 Once again, I welcome you all to MSP lecture series on interpretive spectroscopy. So, we really started looking to the problems to make ourselves more familiar with interpretation of the data obtained from these important for spectral methods. This helps in understanding what if we are familiar with interpretation, it would be very easy to elucidate the structure and also look into the properties and reactivity and also applications in many avenues. So, let's solve more problems in this lecture as well. Let's look into one problem here. The following peaks were from a 1 H NMR spectra from a 300 megahertz spectrometer, convert to PPM unit, Delta unit. So, for example, one is 693 hertz, another one is 1060 hertz. Again, we know the instrument frequency and simply divide chemical shift in hertz by the spectrometer frequency you get in PPM. For example, here, 2, 693, 693 divided by 300 into 10 raise to 6 would give you 2.31 into 10 to the power of minus 6 or 2.31 PPM. And similarly, when you take this one, it becomes 3.53 PPM. So, that means when in a 1 H NMR spectrum recorded 300 megahertz, CH OH appears at 2.31 whereas CH CL appears at 3.53 PPM. And you should remember this value, chemical shift value in PPM doesn't change with the field strength. So, whether I take 500, 600, 700, it still remains 2.31 only. The corresponding chemical shift in hertz increases, but once when you take the ratio, it is independent. So, that's the reason we represent always chemical shift in PPM to avoid any confusion and minimize complexity. Now, let's look into another interesting example here. Butte, to own shows a chemical shift around 2.1 PPM on a 300 megahertz spectrometer in the 1 H NMR spectrum. How far downfield is this peak from TMS in hertz? TMS always be considered as 0. So, if the spectrum was done with a 400 megahertz instrument, would a different chemical shift be seen on 400 megahertz spectrum? What would be the difference in hertz from chemical shift and TMS? So, there's three questions are there associated with this question. First one is how far downfield is this peak from TMS in hertz? For that one, what one should do is, you have 2.1 PPM into 300 into 10 raise to 6 and divided by this one, it will give 630 hertz. So, the peak at 2.1 PPM would show at 630 hertz and a 300 megahertz spectrometer. So, we converted from PPM to hertz. And then, since TMS is always at 0 hertz, so the peak would be 630 hertz on the downfield from the TMS. If the TMS is here, 0 for TMS, so it will be somewhere here, it is 630 hertz. If the spectrum was done with a 400 megahertz instrument, would a different chemical shift be seen? The question, the answer is no. A different chemical should not be seen in PPM. The point of using the PPM scale is to be able to discuss the same peak at the same chemical shift unlike when using hertz. Hertz, it will vary, but when you convert that hertz into PPM, it doesn't vary and it is independent of magnetic field strength. And then, on 400 megahertz, what would happen? Simply multiply a 2.1 PPM with 400, so then you get 840 hertz. So that means, on a 400 megahertz, the frequency chemical shift will be appearing at 840 hertz. And again, if 840 hertz, if you divide by 400, you get back to a 2.1. So this itself shows how chemical shift when measured in PPM is independent of magnetic field strength. So just to show this fact, I have taken this example here. Now let's look into another example, focusing on one HNMR or NMR spectroscopy. If you look at a spectrum of paraxylene, common name, or one for dimethylbenzene is called paraxylene. And 1-2 is orthoxylene, 1-3 is metaxylene. And how many types of protons are there and what does the ratio mean? Ratio means we are referring to equivalent number of different set of hydrogen atoms or nuclei, that's what we refer to. So let's look into it. 1-4, this is 1-4 dimethylbenzene. Just if we look into it, how many different type of hydrogen atoms are there? So these four are equivalent and these two are equivalent. That means we have two different types of hydrogen atoms. As a result, when we look into one HNMR of paraxylene, we get two type of signals here. One for methyl groups of six protons, another one is aromatic protons of four in the ratio 6 is to 4 or 1 is to 1.5. I have taken one HNMR for this 1-4 dimethylbenzene. I will show you here. So this is how it looks. As I mentioned, we have two signals are there and two singlets are there. And this singlet one can readily identify for methyl group here. And this one is for aromatic hydrogen atoms, aromatic protons. And if you look into the ratio, of course, with integration if you take, it is for six protons and it is for four protons. And then if you take this ratio, it will be 3 by 2, 1 is to 1.5. It