 I once again welcome you all to MSP lecture series on interpretative spectroscopy. And now let us continue again discussing more interesting examples of phosphorus compounds and then see how easy it is to interpret the data obtained from 31 Panama spectra of various very interesting phosphorus compounds. So in my previous lecture, I showed these three molecules here. And then I could show you how to interpret or how to sketch NMR spectrum of both 195 platinum as well as 31P phosphorus in case of platinum compounds. So let me now discuss the splitting pattern and also sketch NMR spectrum corresponding to rhodium complex here. First let me look into 31P NMR spectrum of this rhodium complex. As I said earlier, rhodium 103 isotope is 100% abundant with I equals half. So it couples equally. And then we have phosphorus is there. And the moment you look into the molecule, you should be able to tell that there are two different type of phosphorus atoms are there. One is directly attached to rhodium with two fluorine atoms on it and one oxygen. Whereas the second phosphorus is doubly bound to sulfur and then it is not directly connected to rhodium. So we have two phosphorus atoms are there and also fluorine can also couple with both the phosphorus atoms and rhodium can also couple with both the phosphorus atoms. And rhodium phosphorus coupling constant is in the order of 180, 180 it can go up to 300 or 350 hertz. Now let me go to the spectrum of rhodium complex here. In rhodium, I have labeled as PA and PB here. And of course, I can also label them as A and X. This is a typical AMX spin system, AMX spin system. But we are looking into A and X signal because rhodium is a heteronucleus here. So now first, if you look into PA, it is first coupled with rhodium because rhodium to phosphorus coupling is larger than phosphorus to fluorine. So it first couples with rhodium to give a doublet here and this is 1J rhodium phosphorus coupling. Now it is coupled with fluorine since two fluorine atoms are there and each of the doublet lines are spread further into triplet, something like this. And this is 1J PF coupling. And next we have is two bond apart P, P5, PB so that P also couples with PA to give a doublet. So each one is split this one. Here it is 2J PP coupling. So now the spectrum should look like a doublet of triplets of doublets, a doublet of triplets of doublets. So this is how we have to pronounce. So initially start from doublet of triplets of doublets. So this is how we can pronounce the multiplet related to PA signal. And now let us look into PB. When we look into PB, first it will couple with phosphorus because that 2J phosphorus to phosphorus coupling is a little larger. It couples with 2J PP and then each line will be split by rhodium that is two bond apart. So this is rhodium coupling. This also 2J rhodium phosphorus coupling. Now we have 1, 2, 3. So three bond apart two fluorine atoms are there. So they are also likely to show some coupling, long range coupling of three bond coupling. So each line here will be split into triplet here due to the presence of two equivalent fluorine atoms. So this spacing, what is there? This is 3J PF coupling. So that means basically what we have is a doublet of doublets of triplets. And this is how the spectrum would look like. So this is about 31 PNMR spectrum. And probably you can try, try to sketch 19 F NMR spectrum for this molecule. And also you should try to write 103 rhodium. Try very simple, try to sketch 19 F NMR spectrum for this molecule. And also try to sketch 103 rhodium spectrum also this molecule. I shall tell you in case of 19 F fluorine NMR, what happens is first it splits into a doublet due to PF coupling. And then each line will be split into another doublet. Each one will be a doublet because of fluorine to rhodium coupling. And then each line will be split into again doublet because of three bond PF coupling. So I have already given you hint. You can do in the case of rhodium 103 rhodium which is exactly very similar. First it will be coupled with phosphorus one jet to give a doublet. And then each line will be split by fluorine to give a triplet. And then each line in the triplet will be further split into doublet because of PB. So now you try to sketch the NMR spectrum of both 19 F as well as 103 rhodium. If you have any difficulties, let me know so that I can provide you later. So now let's look into this molecule here. This molecule we have three cases as I mentioned. And when spectrum is not given, if somebody asks you to sketch the 14 NMR spectrum of this molecule here, this is bis-definite phosphine amine, this is called. And then you have to look into all the three cases. One is NH coupling is larger than NP coupling. Other one is NP coupling is larger than NH coupling. And other one is NP coupling is equivalent to NH coupling because all of them are one bond apart. So first let's look into PN coupling is larger than one H coupling. Since we have two phosphorous atoms are there, what happens first it would be split into a triplet and each line will be split into a doublet. Something like this. So we have doublet of triplets we have. And next when we look into one J PN is less than one J NH coupling. In that case what happens first it splits into a doublet and then each line will be split into triplet because of two equivalent phosphorous atoms. This is how it looks like in this case. So it is simply something like this. Something like this here it would be something like this. And then in this case the last one where PN coupling is equivalent to NH coupling then we are identifying three nuclei equally coupled to nitrogen. In