 Welcome you all once again to MSP lecture series on Interpretive Spectroscopy. So I've been looking into problems and finding right solutions with the combination of data taken from more than one spectroscopic methods. So let me continue solving more problems in this lecture as well. So I have a problem here. Following four compounds are put in four unlabeled vials and some information is given about that 1-H and 13-C NMR spectra. So now we have to identify by looking into the information that is provided for right or appropriate molecules. So here four molecules are there. They're all chloro derivatives of propane, 1-1 dichloropropane, 1-2 dichloropropane, 2-2 dichloropropane, and 1-3 dichloropropane. It seems with the two chlorine and propane, we have looked into almost all four possible isomers. So now some information is also given here. First, what we should do is, let's draw the structure. Once after drawing the structure, we have to identify how many different type of signals are there in 1-H NMR as well as 13-C NMR spectra in each case. First, let us focus our attention on 1-H NMR of first molecule here. This is 1-1 dichloropropane. If you see here, we have all the three hydrogen atoms are different as a result in 1-H NMR, we expect three signals. And also in 13-C, also we anticipate three signals here. And then if you look into most deshielded is this one. So this one, if you look into this here, it should give a triplet coupling with this one. So that information also, let us write it down here. So the lowest field signal is a triplet here, this one. This information, let it be there. Now we shall move on to the second one, 1-2 dichloropropane. Again here, three signals are there in case of both, 1-H as well as 13-C. And if you look into this one here, this is the lowest field signal. This one is coupled equally with these two. As a result, it would show six lines. Or it is, okay, six lines. So now let us look into this molecule here. Of course, you can see rotation is there, C2 axis of rotation. As a result, only half we should consider. So these two are identical. Only one signal will be there. And then two carbon signals will be there here. Only one signal. This is unique compared to the rest of them because 3-3, 3-3, here 1-2. Now let us look into it. Again, we have C2 axis of rotation. And then here, we have two type of signals. Methylenes with chlorines on both sides and this methylene. And then of course, if you look into the NMR of signal of this one, this would show a quintet. And then these two will show a triplet. This should be like this. And then two signals will be there. So this is also, now this one is the lowest one is also a triplet now. So now with this one, we have to identify and take the corresponding chloro derivative from these four columns. So now, for your information, I have provided one HNMR that had simulated as well as 13 CNMR for each compound. You can see here, a triplet is there. And for this one, quintet is there. And then we have a triplet here. This is for 1-1 here. So this one would show, both are coupled equally four. So five lines it is showing. And then this is showing a triplet. And this also showing a triplet here. And then we have 1, 2, 3 signals are there. And then we look into here. This one, of course, this one is the lowest one. This will couple with both of them. As a result, it shows six lines. And then this geminal coupling is there. As a result, you can see here, it is going like this. So there is first second order splitting is there. And then we are seeing for this one, a doublet here. And then, of course, here three are there. One, two, three. And here also, one, two, three signals are there. So now, three signals between zero to five, lowest field signal is a triplet. So that information given is for this molecule here. And similarly here, this gives us six lines. So information provided is this one. And in case of this here, this shows a singlet, one HNMR. And then we see two 13C signals. And in this case, this one show a quintet here. And then these two show triplet here. And then for this one, the lowest one is a triplet. And then two signals are there. That means here, one singlet at 1.2, this is 2 to dichloropropane. And here it is 1, 3 dichloropropane. So now, we shall come back to this. The answer for the first one is 1, 2 dichloropropane. Answer for the second one is 2 to dichloropropane. Answer for the third one is 1, 3 dichloropropane. And then for the last one answer is 1, 1 dichloropropane. So this is how we can use this information, draw the structure, look into the symmetry, and identify how many different distinct groups are there, and then arrive at the answer. It is very simple, right? Now let us look into another example here. With respect to the 1HNMR spectrum of C10H12O2, answer the following questions. So now just look into carefully here. We have 1, 2, 3, 4, 5 signals are there here, apart from TMS. TMS is a reference that is at 0. We have five signals are there. So now, how many discrete groups of proton signals are there? Except reference is, answer is five here. So then what is the multiplicity of the highest field signal? Highest field signal is here, this one highest field signal, most shielded one, close to TMS, that is triplet. So this