 It is a pleasure to welcome you all once again to MSP lecture series on interpretive spectroscopy. In my last lecture I showed you how very nicely we can use phosphorous NMR to monitor a reaction and also we can gauge when the reaction is going to end and also whether any products are formed or not other than the expected product. So now let us continue looking into more such examples of course using our laser NMR active nuclei as well. Now let us look into 19 F NMR and of course about the abundance, gyromagnetic ratio all those things I have given in the beginning. So now let us focus our attention to the chemical shift values as far as 19 F NMR is concerned. 19 F is 100 percent abundant and I equals half. Then it is as simple as 1 H R 31 P to measure and understand. So here if you just look into CF here when we have something like this on a alkane chain reference we are using is CF Cl3. So relative to trifluorocytic acid so it will show around minus 131 or with respect to fluorobenzene, monofluorobenzene it comes around minus 96 but with respect to CF Cl3 it comes at minus 210 different reference are used that is the reason I have given the value relative to different standards. And then in case of 2 fluorogrups present on same carbon we see a peak at minus 140 in case of CF Cl3 and minus 69 with trifluorocytic acid and minus 26 here in fluororomatic compounds the range is minus 140 or minus 60 or minus 26 and then in case of fluorine atoms present next to carbonyl group the ranges are minus 125 or minus 46 or minus 11 and again we have a CF3 group next to a CH minus 75, 4 or 39 and then a CF3 group next to carbonyl group would appear at minus 81 or minus 2 or 33 and if you have a sulfonate group it comes at minus 50 or 129 or 164 these are some of the important ones besides that we have numerous compounds. So, for example, if somebody asks you to sketch the 19 F NMR spectrum of Cl3 simply by looking into valence shell electron peripersion theory you should be able to arrive at the right kind of geometry for ClF3 and then again let us recall your basic understanding of VSEPR theory. So, valence shell electron peripersion theory the abbreviation is self-explanatory and now if you just look into the valence electrons present on chlorine and fluorine atoms we are talking about 7 plus 3 here the 10 electrons are there 10 electrons means steric number equal to 5 are there and in case of 5 here 3 bonded pairs are there and 2 lone pairs are there. So, the preferred geometry for steric number 5 is trigonal bipyramidal and then we may have trigonal bipyramidal with 3 bonded pairs and 2 lone pairs all these possibilities are there and then we have to look into lone pair, lone pair, lone pair, bond pair and bond pair, bond pair repulsion and from this basic knowledge we will arrive at the right structure where we have 2 lone pairs in the plane and then 1 fluorine atom also in the plane equatorial and other 2 fluorine atoms being axial. So, this is the right structure and also we say that it has a T shape and then we have to sketch the 19 FNMR spectrum of this one here and assuming this molecule is static we can anticipate 2 type of fluorine signals one is due to axial one other one is due to equatorial one equatorial one will be coupled with 2 axial ones to give a triplet and then 2 axial will be coupled with equatorial one to give a doublet the spectrum should be a doublet and a triplet something like this. So, this is how the 19 FNMR spectrum of CLF3 looks like. So, now let us look into another interesting molecule again some of these molecules are really meant for explaining VACPR theory SF4 very interesting to look into the shapes of the molecules are species and also the geometry and you should remember the difference between the shape of a molecule and geometry of a molecule when we define the geometry of a molecule like this we have to consider both lone pairs and bonded pairs that means the entire steric number has to be considered while defining the geometry, but while defining the shape we can ignore the lone pairs that one should remember sketch the 19 FNMR spectrum of SF4 of course SF4 also same thing if you look into SF4 we have 6 electrons are here on sulfur S2P4 and we have 4 electrons coming. So, we have 10 electrons are there and steric number is 5 and here 4 bonded pairs are there and 1 lone pair is there. So, that means again trigonal bipyramidal geometry with trigonal bipyramidal geometry what basically happens is one something like this if S is here what we have is one clue in here one clue in here and lone pair is there. So, lone pair will be occupying and then this is a trigonal bipyramidal molecule, but the shape is CSA shape and also to minimize these repulsions what happens they are further bent little bit in this fashion. So, it is this fashion. So, that means here we have just by simply looking into the molecule we can say that there are 2 types of fluorine environments are there and 2 axial ones and 2 equatorial ones. So, when we look into the signals we anticipate 2 type of signals and each signal will be a triplet. So, that means we can see presence of 2 triplets for SF4 here. So, it should something like this this is a 2 triplets are there one for so axial one for equatorial and then this is FF coupling the spacing what we see here or here it is FF coupling. So, this is how we can sketch the 19 FMR spectrum of SF4 let us look into another molecule here. Now, we have taken SCF4 you know that the 77 selenium is NMR active with I equals half and it is only about 7.6 percent present rest 78 is