that case what happens it would be a quadrate. When we have a quadrate it would be something like this. So all the three cases we should draw. If we plot NMR 14 NMR and it should correspond to one of these. So this is how you should be able to sketch and compare and analyze and interpret the data. It's not really complicated. We have another interesting problem here. Consider all possible isomers that could be obtained for the eight membered ring compound P4 N4 CL6 F2 and indicate the ideal 30 NP and the 19 F resonance spectrum expected for each. Of course the moment you look into this one it's already stated that it's a eight membered ring. We have alternate phosphorous and nitrogen with alternate P and double bonds are there. They are called cyclophosphazanes. This is cyclo-tetraphosphazane. And then each phosphorous having two halogens. That's the reason each P is pentavalent and tetra coordinated. So we have each one has two nitrogen bonds. One is double bond, one is single bond and it has a combination of R. It can have either chlorine two or it's chlorine and chlorine. So now what we should do is we should try to sketch all possible isomers and then we should speculate or we should try to sketch NMR 30 NP NMR spectrum and also 19 F NMR spectrum for each isomer. I have already written all possible isomers. Totally five isomeric structures are possible for this composition of N4 P4 CL6 and pH of course here I have taken a different molecules here and in the problems I showed you P4 N4 CL6 F2 but whereas here I have taken a different example here to make it a little simple. I have considered four chlorine atoms and phenyl atoms maybe in my next lecture I can sketch all possible isomers for CEL6 and F2 combination and then sketch NMR spectrum for each case. So now let's look into the simple one. In simple one we are looking to own the coupling between non-equivalent phosphorus atoms. That's the reason I thought it is to begin with. Let's go for a simple one. And this is one isomer. You can see in isomer one we have two different type of phosphorus are there. Two of them on opposite sides are doubly substituted with phenyl whereas other two with chlorine and is very symmetric molecule it is. We have two different type of that means basically we have A2 X2 type system. And then we have here again we have identical ones. If you just look into it we can have a axis of rotation. It looks identical from one portion here. So these two and these two again it is a A2 X2 type system. And then if you go for the third molecule here third molecule we have three different type of phosphorus atoms are there. One having two phenyl groups, two phosphorus having one each of chlorine and phenyl group and other one having two chlorine groups. That means here we have some sort of AMX system. And in this one it is very symmetric. We can anticipate a single resonance for this molecule here single. And then there is one more possibility. And then here also we have four different type of phosphorus atoms here. One, two, three, four and we can also see here. Each one is having of course one, two, three, four couplings are there how they're going to look like. We can look into it. So first one I have shown here. First one as I said A2 X2 spin system. We get these two in blue color would couple with two red colored phosphorus to give a triplet. And same thing is true for the other set. So we get two triplets we are getting here. And of course in case of two also it is identical we can get two triplets A2 X2 again in case of three. What we have is AMX spin system. So basically what we have is this one is coupled with this one to triplet. And each triplet is split into a doublet. So we get this one. And then these two will be coupled with this one first give a doublet. And then this corresponds with splits for another one doublet so that we get doublet of doublet for this one. And then this is a triplet and this is a triplet here. So you can see this is given here. And then the four all phosphorus atoms are identical. We are getting a signal here. And in case of five as I said it is we have one, two, three, four signals are there. All of them are separate. We are getting well further coupling is not possible. One, one, two, three, four coupling is not possible. Only they are coupled to two. As a result what happens? Each one will show a doublet of doublet here. So all four of them will be showing different ones. And with color code I have given this color corresponds to the color given for the phosphorus atoms in the space so that understanding would be rather easy. So this we get four quadrates we are getting. Whereas in case of five because all four are identical. And they are coupling with only the nearby ones. The fourth one is further away so that they are not coupling. So P shows coupling with only these two P and red and pink. And similarly this one will couple with blue and green. And this one will couple with red and pinks. It goes like that. So we have four of them. We'll be showing each one a doublet of doublet. So now I have listed compounds. I have given the molecular formula here. With molecular formula is there. And also I have given the corresponding chemical shifts I have given. If you recall the extensive list I gave you in my previous lecture, one of my previous lectures, if you compare from that one where exactly the chemical shift falls based on that one, whether it's downfield shifted or upfield shifted, you have to write the spectrum. For example, if you just look into the molecule given here, one signal is there. It means only one phosphorus there obviously. And then we have C3H9O3. So that means basically you can assume this may be P O CH3 three times. So this will be having composition like I have shown here. In the same way, you identify the type of molecule here and then try to write the structure in the blank given here. If you have any difficulty try, it's very interesting. It covers most of the phosphorus compounds we come across. And also the extensive list I gave you in one of my lectures, you can use that one and write the structure for all of them. Once you write the structure, it is very easy to understand. And of course here only this one is missing. Further information is also giving how each signal would look like, the multiplier would look like. Here we have 10 lines. Why it is 10 lines? Because 9 equivalent protons are coupled with phosphorus to give 10 lines. 