 I once again welcome me all to MSP lecture series on interpretive spectroscopy. In my previous lecture, I showed you how nicely phosphorous NMR can be used to detect or determine the number of epoxide groups present on carbon nanomaterials. It is very, very important especially when you want to use pure nanomaterials for electronic and other semiconductor purposes, we have to ensure that there are no such parasites are sitting on graphene. From that point of view, this NMR spectroscopy especially phosphorous NMR comes very handy. So, let me continue from where I had stopped. So, now let us look into some important information regarding instrument operation as far as measurement of 30 NMR spectra are concerned. The instrument operation will vary according to instrumentation and software available. It varies with different company instruments and the software is employed there. However, there are a few important aspects to instrument operation relevant to 31 PNMR one should know before they put their hands on instrument to measure 31 PNMR spectra. The instrument probe which excites nuclear spins and detects chemical shifts must be set up appropriately for a 31 PNMR experiment. This is very, very vital of course, this guidelines will be given. For an instrument with a multi nuclear probe, it is a simple matter to access the NMR software and make the switch to a 31 P experiment. Again technician will guide you in this one. This will select the appropriate frequency for 31 P for an instrument which has separate probes for different nuclei. It is imperative that one be trained by an expert user in changing the probes in the spectrometer. So, this is a typical NMR tube we are using and this is a capillary tube containing 85 percent phosphoric acid and this can be inserted and also we can measure. So, for example here before running the NMR experiment consider whether the 31 P spectrum should include coupling to protons or not and you should remember the fact that 31 P NMR spectra are typically reported with all protons decoupled. That is a default setup and this is usually the default setting for 31 P NMR experiment to change the coupling setting follow the instructions specific to your NMR instrument software and chemical shifts in 31 P NMR reported relative to 85 percent phosphoric acid. This must be an external standard due to high reactivity of phosphoric acid. So, one method for standardizing an experiment uses coaxial tube inserted into the sample NMR tube I showed you in the last slide. So, 85 percent phosphoric acid signal will appear as a part of the sample NMR spectrum and can thus be set to 0. So, when you are measuring you can also see signal due to phosphoric acid that you can set as 0 and accordingly with reference to that one other chemical shifts can be set automatically set. So, this is the one as I mentioned this the capillary with phosphoric acid a sealed capillary and then this can be inserted nicely. So, another way to reference an NMR spectrum is to use 85 percent phosphoric acid as a standard sample these can be prepared in the laboratory or it can be purchased commercially it is very easy to make provided we have a glass blower in our laboratory to allow for long term use these samples are typically vacuum sealed. So, and what is the procedure for using a separator reference. So, insert NMR sample tube into the spectrometer tune the 31 P probe and shim the magnetic field according to your individual instrument procedure remove NMR sample tube and insert phosphoric acid reference tube into spectrometer begin NMR experiment as scans proceed perform a Fourier transform and set the phosphorous signal to 0 PPM continue to reference spectrum until the shift stops changing and stop experiment and remove phosphoric acid reference tube and insert NMR sample tube into spectrometer run NMR experiment without changing the reference of the spectrum. So, this is how you can do alternately referencing phosphoric acid when you are measuring 31 P NMR signals and when we are using 31 P NMR it gives distinct sharp peaks that helps in identifying phosphorous containing new products and unreacted starting compounds and any other phosphorous impurities. So, that is the advantage and so 31 P NMR is very simple technique for assigning sample purity readily a clean 31 P NMR spectrum does not necessarily suggest a pure compound, but it can tell you that it is free of phosphorous containing impurities there may be other organic impurities are there, but that can not tell you any further information, but only if the spectrum is clean it would indicate you definitely that there are no phosphorous impurities or any other minor products containing phosphorous. 31 P NMR can also be used to determine the optical density of a chiral sample adding an ansomer to the chiral mixture to form two different diastereomers will give rise to two unique chemical shifts in the 31 P spectrum that probably at the end I will show you some examples the ratio of these peaks can then be compared to determine the optical purity. And now let us look into couple of more examples considering the reactions carried out with organometallic compounds with phosphorous containing compounds especially phosphines. So, 31 P NMR can be used to monitor a reaction involving phosphorous compounds no doubt in it consider the reaction of nickel chord twice is a nickel 0 compound with a slight excess of bisphosphine. So, I am going to show the chemical reaction here. So, this is the bisphosphine we are considering and then when it is treated with nickel chord this type of nickel 0 compound is formed with some interaction with ortho carbon present here. So, still nickel is in 0 valence state and it is a 14 electron species and then this can be readily monitored how to monitor it is very simple take the ligand and add that one to the nickel chord twice in an appropriate solvent. And then initially the signal when you just begin the reaction you can see the signal due to only phosphine here with the time for example, after 1 hour you can see another peak started developing here much down field. This indicates the beginning of the formation of this compound and then a 2 hours it is gradually concentration of ligand is decreasing and then this is slowly increasing after 3 hours you can see steadily it goes and after 5 hours