 Hello everyone. Once again, I will welcome you all to MSP lecture series on interpretive spectroscopy. This is the ninth lecture in the series. In my previous lecture, I was discussing about the second-order coupling. Second-order coupling we come across more often with 1HNMR because of small chemical shift range we observe in 1HNMR. It's 1 to 10. As a result, what happens? Especially when we have chemically equivalent and magnetically non-equalant protons present on same carbon, that is, geminal coupling arises. Whenever geminal coupling arises, we come across second-order splitting and also when we have two methylene groups on either side, when they are substituted in a different way, we come across, again, difference in magnetic properties of hydrogen atoms present on same carbon. As a result, we come across second-order splitting. So, let me continue from where I had stopped. So, this is the spectrum of styrene recorded at 60 MHz. You can see the peaks are little different compared to what we see when the same spectrum is recorded higher field. So, here you can see some sort of pyramidal effect. Pyramidal effect means when you see multiplets of different lines, if they are in the pyramidal shape, increasing intensity on both sides, resulting in some sort of tower type. So, that is called pyramidal. That can give you hint that this is not a simple first-order spectrum. This is a second-order spectrum. One such example is shown here in case of styrene, olefinic region I'm referring to. Here, if you just see the interaction of HA is not same with the interaction of HB and as well as HC. As a result, we come across one is cis coupling, one is trans coupling and also the coupling between B and C. So, here it is a typical ABC type spin system and the advantage with higher field is the chemical shifts move further away. If you increase the magnetic field as a result, complicated second-order spectrum would become simplified first-order spectrum. There is the advantage of recording NMR at much higher field. So, in this one the peaks are pyramidal in shape, can be seen in all ABC and ABX spectrum. Olefinic region of monosubstituted alkenes, monosubstituted alkenes exhibit often ABC or ABX spin system with peaks having pyramidal shape, but spectra taken at higher field strength resemble AMX spin system. Also in my earlier lecture, I showed you how a ABC become ABX and eventually it becomes AMX spin system in the case of styrene itself. For example, you can see here, this is recorded at much higher field. Just focus your attention to this olefinic region. All are having very clean doublet of doublet pattern because of coupling with different protons in a different way with different magnitude of coupling. For example, if you consider HA and HB, this is the higher coupling and first that will show a doublet and then this will be coupled with cis one and to see doublet of doublet and correspondingly HB and HPC also showing the same fashion. So, this is for HB and this is for HC. You can see here the values also is presented here 5.76, 5.25 and 6.72. You can see clearly 6.72 corresponds to this one where we will see this trans coupling first doublet and then it is further split by cis coupling to give another doublet. So, it appears as a doublet of doublet in the same fashion this one is at 5.76. This will show first couple with this one and then it will couple with this one. This magnitude is very small. That is the reason you can see the space between the two lines is much smaller whereas this one is due to this. This coupling is first appears as a doublet and then it further appears like a doublet of doublet. And if you see the coupling whether you measure from the chemical shift of this one to this one or this one to this one that should be identical. And similar spectrum is also shown here recorded at 90 MHz. Again you can see pyramidal effect. This clearly shows that it is also a ABC spin system. But if we record this one at 400 MHz it can be further simplified to appear as AMX spin system like what I showed in my previous slide. So now I have another interesting molecule here. In this one you can see again HA and HBR there and HC. This is a typical ABC spin system. You can see I have three figures listed here. The figure once is simulated first order spectrum. This one the second one is simulated spectrum with different coupling values are given. JAB equals 18 and JAC equals 2 Hz and JBC equals 10 Hz. And then actual ABC a portion of this molecule is shown here. This is pretty second order is there. But when you take it to higher field it appears like a simple AMX spin system showing doublet of doublets in each case here. And of course here you can see again pyramidal effect is there and it is much simplified when you go for higher field. That means the figure one is simulated first order spectrum whereas figure two is simulated second order ABC spectrum and figures three is the actual spectrum. So among different types of second order spin systems we come across that is A2X2, A2B2, AA prime, XX prime or AA prime, BB prime or A2X2. So among all those things A2X2 is the least common spin system and one such system is that of difluoromethane. And each hydrogen is coupled equally to each fluorine and hence it is A2X2 in case of difluoromethane is a typical example for A2X2 spin system or coupling system. And one should remember is if we have a molecule having two methylene groups adjacent to each other in the middle and substituents with similar electronegativity at both the ends. That means here as well as here if we have groups with similar electronegativity that would show A2B2 spin system. One should remember that. Now let us try to understand few terms so that understanding their futures would become rather easy. Diastereotopic protons in organic chemistry must have studied all those things but nevertheless let us refresh our knowledge and understanding about these typical terms. Diastereotopic protons occur when there