 Hello everyone, once again I welcome you all to MSP lecture series on interpretative spectroscopy and I am sure you are all enjoying the interpretation of very interesting phosphorus compounds and also the metal complexes having several other NMR active nuclei. So let me continue with some of the molecules that I showed you in my last lecture. I showed you in my last lecture this molecule and then three spectra here and the first one we concluded that it is due to 31 PNMR and the splitting pattern also I showed you and just remember now we have another set of signals here in the second one. Second one if you just see 1, 2, 3, 4, 5, 6 are there that means 4 multiplets of 6 lines each and again in the last one also we have 6 lines and 4 multiplets are there each one having 6 lines, but the spacings are little bit different that means they are not due to the same nuclei they are due to the different nuclei. So just let us look into them now it is very clear that 31 PNMR spectrum can be interpreted in this way by writing this coupling tree here you can see first it is coupled with 3 fluid atoms to give a triplet and then each line is coupled with 15 N to give a doublet here and then each line of this doublets again split into doublet because of 2 bond hydrogen coupling something like this and then each line is coupled with 3 hydrogen atoms present on silicon to quadrate in 1 is to 3 is to 3 is to 1 ratio and if you count all of them we have totally 48 lines and that 48 length spectrum looks something like this the original 1 is to 2 is to 1 ratio what we saw here is retained here and then again if you see here this doublets are 1 is to 1 is to 1 that is also maintained here. So now let us look into 19 F NMR spectrum it looks little bit complicated, but it is not really. So then first of all we have to analyze how each multiplet has 6 lines here before that let us start looking into 19 F NMR assume it is 19 F NMR and at the end we can conclude about that one and if you take here if you consider 19 F NMR first that will be split into a doublet because of phosphorus coupling. So, this is the PF coupling here and next this is coupled with 2 bond 15 N that is again each one will be a doublet here this is F N coupling here this is F N coupling and then each signal of a fluorine is coupled with hydrogen to give another doublet here. So, this is your 3 F H coupling due to hydrogen that is present on nitrogen now we have to look into 1 2 3 4 bond coupling here. So, 4 bond coupling would be here so that would split each line into a quadrate because of the smaller spacings of F H with the magnitude of that one is much smaller what happens first 2 lines of the second signal and last 2 lines will merge as a result what happened the intensity of this will go up and then same thing happens in case of all the 4 signals here and you can see here they are merging little bit because of closer spacings and very marginal difference in their chemical shifts it can show now 1 2 3 4 5 6 are there, but intensity varies now as a result what happens it appears like a sextet in each case and then if the spectrum is given sometime often we get confused why sextet is there each line should have been a quadrate very similar to what we saw in case of phosphorus, but here it is sextet because these 2 are merging here. So, as a result we get something like this this is a typical 19 F NMR spectrum shown here. Now, let us look into 15 N NMR spectrum again we have 2 options are there whether which one is larger whether N H coupling is larger or P N coupling is larger let us assume P N coupling is larger getting some hint from the previous 31 P NMR spectrum we looked into first this N will be split into a doublet this is P N coupling here and then this will be coupled with hydrogen. So, something like this so, now if you just look into it now we have 6 lines are there each one that means there is no individual splitting is there that means basically 1 2 bond 1 2 bond. So, fluorines also 2 bond apart from 15 N and also silicon bond hydrogen atoms also 2 bond apart from 15 N that means totally if you assume the coupling constant is of same magnitude in case of both F N coupling as well as 2 bond N H coupling then we have to count together. So, we have 1 2 3 4 5 equivalent nuclei are there then if you just use 2 N I plus 1 rule here of course, all of them are with spin half when we have 5 of them half plus 1 it goes it gives 6 lines. So, each one should be a sextet so that means we have something like this 1 2 3 4 5 6 1 2 3 4 5 6 and the spacings are little bit different compared to what we saw in case of 19 F N M R. So, it is the spectrum should look something like this yes now we have the spectrum. So, very easy so that means basically the 3 spectra that is given are for 19 F and 31 P and 15 N. So, what is left is 1 H. So, 1 H I had left it for you people. So, you people try to sketch N M R spectrum of writing the coupling constant tree a splitting tree and then just see how it looks like and you can take about the couplings you can take hint from these 3 spectral data. Now, I have given another compound here this is cyclo-diphosphazene and cyclo-diphosphazene are 4 membered inorganic ring system saturated ring system with alternate arrangement of phosphorus and nitrogen with phosphorus in rivalant state and they can exist in cis and transforms cis is more stable kinetically more stable isomeric form conformation and then the nitrogen is planar here because of the 2 pi electrons interacting with the sigma star of