 Hello everyone. Once again I welcome you all to MSB lecture series on interpretative spectroscopy. In my last lecture I was discussing about spin-spin splitting. So that means often when we run NMR spectrum for a molecule we see not a single line and single line consists of many lines with different heights that means different intensity. Why that happens? So that I started discussion on that one. Let me continue from where I had stopped. As I mentioned another useful property that allows NMR spectra to give structural information is called spin-spin coupling which is very very important and it is very vital in understanding and diagnosing the molecule while elucidating its structure which is caused by spin coupling between NMR active nuclei. You should remember NMR active nuclei that are not chemically identical. Different spin states interact through chemical bonds in a molecule to give rise to this coupling. That means if you have two carbon atoms, two carbon atoms are connected by a bond or two. In that case protons present on both the carbon atoms interact provided they are in different chemical environment. In NMR spectra this effect is shown through peak splitting that can give direct information concerning the connectivity of atoms in a molecule. Not only that one and also they can precisely tell you how many protons are there, equivalent protons are there on the carbon, the adjacent carbon atom. Nuclear which share the same chemical shift do not form splitting peaks in an NMR spectra. This is very very important. Nuclear which share the same chemical shift do not form splitting peaks in an NMR spectra. In general neighboring NMR active nuclei three or fewer bonds away leads to this coupling. In special cases even if they are 4 to 5 bonds away they can still interact and influence in splitting the signal. The splitting is described by the relationship wherein neighboring nuclei results in N plus 1 peaks. That means if we have N number of proton atoms are there on adjacent carbon atom that leads to the splitting of the adjacent one by N plus 1 peaks. If three CH2 are there then in CH2 so that two are there then it will make the neighboring one into three. If three are there then it appears as four something like that and that information directly comes from Pascal triangle for a particular I value. So, you should remember Pascal triangle will be varies with I values, I equals half we have a different Pascal triangle, I equals 1 we have a different Pascal triangle and we have to use a different totally different Pascal triangle when I equals 3 by 2 or 5 by 2. So, however being adjacent to a strongly electronegative group such as oxygen that can prevent spin-spin coupling between two neighboring groups where we have protons. For example, a doublet would have two peaks with intensity ratio of 1 is to 1 while a quadrate would have four peaks a relative intensity of 1 is to 3 is to 3 is to 1 or if we have a triplet the relative intensity will be 1 is to 2 is to 1. So, the magnitude of the observed spin splitting depends on many factors and is given by coupling constant J which is in units of hertz. So, when we see this triplet, quadrate or doublet the spacing between the two lines we call it as coupling constant J, it is always represented in units of hertz. Also we get information about peak intensity. So, the size of the peaks in the NMR spectra can give information concerning the number of nuclei that give rise to that peak. So, this is done by measuring the peak area using integration, that information again comes from NMR spectra yet even without using integration the size of different peaks can still give relative information about the number of nuclei. Even if you do not have that integration simply by looking into the height of the peaks we should be able to judge, but sometime it can be misleading and we should not just go by the height sometime what happens is the broader one the height is small that can occupy more larger area and still that can be having considerable peak height. So, one should be very careful about that one and it is better to integrate rather than looking into the height to decide the intensity. For example, a singlet associate with 3 hydrogen atoms would be about 3 times larger than a singlet associate with a single hydrogen atom. What are the limitations of NMR there? Despite its ability there are several limitations that can make NMR analysis difficult or impossible in certain situations. One such is that the desired isotope element that is needed for NMR analysis may have little or no natural abundance. For example, I am interested in a particular isotope of a nuclei which is NMR active, but that may be in very insignificant quantity in the natural resources. So, that means for example, if you consider the natural abundance of 13C the active isotope of for carbon NMR with i equals half is about 1.1 percent, whereas rest is 12 carbon which is NMR inactive because i value 0. Still 1.1 percent is quite sufficient to plot 13C NMR and look into NMR spectrum and it can still help to great extent in illustrating the structure and getting lot of information. However, in case of oxygen the