 Once again welcome you all to MSP lecture series on interpretive spectroscopy. This is 54th lecture. In the last couple of lectures, I started solving that of problems and giving different kind of solutions to understand interpretation. So, let me continue from where I had stopped. Let me take some more interesting problems associated with IR mass spectrometry as well as NMR. Now I have a question fully based on IR data. You can see here determine the structure or possible structures for a compound with formula C5H10N2 and the spectra spectrum is given here. And as I mentioned first you should look into the molecular formula here and then once after looking we should find out hydrogen deficiency index. So, that formula also I am sure you are familiar with it. So, if you just consider here 6 this will be 5 here nothing is there plus 1 here. So, now 2 is the deficiency that means possibly there can be a ring or 2 double bonds or there can be triple bond. So, let us look into it now. First let us try to analyze there is a signal in the region 23 that means N is also there and there is a signal means you can think of something like C triple bond M nitrile group and then this would also give you about presence of CH and then we have several peaks are there fingerprint region it is very difficult to identify without much knowledge and also we can see nothing is there in the 1600. So, that means possibly there is no aromatic group here if the aromatic group is not there then possibly 2 or there means there should be a triple bond as well that information has come from here. So, now you read out some of the points I have made about this problem when only minimum data is given the index of hydrogen deficiency is calculated just I showed you it is 2 a glance at the spectrum shows a prominent peak near 2250. So, it this has to be for nitrile C triple bond N or an alkane C double bond C and alkane would appear closer to 2150 centimeter while a nitrile would appear at the value of observed in the spectrum. So, that means now C double bond C is ruled out. So, it should be a C double bond N or triple bond N since the index of hydrogen deficiency has a value of 2 that it fits a compound with a triple bond triple bond is there the value will be 2 the remaining nitrogen atom would like to be an amine. So, in that case what happens if 1 is going as a nitrile group say triple bond N other one should be yes NH 2 should be there in it unfortunately the region between 3600 to 3200 does not reveal easily what type of amine may be present. So, that means if NH is there we could see a prominent peak between 3200 to 3600. So, that is not there. So, the region is often obscured by the presence of water in the same or from weak volt on peaks from other parts of the molecule. However, a primary amine should show a prominent doublet while the secondary amine should show a singlet. So, since no prominent peak appear between 3600 to 3200 we can conclude that this is a nitrile with an attached tertiary amino group. So, that means possibly it is not primary amine or secondary amine it should be tertiary amine. So, in that case the structure is going to be something like this. So, now we have dimethyl amino group is there and then we have a CN is there and now for curiosity I have also taken 1 H NMR for this one and 1 H NMR you can identify 1 2 3 different type of signals here and we have 3 signals are there and then if we just look into these 2 these 2 should show a singlet that comes here and then this would show a triplet and this would show a triplet. So, CH2 both are CH2. So, both of them will show a triplets 2 triplets are there. So, that is it. So, problem is solved now this is a structure and this nicely fits into this 1 H NMR spectrum shown here and also for curiosity I have also taken 13 C NMR spectrum for this molecule here and if you look into 13 C you identify how many non-equivalent signals are there. So, these 2 are integral 1 2 3 4 should be there we have 1 2 3 4 and 4 can be assigned very nicely and this is 1 1 9. Astronaut this comes here and then these 2 will be here because they are on N and then this is close to N. So, it comes here and then this is coming here we have confirmed with the same molecule when you take. So, it gives the expected spectrum in case of 1 H as well as 13 C. So, this is 3-dimethylaminopropane nitrile I have 4 different molecules are there for each of the compounds indicate the number of actually distinct groups of carbon and hydrogen atoms. It is not to specifically tell the chemical shift values very difficult ok. Nobody would ask you all the information only thing is you should be able to tell how they are going to split or at least as simple as how many different type of you know distinct groups of carbon and hydrogen are there. For this one what we should do is we should go for symmetry here. For example, if you consider a C 2 axis of rotation. So, that would make this one this one identical and this one this one identical and here we do not have any hydrogen. So, what would