 Once again, it's a pleasure to welcome you all to MSP lecture series on Interpretative Spectroscopy. In my previous lectures, I started discussing about NMR spectra of various phosphorus containing compounds and also how to analyze if you get more than one product in a reaction. For example, if you get cis and trans isomers or if you get some other isomers, we should be able to elucidate and understand the structures of those molecules by simply applying phosphorus NMR spectroscopy. So, let us continue from where I had stopped. Let me take another interesting reaction here. This reaction, you can see I have used two entities. One is cyclopentadienyl bis-tryphenylphosphine chloro-ruthanium 2 plus that is reacted with bis-tryphenylphosphine amine ligand. So, that means in this reaction, although we know that when you are mixing in a no-starch geometry, we are supposed to get one product, but on the other hand due to some reactivity difference that we gauge and that is what happens later, we end up getting more than one product. So, that means we may get a mixture of products as well. In that case, if we analyze by looking into the 2NPNMR spectrum of that reaction mixture that can tell you lot more details and also that can also guide you how to plan our reaction by altering reaction conditions. For example, in this one as I said, this ruthanium 2 compound was treated with this bisphosphine. In the above reaction I have shown here, we are getting 3 products 4, 5 and 6 in different stoichiometrics although the reaction is carried out in 1 is to 1 ratio, the expected product is 4, but how other 2 compounds are formed and how did we got this information that one can see by simply looking into the 3NPNMR spectrum of the reaction mixture here. So, these are the 2-triphenyl chemical shifts are there, but if you recall the first compound 4 where 2-triphenyl phosphines are replaced by bisphosphine. So, here both the phosphorous environments are identical as a result one can expect a single resonance here that means this is assigned to the product 4. Now let us look into other products. Let us go back to the reaction. In this product 6, one triphenylphosphine is there and one bisphosphine is there, bisphosphine chemical environments are identical and we have PPH3. So, by simple thumb rule by looking into 2Ni plus 1 rule, we can anticipate a triplet for triphenylphosphine and a doublet for bisphosphine because bisphosphine will be coupled with both of them together coupled with triphenylphosphine whereas triphenylphosphine will be coupling with the 2-equivalent phosphorous centers from bisphosphine. So, we expect a triplet and a doublet for 6. So, you can see here a doublet is there and a triplet is there this is what triphenylphosphine and this is for chelated 2 phosphorous mites of bisphenylphosphine I mean. So, then we have left with one here, one here, one here for product 5. So, what is that product 5? So, product 5 you see we have one dangling phosphine is there acting as a monodentate ligand. So, this is different and this is different and these two are identical. So, that means, we are anticipating 3 signals and here this phosphine is further away from this phosphine as a result this can just couple with coordinated phosphorous to show a doublet, but on the other hand the chelated phosphines also can couple with coordinated phosphorous from the monodentate ligand to show a doublet whereas, this one would show a triplet with this one and a triplet is further split into doublets from this one. That means, we are seeing a a multiplied for mono coordinated phosphorous and a doublet for biscoordinated or chelate compound and then again doublet for this one. So, let us see whether we have those things here yes we have a doublet and we have a doublet with very small coupling constant value and also we have a multiplet here it is a triplet of doublets. So, we are seeing that one so that means, this is giving some idea that three compounds are formed, but again looking into the integration we can gauge how much quantity of each one is formed in what ratio and then looking into other two compounds where ionic products are there yes that means, probably instead of using let us say I have used dichloromethane here instead of using dichloromethane if I use a polar solvent probably the formation of four can be diminished and formation of these two compounds can be enhancer yields on the other hand again by taking two stoichiometry two equivalents of bisdefinifene or phosphine probably I can enhance the possibility of formation of 5 even more. So, this vital information comes for using those information so this reaction was carried out and this pure compound was obtained carrying out at low temperature and then by using two equivalents of this phosphine in polar solvent resulted in this compound and then using a polar solvent, but one equivalent of bisdefinifene and phosphine resulted in this compound of course, when once after making this compound if you bubble CO here this can form a carbonyl compound. So, this all vital information comes very nicely by analyzing the 31 PNMR spectrum of the mixture of this reaction product that means it is not just to analyze and understand what kind of compounds formed it can certainly help us in altering our reaction conditions so that we can refine our methodology and later all these compounds were prepared in pure form and analyzed you can see here so these compounds are