active isotope for NMR is 17 oxygen which is only 0.035 percent naturally abundant. This means that there are certain elements that can essentially never be measured through NMR. So, no matter how much quantity we take it is very difficult to identify and run 17 O NMR spectrum for a molecule where we have insignificant quantity of 17 O isotope. So, that means, in case if you want to run 17 O NMR it has to be enriched. So, another problem is that some elements have an extremely low magnetic moment that means, the sensitivity is very very low. So, the sensitivity of NMR machines is based on the magnetic moment of the specific element, but if the magnetic moment is too low it can be very difficult to obtain NMR spectra with enough peak intensity to properly analyze. So, these are the limitations of NMR. One is less abundance and another one is the low magnetic moment value. What are the basis for spin coupling? Well, all the day I discussed some of those in the couple of slides earlier, but let me go back again and repeat some of these facts so that understanding would be much better. So, NMR signals arise when nuclei absorb a certain radio frequency and are excited from one spin state to another one. So, this whatever the energy we are supplying is equal to the normal frequency of this one then flipping of this one takes and that we call it as electronic transition and exact frequency of electromagnetic radiation that the nucleus absorbs depends on the magnetic environment around the nucleus. So, that means, whether it is given nucleus is shielded or deshielded also should be considered. This magnetic environment is controlled mostly by the applied field, but is also affected by the magnetic moments of nearby nuclei. So, nuclei can be in one of many spin states giving rise to several possible magnetic environments for the observed nucleus to resonate in. So, this causes the NMR signal for a nucleus to show up as a multiple or multiplied rather than a single peak. So, because of the interference of magnetic field generated by neighboring nuclei what happens the signals will be split into multiplets 1, 2, 3 or 4 depending upon the magnitude of the field and also number of such nuclei that are generating induced magnetic field. In NMR experiments only nuclei with I equals nonzero will show up NMR spectrum again I am telling you. So, when I equals 0 there is only one possible spin state and hence nucleus cannot flip between states. So, why? So, it has only one state so that it cannot flip. Since the NMR signal is based on the absorption of radio frequency and the nucleus transition from one spin state to another one plus half to minus half or 1 to minus 1. I equals 0 nuclei neither show up NMR nor cause splitting of other NMR signals because they only have one possible magnetic moment. This simplifies NMR spectra. In fact, this also comes very handy in simplifying the NMR spectra in particular of organic and organometallic compounds greatly since the majority of carbon atoms are 12 carbon which have I equals 0 value or we have to consider only 1.1 percent of 13 C that has I equals half value. So, here I have given a typical spectrum here chloroethane and you can see here we have a triplet of 1 is to 2 is to 1 intensity and we have a quartet of 1 is to 3 is to 3 is to 1 intensity for this methylene protons and a triplet for methyl protons. How we get triplet here and how we get quartet here I will show you after a couple of slides. So, now for a nucleus to cause splitting it must be close enough to the nucleus being observed to affect its magnetic environment. For example, if a particular group protons present on a particular group are influencing the splitting they should be on the adjacent carbon atom or at least they should not be beyond 3 or 4 bonds that is what it means. The splitting technically occurs through bonds and often we can also come across through space coupling also that I would take up at the end. So, here the splitting technically occurs through bonds not through space. So, as a general rule won't be nucleus separated by 3 or 4 bonds can split each other. However, even if a nucleus is close enough to another it may not cause splitting for splitting to occur the nucleus must also be non equivalent. So, they must be non equivalent as I mentioned in case of ethane and we have CH3 group and CH3 group are there and both the CH3 groups are not splitting each other because both are chemically equivalent. That means, there must be some non equivalents there to split the neighboring signals. To see how these factors affect real NMR spectra consider the spectrum for chloroethane here there are two groups of peaks in the spectrum as I mentioned a triplet and a quadrate these arise from the two different types of I equals non-zero nuclei in the molecule. Yes, the protons on the methyl and methylene groups methyl and methylene groups the multiplied corresponding to CH3 proton appears as a relative integration the peak area comes here of 3 1 for each proton and is split by the two methyl protons N equals 2 which results in N plus 1 peaks that is 3 which is a triplet. So, the multiplied corresponding to CH2 protons has an