it is completely reacted and then if you just compare the integration you know that you have used a slight excess of this one only the slight excess of bisphosphine usually is left unreacted that indicates the completion of the reaction. So, that means by using variable time 31 PNMR spectrum you should be able to monitor the reaction in a continuous process and see you can stop as soon as you come to know that the reaction is completed. Not only in this reaction when we are doing catalytic reactions also we can look into the consequences that are observed on phosphorus in a catalytic reaction can also be very nicely followed using 31 PNMR spectroscopy. So, now one more interesting example is there here that is with respect to cyclo-tryphosphazine. So, indicate the number of isomers of cyclic compounds of formula and sketch the 31 PNMR spectrum for each. So, in this case some hint is given assume delta is much larger than J and J pH is small if J pH is small J pH can be ignored for phosphorus atoms which do not contain methane groups. That means when you are writing like this we have two methane groups are there and four chloro groups are there. If we write all possible isomers it so happens some of the phosphorus atoms will not be having chlorine in that case they will be little further from the methane groups in that case what happens the pH can be ignored. That means we are plotting not decoupled one we are plotting coupled one. What would happen if we take a decoupled one it is straight forward we have to identify how many different type of phosphorus environments are there in a given isomer. Here one the two isomers are possible either having both the methane groups on one phosphorus or two methane groups are distributed at two different phosphorus since only we have two. So, either we can have this possibility or we can have this possibility and in this one if you are considering protons also it would resemble A2MX6 spin system or here what happens it would be A2MX3 spin system. So that means let us look into how the NMR is going to look like ignore AX coupling in both the cases. So AX coupling because A and X are far apart you can ignore the coupling only you can see the direct coupling. So here for example it is 2 bond pH coupling can be considered and rest of the long range coupling can be ignored. Now if you just look into this one these two are identical and this is different. So that means basically when we look into if I say PA here and this is M and this is AX here if you look into delta A first it will be split into a doublet this is 1, 2, 2 j P P coupling PA P B and then here this is coupled with six of them are there. So if six of them are equally coupled so you should get six means seven lines and plus one line should be there because I equals half. So what you get is 1, 2, 3, 4, 5, 6, 7 similar here also 1, 2, 3, 4, 5, 6, 7 so something like this we get it and for PM so this will be coupled with this one into triplet and then each triplet will be again a separate here so you can see this one. So this is the NMR spectrum for this one and of course here you can also ignore the coupling in this case but anyway I have considered both small range coupling then if you go for this one here again here these two are identical if you put A or if I put A here this is M and this is X I will show you spectrum here you can see A to MX 6 system as I said these two do not show proton coupling I showed you there but it will be something like this simply shows a doublet coupled with this one we are seeing here this if you say A and this is M this is delta A and then this is for delta PM. So here first it shows a triplet and each line will be split and what you see here this is your pH coupling so 2 1, 1, 2 1 coupling and this is 1 is to 2 is to 1 triplet and then we have each one is accepted. So and AM to X 3 you can see here and this triplet is for this one if I say A, M and X so A will be delta PA will be a triplet here and then this coupling is 2 J P P so you are not seeing long range if you see long range again this will show a accepted again. So now the other one since both are identical here not we are going to get accepted we are going to go to the only quadrate. So each line in this one so if you take these two are coupled with this one to a doublet something like this and each doublet now since this is identical with this one only a quadrate will be C you see quadrate and this is J pH coupling. So this is how you can interpret and you can identify if both the isomers are present in say 1 is to 1 ratio or 1 is to 3 ratio or whatever so we should be able to clearly distinguish them by using simply phospho-cinema spectrum. Now let us look into another interesting compound here the reaction of zirconium tetrachlorate DPP DPP is this diphenyl phosphinoethane with magnesium dimethyl magnesium here gives a compound of this type where we have four methyl groups are there and DPP is there acting as a bidentate ligand say NMR spectra indicate that all methyl groups are equivalent draw octahedral and trigonal prism structures for the complex and show how the conclusion from NMR supports the trigonal prism assignment. So this is very simple and it shows that this compound with coordination number 6 can have both octahedral geometry and trigonal prismatic geometry so this is octahedral geometry here and 4 in the plane and 2 axial and also this is octahedral compound geometry is also known as trigonal antiprismatic geometry trigonal antiprismatic geometry. So in this one if you try to arrange four methyl groups and bridging diphosphine you cannot really come up with a isomeric form that has all four methyl groups equivalent. For example, to make them equivalent I have to put all of them in the plane when I put all of them in the plane what happens these two what is it will be far apart it is very difficult to fix this diphenyl phosphinoethane in this way because the ring size is not sufficient. So it can either go here when it goes here again it will not be very symmetric. So from that point of view what happens it is ruled out on the other hand trigonal prismatic geometry all four can be for example this is a trigonal prismatic geometry that is trigonal antiprism and of course how to why you can say prism is something like this you can see and then if I turn something