 I once again welcome you all to MSB lecture series on interpretative spectroscopy. So today again, let me continue discussing some of the important problems pertinent to all these spectroscopic methods. How to analyze the data and how to interpret the data and elucidate the structure of unknown sample. So now problem 40 is here. UV spectrum of this compound shows only end absorption. You should remember that. Determine the structure of the compound using the following data. So here what are the data we have given? Mass is given. Mass parent molecular ion peak is given and also 1HNMR is given. And then IR spectrum is also given and also 13C is also given. So with this, we should find out what molecule we are referring to. First let us look into 102 here molecular mass. From this one, we have to try to arrive at tentative molecular formula. And then we have to look into saturation and unsaturation if there are any. And we have to proceed that way. So let us begin with 102 here. So if 102, first we shall divide it by 13. So it becomes 91 plus 11. So that means basically C7 and H will be 7 plus 18. And then if you go back to the spectrum here, you can see we have a very strong peak at 1740. And also we have peak around 1200. That also we should remember. This indicates we have both CO as well as C double bond O. That means carbonyl is also there and yester group is also there. So that the information comes with this. Now let us add one oxygen here. When we add oxygen one, we have to take out one CH4 group. So here it becomes C6H14O. Since both carbonyl and yester group is there, it has to be COO. So we have to account for one more oxygen. So we will take one more 10O2. This is going to be, let us assume this is going to be the composition, molecular formula here. And now with this, assuming this is correct, let us look into hydrogen deficiency index. That we know the formula C plus 1 minus half H because other heterotoms are not there, nitrogen as well as heterogens are not there. So if we take here, this is 5 plus 1 6 minus 5. So it shows 1. So hydrogen deficiency is 1. That indicates there is only one double bond is there. In that case, it looks like a linear hydrocarbon. So now we have to write the structure. Before we write the structure, we should go back and analyze again. We have two quartets are there and two triplets are there. And that means if we are seeing a quadrate that has to be next to a methyl group. And same thing, that means we have two methyl groups are there here. In that case, and again if you look into 13 CNMR, we have 1, 2, 3, 4, 5 distinct carbon groups are there. That means we have to think of something like this. So this 5 are there. And then extra group is also there. So we will put here something like this. And then you put another one here and see whether this matches. If this is a correct structure, then this would show a quadrate. And this will show a triplet. And then this will also show a triplet. And this will also show a quadrate. But this is little bit more D shielded compared to this one here. So let us see whether that one is D shielded or not. We can see here. This is D shielded. And then this one is here. And we two triplets are there. And then of course we look into this one. We also saw this peak as well as this peak. And then we saw five of them. And then one of them carbonyl is here. We can see quaternary carbon. So from this one, we can say that this is the correct structure. So this is how we can use all available information starting from rule number 13 and then hydrogen deficiency index. And then looking into any heterotoms are there and looking into functional groups such as CO, NH or OH. And then we can arrive at the structure and interpret the data what we have obtained from actual spectrum. Now let us look into another example here. So here a compound, composite of carbon, hydrogen and oxygen has a molecular ion at MZ decos 90. This is important in its mass spectrum. The base peak is at MZ decos 45 atomic mass units. The infrared spectrum shows a strong absorption in the 2840 to 2980. And a very strong absorption from 115 to 125. And one HNMR shows two sharp signals at 3.40 and 3.55 in the intensity ratio of 3 is to 2. And then 30 NMR also has two signals. So we have to identify what is this one. First let us take 90 here divided by 13. We take 78 and 12 will be there. 12 means it becomes here what we get is 18 here. So from this one oxygen is there. Let us remove one oxygen here. It becomes 14 O because we have to remove CH4 that is equivalent to 12 plus 16 oxygen. Let us try to remove one more to see what would happen. See 4 H10 O2 we got here C4 H10 O2 here. And then when we look into these two chemical shifts here we have 3.40 on 3.55 3 is to 2 ratio. So that means basically we should divide this into 6 and 4. So that means probably we have two type of hydrogen atoms are there. And of course here if you look into hydrogen deficiency, let us look into see whether we have any hydrogen deficiency here. So 5 minus 5 equals 0. So no unsaturation is there in this one because it is a hydrogen index is 0. So now we have to look into the spectrum. So it has two carbons are there. 