integration of 1 for each proton and is split by 3 methyl protons N equals 3 which results from N plus 1 peaks that is 4 peaks which is a quadrate and each group of nucleus splits the other. So, in this way they are coupled that means, both are coupled here and if we measure the spacing between the lines here and the lines here that is similar this the spacing what we call between the two lines in a chemical shift is called coupling constant and is represented in hertz. To see how these factors affect real NMR spectra again I am considering this chloroethane there are two groups of peaks in the spectrum a triplet and a quadrate I am just repeating again these arise from the two different types of I equals non-zero nuclei in the molecule the protons on the methyl and protons on the methylene group the multiplied corresponding to methyl protons has a relative integration of 3 for each proton and is split by two methylene groups protons N equals 2 which results from N plus 1 peaks here N plus 1 peaks are coming because of CH2 CH2 plus 1 3 which is a triplet and the intensity ratio is 1 is to 2 is to 1 the multiplied corresponding to CH2 has an integration of 2 is this entire integration corresponds to two protons, but it is split into 4 with a intensity ratio of 1 is to 3 is to 3 is to 1 because of N plus 1 coming from here 3 are there 3 plus 1 4. So, this methyl protons are split this into quadrate methylene protons are split this into triplet that is what it is. So, now let us consider now 1 1 2 dibromoethane here the protons and adjacent carbon are non equivalent. So, why this doublet and a triplet? So, that means, this one shows a triplet whereas, these two show a doublet here. So, how that happens let us look into it and then values are given here we can see 4.26 4.26 this one these two are showing a doublet here whereas, at the 6 point centered at 6.51 for this one we give only one value the middle one we are giving or if you have 5 peaks 5 peaks you know the middle one we are taking and if 6 peaks are in between 3 and 4 we take the center point of this one and it is represented as chemical shift and here chemical shift is 6.51. So, now we have to see why this is doublet and why this that triplet in a more let us first look into this HA signal HA. So, first this is split by these two into triplet how we will see. So, now basically what happens? So, if you just see here we have two protons are here now. So, these protons both the protons the magnetic field generated by both the protons can be aligned with the magnetic field this is more towards the shielded region and then in this case what happens one can be upward one can be downward this is one possibility. So, we can also have this kind of orientation if you represent this 1 and 2 and 1 and 2 1 and 2 both are upwards and 1 is up and 2 is down and now one can be down and 2 can be up this is another possibility and then we cannot have any other possibilities. Now, the third one will be something like this 1 and 2. So, that means now if you just see these two are degenerates. So, this is one set and this is two set and this is one set. So, this is how the signal due to HA HA appears as a triplet something like this something like this this is for HA. So, now let us look into HB HB signal we have to see. So, HB signal. So, now these two will be split based on the magnetic field generated with this one. So, now one can be something like this only one is there or one can be something like this only two possibilities are there as a result what happens either this one can influence its chemical shift as a result what happens it will come little bit D shielded and this one will be little bit shielded. So, we get two peaks two lines of equal intensity. So, this is how the splitting can be seen and overall one can write something like this something like this and this whatever the spacing is there this is spacing are identical. So, this we call it as J and if here if you see 1 2 3. So, this we call it as 3 J H H ok even if you want to be more precise you can say that one. So, this is how this is your superscript and this is subscript and this should be in italic this is how the coupling constant is presented. This is italic and then here you see the bond separation and here the coupled nuclei H A H B. So, this is how a typical coupling constant is represented. So, now what is this n plus 1 rule? So, if a signal is split by n equivalent protons n equivalent protons it splits into n plus 1 peaks that means the relative peak intensities of symmetric multiplets of first order and then whatever this Pascal triangle that predicts number of peaks with i equals half it is very unique for this one and also this holds good only for first order spectrum. That means if I say first order spectrum that means something else can also be there yes that is called second order spectrum second order will be totally different. So, that I shall discuss after completing about first order spectrum and looking into more examples. So, if in a typical first order spectrum and if the i value is half then we can use very effectively this Pascal triangle. For example, when number of equivalent protons causing