NMR in active. So, when we look into 19 FMR we can see again very similar to SF4 it has a CSA type structure. So, with color I can distinguish them if you look into 19 FMR again we are seeing a triplet in each case of course, in each case triplet is there and again because of selenium being NMR active that is present is 7.6 percent or so what happens we can see satellite peaks that means basically if you see a triplet something like this. So, let me make it little bigger now we can see this coupling will also have this will give you 1 j FSC coupling and then what you see here is this spacing is j FF coupling 1 2 j coupling. So, same thing happens in both the cases both the cases we get a triplet of a triplet and also a triplet of doublets that would be here. So, this happens in both the cases, but when we look into 77 selenium NMR 7 different selenium NMR what basically happens is you get a triplet first this is coupled with axial ones and then each one will be split further into triplet into equatorial ones. So, if I plot this one it should be 1 is to 2 is to 1. So, this how we can sketch both 19 FMR spectrum as well as 77 NMR spectrum for SCF4. So, you can see here triplet of a triplet it is pretty easy. So, now let us look into another example here SCF4 SCF4 is a tetrahedral molecule and if you look into fluorine NMR this is fluorine NMR fluorine NMR would not show much difference because all fluorine are in a single environment and if it is tetrahedral molecule here, but you should remember the fact that silicon has 3 isotopes 28 silicon 29 silicon and 30 silicon and then 28 silicon is 92.23 percent 29 is 4.67 and this is 3.10 30's and now this is i equals 0 this i equals half and then again this i equals 0. So, that means basically we should think of a signal coming here without involving any coupling in case of 19 F this corresponds to 92 means 95.33 which is NMR in active i equals 0 and this one whatever we are seeing is 1 j 29 to silicon coupling that is about 178 hertz and if you look into intensity this corresponds to 2.32 or 2.32 or something like that. So, this is how this again satellite peaks, but if you look into 29 silicon NMR. So, you can see here then it looks all fluorine atoms being identical all 4 fluorine atoms would split silicon signal into a pentate a quintate we do not use the term pentate for 5 we use the term quintate. So, we have 1 is to 4 is to 6 is to 4 is to 1 we are getting something like this and here again if you look into the coupling the magnitude of this one should be same as this one. So, again double verification we can do. So, silicon NMR looks like this 29 silicon NMR for Si of 4 would be a quintate you see now how simple it is to look into various other NMR active nuclei other than 1813 C and also 31 P. Let us focus our attention to two more NMR active nuclei that is nitrogen 14 N and 15 N. The 14 N NMR experiment is much less sensitive than one edge, but has a much larger chemical shift range and signals are broadened due to quadrupolar interaction of 14 N having I equals 1. So, the larger the molecule and the more symmetric the nitrogen environment the broader the signal this one should remember if running 14 N or 15 N is inevitable 15 N there should be any problem, but if you have when you are running 14 N what we should remember is the larger the molecule and the more symmetric the nitrogen environment the broader the signal. The small and highly symmetric aqueous ammonium ion gives a very sharp line less than 1 hertz wide. So, that means if the molecules larger and more asymmetric around the nitrogen environment the broader will be the signal and small and highly symmetric aqueous ammonium ion gives a sharp line with less than 1 hertz width line width. So, on the other hand if you look into liquid ammonia being less symmetric here the width is 16 hertz on a 400 megahertz spectrometer at room temperature and urii is larger and asymmetric. So, the line width is approximately 1 kilo hertz you see very broad signal. So, molecules that are significantly larger than urii yield signals too broad to be observed with a higher resolution NMR spectrometer even with a 600 it is very difficult to observe. Since the nitrogen chemical shift range is wide may be readily used for distinguishing nitrogen species in very small molecules and if you want to run 15 N it has to be preferably enriched otherwise the percentage is so low it is very difficult you can see how much percentage is there. So, for example, natural abundance of 14 N is 99.63 and chemical shift range is 900 ppm from 0 to 900 ppm and frequency ratio is also given for ammonia and reference compound is nitromethane in CdCl3 and line width of reference is 19 hertz and T1 black station time is also given and then all this vital information is given and one should keep distance in mind especially when you want to run 14 N and of course 15 N is very very important from biological point of view because we have lot of nitrogen compounds. Now, as I mentioned if the molecule is more asymmetric the line width will be more you can see here the spectrum of uria it is very huge and then if you see here NH3 in liquid very small and less symmetric. So, 16 hertz is there and then if you see NH4 plus very small and highly symmetric you can see the line width is 0.8 it looks like 13 CNMR or phosphorous CNMR very simple. So, that means if you have a smaller molecules and symmetric looking into 14 N NMR would be rather easy. So, this plot essentially to show how the line width varies with the molecule symmetry and also the size. Now, this