like this then it becomes trigonal antiprism staggered trigonal prism. So now with this one what happens let us say it has four rectangular faces are there if four rectangular faces let us say if I can put here very conveniently 1, 2, 3, 4 methyl groups are there and then if it is the bisphosphine I can fix in this way. If I fix this way what happens now we can see if I cut a axis here this can divide and now both of them can be magnetic equivalent or if I do some rotation from this one C2 axis of rotation they will be identical. So that means with trigonal prismatic geometry with this kind of conformation what happens all the four will be identical that means when we look into 13 CNMR spectrum of this compound it shows only one signal for all four carbon atoms that clearly indicates that this compound has trigonal prismatic geometry with all methyl groups are in one plane something like this. So we can conclude without any problem so that means whatever the statement that is made in the question can be fulfilled by putting this kind of geometry here and then of course these two phosphorus are identical. If you look into phosphorus NMR it shows only one signal and if you look into 13 CNMR and if you focus our attention to methyl groups you will get only one signal. So I have shown here for example here it has a different one and two type of we will come back to axial and equatorial again same thing is true here and whereas here trigonal prismatic either you can see here or here they look identical and then you can see what I showed through this models. So this is how once again NMR can be used to understand the geometry and the preference of groups very readily by simply looking into the corresponding NMR spectra. So now I have another interesting reaction here this is with tris-trimethyl phosphine chloroeridium. So when the four coordinate square planar complex IRCL PME3 thrice where PME3 is trimethyl phosphine you all know reacts with chlorine to six coordinate products of formula IRCL 3 PME thrice are formed. So that means we are adding oxidatively chlorine to iridium 1 to form iridium 3 compound having six coordination from square planar 60 electron it forms a octahedral 18 electron complex iridium 1 changes from plus 1 to iridium plus 3. So 31 PNMR spectra indicate one phosphorous environment in one of these isomers and two in the other what isomers are possible. So when you look into MA3 B3 type system so we have MA3 B3 system and of course when we have MA3 B3 system and when we have MA4 B2 system. So we know that two isomers are possible in case of octahedral compounds of course here coordination number is 6. So in case of this one we get facial and meridional isomers whereas in this case we get cis and trans isomers. So since after oxidative addition we are getting MA3 B3 possible isomers are facial and meridional. So that means here and let us assume A is CL and B is PME3. So now we have two isomers are there and you can clearly distinguish them in this one all A's are on one phase. So that all trimethyl phosphines are one phase and all chlorines are in another phase. So this is a facial isomer and this is meridional. So we have trans as well as cis. So in this case what happens if we just look into we have two different type of phosphorous atoms are there and then if you look into the spectrum here this will give a triplet and then whereas these two will show a doublet. So we get a pattern something like this whereas in this one we get a singlet. So that means what it says that even enema spectrum indicate one element in one of these isomers this is facial and two in the other two will be in case of meridional one doublet and a triplet. So this is how we can identify simply by even not even plotting simply by looking into the number of signals they are present. Of course when we look into signals we know that one has to be a doublet and one has to be a triplet and then this doublet intensity should be it should corresponds to two and it should correspond to one. So now there are a number of advantages for using 32 MP NMR for reaction monitoring when available as compared to one HNMR. It is very simple you saw with several examples how simple it is to use phosphorous NMR as a tool for our advantage to understand the type of reaction we are doing and how to change the course of the reaction and all other details. Another advantage with phosphorous NMR spectroscopy is we do not need a deteriorated solvent which simplifies the sample profession and save time and resources. That means when we are doing a reaction say continuous reaction especially in homogeneous catalysis. Let us instead of doing batch reaction we can take a regular intervals a liquid and they can look into if we have provided we have phosphorous might is in it we can readily monitor without going to expensive deteriorated solvents. So 32 MP NMR spectrum is simple and can be analyzed very quickly much easier. So the corresponding one HNMR spectra for the above reaction would include a number of overlapping peaks for the two phosphorous species as well as peaks for both free and bone cycloactra in ligand. So that means it is very simple here instead of going for one H one can conveniently go for phosphorous 31 P NMR spectra. So purification of product is also easy and that means how we can purify of this information also comes from a spectroscopy 31 P NMR does not eliminate need for one H NMR characterization our impurities lacking phosphorous will not appear in 31 P NMR experiment. So basically what happens 31 P NMR does not eliminate the need for one H NMR characterization but as far as 31 P NMR is concerned if the spectrum is clean that indicates that there are no phosphorous containing minor products or impurities but on the other hand there can be something else. So that means one should also confirm purity by going for one H NMR or other nuclei NMR as well. So however at the completion of the reaction both the crude and purified products can be easily analyzed by both one H NMR and 31 P NMR spectroscopy. So let me stop and continue in my next lecture about many other NMR active nuclei for example 19 of 14 and 15 and other nuclei such as even boron. So let us continue discussion on spectroscopy especially NMR spectroscopy in my next lecture until then have an excellent time. Thank you.