splitting is 0 if you have nothing then only singlet will be observed because n plus 1 n is 0. So, only one will be there and if only one is there then we will see n plus 1 2 lines will be there in 1 is to 1 ratio and if we have 2 neighboring protons are there they cause this one into a triplet then intensity will be 1 is to 2 is to 1 and then if 3 are there we get a quadrate that is 1 is to 3 is to 3 is to 1 and then if 4 are there quintet we get 1 is to 4 is to 6 is to 4 is to 1 ratio and then if 5 are there then n plus 1 will be sextet we get 6 lines the relative intensities are 1 is to 5 is to 10 is to 10 is to 5 is to 1 and then if 6 lines are there we get a separate then intensity ratio will be 1 is to 6 is to 15 20 15 6 and 1. So, this how even simply if you know 1 second one can be written and then you just 1 do this 2 and this 1 and next 1 this 2 3 and this 2 3 and this 1 1. So, you should be able to write once you know only this one there is no need to remember this one next one you can write 1 4 6 4 1 you can write here and next one is 1 5 and then 10 and then 10 and 5 1 you should be able to write. So, now it is 1 is there 1 6 is there and then 15 is there and then this is 20 and then again 15 is there and 6 is there. So, you can keep on writing like that. So, once we know up to here you know extrapolating that one and writing for higher number of lines should be very easy using this Pascal triangle. So, using this Pascal triangle we should be able to predict the number of peaks and this equals i equals half and another equation we use is n i plus 1. So, then if you do not know or if you have different i values still this can also be used here and eventually it will if you take half is there for example, 2 ok n again is half plus 1 this will cancel that is the reason we simplify it and use n plus 1 peaks. So, this Pascal triangle is for i equals half value. Similarly, when i equals 1 is there it changes and now if you see for example, if we have 0 you know the nuclei which are influencing still we get n plus 1 we get 1 peak and then when we have 1 we get 3 peaks because n plus 1 here we have to use 2 n i plus 1 2 n i plus 1 we should use i is the value 2 into if say 1 is there and then this is 1 plus 1 i value is 1 i value what we are considering is 1 we get 3 lines ok this is what is there and if 2 are there 2 into 2 plus 1. So, we get 5 this is how you can calculate and if 3 are there 3 into 2 plus 1. So, we get 7 lines. So, you should be able to calculate. So, it 5 level and 630 and this is the corresponding intensity ratio for number of peaks goes like this. So, this is this Pascal triangles holds good for those nuclei which have i equals 1 value. Now, I have written another Pascal triangle for i equals 3 by 2 for example, if we are using say 11 boron is i equals 3 by 2 and if we come across any spectrum and if we have more number of lines we can conveniently use this Pascal triangle to calculate the relative intensities. So, here again if you simply take 0. So, here what happens 2 n i plus 1. So, then of course, 0 means it becomes only 1 if you have 1 is there then 2 into 1 into 3 by 2 plus 1. So, we get 4 lines here 4 lines are there what would happen if 2 are there 2 into 2 into 3 by 2 plus 1. So, it becomes 7 lines 7 will be there and similarly if you go for 3, here 3 into 2 into 3 by 2 plus 1. So, we will get 3 into 3 9 plus 1 10 lines. So, 10 will be there and similarly you can go for 4 we get 13. So, that means, if 4 equivalent nuclei with i equals 3 by 2 are splitting then 13 lines will be there and the corresponding intensity will be this one. We have some rough idea about coupling constants what we have to understand is equivalent protons do not split each other. So, that means, if you consider a CH 3 CH 2 OH and these are equivalent protons they do not couple each other. That means, if I expand them we write like this these are all geminal that means, when they are equivalent when they are geminal coupling is not observed in first order spectrum that is what it is equivalent protons do not split each other and then protons bonded to the same carbon atom will split each other if they are not equivalent due to unsymmetrical nature of the molecule. Due to some reason if the 2 hydrogen atoms present on a carbon are non equivalent because of unsymmetry present in a molecule or molecule is differently substituted on both the ends in that case what happens geminal coupling can be observed in that case invariably we come across second order spectrum. So, protons and adjacent carbon atoms normally will couple. So, always the protons and adjacent carbon atoms provided they are not chemically equivalent they normally couple. Protons separated by 4 or more bonds will not couple or show very weak coupling that means, to see very effective coupling they should not be more than 3 bonds apart. If they are more than 3 bonds apart we may or may not observe coupling and in case if you observe coupling the magnitude of the coupling is very small and hence that is very weak. Considering more examples and discuss more interesting aspects about coupling constant in my next lecture until then have a good time. Thank you.