is 1H NOS spectrum of ammonium chloride this is a tetrahedral molecule you can see NH4 plus and then this 4 protons are identical and they are coupled with nitrogen I equals 1. So, you see a triplet of equal intensity 1 is 2 1 is 2 1 here and then the coupling is 1J 14 N to hydrogen coupling is 52.4 hertz. So, what is interesting is in this one we are also seeing very very tiny peaks here this is due to 15 N less than 0.4 percent or something this is about 1J 15 N H coupling is 73.5 hertz probably they have taken much longer time to identify these two peaks here this is the 15 N satellites here this was observed in case of ammonium chloride. You know that 14 N coupling constant is smaller compared to 15 N coupling because 15 N lower resonance frequency. Now, if we look into 14 N NMR spectrum of ammonium chloride now 14 N this nitrogen is equally coupled to 4 equivalent hydrogen atoms as a result what happens it shows a quintet of this type. So, 14 NMR spectrum of ammonium chloride consists of quintet whereas, 1H NMR spectrum of ammonium chloride consists of a triplet of equal intensity 1 is 2 1 is 2 1 line and of course, this also you can say it is a EX4 spin system. Look into this molecule here 15 N enriched cyclosporine A in CdCl3 this spectrum was acquired in 1 hour very interesting we got a beautiful spectrum for this 15 N because it is a 15 N enriched molecule here still it took 1 hour for acquiring the entire data to identify all this 1 2 11 nitrogen atoms here to obtain similar sensibility without enriching would take about 10 years that means basically we have to take this sample put into NMR spectrometer be 600 megahertz or something run continuously for 10 years run 15 N NMR for 10 years continuously without stopping to see something like this. So, you can imagine the presence of 15 N so low and then how much sample is needed and how much time it takes but on the other hand when you enrich this one with 15 N it appears like any other simple molecules and measuring 1 H NMR beautiful spectrum you there you can see all 15 N are here 1 2 3 4 5 6 7 8 9 10 11 in biological studies 15 N plays a very very major role. So, now this one is for sodium azide sodium azide of course you have 2 different type of nitrogen atoms are there one is cationic and other 2 are anionic and of sodium you say so sodium plus so it is neutral and then the middle one is appearing here and these 2 anionic are appearing here and the ratio is 1 is to 2 here. So, this is a 15 N enriched sodium azide taken in T2. So, let me show you another very interesting example and this is about 13 C NMR the structure of tertiary butyl ethium is very similar to the top methyl ethium we know that methyl ethium has a Cruban structure where alternate kernels are occupied by methyl group and lithium with 4 lithium atoms having tetrahedral relationship but with each hydrogen atom replaced by a methyl group in case of methyl you get tertiary butyl ethium that is it. So, in methyl we have 3 hydrogen atoms are there if you replace each one with a methyl group we end up with tertiary butyl ethium that also has very similar structure to that of methyl ethium Cruban structure. The 75 megahertz 13 C NMR spectrum of a sample of T butyl ethium prepared from 6 lithium metal consists of 2 signals one for the methyl carbons and one for the quaternary carbon atom. That means if you look into 13 C NMR spectrum we get 2 signal sets one is for quaternary carbon directly interacting with lithium other one is methyl groups. So, now if you focus our attention to the quaternary carbon the spectrum look like what I have shown here at 185 Kelvin that is at low temperature it shows 7 lines here a typical septet on the other hand at 299 Kelvin that is at room temperature it shows 9 lines. So, now we have to analyze and explain. So, why at low temperature 13 C NMR spectrum for quaternary carbon shows a septet whereas at room temperature or 299 Kelvin it shows a multiplied having 9 lines. Now, let us look into the Cuban structure this is how you can show the relationship of 4 lithium atoms in methyl ethium or T butyl ethium this heteromeric and then in each phase we have a carbon is sitting like this it is a methyl or T butyl group and that means at 185 Kelvin what happens and this the molecule is very static. When it is very static this carbon present in one of the triangular phases would see only 3 lithium atoms it is seen only 3 lithium atoms because the whole molecule has no flexibility and it is very static and it is sitting something like this overlapping with orbitals of 2 S orbitals of 3 lithium atoms and it is a 4 centered to electron bond system. So, in this case what happens it identifies all equally that means the carbon is split by 3 equivalent lithium atoms if we use 2 Ni plus 1 rule at low temperature we have 3 and i equals 1 plus 1 it shows 7 lines. So, 7 lines so this is carbon but when you look into room temperature what happens it is no longer static. So, all lithium atoms are exchanging as a result what happens carbon sees 4 equivalent lithium atoms on 4 they will be exchanging as a result what happens you will see this carbon see observes are in contact with 4 equivalent lithium atoms now. So, 4 equivalent lithium atoms are there at 299 Kelvin. So, what would happen is if you use again 2 Ni plus 1 rule we have 4 identical 1 i equals 1 so this is 9 lines. So, we will see 9 lines here. So, that this how even the dynamic process can be understood readily using NMR spectroscopy. So, let me continue in my next lecture more details and more examples thank you. you