is interesting now. How many different types of carbon atoms are there in this molecule? So now this is one. This is both are one type. And this is two. This is three. This is four. We have four different types of one, two, three, four. Different types of carbon atoms are there. That means we should anticipate four different types of signals in this molecule. So this is the one. As I mentioned, you can see 2.19 for these two methyl groups. And around seven, we are seeing a singlet for these four hydrogen atoms. Now here we are seeing three. As I mentioned here, one, two, three. We are seeing here, no, not four, because this is very same. Because you can do rotation like this or you can do rotation like this. So this brings similarities between these. So we are expecting only three different type of carbon atoms. And we are seeing three carbon signals here. And you can see here, 21.3 for methyl carbon here. And then 128 for all of them are at 128. And then this one, 135 we are seeing here. And again the ratio will be one is two. One, this is two. Of course, you cannot get the ratio here. But now the rest you can get two is two, two is two, four we are getting here. And four is there, you can also see intensity is very high. The two signals, as there are two types of hydrogens, the ratio is one is two, 1.5. So that means basically if we have to write in words, this is how we have to write instead of using telegraphic shorthand or something like that, it should be clearly spelled out. The area under the peak at 2.19 is 1.5 times greater than the area under the peak at 7.0 or 7.01 ppm. This molecule has two sets of protons, six methyl protons and the four aerobatic protons. The ratio is one is two, 1.5. So completes the solution. And then in carbon, we have three types of carbon atoms are there. As a result, we see three signals in its 13C NMR spectrum. So now we have two more molecules are there. How many signals the following molecule would have and the integrations in 1H NMR, skid the spectra and estimate the integration of the peaks. So now if you just look into the molecule, you can clearly tell that there are three types of signals are anticipated in 1H NMR for this molecule here. And also if you look into 13C, also shows three type of signals here. All three are different. And then how it is going to look like? This would be split by this one to give a quadrate. So this will give a quadrate. And then this will give a triplet. And this will also give a triplet, but this is slightly downfield. In the same range, if you look into the energy here, they appear like this. So this is how the spectrum is going to look like. And then this is not coupled. Or this is not coupled. You should be able to get a single resonance here. In this case, this dibromo compound, also we have two different types of hydrogen atoms are there. These two are identical and this one is there. And these two will be coupled with this one to show a triplet. And then this will be coupled with these two to show a triplet. So we anticipate two triplets here. Let's look into NMR. And carbon also, we anticipate two signals in this case, three signals in this case. So here, 1H NMR, there will be three signals in the ratio 3H to 2H to 1, 3H to 2H to 1. That's what I have written here. There will be two signals in the ratio 1H to 1 here, because 2, 1H to 1. And 13C NMR, there will be three signals. And 13C NMR, there will be two signals here. I have also taken 1H as well as 13C NMR spectra for these molecules. I will display now. You can see here. We have a triplet and a quadrate. And then a singlet. The singlet is for this one here. And then the quadrate is for this one. You can tell it is a quadrate. And then 3.53 and then for this one, a triplet is here. Beautiful spectrum. And then I have also taken 13C. 13C also, as expected, it should show three signals. One is here, 14.9 for this one. And then 67. 67 for this one here, because this is next to this one. And also both are next to oxygen. So they're a little bit downfield. So it appears around 59. And this appears around 67. So this is 13C NMR spectrum shows three signals. And 1H NMR also shows three signals in the ratio, 3 is to 2 is to 3. And now let's look into the spectrum, 1H NMR spectrum of this tetra bromo propane. Here it is. It shows two triplets. So the middle one, this is 3.21 is here. And then these two are here, triplet. And then 13C NMR would show only two signals here. This is 54.9. And then these two are 40.3 here. Ratio also you can see here. This is 1 is to 2, almost intensity wise you can tell. Now let's look into another interesting molecule here. How many proton signals are expected in the 1H NMR spectrum of the compound shown below? Also predict the number of 13C