 to today's class. So we will be continuing with high resolution NMR spectrum of molecules. So let us now using this concept of coupling constant and chemical shift, can we start interpreting NMR spectrum of few molecules and these can be used for distinguishing say isomers. So let us take an example. So if I have suppose a molecule which is something like C11, C11, H16 and O, this is a molecule which has two isomers. Now isomer suppose is something like this CH3, CH3, CH3 and O, CH2, C6H5 and the other molecule suppose we have CH3O, CH3, CH3, C6H4, CH3. So if these two are isomers of the same molecule, how we can distinguish based on the NMR spectrum that which molecule belongs to what. So what will be the NMR spectrum for say first molecule? So let us look at the groups that are present closely here. We have a benzene ring. Now as you know benzene ring comes around say 7 ppm. Then we have three methyl groups and these methyl groups comes quite like uphill shifted that is like near 0 ppm and then we have a methylene group. So three types of carbon mostly we have one here of benzene ring, methylene group and methyl group. So three types. So essentially we should see three peaks. What are those three peaks? And these three peaks are essentially like this. So here say our 0 ppm reference is there. Now we have the equivalent nine protons that are contributed by these three methyl group. So we have a peak which will be like this methyl groups here and that will corresponds to nine peaks. Then we have another proton which is methylene group. So this is say 0.09 ppm. Then we have another group which is methylene group and here it is contributed by two protons. So that will be around 3.5 ppm. And then we have far down here around 7 ppm that is one peak which is given by the five protons and that is 7.2. So this is our spectrum for this molecule. What happens to here? This molecule which is there is a benzene ring, one methyl group here, two methyl groups are here and one methyl groups are here. If you look at among these methyl groups the chemical shift is not same it is going to be different. Therefore we have a different spectrum for this molecule and this molecule will give a spectrum something like this. So the most downfield shifted will be these two methyl and that will be slightly downfield than this 0.9. So here is our TMS. So these six protons will be around 1.2 ppm. So this is six. Then it comes like other protons which we have like here and here. So even in these methylene protons we have three kind of chemical shift. For this benzene ring it will around 7.2 ppm. Then methyl will be around 2.2 ppm this methyl and then we have 1.2 ppm. And there is one more methyl which comes around here that is 3.5 ppm. So we have a six protons that will be around 1.2 ppm. Then three protons, three protons and this will be four protons. So let us go to next page and then I will write it clearly. For molecule like this that we have here is 0 ppm. We have the most downfield are six protons coming from these two groups. If you look at CH3 group and CH3 group. Now then we have one CH3 group which is attached to benzene ring and that comes around to say 2.2 ppm and these corresponds to three peaks. Then comes so this is what we are talking about this CH3 which is attached to benzene group. Then we have 3.5 ppm CH3 corresponds to three protons and that we are talking about OCH3 group. So even in the methyl depending upon what is attached here we are getting that different chemical shift and most downfield shifted for 4 protons that is coming because of benzene ring. So benzene ring has 4 protons because two groups are attached at two positions. So that is around 7.2 ppm. That is how we can distinguish. So if you go back, so spectrum of the previous proton here we have different than the current one and that is how you can distinguish between two isomers. Let us take the another isomer. Say its chemical formula is C7H12 and two structures that we have is CH3CH3CH3C triple bond CH3 one is this and suppose we have another CH3CH3CH3CH2 triple bond CH. So if we look at here now what we have is all these are in the same chemical environment. So 9 proton and they will be quite affected near 0. So we have a 9 proton corresponds to these three moiety and we have here 0.9 ppm. Then we have another type of chemical shift which is coming from proton chemical shift which is coming from here and that is we have one type of proton. So that is 3 and this comes around 2. So if you take ratio 3 to 9 or 1 to 3. So here we have ratio of signal 3 to 9 or 1 to 3. For this molecule what we have here we have a different kind of proton. So here are 3 one kind of proton then CH2 we have another kind and then CH we have that is third kind of protons. So this CH3 again say all 9 will come at 0.9 ppm. Then we have another proton two protons which will come around 3 ppm and this one proton which will again come very closer to that. So we have two protons and one proton that will come very close to each other. This is three proton and these are methylene protons. So that comes quite down full compared to this