formed and then this compound was again prepared in a separate reaction by getting that information from the reaction mixture and then this was obtained in pure form. And now let us look into few more examples here I have given three examples here one is a bisphosphine with both the phosphorus oxidized by selenium that is bis selenide derivative here and then I have here analogous to Vasca's compound both cis and trans chlorocarbonyl bisdefinyl phosphine bis triphenyl phosphine rhodium and also the its cis analog and they they differ significantly in their 31 PNMR spectra. First let us look into bisphosphine here of course when we look into bisphosphine with both the phosphorus oxidized by selenium here 31 PNMR spectrum would show a singlet and then it is coupled to selenium 77 selenium and that shows satellites here separation is called phosphorus selenium coupling constant so these are called satellite peaks and then if you look into selenium NMR again both the selenium centress are chemically and magnetically equivalent as a result they are equally coupled to phosphorus and we will see just a doublet and then this whatever the separation is there here in case of selenium NMR should be same as what we obtained in case of 31 PNMR that means coupling constants remain same no matter which nuclei you are using. Now let us look into trans compound here in case of trans compound you can see both the phosphorus are identical by performing a C2 axis of rotation passing through C1 CL as a result you can see only one type of phosphorus and when you look into 31 PNMR they are simply coupled to rhodium to give a doublet here something like the very simple doublet is here and then this separation is called 1JRHP but when you look into this one here one triphenyl phosphino moiety here it is triphenyl phosphine one triphenyl phosphine is trans to CO whereas the other one is cis to CO so we cannot have C2 axis of rotation as a result two phosphorus atoms are chemically equivalent but they are both chemically and magnetically non-equivalent as a result it resembles AMX spin system so PA is coupled with rhodium and then coupled with phosphorus it shows a doublet of doublet in the same way PP is also coupled first with rhodium and each line is further coupled with phosphorus so first this is rhodium coupling and then this is phosphorus coupling whether you consider PA, delta of PA or delta of PB they are very similar but slight marginal difference is there in chemical shifts and coupling constants so this is rhodium phosphorus coupling and then here it is 2JPP coupling so this is how we can distinguish cis and trans isomers in case if they are obtained in a reaction let us look into two more examples here and we have here we have a mixed ligand complex platinum zero a tetrahedral complex where we have two trimethyl phosphine and two diphenyl phosphine so that means platinum is in plus two state here two trimethyl phosphines are neutral ligands and here we have two diphenyl phosphines phosphides are there they are anionic so platinum is in plus two state and it is a 16 electron complex so now if you look into phosphorus NMR spectrum we can anticipate two type of signals so these two are identical they are coupled with these two to give a triplet these two will couple with equally to this one give a triplet and then these two will couple with this one to give a triplet so what we should get is a triplet something like this and then platinum coupling would come so basically platinum coupling would come with platinum satellites because of 195 platinum is about 34 percent NMR activity equals half so as a result this is for 196 platinum which is i equals 0 so then what we get is this is split into a doublet here something like this and then something like this so if we look into the spectrum it should look like this distance is called 1j ptp coupling this is similar for both of them this is similar for both of them whether we consider a signal for trimethyl phosphine or we look into that one they'll be identical but they'll be having different chemical shifts here this one is for 196 platinum and then if you look into the integration this would account for 66 percent and then this would be 17 percent and then this is 17 percent so this how we can analyze the spectrum and you can ignore this spectrum here you can ignore this one and of course you can also look into one HNMR spectrum here because trimethyl group is there and this trimethyl group what happens this is of course here you can see both are identical both the trimethyl phosphine groups and CH3 first it will be coupled with phosphorus to give a doublet and then each doublet something will be there rhodium to hydrogen coupling and then we will see a hydrogen to platinum coupling will be there something like this again so this is how the one HNMR would look like and most of the time by default when we are looking into phosphorus NMR what we get is proton decoupled one and of course if we are interested in looking into phosphorus hydrogen coupling then we can go for coupled one in that case it should be represented something like this the moment we write in flower bracket next to phosphorus 31p it indicates that one H is decoupled and that means interaction of hydrogen protons in the magnetic field when we are measuring phosphorus NMR is nullified so now I have another interesting compound here in this one it's a dimetallic compound here and two platinum atoms are