in case of fourth molecule one of the four phosphorus atoms has three PS bonds. So others they keep PP bonds intact just by looking into the molecule you should be able to tell that we have two different type of phosphorus atoms in a ratio 1 is to 3 and this is one type if I say this is A and this are all B or if I say A this can be X. I think it is appropriate to give use letters from farthest because one is very different from the rest of three. So if I put A here they should be X here that means we are talking about AX3 spin system. AX3 spin system A will be coupled with 3X to give a quadrate delta A if you look into it it would be something like this quadrate and if we look into delta AX so this will be A doublet. So these three are identical they split with A to give a doublet and then A will be split by 3X to give a quadrate. So we have here so this is for four. So that means basically when we take white phosphorus and react with excess of sulfur and it is likely that if you abruptly stop this reaction it is likely that we can get all these products along with unreacted white phosphorus also in that case without any problem we should be able to distinguish then how to distinguish between this one and this one for that one we have to take the spectrum of pure white phosphorus then we should be able to identify or we do excess of reaction for this isolated molecule you take and then we should be able to distinguish between this one and this one. So this is how when we get more than one type of product in a reaction mixture and if we have phosphorus in it it is very easy to distinguish these and identify the corresponding products. So here that is the explanation I have given here. This one you can call it as A2 X2 spin system and this is AX3 spin system. Now let us look into the NMR spectrum of PF5 both 31 PNMR and 19 FNMR. PF5 we all know that it is highly flexional molecule and it is trigonal bipyramidal and because of flexionality what happens we may not be able to distinguish axial fluorine with equatorial ones. So at room temperature if you record 31 PNMR so all are identical as I said we cannot distinguish between equatorial and axial ones as a result all the five would be coupled with phosphorus to give a five or the sextet it should give. So six lines will be there and it should appear something like this and this is 31 PNMR spectrum. Now if you look into fluorine NMR fluorine NMR again you cannot distinguish between axial and equatorial as a result you will see a single resonance for all the fluorine atoms but that is coupled with phosphorus so it appears as a simple doublet something like this something like this but if the molecule is static what would happen the next question is if the molecule is static then let us look into 19 F spectrum if the molecule is static then this is axial and this is equatorial. So let us say then there are two possibilities first it can interact with the axial ones or it can also interact with equatorial ones that means unless we know the magnitude of the coupling constant of phosphorus with equatorial fluorine and axial fluorine we may not be able to tell it but on the other end if the molecule is frozen say let us say minus EIT degree centigrade when we take it we assume that all dynamic process comes to an halt and it is static so that there is no very pseudo rotation to exchange axial ones with equatorial one in that case if we take record the spectrum we should be able to tell simply by looking at the spectral pattern whether the fluorine to phosphorus axial is larger in magnitude or phosphorus to equatorial fluorine coupling is maximum. So let us look into both the cases here first let us assume 1 J P F axial is greater than 1 J P F equatorial so now first what would happen is the phosphorus signal will be split into a triplet here and then each line will be split into a quadrate something like this that means it is basically a triplet of quadrates we will see a triplet of quadrates we will see on the other hand if we see P equatorial is greater than 1 J P F axial first they split this into a quadrate and then each one will be split into a triplet here because of two axial ones and this one is J P F E and now each one will be split into triplet so this coupling will be here axial in this case it appears like something like this so now by simply looking into the spectrum we should be able to tell whether magnitude of phosphorus to fluorine axial one is larger or phosphorus to fluorine equatorial is larger so this two this one will be something like this this is a triplet something like this it shows of course here is 1 is to 3 is to 3 is to 1 is maintained and overall it is 1 is to 2 is to 1 is maintained so this is how we can sketch NMR spectrum of looking into the coupling tree we call this as coupling tree now I have three more examples here sketch the 19 F NMR spectrum of P5 that I have already done sketch the respective NMR spectrum for the following compound showing all reasonable couplings here so let