phosphorus that we call it as negative hyper conjugation as a result what we have is 2 pi sigma star interaction because of that one nitrogen lone pairs are delocalized here as a result the P N has multiple bond character and hence this appears like a planar molecule having S P to hybridization nitrogen. So, when you react to this one with appropriate metal reagents either you can form binuclear bimetallic complexes or monometallic complexes, but it cannot kill it because of the planar structure, but it can only act as a bridged bident teteligant or if use one equivalent it can act as a monodent teteligant leaving either phosphorus uncoordinated we have one such case here rhodium chlorocard dimer when it is reacted with the two equivalents of cyclo-diphosphosine what we get is a mono substituted mono coordinated compound of this type here we have chord cycloctadiene is there. And now if you just look into it this is a typical AX spin system in case of 31P and now if you see this PX PX is directly attached to rhodium and first it will split into a doublet and then of course, each line can further split into a doublet because of P P coupling in that case what happens it will be the values are very small and it would appear something like this and whereas, P is coupled with PA is coupled with X to give a doublet and then it does not show in a rhodium coupling otherwise it would have split into a doublet of doublet here again you can see here that is rhodium phosphorus coupling of 229 hertz and then see whatever the coupling we anticipated is not there, but when you since we have one free phosphorus is there you can again coordinate that one with another heteroatom for example, when it is treated with AUCL SME 2 here this is reacted with AUCL SME 2 SME 2 is a very labile ligand and SME 2 goes off and then what you get is here AUCL is bound now again we have AX pin system both are coordinated earlier we have one free uncoordinated phosphorus now you can anticipate some set of P P coupling comes now P P coupling is coming here in the previous one we did not see now the P P coupling is r of 30 hertz that you can see here and also surprisingly now phosphorus is coordinated to gold it also shows rhodium coupling that is 3 J rhodium phosphorus coupling of 5.58 hertz and this one what we see here is due to the complete substitution of both the phosphorus with gold that means basically when this compound is treated with AUCL SME 2 apart from forming heteronuclear compound partly it has replaced rhodium also to form di gold complex that is what it indicates here some of this vital information you can extract when you look into NMR spectrum carefully so it gives much more than what you are looking for this is where the importance of spectral data comes into picture in understanding the reactivity and other things for example now if I start using excess of AUCL SME 2 probably I should be able to replace rhodium also to form a di gold complex in fact we have done that one in our laboratory sometime back so interpretation is easy and also we came to know that suddenly phosphorus phosphorus coupling is increased when both the phosphorus are coordinated and it is now very interesting to look into the electronic situation at both the phosphorus with some theoretical background or calculation DFT calculation and other things that might tell you why now P P coupling is very intense compared to the compound where only one phosphorus is was coordinated so this is for this one as I mentioned here so now one more interesting molecule is there again this is a rhodium chlorocarbonyl complex having phosphine immune ligand so one phosphorus is directly attached to rhodium and other nitrogen has phosphoryl azide is bound one so we have three phosphorus atoms are there now so that is like almost AMX spin system and in this one in first case where we have P this phosphorus here this is directly attached to rhodium so it can first couple to give a doublet and each line can be further split into a doublet because of this two bond P P coupling so this one you can assign for P here so this is the rhodium coupling here 1 j rhodium phosphorus coupling and this is 2 j P P coupling here and this is further so this is not showing any coupling with this one now if you look into this one it's a doublet of doublet this may be due to this one here this first couples with this one and then this is coupled with this one also it's a doublet of doublet here so this phosphorus doesn't show rhodium coupling whereas this one shows now only this one is left now this one only coupled with this one so it gives a simple doublet so that means now very nicely interpret the data and also check whether the compound is formed or not for example this ligand is taken when it's treated with half equivalent of rhodium chlorocarbonyl dimer so this is taken and then this is reacted with so this is two equivalents we had taken and then it gives the compound shown here this compound is shown here yes now we can conclude that this bond is broken and it is formed a mono rhodium compound in this case now we have another interesting molecule here this is a platinum compound of bisphosphine where it is a unsymmetrical bisphosphine one side it is OPPH2 other side it is NPPH2 and then if you look into the 31 PNMR spectrum of this one it shows two signals without any PPP coupling as they are 1 2 3 4 5 bond apart so that means it doesn't show when we make a compound with this one say platinum or something now through metal they can interact because