 Hello everyone, I once again welcome you all to MSP lecture series on interpretive spectroscopy. This is the 10th lecture in the series. In my previous lecture, I was discussing about 13C NMR spectra and related aspects and how one can compare the data obtained from 1H NMR with 13C and how to simplify 13C. If you have large number of hydrogen atoms, we use decoupling system. Let me continue from where I had stopped. Just if you recall the spectra I showed in my last lecture, you can see here 13C NMR spectra is recorded both with coupling and without coupling. This is without coupling, I would say 13C like this. Once if you write in the flower bracket any nuclei next to the one we are looking into that indicates any signal due to this one is decoupled. And this one is coupled one. So in the coupled one, you anticipate interaction of carbon with adjacent hydrogen atoms. For example, here when carbon interacts with 3, we should get a quadrate we are seeing here. And then again this one is also this carbon also coupled with these 3 and it will show another quadrate here. And then this one methylene is split by 2 hydrogen atoms. So we are getting 3 here. If you are not understanding how this is giving for example, here we can use simply 2 Ni plus 1 rule here. So here carbon is attached to 3 hydrogen atoms 2 into 3 and its spin is half. So it gives 4 lines. This how you can calculate. So we are seeing 4 lines and in case of this one also we are seeing 4 lines whereas in case of methylene we are seeing 3 lines. And of course here what happens we have 1, 2, 3, 4 different type of carbon nuclei are there as a result we are seeing 1, 2, 3, 4. So this is much better. This is simplified one. In case if we want to have further information to see how it is interacting and one can always record coupled spectrum whether it is 13C or whether it is phosphorus NMR or it is fluorine NMR we can always record coupled as well as decoupled ones. An example is shown here. We have a para position, we have fluoro group is there. So para, fluorotoline it is and if you just look into this molecule here this one and this one are identical and this one and this one are identical. This is unique, this is unique and this is unique. That means 1, 2, 3, 4, 5 peaks should be there. So we have here 1, 2, 3, 4, 5 signals are there. And then it is a simple but fluorine is also there. It is likely that fluorine can couple strongly with this carbon atom here, this carbon atom here and then it can weakly couple with these two as well but here it may not show any coupling because 1, 2, 3 bond coupling. Let me see how it looks like. Yes, you can see here the one with larger coupling what we see here this is due to this one bond carbon fluorine coupling whereas these two will be coupled with this one that is that long range you can see here. And then these two are coupled weakly and that can be seen here and then this does not show any coupling it comes here and this one is coming here. So that means basically coupling also helps in understanding the position of signals in a given spectrum. Now I am presenting another interesting molecule here. You can see dichloro trimethysyl methane here and if we look into 13 CNMR spectrum we are expecting two signals here. One is for this methane carbon and then other one is methyl group present on silicon. So here and simply by looking into the coupled spectrum the C 13 C this one will couple with H 2 show a doublet and then this carbon here all three are equivalent as a result we can anticipate only one carbon signal in each carbon signal. So you can see the coupling with three protons so it will show appear as a quadrate. But if we look into the fine spectrum of this one of expanding this quadrate on expanding each line would be consists of a septate and each septate is again each line is a doublet. And similarly if we look into the signals here in the doublet each signal is consists of 10 lines here. So it is interesting to analyze how this 10 lines are coming here and how each one is a separate of doublets how it is appearing here. You can see here first I have expanded version of this one I am showing here in why each line is showing 10 lines here and of course if I just write the structure here. So first this carbon will couple with hydrogen to give a doublet certain larger coupling one can anticipate. Now this carbon is coupled with 1 2 3 bond coupling 3 bond coupling with 9 hydrogen atoms are there 9 hydrogen atoms are there if I just again use this 2 ni plus 1 rule here 2 into we have 9 so 10 lines you can expect 10 lines that is what we are seeing here. So that means you can also see 1 j C H coupling plus we are seeing 1 2 3 3 j C H coupling. So first it will appear as a doublet and each line will be split into so each one so you will see 10 lines here you can see here 10 lines are here. So this is how you can understand the complex pattern appears although it appears like a doublet of the expanding you can see the fine splitting here showing both 1 j as well as 3 j coupling with hydrogen atom. Now let us look into the quadrate here and in the quadrate it is very interesting let us just look into try with a select group here. So once when we are considering one of the signal here 13 C others may be 12 not necessarily there will be 13 in this case what happens first it will split with this one into a quadrate and then since now these two are non equivalent they can split protons will split this one into each one into a separate the same thing happens in all cases after that one you can see again 3 j coupling this one will be coupled with this hydrogen 1 1 2 3 coupling. So 3 j coupling