should be triplet. And then the sample has a singlet at 3.8 ppm. Its value in hertz is, if the ppm to convert into hertz, what we have to do is we have to multiply that by field strength 90. So 3.8 into 90 would give you the hertz, that is 342 hertz. So now what structural feature is suggested by the singlet at 3.8? So 3.8, if you look into it, so this is 3.8. So this one is not coupled with anything. And we have oxygen, probably the options are given here. We have to identify which one. So now, since it is not coupled with any other proton, that means there is some electronegative autumn comes in between two carbon atoms. And here obviously we have other heteroatom is oxygen. So it has to be OCH3. Next predict which of the other signals is coupled to the quadrate at 2.9 ppm. So let's look into quadrate at 2.9 ppm. So we have a quadrate. And if quadrate is there, in the neighborhood, there should be three protons. So for example, if you consider CH2CH3, and if you are looking into this one, this would be a quadrate. And this will be a triplet. So then we have to find out which is coupled with this one. Here coupling constant values are not given, but simply looking into the signals and the separation of the individual lines in these multipliers, we should be able to tell that it is coupled with this one, that is 1.2. So next, predict the number of protons present in each signal, 7.9, 7.92 are there. So that means number of protons here is one, and here also one. And then in case of 6.9, this is 6.9. And then 3.8, 3.83, by integration we should be able to get to 1.3, and then 2.92, and then 1.23. So now with this information, we have to identify what molecule it is. What we can do is, hydrogen index deficiency, we can look into it. So that would say 11 minus 6 equals 5. So 5 indicates one ring is there, plus four double bonds are there. So this information is there. So with this information, and by looking into these five different type of signals, then one ring is there, it has to be aromatic group, something like this. And then we have two identical ones. In the aromatic region, if they are coming, two identical ones are there. That means probably this one, this one, this one, we should consider. These two are one type, and these two are another type. So it goes around four protons. And then CH3, CH2, CH3 is there. So that means this one here, 3. And then here, this one is there. That indicates there may be something like, so another one should be something like this. If you count now, so 12 are there, H12, and then O2. And then we have C10. So that means here, 1, 2, 3, 4 double bonds are there. And then one ring is there. So this also satisfies rule whatever we are talking about hydrogen deficiency. Four is satisfied. And then these are identical now. And these two are coupled with this to show two doublets. This one is doublets, and this one is doublets. They are very minute differences there in the chemical shift. They are coming together. And then this is showing a singlet here. And then this is showing a quadrate. And then this is showing a triplet here. So this is a molecule. So this is how we can identify the molecule. And then we can also answer simply by looking into even before identification of the molecule, we should be able to answer the questions that are listed here. Now let us look into another problem related to phosphine complex of rhodium, coordination number 5. And rhodium is in plus 1 state, because 4 trimethyl phosphines are there, neutral ligands. And then we have one methyl is there. So rhodium is in plus 1 state. So for this one, very interesting. Two spectra are recorded, one at room temperature and another one at minus CIT. Interesting is at room temperature, a doublet is observed. Simple doublet is observed. And at minus CIT, a doublet of doublets and quadrate of doublets. That means two signals are observed. So now we have to see what are the right geometry that explains these two kind of situations in case of 31p NMR. Well, when we have coordination number equals 5, one of the two options are there. One is it can be square pyramidal. And other one is trigonal bipyramidal. So now since it shows a doublet, and also we should remember the fact that 103 rhodium is i equals half. And it is 100% abundant. And then usually rhodium to phosphorus coupling comes in the range of 152 even 300 hertz. That means much larger coupling. This doublet explains probably all four trimethyl phosphines are equally coupled to rhodium resulting in the form signal, which looks like a doublet. So that means we have to write a right kind of structure in which all four trimethyl phosphines are identical. We just look into it. This is a square pyramidal in which all four phosphines are in the plane, and they are identical. So here they couple with this one to show a doublet here. And now let us look into another example. Now doublet of doublet means probably it is not retaining at minus 80. It is going to geometrical isomerism. And other alternative is trigonal bipyramidal. When we go for trigonal bipyramidal, we have several options. Let us write one or two options to begin with. So this is trigonal bipyramidal. I have drawn one isomeric form where all three trimethyl phosphines are in plane and one in axial position. So now let