1, 2, 3, 4, 5, 5, 6, 7, 8, 9, 10. Looks something like this. And then next one is separate is given. If the separate is given and PF is 80.6 hertz, then you should be able to analyze and then write the spectrum here and also write the structure. So make an attempt for all of these molecules. It's very simple because all the data is given. It's almost like providing NMR, 30 and P NMR spectrum here. And you have to molecular formula is given. You should be able to write appropriate structure for in each case. You can see here chemical should 159. And then we have 144. And then 2.3 minus 27 plus 11. 22.7 minus 2.8 plus 9 minus 66.4 and 53 and minus 45 and minus 144. I have very interesting molecule here. You can see here this molecule has a silicon and three hydrogen atoms are there on silicon. And then nitrogen 15 N enriched molecule this is. And then H is there and phosphorus is there. That means it's a very interesting molecule. You can sketch NMR for almost all nuclei here. You have 19 F, 1, 2, 3, 4, 5. Five different type of NMR active nuclei are there. Of course, silicon abundance is very less, 29 silicon. But nevertheless, when you are looking to fluorine NMR, phosphorus NMR, 1H NMR, and 15 NMR, you can ignore the coupling due to silicon. And of course, in case of silicon NMR, you can consider all other couplings. So, let us try to do one at a time here. First, let us look into 31 PNMR here. 31 PNMR, let us look into it. In 31 PNMR, you can see two fluorines are there. And you know the magnitude of PF is 1J PF coupling is much larger. So, what you can do is first you split into a triplet. This is your 1J PF coupling. So, next it is coupled with the nitrogen here. This is your 1J PN coupling. Next we have 2J pH coupling is there. Each line will be, so this is 2J pH coupling. So, next we have 1, 2, 3, 3-bond coupling with the silicon-bound hydrogen atoms. We have three equivalent hydrogen atoms are there. Each line will be split into a quadrate here. So, this many lines will be there. If you count 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 12 into 4, 48 lines will be there in it. And then the intensity should be maintained. So, 1 is to 3. Overall intensity should be 1 is to 3. And here it is 1 is to 3 is to 3 is to 1. It is 1 is to 1, 1 is to 1, 1 is to 2 is to 1. So, when we sketch spectrum of looking into this coupling tree, we should ensure that the intensity whatever is there that is maintained here. So, now with this one you can interpret which spectrum. The spectrum, which spectrum it is not mentioned, but by sketching say 31 PNMR, 19 FNMR, 1H NMR, 15 NMR you should be able to tell which spectrum is for which NMR nuclei we are talking about. So, this one should be phosphorous NMR. So, this one if you just see it looks like this here. Yes, this is for 31 P. So, let me show you in that one in the next slide. So, here I have taken 31 PNMR. You can see here the whatever the coupling I showed you. Here it is there. You can see first phosphorous is coupled. You know split by three fluorine atoms into triplet. And then each line in the triplet is split into a doublet because of 1J PN coupling doublet. And then each line is further split by 2J pH coupling with nitrogen bond hydrogen here. And each one is a doublet here 1, 2, 3, 4, 5, 6. And then each line here will be split by SiH3 hydrogen to a quadrate something like this. Now, we have something like this here. We have 48 lines are there. So, this is how you can show. And then the corresponding spectrum would look like this. Very interesting molecule, right? And also spectrum looks wonderful. Is that right? So, very nice one. It's, as I said, once you understand, once you write clear structure, even if the molecular formula is given or structure formula is given, try to write the full structure showing all the bonds and then start analyzing all NMR active nuclei, how far they are from the one we are looking into. And then with little bit of information you are having about the magnitude of the couplings. Of course, no matter what happens, 1J coupling is always larger than the 2J coupling, then the 3J coupling, then you see, you start analyzing. And due to some reason, if we have two different nuclei, which are about three or same number of bonds apart, then it's likely that they are coupling equally. So, there's another possibility. So, that also one should look into it. That we saw in case of this diphenyl phosphinoamine case, where we saw two phosphorus and one hydrogen were coupled equally to nitrogen. One should look into all those things and analyze. It becomes very easy for interpretation. So, one that I looked into 31 PNMR for this molecule. In my next lecture, I shall discuss about other two spectra that is yet to be analyzed. That I shall do it in my next lecture until then have an excellent time. Thank you.