be? So, H 1 if you consider 1 H NMR 2 signals will be there and then if you look into 13 C for this molecule again with the same analogy. So, there is 1 here 2 here and 3 here and this side it is identical. So, that means, in 1 H NMR spectrum of this molecule you expect 2 different type of signals for hydrogen atoms and then in case of 13 C one should expect 3 different type of single resonances. So, here. So, this is the answer for this one. Now we will look into this here we have CH 3, CH 2, CH 2, CH 2. So, CH 2. So, if you just look into this one it appears like only 2 type of signals are there that is not the case here. For example, this one if you see here it is surrounded by 2 methylene groups and this one is also surrounded by 2 methylene groups whereas, this one is surrounded by 1 methylene group and 1 methyl group and same thing is true for this one. So, that means, these 2 are of 1 type and then this and this will be of another type and this one and this one will be another type. That means, 1 H NMR should show here 3 signals and also similarly the same rule appears here 1 2 3 identical ones are there 13 C also shows 3 distinct signals for this molecule here and of course, if you are curious you can just see here it can show a quintet and this can also show a quintet identical whereas, this one will show a triplet further coupled with this one this would show a quintet here. I have not plotted NMR for these things because I am asking a different type of question here. Now, let us look into this molecule here. So, of course, this one one can do like this 3 C 2 axis of rotation one can consider this is a symmetric molecule here. So, here if the CH 3 are there this is 1 type of group here and then we have another one here that means only 2 type of signals you can expect for this molecule here all CH 2 methylene groups are identical and all methyl groups are identical here. So, you get 2 signals and in case of 13 C 1 2 3 13 C we expect 3 because 1 here 1 here and 1 here. So, that means, quaternary carbon methyl carbon and methylene carbon 3 signals are anticipated for this one then here again you can see C 2 axis is there and this will be identical. So, this will be identical this will be identical and then this is one more here. So, this one so that means, here also you can see 3 here and then if you look into carbon 1 2 3 4 5 different type of carbons you can expect here. So, 1 2 3 4 5. So, 5 different carbon environments are there here as a result we are expecting 5. So, this is the answer for these questions the first one will show 2 distinct group of hydrogen atoms and 3 distinct group of distinct carbon signals and the second will show 3 hydrogen and distinct multiplets and then 3 for different carbon atoms and in this one 2 and 3 here 3 and 5. So, minimum knowledge of symmetry and looking into simple operation rotation rotation reflection you should be able to guess the magnetically non equivalent or magnetically and chemically equivalent nuclei very readily and assigning and understanding would be very easy. Now, one more example here let us look into this example here again similar problem here. So, this molecule is there again for this molecule also you can look into C 2 axis of rotation is there here C 2. So, as a result what happens this would be identical and this one and this one. So, you you can see here 3 hydrogen atoms 3 will be there and in case of carbon 1 2 3 4 4 signals are anticipated here 1 2 3 and 4 4 carbon signals are expected and 3 hydrogen signals are expected here. Whereas in this case again very similar here in case of this one this is very similar to the previous one we discussed here. So, we have H here and then this one here. So, that means 2 signals are anticipated in case of 1 H N M R and also here 1 2 3 signals are expected in case of 13 C 3 signals are expected in case of 13 C. And now if you look into this here hydrogen this is distinct and this is distinct this is distinct this is distinct. So, we expect 4 all are different we expect 4 different signals for example this will show 2 doublets this will show a singlet and this will show a singlet. And then 13 C if you look into it 1 2 3 4 5 6 7. So, we expect 7 distinct 13 C signals for this molecule here. If you take out 1 chlorine here let us say if it is not there then it is much simplified. So, let us look into another set of molecules here same analogy here for each of the compound we have to identify distinct group of carbon hydrogen atoms in their respective spectra 1 H here and 13 C again. So, here these two are identical and then we have 1 H and then these are CH 3 they are all identical. So, we can expect for this molecule 3 and then if you look into carbon 1 2 3 and 4. So, 4 we are expecting. So, quaternary carbon is there and here we have another carbon and we have 2 carbons are identical. So, methyl all methyl's of this one is identical and then we have 1. So, we have 1 2 3 4 4 signals are expected for this one and in case of here we expect again these two are equivalent and then these