methyl protons. So this is how you can distinguish the isomers. So now how like we can do the analysis just I will go forward and try to explain again. So first thing you need to know if you have a chemical formula how many kind of protons we can have. So that will be simple given by how many double bonds that we have. So there is a simple calculation which is called double bond equivalence that you can find it out by a simple formula called 2A plus 2 minus B divided by 2 for a compound which is like CA HB OC HB OC. So here A is the number of carbon B is the number of proton. So like whatever we had earlier if you look at C11 H16 O we have. So how many double bond we are expected to have here? So if you look at here 2A 22 plus 2 minus 16 divided by 2. So 24 minus 16 8 divided by 2 means 4 and therefore we had a one benzene ring there. Now that is how this can be used let us take the another example that we had in the previous example C7 H12. How many double bond we are going to expect here? So 7 into 2 14 right 14 plus 2 minus 12 divided by 2. So that is 16 minus so that will be 2 number of double bonds. So DBE is 2 and therefore we had one triple bond there. So in the previous example. So that is how we can calculate the number of double bond equivalent. So now let us see if I give you chemical formula and now we have to predict the spectrum for that. So let us see we have a molecule like CO2 H4 and Cl2. Now if you calculate using that then we do not have a double bond because 2 multiplied into 2 plus 2 minus 4 divided by 2. So 4 plus 26 minus 4 that is 2 divided by 2. So DBE is like 1 so we do not have any double bond here. So that means all carbon are saturated. So probably the structure for this can be something like this CH Cl HCl and Cl or it even can be like this CH3 CH3 sorry CH and CH Cl and Cl. So how do you distinguish whether the spectrum that we are getting belongs to here or not? Now we record this spectrum for this molecule and suppose we are just getting only one peak here is our TMS this is only one peak. So now only one type of proton we have. So only one type of proton can be only this. So like where two protons on each carbon are attached and two chlorine are attached. If we have this case then we have two different kind of protons and this will generate two peaks. So if you are getting only one peak that means this is the correct molecule. That is how we actually interpret this spectrum. Now I will give you an a spectrum and a chemical formula and then we have to find what could be the structure of that. So let us see if I have a chemical formula of a molecule which is C9 H12 and the spectrum that I am getting is something like here 0 and one peak here one peak here ratio is 1 to 3. Now to find it out what could be. So you can similarly calculate the double bond equivalent and then two things are visible if I give you chemical shift. So chemical shift suppose for this is around less than 2 ppm say 1.5 something like that and here we are giving you around 7 ppm. So it is now clearly evident that we have two kind of proton one belongs to say methyl group another belongs to and benzene ring. Now so and ratios are 3 to 1. Now we have a total of 9 carbon and 12 protons. So 9 carbons and 12 protons if you calculate your double bond equivalent we can find it out that these three protons can only come if we have a methyl groups attached to a benzene ring and those methyl groups seems to be equivalent to chemical equivalent therefore something like this here one group attach another group attach here and here. So that takes care of our 9 protons remaining three protons are these and therefore we have two kinds of proton one attached to the directly from the benzene ring and 9 protons attached to the methyl group that is why we have 1 to 3. Let us take another example now here suppose our chemical formula is say C4 H10 O2 and our spectrum is say something may be like this here we have a 6 integration here we have a 4 integration and this say chemical shift is 3.2 and this chemical shift is say 3.6 something like this very close now can we predict the chemical structure of this compound. So we have a 0 double bound so one can calculate it. Now one thing is obvious that even there is a methyl group it is quite down field compared to the typical methyl group and that is because two oxygens are attached and we have two only two kind of protons one looks like from methyl group another looks like from methylene group let us put two oxygen here and then we have two methyl attached to these. So let us attach those two methyl and then we have remaining two carbons and then we can balance the four protons. So now these all four protons are of equivalent chemical shift and that belongs to these and this and these two methyl groups belongs to these. So we have two kind of proton one contribute to 6 integration another belongs to 4 integration this is from methyl group this is from methylene group that is how we can solve these structure of few of the chemical compound rest you can practice and then we can discuss in the class if required if you have any doubt. So