 Once again, I welcome you all to MSB lecture series on Interpretive Spectroscopy. Since my last lecture, I started really looking into lot of problems and giving the right kind of solutions, real interpretation I started considering problems, although I was discussing many problems in between while discussing many of the spectroscopic methods. Let me focus your attention entirely to only looking into the problems and different ways of finding solutions when some data is given from any of these spectroscopic methods. So let's begin today with another problem, solve the structure using the following spectral data, molecular formula is given here. So I'm going to show you one HRMR spectrum in a minute or so. Now first you look into the formula and try to make familiar yourself how many different type of elements are there in it, in this one carbon is there, hydrogen is there, bromine is there and oxygen is there. So molecular formula is given but structural formula is not given that is more important. And now when this compound is given we also know whether the compound is saturated or unsaturated whether double bond, triple bond is there by looking into hydrogen deficiency index. So that formula I'm sure you are familiar. Let me write again C plus 1 minus half H minus half X, X is halogens and half N. So this one should remember this is the formula we use to identify hydrogen deficiency index. From this one we can get information about possibility of the presence of a ring double bond and triple bond. So if I use here 8 is there 8 plus 1, 8 plus 1 minus 7 by 2 means 3.5 and minus half 0.5 equals 9 minus 4 equals 5. So that means here the deficiency is 5. 5 means certainly there is a ring and plus 4 double bonds. Now we know that ring is there means probably aromatic group a benzene ring is there and then that will take care of 3 double bonds and the ring. So now we have to account for one more double bond. So that means we have O possibility of a ketonic carbonyl group or omethylxy group or something would be there. Now let us with this information let us look into one H NMR spectrum. You can see here and in NMR spectrum first identify number of groups of identical protons are present. So that you can just see by simply counting the number of multipliers we see here. We have here 1, 2, 3, 4, 5, 6, 6 groups are there. That means you can clearly tell 6 different type of hydrogen atoms are there. When you look into a doublet is there which is around 3.7 or something. There is a doublet here and then we have 2 triplets closely spaced. That means they have very similar chemical shifts and similarly we have 2 more doublets. Again they have very similar chemical shifts and then we have a triplet far away much deshielded. So that means probably this could be due to a aldehyde or acidic hydrogen means it does not show any coupling. So it must be aldehyde hydrogen. So that means we know now. And then by looking into this probably it looks like they are from aromatic group. Now let us try to write down the structure here. First I would write a benzene ring something like this. Then bromogrupe is there. So let me put one bromogrupe here and then by just looking into this probably there is a something like this I can write. Now we look into how many different type of proton signals can be expected for this one. So certainly 1, 2, 3, 4, 1 to 4. So if you just see here because you can do free rotation C2 axis. As a result what happens these two will be identical. That means 1, 2, 3, 4. So here you can expect 4 different type of proton signals. But we have 1, 2, 3, 4, 5, 6 are there. So this is ruled out. This not. So next let us consider another one. So now let me put it here. If I put here now what happens 1, 2, 3, 4, 5, 6 all are different here. And then once after identifying now we have to check whether we get triplet pattern or not. For example if I consider this hydrogen here this is equally coupled to these two hydrogens. As a result this can show a triplet. And similarly this hydrogen can also show a triplet. And this can also show a triplet. And then this one is coupled only to this one show a doublet. And similarly this can show a doublet. And now this one would show a doublet here. This one and this will show a triplet. This triplet is here. This is. So now we have identified the compound is this one. So now let us see what information yes. And now to further conclusion we have 13 C NMR also. 13 C NMR again it will tell you if I look into 13 C NMR of first compound parabromo substituted aldehyde benzyl aldehyde. So we can see here 1, 2, 3, 4, 5, 6, 6 carbon will be there. But again here we have 1, 2, 3, 4, 5, 6, 7, 8 are there. That means all 8 carbon atoms are non-equivalent. That means here if I just look into it in this artobromo 1, 2, 3, 4, 5, 6, 7, 8 all are different we are getting. So that means without any hesitation one can assign this spectrum and write the structure for this one should be like this is the compound hydrogen deficiency 5. So now yes this is the compound here. Now we can identify signals yes 7.33, 7.33 here and these two are showing triplets. And then this is higher and this is slightly higher