6 plus 4 the reason why I have taken this one is if you just look into the spectrum is not given. If you see both are much more deshielded. If simple CH3 is there next to carbon that should come around 1 to 2. So since they are more deshielded that means they are probably next to electronegative atom. Here electronegative atom is oxygen. So now if you take here CH2, CH2 is there and now they will show a singlet and then this this will also show a singlet and the ratio is 3.40 on 3.55 and this should be the structure of the molecule here. And then this would corresponds to 90 here. So this is called dimethoxy ethane 1 2. So first what we should do is we should look into the mass. From mass we can calculate like this and whatever the reminder is there add reminder to the quotient that would give you total number of hydrogen atoms present. What we get is C6H18, C672 plus 18. So that is correct 90. And then since oxygen is already informed that oxygen is there one oxygen can be added and it becomes C5H14O. Add one more to see what would happen C4H10. Now the 10 is there. 10 if you see if you can make into 3 is to 2 ratio automatically it becomes 6 and 4. 6 and more probably it can tell you 2CH3 and 2CH2 units are there in it. You can analyze like that. And then we look into here we have two signals are there one is 72, one is 59. So that means two type of carbon atoms are there and then they are all much deshielded. As a result we can conclude that this is the structure here. This actually the all the information that is provided is about 1 2 dimethoxy ethane. Yes it is here. So even I have given one HNMA spectrum this one here you can see two signals are there one is at 3.4 and another one is at 3.55. 3.55 is for methylene and then 3.40 for methyl group. And then 13C also shows two signals one is around 59 and another one is about 72 ppm. So this proves beyond any doubt that the data pertinent to 1 2 dimethoxy ethane is given here. The question is about identifying 1 2 dimethoxy ethane. Let us look into one more example here. A compound used as a moth repellent has three molecular ion peaks at M by Z equals 146 that is 100 percent and then 148 65 percent 150 10 percent atomic mass units in its spectrum. A pair of smaller peaks are seen at 111 and 113. The infrared spectrum shows sharp absorption just above 3000 and also at 1480 characteristic of an aromatic group you should remember and 1 HNMR shows a sharp singlet at 7.2 ppm. So 7.2 ppm can be assigned without any ambiguity to aromatic hydrogen atoms sharp singlet. That means we have only one type of hydrogen atoms are there and 13C NMR signal shows two signals at 133 and 130 almost very close with this information we had to find out what it is. So molecular formula is 146 again we can divide by this one one. So another 113 that means we get 11 and quotient will be 3 11 plus 3 will be there equals C 11 and H 14. And since we are referring this molecules to be moth repellent that means usually moth repellents have chlorine in them they substitute a chlorobenzene or something like that. So then let us try to remove one Cl, one Cl to remove one Cl we have to look into C 2 H 11, C 2 H 11 means it will be 35 or one can also C 3 H minus 1 or add Cl H to remove 3 carbon atoms. So first let us try to remove one fragment C 2 H 11, so on C 9 H 3 now it goes 11 and then we add Cl. Let us take out one more Cl because we get m plus 2 m plus 4 peak is also there and also this gives this information again. Now let us take this one C 3 H minus that means C 6 and H 4 Cl 2. Let us assume this is the formula. Now let us try to find out hydrogen deficiency 6 plus 1 7 minus 2 minus 1 equals 4. 4 means one ring plus 3 double bonds. So 3 double bonds and one ring fits into an aromatic group here very nicely and now we get 13 C signal only 2 that means we have to make sure that it is highly symmetric in nature. If it is highly symmetric and in what way we can get by putting chlorines on benzene ring to get two signals that is only possible if you put both of them at relatively para position that means 1, 2, 3, 4. 1, 4 dichlorobenzene or para dichlorobenzene we can say 1, 4 dichlorobenzene. So after putting this one let us now look into 7.2 yes now this is highly symmetric because you can do C 2 rotation here along Cl Cl axis and that make this all 4 chemical and magnetically equivalent as a result we get only one signal that is at 7.2 and then this one is there here and this one 1 and these 4 are identical 2 13 C signals we get it and the compound here is 1, 4 dichlorobenzene. See this is the one here. So I have also taken 1, 8 and 13 C NMR to compare whatever the analysis we made along with mass also I have taken here and you can clearly see here 4 peaks are there 146, 148, 150 relative intensity also you can see here and then in this one it shows only one signal around 7 point simulated one shows around 7.4 nevertheless it comes at 7.2 actually. So it is here and then in case of this one 13 C we have two signals here 132 and 130. So you can see those and also this is in 1 is to 2 ratio 2 is to 4, 1 is to 2 ratio. So now our assignment is correct this is