let us move back and let us try to do some of the splitting pattern. So what happens when J coupling is involved here we have not considered any splitting let us consider some of the splitting. Now suppose we have a molecule which has a chemical formula like C3H7 and Cl and we have a spectrum something like this here it is 0 ppm we have two closed peaks and integration of these together comes 6 and here we have a peak which looks like multiplied and here we have integration one. So C3H7Cl what this could be so one thing is obvious that we have two kind of proton 6 and 1 so here we have 7 so 6 and 1 very clear. So then probably we have 6 protons so that might be coming from two methyl group here let us put two methyl groups and there is one chlorine so let us attach that chlorine here sorry Cl here now then there is only one proton which is attached here now what is happening here this proton is splitting this methyl group into two and therefore we have two equivalent like height methyl protons and integration total is 6. Now these six protons are splitting these single proton into septate and that is why we have a multiplied and that comes to be 1 that is how you can find it out that this is the chemical structure even the chemical formula is this. Let us go to next example, suppose we have a chemical formula like C6H5CH2SH. So now the spectrum is something like this here is 0 ppm here we have a triplet of 1 here we have again 2 and here around 7 ppm so here is 7 ppm we have a 5 protons here 2 protons which is less than 4 so 3.8 ppm and this is say 1.8 ppm something like this. So now we have to find the structure for this okay if we have this compound so we can clearly identify that these peaks 5 protons coming from this one C6H5 and then so these triplet this is CH2 so that will split SH proton into triplet therefore we have 1 SH triplet at 1 and then this H will split this CH2 into doublet so we have here doublet structure. So that is how looking at the spectrum and looking at the chemical structure one can identify few of these compounds. Okay let us take an complicated one now you can maybe you try to do it by yourself if we have something like C5H9O4 and N and suppose a spectrum I am getting something like this so here we have two kind of protons that shows doublet and here we shows triplet and this is say 1 ppm and this is say 1.4 ppm this is 1.6 ppm integration of this is 3 or integration of this is 3 and then we have 1 around 4.2 ppm and here we have a quartet like this and then around say 5.2 ppm we have again a quartet something like this. So integration of this is 1, 2 and here we have a 3 and 3. Now can you identify what this compound can be given the formula is like this. Okay so let us try to solve this as well so we have two, three protons that looks like there will be two methyl groups. So looking at these there is no equivalent so there will be two methyl groups we can write it two methyl groups at two end and then we have two other protons which are equivalent to so like CH2 and here CH. Now we have we can have one NO2 and there will be some CO2 which will be interact here so CO2 so our compound something like this we should have a so get down the exact chemical formula I am giving you a hint we have two methyl group there will be one CH2 group one CH group that will corresponds to the spectrum that we just now draw it. Here we have a triplet corresponding to three protons here we have another triplet corresponding to three protons here we have a quartet corresponding to two proton and here we have again quartet corresponding to one proton 5.2, 4.5, 1.8 and 1.6. So identify how these molecules corresponds to this kind of a spectrum and I will suggest you guys to do practice take any organic chemistry NMR spectroscopy book practice it take the spectrum and then try to deduce the chemical structure of that or take a compound and try to draw the spectrum for that but to understand how these lines comes we have to go into little digress details of explanation for their intensity for their resonance frequency all those comes if you understand little bit of quantum mechanical analysis. So what I will do today for this explanation I briefly venture into quantum mechanical description of these spins so I will just like you to go through the basics of quantum mechanics but we will be wherever required we will introduce it some of the phenomena. So let us start with a basic principle of quantum mechanics. So as you have studied your in your undergraduate or so for basic principle of quantum mechanics we have a operator which operates on a on a eigen function so we have a operator and we need to define an Hamiltonian function and operator operates on eigen function and it gives its eigen value. So these are basic terms like operator Hamiltonian eigen values and eigen function in NMR we particularly are interested in something called angular momentum operator and this is defined as i. So angular momentum is of our particular interest and we define that i, i is a operator of total angular momentum i has a three direction x, y, g and so here i z direction