NMR signals. If you just look into it, this looks like a very symmetrical molecule here. You can do C2 rotation. Both the portions, left and right portions, bridged by methylene, looks identical. And these two are identical. And then this one, and this one is identical. This one, this is, of course, no hydrogen here. These two are identical, and these two are identical. That means basically 1, 2, 3, 4 signals are expected, 1, 2, 3, 4 signals are expected. As this one, 1, 2, 3, 4, and 5 signals are expected for this molecule in its 1H NMR spectrum. And about carbon, carbon if you see, this is 1, this is 2, this is 3, this is 4, you can see here. So, 1, 2, 3, 4, 4 different, of course, this is one, 5 different type of signals are expected for this one in its 1H NMR spectrum. Let's look into it now. See, 1, 2, it is merged here, because these two are coupled to each other. This also shows a doublet, and this also shows a doublet. They are merged here, and then there is one, and this one is for this OH, and then this four is for methylene here, and this one. And then this one is here, here, this is singlet. I have shown here values, 3.96 here for this one, and then 6.90, yes, here 6.95, we have two doublets here, something like this, and then 7.14, and here it is, this is a singlet, these are identical. So, this is how you can see, 1, 2, 3, 4, 5 signals here, ok, so 1, 2, 3, 4, 5 signals are there. And if we look into 13C, you can see 1, 2, 3, 4, 5, 6, 7 signals are there. What are those 7? 1, 2, 3, 4, 5, 6, 7, rest are identical, mirror image, so that means you can see 7 signals are there, and you are seeing here 1, 2, 3, 4, 5, 6, 7, so all these are clearly identified. One is done here, you can see here, it's pretty simple, right, let's look into one more example here. Thus all the structure using the following spectral data, molecular formula is C6H12O3, IR spectrum is shown below here, and of course I have also taken 1H and MR spectrum also here, and then what you should notice here, here we have a carbonyl peak is there, we have to identify, and then you have CH, and then of course you have OH is also there, so that means these things we have identified, and then we have 1, 2, 3, 4, 5, 6, 6 signals are there, and some of them are much deshielded, and then there is 1 with 5 splitting, that means you can think that yes, here 5 lines are there, and the 5 lines is probably it is coupled with 2 equivalent methylene protons, and that means you have a chain of CH2CH2CH that immediately comes to your mind, and then 2 methyl groups are there, and those methyl are much more deshielded probably next to oxygen or something like that. So you can speculate, and once if you speculate that one, you can even write the structure, so this is the structure, and of course the formula is CH6H12O3, and this is CH3CH2CH2CH2CH2CH2. The moment you see here a quintet here, the quintet can be assigned to 1.88 here, so that is there, and then 3.69, and this one at 3.69 is here, and 4.13 this one, and here CH2 is there, and then this is a quadrate 2.41, and then this one is like a typical ether one, so this is here. So all of them are and water is there, and this OH is there here in this one. The molecules CH6H12O3 has 3 oxygen atoms, and it has a hydroxy group is there, ketonic carbonyl group is there, and we have identified those things. This is the spectrum recorded for this one, EMS shows, and the same in a different way mass data is also added, mass shows 132 if you go back here, so this 72 plus 12 plus 48 132 mass. All this information is not given, IR is not given, and then if IR value is also given here, IR shows a prominent peak at 1750 means you can immediately think of having something like this, and then 132 is there, and formula should be worked out by looking into these things. When you look into these things here, we can readily identify 1, 2, 3, 4, 5, 5 different type of carbon atoms are there. You can see here 1, 2, 3, 4, 5 different type of carbon atoms are there, and some of them are quadrates, only one quadrate is there, one quintate is there, and 3 triplets are there. That means you should be able to judge very wisely where exactly they fall. So you have to see quadrate means 1 and 2 quadrates means here. This one will be a quintate here, and then this one will be triplet, this one will be a triplet, and then this will be a quadrate, and this will be a triplet again. So by just looking into the positions of the peaks and their splitting pattern multiplicity, we should be able to assemble the entire molecule in this fashion with little information from IR as well as mass. I think let me stop here and continue discussion on more very interesting examples in my next few lectures. Until then have an excellent time.