two are equivalent. So, we expect 2 signals and then 13 C 1 2 3 we expect 3 signals here next for this molecule here again very nice you can see C 2 axis of rotation is there. So, this one this one is same this one this one is same and this one this one is same. So, we are expecting here also 3 signals here and then if you look into carbon 1 2 3 4 signals we can expect here ok. So, 4 signals only thing is little bit of symmetry and understanding symmetry operations if you are familiar with we can see similarities between different fragments here and then anticipating or identifying distinct groups should be a problem here. In the first one we have 3 hydrogen signals in its 1 H N M R and 4 signals in 13 C whereas, in case of 2 here 2 and 3 each and in case of this one 3 and 4 each we can expect here. So, now another interesting question is there following 3 compounds are put in 3 unlabeled vials and some information is given about their 13 C N M R spectra match the spectral futures with corresponding compound from the list. So, what are those compounds provided 1 2 3 trichlorobenzene 1 2 full trichlorobenzene and 1 3 5 trichlorobenzene all are trichlorobenzene, but positions of chlorine atoms are different. So, I have written all 3 substituted chlorobenzene here it is 1 2 3 1 2 3 here it is 1 2 4 here it is 1 3 4 1 5 here 1 3 5 here. So, now what is the question here 2 peaks between 125 to 140. So, we have to identify now if we just look into the different distinct carbon signals are there for each one then answering this question should be a problem. Now, if you look into this one of course, you can see here a rotation is there. So, as a result what happened this is 1 this is 2 this is 3 and this is 4 signals can be seen here and in this case what would happen. So, 1 2 3 4 5 6 all 6 are different all 6 are different all are different whereas, in this case we will see only 1 2 2 signals here. So, that means, 2 peaks between 1 2 5 125 and 140 will be for 1 3 5. So, you can take this one here next 6 peaks between 125 to 140 ppm is for 1 2 and 4 highly unsymmetrical this one the last one is very easy now identifying, but still we can analyze. So, 4 peaks 4 peaks should be 1 2 3 here this is how we can answer these questions very easily provided we write the structure and look into how symmetrical the molecule is to identify equivalent magnetically and chemically equivalent nuclei. So, now another one now following 3 compounds are put in 3 again unlabeled wires and some information is also given about the 13 CNMR spectra. Mat the spectral features with corresponding information provided about each one. So, now we have 4 tetrachlorobenzene. So, tetrachlorobenzene also we are considering 3 molecules having different positions of chlorine in these molecules. So, first let me write 1 2 3 4 this is 1 2 3 4 next 1 2 3 5 1 2 3 5 here next 1 2 4 5 say 1 2 4 5. So, now let us look into the symmetry where any symmetry elements are there yes we can consider here C 2 is there in this fashion, but it does not really degenerate whether we consider like this here yes it is possible. So, this is ruled out whereas, this one is possible here. So, that means basically 1 2 3 we have 3 identical ones are there here 3 as far as 13 C is concerned 1 2 3. So, 1 2 3 will be so 3 will be there here next if you consider this one here still we can see some symmetry here. So, with this one we can see 1 2 3 4 are there here 4 identical ones are there 4 13 C signals will be there and then in this case you can rotate this way. So, 1 2 are there here so 2 are there here. So, that means basically the 2 peaks between is for this one 1 2 we take 1 2 3 4 5 1 2 4 5. So, this one we can say and next 4 are there 4 means this should 1 2 3 5. So, this one and then with this one 1 2 3 4 signals are there. So, this one is 1 2 3 4 are there so this is 1 2 3 5. So, this one will be 4 signals it should be 2 3 5 that is done. So, now for this one 1 2 signals 2 signals will be there this is for 1 2 4 5 the last one is this one is left here obviously 3 peaks here this will be 1 2 3 4 1 2 1 1 2 3 4 will be this one. So, we can identify in this fashion here. So, let us look into one more example here free the one head splitting pattern for the hydrogen in the red color in the following compounds. So, here this one is the red one. So, this one should show a septate here because 6 equivalent will be coupled and whereas, this one because of one side nitrogen is there other side oxygen is there it is not coupling it will show a singlet and whereas, in this case this will be showing doublet and then here they will be showing both are coupled to equivalent one this will be quintet and then here this will be showing a quadrate and then this will show a triplet. So, let us look into it now a septate yes and then a singlet this is a doublet and then this is again a triplet and this is a quadrate and this is a quintet. So, like that we should be able to predict it. So, let me come up with more examples in my next lecture until then have an excellent time.