than this one this is a doublet again and further high because next to bromo this is one and then 3.66 it should be a doublet here it is for this one and then this hydrogen here this one it is showing get triplet because of coupling with this two ok. So very easy right. So this how one can do very nicely interpretation of course when we have molecular formula we should first find out hydrogen deficiency index by simply looking into this formula here and then you should be able to identify. So now for curiosity sake let us try to analyze the spectrum of parabromo. As I said in case of parabromo we should we are expecting 1, 2, 3, 4 different type of hydrogen atoms ok 4 different type of proton signals and then 1, 2, 3, 4, 5, 6 different type of carbon signals I have also simulated that one let us look into that now ok. So this is again for 13C all the signals chemical shifts are assigned here ok. So 1, 99 you know again Keto aldehyde carbon will come here and then we have here 6 and then 1 here this is due to this one here. So now this is the one the 4 signals as I mentioned here we should get 4 signals for this one 1, 2, 3, 4 yes 4 signals are there these two will be showing doublet and this will be again a triplet this one will be again a doublet here and then this would show a triplet here very similar to Arthobromo 1 except for we have 2 more signals here because of non equivalence of all the hydrogen atoms. And as I mentioned in case of 13C we also expect 1, 2, 3, 4, 5, 6 different signals I have here 1, 2, 3, 4, 5 and then 6 are there yes this is for parabromo benzyl aldehyde. Now let us look into another problem here an organic molecule shows 2 absorption peaks at 870 and 975 hertz in a magnetic field of 3 tesla. What are the corresponding chemical shifts in PPM first yeah it is given in hertz we have to convert into PPM. Before you convert this into PPM we should know this is a 13C NMR we should know first the frequency corresponding to 3 tesla. To solve this problem what we should do is first we should find out the frequency corresponding to 13C nucleus with respect to magnetic field strength of 3 tesla for that one we should use this equation h nu equals gamma h over 2 pi b naught. So, we know now nu equals gamma over 2 pi to b naught simply if you use the formula here and then apply here we have gyromagnetic ratio is given here 6.7263 into 10 raise to 7 to b naught over 222 by 7 by simplification of this one what we get is 32.1 or approximately 32 hertz. That means the frequency corresponding to 13C in a magnetic field strength of 3 tesla is 32 hertz. So, now we know now we have to convert the given peaks in hertz to PPM first it is considered 87. So, to convert delta we know that delta in hertz or the frequency. So, here this is 870 by 32 is 1 and another one is 975 by 32. So, this will give you 27.1 PPM and this would give you 30.4 PPM. So, that means this one is 27.1 PPM and this one is 30.4 PPM. Now, you know that this chemical shift presented in PPM is independent of magnetic field strength. So, now let us go to another problem here very similar problem in a magnetic field of strength 2.349 tesla the resonance frequency of 15 N nuclei is 10.13 megahertz. What is the resonance frequency of 15 N in a magnet of 11.745 tesla? So, that means we are considering two magnetic fields and also we are focusing on 15 N nucleus. So, in 2.349 tesla magnetic field the resonance frequency for 15 N is 10.13. So, now we have to find out what would be this value with a magnetic field strength of 11.745 tesla. So, it is very simple here. So, again you use the same equation h nu equals gamma over 2 pi to b naught. So, h we can remove this one. So, we can remember this way. So, now what we can do is 1 equals gamma. This is constant we can ignore this one time being and b naught 1 and then nu 2 equals gamma remains same it is not given and it is not necessary now. So, we can cancel this one. So, nu 1 by 2 equals. So, let us say this one is given value here 10.1 then we have to find out this one in this. So, nu 2 we have to find out nu 1 b 1 is there and this is also known now nu 2 equals. So, here this would come around approximately 5. That means if you just simply look into the ratio of this one this is 5 times more. So, that is what we are getting here and then 1.3. So, this will give you 50.65 hertz. So, this is megahertz. So, this is how you can calculate it is very simple now even if you take the ratio of two magnetic field and we know this 5 times. So, that obviously if it is 10.13 into 5 it will be 50.65 will be the corresponding frequency for a 15 n of magnetic field strength 11.745 tesla. Let us see one more problem in NMR spectrometer commonly used in medicine the resonance frequency for the protons in water is 60 megahertz. If such an instrument was to be used to observe 31 p what frequency of radio radiation would be required. So, this is again a simple question. So, here water it is 60 megahertz and then we have to find out what would be the frequency for phosphorus if the field strength is 60 megahertz. If the frequency corresponding to hydrogen is 60 megahertz and then if such an instrument was to be used to