through metal they are only two bond apart so if through the ligand framework they are 1 2 3 4 5 bond apart so when they are 4 or 5 bond apart it is less likely that they couple with each other instead they will show a singlet but on the other hand when a complex is formed what happens now they are coming closer through metal you can anticipate through metal coupling with this one it is a very nice example where the phosphorus phosphorus coupling is absent in the free ligand whereas when it forms a chelate complex now they are coming closer and they can show coupling here that means it also gives some idea about when such molecules are there how much the coupling one can anticipate that information comes directly from this one initially this free ligand shows one the two signals singlets and then when you make a platinum complex now both the phosphorus are coupled apart from coupling to platinum I have shown here you can see here pn bound one is shown here and whereas this is for p o bound one and then they are also showing the satellite as expected because this whatever the doublet is their major doublet this is for 66% 196 platinum i equals 0 and then the corresponding one is here and one is here this this coupling is 3933 hertz and this is 1j pt p coupling here and 1j pt p coupling is 3933 hertz and in case of p o it is 4066 hertz are there and then the pp coupling is 13.3 hertz here and this is a again interesting molecule here you can see the structure it is it has a puckered chelate ring you can see here now they are coupled through space one more example is there here just look into this molecule here this is a cationic complex we have three different type of phosphorus atoms are there and all of them are coupled with each other because they are all two bond apart so now again interpretation is also easy here just to keep you see here all are different so you have something like a m x spin system here this phosphorus couples with first this one trans and then it gives doublet of doublet and then similarly this one would couple with this one or this one doublet of doublet and this also couples with doublet of doublet because all are in different chemical environment is there you can see each one is showing a doublet of doublet a typical a m x spin system so in case of a m x spin system and also the field strength is very high it doesn't show any second order splitting here second-order splitting is invariably observed in case of one HRMR where we have two hydrogen atoms present on same carbon the geminal coupling will be seen the carbon is substituted with different groups on either side so now let us look into an interesting reaction here this ruthenium compound this ruthenium compound I'm sure you know the structure of it this is a tetrahedral molecule with ruthenium in plus 2 state you can also do electron counting for this one if you do by neutral method this is 8 plus 5 for CP group and this is for 1 and this is for 2 and this is for 2 this is a 18 electron complex 14 plus 16 plus 18 and this is by neutral method by ionic method also we shall do this ruthenium 8 means 6 are there and this is giving 6 electrons now and this is giving 2 electrons now and both of them are giving 2 electrons each this is 18 electron so this is the 18 electron system when it is reacted with this PNP ligand what happens we are getting lot of different type of compounds are there here for example you can see here when it is reacted it give a mixture of compounds here it give a mixture of compounds one is a chelated compound replacing two triphenyl phosphines whereas here it becomes a cationic complex here two PpH3 and Cl also has come out and then one chelation and one acting as a dangling monoid ligand in other one PpH3 is there and so all these three are formed here once all of the how do we know that all the three are formed simply by looking into the 31 PNM spectrum of the reaction mixture that reaction mixture also gives in what ratio these compounds are formed once we know that in what ratio these compounds are formed whether it is possible to prepare those compounds in pure form by varying the stoichiometry and varying the reaction condition for example if this you take this compound here and then ruthenium to chlorine bond is ionized that means if I use a polar solvent and you prolong the reaction it is likely that I can form a cationic compound something like this or if I use excess of phosphine this phosphine one can get like this but by controlling the stoichiometry if you use one is to one reaction in a polar solvent you can get this one that means this kind of vital information one can get simply by analyzing the spectrum of reaction mixture again using very simple 31 PNM spectroscopy so let me show you the spectrum for this molecule in my next lecture very interesting and also how did we come to conclusion about the spectrum analyzing this particular set of multiplets for this molecule that of course when we carried out individual reactions also separate reactions and isolated compounds we can conform and in some cases it's very simple whereas in this case it's very simple you should get a droplet and a triplet that is there and then in this case you can get a singlet and so there is no scope for any confusion but if there is scope for confusion of having similar pattern for two or more molecules then we have to separately make them and conform their chemical shifts okay let me discuss more about this one in my next lecture until then have an excellent time thank you so much