is also and split only one is there each one will be split into a doublet so you can see now this is how a quadrate is a simple quadrate of expansion it appears like separate of doublets. You can see here I have shown here a quadrate each quadrate is split into 7 lines and then each line in the separate is further split into a doublet because of 3 j C H coupling and then eventually if you just look into it it appears like this it is very simple to understand if we expand and get the fine spectrum and let us look into another example here and in this molecule all are very different as a result we can anticipate 6 carbon atoms in the aromatic region and all the 6 carbon atoms are shown here one is here 2 is here and 3 4 5 6 are there and now if you just look into 1 1 is split into a doublet because this one is about 1 2 bond coupling this is due to 2 bond carbon fluorine coupling and then 2 is directly attached to fluorine so you can anticipate larger coupling as a result it is split further this is 1 j C F whereas, this portion is 2 j C F and in the same way you can see C 5 C 5 is further the coupling is very small because 1 2 3 4 bond coupling is there so you can see here 4 j C F and then 4 4 will be little here 1 2 3 so 3 bond coupling will be there and 3 will be 2 j coupling and then 6 does not show any coupling as it is very far that means if we have another NMR active nuclei is there assigned becomes much more simplified just by looking into the magnitude and how further this nuclei is from the nuclei under consideration we should be able to illustrate the structure very easily this is how one can use other NMR active nuclei present in a molecule for better interpretation. Now, let us look into symmetrically substituted compound here here we have if you consider from here we have 1 3 position we have a substitution is there. So, now these two are identical and these two are identical and this is different and this is different that means one can anticipate one 4 signals for this diphluorobenzene metadiphluorobenzene you can see here we have 1 2 3 4 5 5 are there and 4 and 6 appear together the reason is although we say these two are chemically equivalent and their interaction with fluorine is very different. For example, this one has a 2 bond coupling with fluorine they are identical you we have C 2 axis of rotation is there, but still 6 interaction with fluorine F 1 is different from F 2 and similarly interaction of 4 with F 1 is different from F 2 as a result we get a different pattern here whereas C 2 will show very nicely triplet and then C 5, C 5 will also show a triplet here and then C 1 and C C 1 I will show a doublet and C 3 will also a doublet of almost same coupling value whereas here this portion we have a different spin system is there it appears something like this here. So, this we call it as virtual coupling. So, now let us try to analyze 1 H NMR data for better understanding again the presence of a paramagnetic center with one or more unpaired electrons in a compound has significance on 1 H NMR spectrum chemical shifts. First the local magnetic field at each of the 1 H nucleus is affected when NMR active nucleus is placed in an external magnetic field the energy difference between nuclear spin states arises due to the interaction of the magnetic fields of the spinning nuclei with the applied field. However, the local field experienced by the nuclei is not the same as the applied field because the electron pairs in the vicinity of the hydrogen nucleus generate small local magnetic fields and these magnetic fields as I had mentioned earlier they can align with the magnetic field or they can oppose the magnetic field depending upon their alignment we will see some sort of shifts and also the further splitting. The local magnetic field is the sum of the applied field and all the smaller fields the later depend on the chemical environment of the hydrogen nucleus this is very important. So, all these chemical shifts essentially depend on chemical environment of the hydrogen nucleus typically the difference in local magnetic fields for protons in the different environments are small and as a consequence the chemical shift range over which the 1 H NMR is not large. So, whatever the chemical shift difference that we come across that is considerably low compared to other nuclei as a result we have a small spectral range of 1 to 10 ppm in case of 1 H NMR. In a paramagnetic compound there is an additional factor a large local magnetic field is arising from the unpaid electrons or electrons on the paramagnetic center. So, when we have paramagnetic center what happens because it generates again another magnetic field that can again oppose or align with the magnetic field and hence we will see unusual shifts in such molecules and this happens especially in case of metal complexes where we have lot of unpaid electrons are there and if it is paramagnetic and if it is diamagnetic it is not an issue if the complex is paramagnetic due to unpaid electrons and if we have some hydrogen atoms attached directly to metal we can see enormous shift will be there towards shielded region. This contributes to the energy difference between nucleus spin states and as a consequence the chemical shift range for the 1 H NMR signals is much larger than in a diamagnetic compound it can go up to minus 15 minus 20 or it can go up to even minus plus 60 sometime. The second effect that is observed in 1 H NMR spectra of paramagnetic compounds is broadening of the signals. This effect has its origins in a significant shorting of the exact state life time the relaxation time is very short that means we have to