observe 31 p what frequency of radio frequency would be required. So, it is very simple the what information that is missing is gyromagnetic ratio is not given gyromagnetic ratio for phosphorus is 10.841 for phosphorus for hydrogen it is 26.752 this is very important this is not given without this it is very difficult to find out. So, let us see B is constant magnetic field is constant this we need at least here. So, now we should be able to solve this problem. So, B is 10.841 and then this is 26.752 into 60. So, this will come around 24.31 hertz. So, that means here if you just consider this will be approximately 0.0 this would come around approximately 0.405 or something it comes and then if you multiply this one by 60 you should get it. So, that means corresponding frequency for phosphorus in a known magnetic field if the what has a 60 megahertz then the phosphorus will be having 24.3 megahertz. So, 1 h as 1 h 60. So, this corresponds to 31 p equals 24.31 megahertz. So, now we have another problem here I have also given the solution here the magneto gyric ratio of the deuterium 2 h R D nucleus is approximately 6.5 times smaller than that of the proton. So, in a magnet where 1 h spectrum can be observed at about 400 megahertz what is the approximate radio frequency radiation you would need to observe the 2 h NMR spectrum. That means in a magnetic field strength known magnetic field strength if the 1 h NMR is observed at 400 megahertz. So, then what would be the corresponding frequency for deuterium the question as for a comparison between the frequencies required for the observation of protons and deuterium in the same magnet under the same magnetic field strength the magneto gyric ratio of 1 h is 6.5 times that of 2 h. That means we also know the frequency I mean gyromagnetic ratio also we know 2 h will be. So, it is simply what you can do is 2 h frequency equals 1 h frequency by 6.5 6.5 is the ratio of gamma 1 to gamma 2. So, that comes around 61.54 h. So, this is how we can look into the problem here. So, let us look into one more interesting problem before I conclude this lecture. This already I discussed I believe while looking into NMR while discussing NMR spectroscopy. The structure of tertiary butyl lithium is similar to that of methyl lithium and you should know the fact that both exist as tetraomeric compounds, but with each h atom replaced by a methyl group tertiary butyl lithium is very similar to methyl lithium where lithium is replaced by tertiary butyl. So, 13 CNMR spectrum of a sample of tertiary butyl lithium 4 times prepared from 6 lithium metal. So, here we are taken exclusively 6 Li isotope consists of 2 signals one for the methyl carbon and one for the quaternary carbon atom. The signals for the quaternary carbon is shown below at two different temperatures one this one is taken at 185 K and this one is taken at 299 K. That means there is a dramatic difference is there in the multiplied pattern at recorded at 185 K and 299 K. So, how these signals arise and then the further information is 6 lithium i equals 1. Now, we are looking to 13 C spectrum. So, for this one one should know the geometry of tertiary butyl lithium I have shown you that one in here. So, let us look into this one here the red ones are lithium and these are all tertiary butyl one carbon having 3 tertiary but something like this. So, now you can see and they are occupying alternate carnos lithium and the tube butyl carbon are occupying alternate carnos of cubane and then you can imagine this one like a lithium tetrahedral similar to white phosphorus and now each triangular faces of 3 lithiums one tube butyl carbon is interacting and then this leads to 4 centered 2 electron bond. Go back now at 185 K we are seeing 1, 2, 3, 4, 5, 6, 7 we are seeing and then at 299 we are seeing 1, 2, 3, 4, 5, 6, 7, 8, 9 we are seeing. So, now this we have to explain. So, that means at this temperature what happens if you assume lithiums like this in one of the place and then we have the structure is static as a result of this carbon is confined to these 3 when it is confined to these 3 what would happen it is interacting with these that means if I use 2 Ni plus 1 rule here. So, 2, 3 into 1 plus 1 so 7 comes 7 is coming here but on the other end at 299 because of flexionality what happens this carbon will be moving to another and then this one this one as a result what happens. So, each carbon will look into 4 equivalent lithium atoms. So, in that case if we use 2 Ni plus 1 this is because of dynamics flexionality now 2 into 4 are there. So, this 9 lines that is the reason we see at slightly higher temperature we are seeing 9 lines whereas at low temperature it has a static structure each carbon is linked to 3 lithium atoms and at 299 K due to flexional process quaternary carbons is 4 equivalent lithium atoms hence we see 9 lines this is a structure one should remember about that one you can see here how this you can visualize each methyl group or tributyl group confined to one of the 4 triangular phases of 3 lithium atoms with this let me stop here and continue you know more problems in my next lecture until then have an excellent time. Thank you.