there at terminally they are held by two trimethyl phosphine each and the middle we have diphenyl phosphide ligand is there so both the platinum if it takes symmetric it will be 1 1 or it can be one side it is platinum 0 and one side it can be platinum 2 also so now we can look into the spectrum of this one here how it looks like let me write the structure again here this is about 195 platinum NMR first we can look into 195 platinum NMR this is a structure so now if you see these platinums are both chemical and magnetically equivalent both the platinums are coupled on either side by two different type of phosphorus units so here let us say first this is coupled with two phosphorus units and if you just look into 2 N A plus 1 it should be a triplet and then this separation or this separation would be 1 J P T P I should say P P H 2 and now each of this signal will be further split by two end trimethyl phosphine so this will be coupled with first these two and then these two so this separation is 1 J P T P M E 3 so it shows a triplet of triplet and if you look into the spectrum it should look like 1 H 2 2 H 2 1 something like this so this is how the 195 platinum NMR looks like but on the other hand when we look into phosphorus NMR phosphorus NMR if you look into it these two are identical and they are equally coupled to 4 primethyl phosphine as a result what would happen delta P P H 2 let me write here it will show first a quintet so it show a quintet and then each quintet is further split into a doublet in the form of a satellite so as a result what would happen is we get something like this so something like this again this is for 66 percent and then this is 17 percent and this is 17 percent and then distance from the middle one to middle one or first one to first one this is called 1 J P T P coupling and this spacing is called P P coupling P P coupling this is how you can write it but on the other hand this would give a this one these two are identical with this one when they are coupled with this one we get simply a triplet and then this triplet is split into another doublet because of platinum coupling so this would be something like this something like this again here this is your 1 J P T P coupling so this is how the spectrum can be analyzed so now let us look into other applications we have several applications are there one very important application I am going to show you especially those who are working with nanomaterials and working with carbon compounds such as phallorines graphene graphene oxides and other things very very important one measuring epoxide content of carbon nanomaterials the presence of epoxide on nanomaterials such as carbon nanotubes and phallorines can be readily monitored by reacting these nanomaterials with triphenylphosphine how it helps in understanding the presence of epoxide so this method involves the catalytic reaction of methyl triaxorhenium so an epoxide reacts with methyl triaxorhenium to form a five-membered ring and in the presence of triphenylphosphine the catalyst is regenerated forming an alkene and then formation of triphenylphosphine oxide so when we add triphenylphosphine along with methyl triaxorhenium to nanomaterial containing epoxide oxygen is abstracted by triphenylphosphine to form triphenylphosphine oxide by simply taking the known quantity of triphenylphosphine and by monitoring the 31 P N M R spectra at regular intervals we can quantify the amount of epoxide present by looking into the amount of triphenylphosphine is converted into triphenylphosphine oxide and how much is left at the end of the reaction so this is the reaction for example we have an epoxide is here and epoxide when it is treated with methyl triaxorhenium it forms a cyclic system like this and when this cyclic system is treated with triphenylphosphine it abstracts oxygen from this oxo cyclic system to form triphenylphosphine oxide and regenerating a methyl triaxorhenium through the formation of this olefin so now exactly this is what happens so for example we have one graphene we have epoxides are there you just treat with this one and then as I said here triphenylphosphine abstract and this one is regenerated and then by simply looking into the reaction mixture N M R we can conclude that yes since the amount of P P H 3 used in the reaction is known the relative amounts of triphenylphosphine and triphenylphosphine oxide can be determined simply by looking into the integration of 31 P signals in the spectrum and that intense gives the amount of epoxide present on the nano cubes so this is very very important very simple reaction one can analyze and also one can also get rid of all these epoxides if they are not read it so you can see here 31 P spectra before and after the reaction before the reaction so full quantity is there and now after the reaction what happens partially P P H 3 is converted into triphenylphosphine oxide depending upon how many epoxide groups are present on nano material so then if you just look into the integration now only 0.4 percent is left let us say if we have used two equivalents 1.6 is converted into triphenylphosphine oxide and now only 0.4 percent is left in the solution so that gives how much is there in it this is a very very important method one can use effectively for determining epoxides present so let me stop here and continue with more interesting examples in my next lecture until then have an excellent time reading spectroscopy thank you.