us take one at a time and again if you look into first molecule here we have a square planar complex and in this one again we have two isotopes are there 196 platinum with I equals 0 that is 66 percent and then 195 platinum I equals half with 34 percent abundance and now we have one phosphorus that is directly connected to phosphorus and then we have phosphine imine we have another phosphorus this pentavalent and this if you say this one A this is X and also this is M so we are talking about AMX spin system AMX spin system differs little bit in the sense AM is not 100 percent abundant whereas in this case we can see very identical but instead of phenyl groups and phosphorus we have fluorine are there and another NMR active nuclei and here we have P double bond yes that means basically you can talk this is one type this is another type and this is another type and this is another type so that means we have in case of 31 NMR we can anticipate two signals and each signal will be split in a different way and then if you just look into this case it is very interesting and if you just look into 14 NMR let us say 14 NMR what would happen is hydrogen is also one bond apart and phosphorus is also one bond apart in this case we have to see different situation here so one is nitrogen to hydrogen coupling is larger than nitrogen to phosphorus coupling on the other end and nitrogen to phosphorus coupling is larger than nitrogen to hydrogen coupling and the third one is both the couplings are identical that means if somebody ask you to sketch the 14 NMR spectrum of this molecule without knowing the difference in the magnitude of these couplings we should be able to sketch all three possible ones and after plotting or recording actual NMR spectrum we can compare and we can conclude what actually it is so let us take one at a time now let us look into 195 platinum NMR spectrum of this molecule here first this platinum is coupled with phosphorus this one bond apart to give a doublet and each line in the doublet will be further split into another doublet because of 2 j platinum to phosphorus coupling here so that is shown here and this corresponds to 2 j platinum phosphorus coupling and this corresponds to 1 j platinum phosphorus coupling so it shows a doublet of doublet now I would come back to phosphorus NMR now let us look into this molecule here here a PP bonded compound with one of the phosphorus is oxidized with selenium so selenium again 77 selenium is about 0.6 percent I think 7.6 percent natural abundance is there and I equals half rest is 76 selenium which is NMR inactive but when you are looking to 77 selenium NMR first this will be coupled with doubly bounded so one bond phosphorus and this coupling comes anywhere between 780 to 1200 Hertz first it splits by this one and then we have two bond coupling also that will split each line into another doublet and this coupling magnitude is much lower it can be anywhere between up to 50 Hertz or 100 Hertz so you can see it also appears as a doublet of doublet and then if you look into platinum compound here phosphorus NMR let us look into the phosphorus NMR here 31 PNMR so in 31 PNMR we just look into 66 percent NMR inactive molecules platinum inactive I equals 0 molecules what we get is we get a doublet here and a doublet here this is simply 2 j PP coupling 1 2 j PP coupling and this is 2 j PP coupling so let us say this is PO and then this is PN something like this we can denote and now what happens this accounts for 66 percent of molecules out of 100 molecules 66 have 196 this corresponds to 196 platinum bound but how about 195 platinum bound they will be split further so something like this a doublet and then doublet something like this now now this coupling and this this coupling so this is for the PO 1 j PT PO coupling and this is 1 j PT PN coupling so this can be anywhere between 2500 to it can go to as high as 7000 Hertz so we should be able to again draw 31 PNMR a sketch 31 PNMR spectrum for this molecule here and if we sketch phosphorus NMR here for this one again both are different AX pin system we can get something like this and now this selenium is also there so what happens the one that is safe it is selenium bound and this is with lone pair this one will be split and it will be splitting something like this and then we will be having satellites here and then this coupling what we call it as 1 j PSE coupling here and whereas this one it is less likely to show even if it shows the coupling magnitude is very small let us say if it is there it is something like this and then in this case it is 2 j PSE coupling so once we know the relationship and how what is the natural abundance we should be able to catch appropriate NMR spectrum no matter which nuclei we are talking about so let me stop here and come up with a few more interesting molecules in my next lecture so I showed you here a rhodium compound let me discuss NMR spectrum of that rhodium compound in my next lecture until then have an excellent time thank you