is a chirality center already present in the molecule. If you see here this already chiral center is present in the molecule. In the molecule diastereotopic protons are chemically non-equivalent and hence they show different chemical shifts. I am referring to HA and HB and replacing them results in compounds which are diastereomers. If you replace them with something else then that results in diastereomers. And what is isotopomers? Isotopomers have the same number of atoms of each isotope but in a different arrangement. Isotopomers are isotopic isomers are isomers which differ by isotopic substitution. For example if you take CH3 OD here H is replaced by D and if you compare this one with CH2 DOH number of hydrogens and deuterium remains same here but the substitution is at different place. So and again you can see here D comes here and again here we have this one among these. These are all called isotopomers. And what are diastereomers? The diastereomers are the stereoisomers that are non-identical and do not have mirror images. As a result they are non-superimposable on each other. If you do not have identical, if you do not have mirror images then they are non-superimposable on each other and such isomers are called diastereomers. Now for example you can see here this portion if you compare and this is non-superimposable mirror image but on the other hand if you consider this portion here that is not. So these two are called as diastereomers. And what are the enantiomers? A pair of molecules existing in two forms that are non-superimposable mirror images that are enantiomers. So now let us look into a typical AA prime BB prime spin system. As I mentioned we give the terms AA prime BB prime when A and A are magnetically non-equivalent but are chemically equivalent. Similarly BB prime is given here which are chemically equivalent but are not magnetically. So this is A and this is A prime, this is B and this is B prime. And although same J values were used in while simulating for ortho, meta and para chemical shift differences was reduced to 40 hertz. They are quite different structures as well one is para and the other one is ortho. For example if you see here simulated AA prime BB prime spectrum for this molecule here. In all cases the coupling constants are taken as 9 hertz, 3 hertz and 1 hertz respectively. And here delta A is 630 and delta B is 700 hertz whereas in this case delta A equals 630 and 700 hertz. But you can see the pattern is very different and again you can see this pyramidal effect can be seen here. So this shows that AA prime BB prime spin system. And another example is there and here again these two look identical yes they are identical but they are chemically equivalent but they are not magnetically. So as a result using AA prime and BB prime. The moment you see this kind of prime, the prime indicates that both of them are chemically equivalent but they are not magnetically. So in this case what we get is AA prime, BB prime spin system. This is another example Bromo Aniso. The reason is if you consider the coupling of this one with this one is different from coupling of this one with this one. And similarly the coupling of this one is different from coupling of this one. Because they are the bonds if you consider they are farther away as a result although they are chemically equivalent. Their interactions are different and hence that results in a different spin system, secondary spin system that we call here as AA prime, BB prime. Another example here for Parabromo chlorobenzene. Now let us look into hydrogen and carbon chemical shifts for better understanding. I have given here for comparison for the same functional groups both 13C and 1H chemical shift ranges I have given here. If you just see here aldehyde group and carbonyl group it comes around 9 to 10 in case of 1H NMR whereas in case of 13C it comes around 200 ppm. Similarly if you take ortho hydrogen where ortho you have halogen it comes in this 7 to 8 and it comes somewhere here. And then if you take olefinic hydrogen that comes between 6 to 5 whereas here it comes between 100 to 150 where it is aromatic carbon as well as olefinic carbon. And then if you see a carbon having hydrogen as well as halogen or oxygen it comes between 3 to 4 ppm whereas in this case it comes around 50 to 90 ppm. And similarly if you see a carbonyl group is there adjacent to another carbon having H it comes around 2 ppm. Whereas here we will see that one around 50 ppm. Similarly all cans would come around 0 to 30 ppm in case of 13C NMR. So let us look into the general information about 13C NMR. When we look into 13C we have to consider carbon has two isotopes 12 carbon that is nonzero nuclear spin is only 13C in case of 12. Overall it has no magnetic spin that is I value equals 0 that constitute about 99% and remaining 1% is 13C that has magnetic spin of half but is only 1.1% of carbon in a given sample. So the gyromaniac ratio of 13 carbon is one fourth of that of hydrogen. The signals are very weak difficult to identify as they often get lost in noise. If enough quantity of sample is not taken then especially quaternary carbons will be merged with noise as a result we may not be able to see such signals. That means we have to take more sample and hundreds of spectra are taken and averaged to get a single refined spectrum. I have shown here 13C and 1H NMR spectra of 1 p roll to carbohydrate. And you can see here the aldehyde proton comes around 9.4 ppm whereas in case of 13C it comes around 180 ppm. And we have here these two are assigned to you can see here for CH and CH and CH these three protons are there. And then corresponding ones are shown here. And then if you look into the sample is record in CDCl3 in case of 13C what we are getting is a triplet of equal intensity. So why that happens because if you just look into CDCl3 here. So d i equals 1 if you just use this formula to identify how many peaks will be there so 2 into 1 so it gives 3. So here i equals 