us look into this one here. These three are identical. And they first couple with rhodium to give a doublet. And now they are equally coupled with the axial one. Another doublet comes here. So this is 1j rhodium phosphorous coupling. And then this spacing and this spacing is 2j pp coupling here. So we get a doublet of doublet. Yes, doublet of doublet is there. It looks like this is the right structure I have drawn. Next, let us look into, and this is for equatorial puttee. Let us look into axial one now. So now this would first couple with rhodium. And then it is coupled with the three. Three means you can use 2ni plus 1 rule again. Four lines should be there. So that means basically, so this spacing, any spacing we take. This is 2j. So quartered of doublets. So this is how you can explain and draw a spectrum and then understand the problem here. At minus 80, it shows two signals having regular bipedal geometry. One is doublet of doublets, and other one is quartet of doublets. So at room temperature, it converts into isomerases to square pyramidal geometry, having all four trimethyl phosphorous in the plane and shows just coupling with rhodium. So it's a doublet. Now let's look into another example. I think I already discussed this one while looking into NMR problems. Let me recall again and tell you about 195-platinum NMR, how to understand and interpret. So a spectrum is already given here, 195-platinum NMR spectrum. And the structure is also given. Only you have to interpret and explain why they're giving two doublets of equal intensity. And here, if you just look into it, 115N is already designated. So for this one, I equals half. This is coupling. Fine. First let us say it couples with 15N. So now we have to see this is split into triplets. If triplets means we don't have anything in the close vicinity. We have carbon. Carbon cannot show. Carbon is 13C is only 1%. So you cannot say if at all if you see only you can see satellites. But with this, that information is not provided. The next target is two bond away from platinum is N. Probably if it is not labeled anything, we can assume this is 14N. And 14N we have I equals 1. If I equals 1, 2NI plus 1 will give you three lines. So this one would be split into three lines each one. And of equal intensity. If you compare this one with here, both look identical. So that means this platinum compound is coupled both with 15N, one bond coupling, as well as two bond coupling, 14N. First it splits into a doublet. And each line will be further split into triplet of equal intensity. So this is how you can interpret and understand the coupling constants and eventually the data that is provided. Now let's look into another example here. So interpret the spectrum below for the compound that is shown here is a platinum compound. Again, four coordinated one. Justify all splitings. So spectrum is given here. And also the question is when four coordination is there and platinum here, we have to see whether it is cis or trans. Let us write trans here. We just look into this compound here. This is a square planar complex. If I do C2 axis of rotation, both the phosphorous in the trans isomer are magnetically and chemically equivalent. And if you look into phosphorous cinema, they just show a singlet that is singlet. And then platinum satellites we'll see. So this will be 1JPTP. So that means it has to do nothing with 13 P NMR spectrum. And then, of course, platinum 195, platinum abundance is 34%. That is NMR active with I equals half. And rest 196 is 66%. And that is 196. That I equals 0. Three signals are there, which are the signals. And also if you look into the chemical shift range, it appears like probably it is a 1H NMR spectrum. If it is 1H NMR spectrum, how it should look like. So first we should look into methyl group. And if the methyl group three hydrogen atoms are there, and they couple with these two to give a triplet here. So they give a triplet. And then it would couple with platinum. When it couples with platinum now, what would happen? This NMR active one, when it is coupled, then it will be split into a doublet. So that means basically you will see something like this here. This pattern is more or less same for all of them. So one is for this methyl triplet. This is for I equals 0. And this is for I equals half. This is, you assume, 196 platinum. And these two are for 195 platinum. This is the species, this PTH coupling. So now the pattern is same for all. But we have to look into number of hydrogen atoms. So this is three. So this is one here. And then here, 6, 6, 12 are there. And this is 12. And then here, 4. We have 4. That means all of them are behaving in a similar way. And all of them are coupled with platinum. 34% showing singlet. And then probably that is split into triplet. And then this one also singlet that split into a triplet here. So this is how you should be able to explain the NMR. This NMR given is for 1H NMR that represents CH3, ethylene, 4 protons. And then the ratio would be, if you just look into it, 3 is 2, 4 is 2. So here it's 12. So you can see 3 is 2, 3 is 2, 4 is 2, 12. R16 is essentially same. So this is 1H NMR data is provided here. So let me stop now and come back with more examples in my next lecture. Until then, have an excellent time. Thank you.