dichlorobenzene parodichloro 1, 4 dichlorobenzene. So now there is one more example here determine the structure of a compound with the formula C 10 H 12 O 2 in addition to the IR and 1 H NMR tabulated data for the normal 13 C NMR depth 135 and depth 90 spectra data also provided here. So this would tell you information about CH proton presence of how many CH are there, how many quaternary are there, how many CH 2 are there and how many CH 3 are there this information comes I would be elaborating little bit later. Now let us look into this formula here. From this formula let us find out hydrogen deficiency 10 plus 1 minus 6. So 5 now 5 is the deficiency that means basically we have a ring plus 4 double bonds 4 double bonds are there and then in this case also what we have is you can see here as I mentioned there is a CO group is there and also ester group is also there here. So this indicates clearly and then we have two doublets and we have three singlets are there here that means they are all disconnected or in between we have oxygen atoms and then what we have is in one C double bond O is there and one CO group is there here. So let us look into it now. So now how to write the structure now with this one 13 C NMR also showing 13 C is showing 13 C information is not there 13 C shows this many peaks here whatever the information we have now it is sufficient I believe to arrive at the structure. So now by looking into 1 H NMR you can see here see two doublets and three singlets are there. So these three singlets are disconnected and they do not have any nearby C C bond at all C C bond having hydrogen atoms. So that means one can think of something like this first let me write ring with this one should corresponds to this formula here 1 2 3 4 4 plus 2 6 this is 10 6 7 8 9 10 are there 10 and 12 O 2 this corresponds to this one there is no problem and now these are identical and both of them will show a doublet and similarly these two will show a doublet and then this will show a singlet this will show a singlet and this will show a singlet three singlets are there and two doublets are there these two doublets are in the aromatic region. So from this one we should be able to tell the structure what it is the structure is this one you can see the label is also done appropriately. So A corresponds to this one CH 3 here and then what we have is B B corresponds to methylene group next to carbonyl and C is there methoxy group and then D and E are two doublets they are coupled with the neighboring CH proton. Now what information this gives so as I said this information is quite vital for example this depth means distortion less enhancement by polarization transfer the experiment is used to determine the multiplicity of carbon atoms that is whether they are quaternary carbon primary secondary at tertiary this information and if we do this depth 135 experiment this gives inverted CH 2 and C group CH and CH 3 groups are upright then if we perform depth 90 that produces inverted C group and upright CH the CH 2 and CH 3 are null for some nuclei with a negative gyromagnetic ratio the depth experiment can provide much higher signal to noise ratio than the standard 1D experiment with 1H decoupled nuclear over as a effect. So now for comparison I have shown what are the experiments one can do in this format. So if you see here when you take 13C you get all the naturally you are getting all carbon identified here all carbons will be coming and then to know the nature of this carbon atoms for example if you just take all intensities are same and you don't know which one is which one in that case you perform depth 45 in all these experiments quaternary carbon is not seen they disappear that indicates one information and then with d45 apart from quaternary carbon we see all CH CH 2 CH 3 with depth 90 we see only CH whereas in case of depth 135 CH and CH 3 are seen whether CH 2 is inverted this vital information. So that means this is used to extract three subs spectra of CH CH 2 and CH 3 signals from a series of experiments. These experiments can be run as either 45 depth 90 or depth 135 the number corresponds to a flip angle of the 1H selection pulse that is experimental manual all these details will be there very easy to perform and depth 45 detects signals of all protonated carbons CH CH 2 and CH 3 with the same phase the depth 90 gives only CH peaks you can see and then depth 135 gives signals of all protonated carbons but CH CH 3 signals are positive whereas CH 2 is negative you can see here the signals of quaternary carbons are absent in all depth spectra by combining these it is possible to determine multiplicity of each carbon signal. So by doing this one we can also assigning the multiplicity how many hydrogen atoms are there and then eventually arriving at the structure would be rather easy very easy to understand this one one can also do provided you have instrument in your institute or college it is a very nice learning process. So with this let me stop here and continue in my next lecture more problems on possibly like this as well as related to UVS bus spectroscopy until then have an excellent night. Thank you.