1 the intensity will be 1 is to 1 is to 1 as a result in case of 13C NMR CDCl3 shows a triplet of equal intensity. Now I have given another example here you can see the same example I discussed in the previous slide. And the corresponding chemical shifts are given here and if you just see here 1, 2, 3 so all are different here. You can see all carbon signals are shown here for example it is 127.1, 111.7, 121.6, 132.4 and then 182.6 so all signals are shown here. You can see the distinct chemical shifts for each carbon atom if they are chemically and magnetically non-equivalent. And if you just look into 1H NMR spectrum you can see typical spectrum is shown here. Here are the differences between 1H and 13C techniques when you are running NMR. So resonance frequency is 1 fourth of 1H frequency since gyromagnetic ratio is 1 fourth and gyromagnetic ratio is directly proportional to normal frequency. As a result it is very easy to calculate by just looking into the ratio of gyromagnetic ratio of 2 individual nucleus with respect to hydrogen. For example if you take 60 MHz 1H NMR the corresponding magnetic field strength will magnetic field will be 15.1 MHz, corresponding frequency will be 15.1 MHz for carbon 13. So unlike 1H NMR peak areas are not proportional to number of carbons here. Carbon atoms with more hydrogen atoms are strongly and if carbon do not have hydrogen atoms then absorption is very weak. As a result it is very difficult to identify such samples especially the nice ratio is very high. That is the reason we have to take larger amount of sample to get to identify those quaternary carbon atoms or carbon atoms without having hydrogen atoms directly attached to it. So now let us look into spin-spin splitting. So it is unlikely that a 13C signal would be adjacent to another 13C. So splitting by carbon is negligible. That means carbon-carbon splitting is not seen unless it is enriched with 13C isotope. Because what happens if we have two neighboring say carbon atoms are there and it is highly unlikely that in a given set of molecule both of them will be 13C, 13C. It is almost like taking in two bottles 100, 100 marbles 99 of one color and one of another color and then if you mix it and if you keep on shaking you will come to know how many times these two colored ones come together. So if you look into that probability that probability is very very low. As a result we do not observe 13C, 13C coupling in a normal molecule where it is not enriched. Of course if you can enrich and record then we can see 13C, 13C coupling. 13C will magnetically couple with attached protons and adjacent protons. However if carbon is attached to hydrogen or other NMR active nucleus such as fluorine or phosphorus you can see coupling between them. These complex splitting patterns are difficult to interpret. If we take an organic molecule we have numerous hydrogen atoms are there and if we see CH coupling with two bond coupling, three bond coupling and other things. The 13C NMR spectrum would look very complicated and then assignment of signals would become very difficult. As a result what we do is we do the decoupling. So protons spin decoupling. What we do in that one is to simplify the spectrum protons are continuously irradiated with noise. So they are rapidly flipping that means we are not giving enough time for protons to interact as different the way they were in the molecule. Because of continuously irradiating them with the Raman frequency corresponding to those hydrogen atoms. The carbon nuclei see an average of all possible protons spin states as a result it does not couple with any of them. Thus each different kind of carbon gives a single unsplit peak and this we call it as proton decoupling often we show in this fashion. If 1H is written in flower bracket next to 13C or anything this means it is 1H decoupled 13C NMR spectrum. So half resonance decoupling. So 13C nuclei are split only by the protons attached directly to them or by any other NMR active nuclei such as 31P or 19F. The N plus 1 rule applies here a carbon with N number of protons gives a signal with N plus 1 peaks. Or you can also use 2NI plus 1 peak if you want to add the spin nuclear value I. Then interpretation of 13C NMR spectra. So the number of different signals indicates the number of different kinds of carbon that way it is very simple. And when you decouple it becomes even more simplified. The location indicates the type of functional group as we have a distinct chemical shift for each functional group. It is very easy to identify what kind of functional groups we have in a molecule by just simply looking into the chemical shifts. The peak area indicates the number of carbon if integrated normally we do not do that. The splitting pattern of half resonance decoupled spectrum indicates the number of protons attached to the carbon. The splitting pattern of half resonance decoupled spectrum indicates the number of protons attached to the carbon. Now we will look into 13C NMR spectra taken with and without proton coupling. For example, for this buton on this ketone NMR 13C NMR is recorded both with coupling without coupling. Without coupling you can see we have 1, 2, 3, 4 different carbon atoms are there. You can see here 1, 2, 3, 4 different carbon atoms are there. In the coupled one you can see this methyl group is coupled with 3 protons. As a result it will be a quadrate we can see here. And then this methylene is coupled to 2 and we can get a triplet here. And then again this is coupled to 3 hydrogen atoms again a quadrate here. And this one has no directly attached to hydrogen as a result it shows one singlet here. So, this of course here it is little less complicated. So, we can take coupled one there is no harm. But when we have a complex molecule a protein molecule or biological molecules. Then it is advisable to record decoupled spectrum for understanding and elucidation of the structure. So, let me continue in my next lecture. Thank you very much.