look into relaxation times we have spin-spin relaxation spin that is relaxation what would happen to this relaxation if we have paramagnetic species anyway. So, since I am more focusing on interpretation I am not going to the details of those things if you want to look into the details of those aspects I am sure there are better courses are there where theoretical aspects are discussed in detail probably you can refer to those or you can read appropriate book for that one. So, in some cases the broadening is so great that no well resolved signals are observed in some cases probably we may miserably fail to record a spectrum because of paramagnetic species present in the molecule. So, let us consider a simple paramagnetic species Tris phenanthraline cobalt 2 plus complex here and cobalt 2 can have one or three unpaid electrons in an actahedral complex depending upon the ligand field strength. So, if you see this is D7 system we have something like this we can have something like this or we can also have something like this if it is no field or no spin. So, we can have either one or three unpaid electrons. So, it is very interesting to examine 1 H N n m spectrum of this complex Tris phenanthraline there are four different aromatic proton environments are there in this one and the chemical shifts of the signals assigned to these 1 H nuclei fall in the range of plus 1 10 to 15. So, you can see otherwise in most of the organic molecules the chemical shifts are confined to 10 ppm 1 to 10 whereas, in this case it can go to as low as 1 1 10 to 15 or in some cases it can also go to minus 15 or minus 20. So, here what are those 4 signals we are getting due to which protons you can see here just I have labeled one phenanthraline molecule here others are also identical. So, you can you can label them as 1 2 3 and then 4 here and similarly 1 2 3 and 4 here. So, 4 signals we are seeing in this 1 10 to 15 range. I would come back again to 13 C and 1 H wherever we come across spectrum or at the end I will be discussing problems where I would take NMR spectra along with IR and UV data to elucidate the structure. So, now let us focus our attention to phosphorus 31 P NMR and phosphorus 31 P NMR is very similar to 1 H NMR the concept of 31 P NMR is very similar to proton NMR here also we have natural abundance of 31 P 100 percent and spin I equals half nuclear spin I equals half the 31 P nucleus is useful NMR spectroscopy due to its relatively high gyromagnetic ratio of 17.235 megahertz per tesla for 1 H if you recall it is 42.576 megahertz per tesla and in case of 13 seats one fourth of 1 H. So, it is around 10.705 megahertz per tesla the spectral interpretation is much easier and also it is very similar to 1 H NMR. 31 P NMR is an excellent technique for studying phosphorus containing compounds such as organic compounds, coordination complexes and molecules of biological importance and also in materials we where we have phosphorus and also especially this is very useful in understanding the intermediates in homogeneous catalysis if you have phosphorus as begins in those metal mediated catalytic reactions. And now let us try to look into the differences between 1 H and 31 P NMR and 1 H NMR spectra is a reference to TMS tetramethylsilane whereas in case of 31 P NMR we are using 85 percent phosphoric acid and we are considering its chemical shift as 0 and it is used as an external standard because we cannot add 85 percent H3SO4 directly to the solution containing phosphorus compounds as it can react with it it is used as an external standard. Another standard is trimethylphosphate or tri methoxyphosphine is also used as a standard and unlike phosphoric acid it is independent of concentration or pH so we can use it without any problem and with respect to phosphoric acid the shift of trimethylphosphate is around 140 ppm. Similar to 1 H NMR positive chemical shifts corresponds to down field shift from the standard chemical shifts in 31 P NMR commonly depend on the concentration of the sample the solvent used and the presence of other compounds. This is because the external standard does not take into account the bulk properties of the sample as a result the reported chemical shifts for the same compound could vary by 1 ppm or more especially for phosphate groups whenever we have a pentavalent phosphorus having P double bond O in this case what happens the chemical shifts for the same sample if you record and there can be some variation it is because of homogeneity problem and other things as I mentioned here since the bulk properties are not considered here so that can show this kind of variation. 31 P NMR spectra are often recorded with all proton signals decoupled very similar to decoupling process we use in case of 13 C here also conveniently we can use decoupling process and we can record 1 H decoupled 13 P NMR that will be very simple and it is very simple to assign. This gives rise to single sharp signals per unique phosphorus and here in we will consider both coupled and decoupled spectra but very nicely we can see two phosphorus couplings up to three four or five bonds they are chemically and magnetically non equivalent. Interpretation of spectra easy very similar to 1 H as I mentioned earlier as in 1 H phosphorus signals occur at different frequencies depending on the electron environment of each phosphorus nuclei. So let us consider some examples here so you can see the extensive list of chemical shifts and the chemical shift range for various type of phosphorus compounds we come across. So let me elaborate on this one in my next lecture until then have an excellent time reading on spectroscopy thank you.