or x direction and say y direction or vice versa. So it has three components one in z direction x direction y direction in all three direction. So for a single spin half system two orientation of spin are possible one is like along the magnetic field here suppose we have a magnetic field b0 one is along the magnetic field another is against the magnetic field you can define one as alpha another as a beta or vice versa whatever way you want to define so these are the two orientation of spins along the magnetic field or against the magnetic field. So now these are eigen function of i z operator as I said there is a operator and in NMR it is the angular momentum operator. So now i z is an operator and it is operating on the eigen function like alpha and beta. So if i z operating on alpha state it gives the value of plus half or if it operates on beta state it gives the value of minus half. So that is what we have. So now if these functions alpha and beta are orthogonal that means if you take the bracket notation alpha or beta beta it is 1 or if you take beta alpha beta it is 0. So that is what we mean by orthogonal function. Now so as we said that these are operators i i x i y i z and these spins they do not commute but obey a cyclic relation. So what we mean by cyclic relation in quantum mechanics that if you take this i p or i q they will give you i r that means if you take i x or i y in our angular momentum term they will give i z. So now two commuting operator with non-degenerate eigen value have a common eigen function what I mean by that if you have a a b phi so we have these functions. So here b is non-degenerate eigen values of function b. So that means a phi must be an scalar multiplication of phi what we mean by this is a phi equal to small a phi. This means phi is an eigen function of a and a is an eigen value. So here is phi is an eigen function a is eigen value. So this is basics of quantum mechanics and as we said that we will come back to quantum mechanics we will reintroduce wherever it is required. So now let us move ahead and then look at the high resolution NMR spectrum. So what all the functions and Hamiltonian we have. So main one is a Gman interaction. What we mean by Gman interaction is interaction of spins with an external magnetic field. So as we see that external magnetic field is defined as a b0 and it is in z direction and this is of like if you look at the magnitude it is quite a bit because magnetic field is of Tesla order or in terms of frequency it is a megahertz like 600 megahertz, 700 megahertz and all those. The next one is J coupling that is another Hamiltonian. So J coupling is interaction between the spins through bond. So like there are say here this is interaction between spin a and spin b and through bond how they are interacting that is what we call it J coupling. So it is a intramolecular interaction and this is of relatively very small magnitude. The third one is called dipolar coupling. So two spins in solution so these are two spins and there is a bond vector between them. So spin a and spin b and this is bond vector connecting them. So the angle between the main magnetic field and this bond vector is given as a phi. Now the dipolar coupling interaction happens through a space. So this was through bond interaction scalar coupling the dipolar coupling is through a space interaction and that if you remember little bit of this dipolar kind of interaction this is proportional to cos square phi minus 1. So phi is the angle between this intermolecular vector of these two spins with the magnetic field. Now what happens in solution the two spins all the time tumbles right. So because of this tumbling these term goes to zero and therefore we do not see any dipolar coupling in solution. So what we have mostly is a Giman interaction and J coupling interaction dipolar coupling is averaged out in solution but that is not case in the solid. Solid means does not tumble. So if you want to make this term zero we have to spin faster and faster and that is that we will discuss later whenever it comes. But in liquid state so these are the two interaction terms that we have mostly coming from the magnetic field external magnetic field this is order of megahertz then we have J coupling which is in order of hertz and dipolar coupling is typically order of kilohertz. So this is the main field and then we have a J coupling. So now let us move ahead if we have these interactions that means these can form the Hamiltonian which operates on the angular momentum operator. So the isotropic nuclear spin Hamiltonian isotropic means spins that tumbles in solution. So as we say isotropic that means spinning is happening all the time it is randomly moving so dipolar coupling is not there we have a Giman interaction and scalar coupling interaction which is coming because of J. So total Hamiltonian is H0 and H1. Now H0 is our Giman term and H1 is essentially J coupling term. So H0 one can define that how different spins like i, z, i, i is different spins and as we said that z component of angular momentum gamma is the gyromagnetic ratio of the each nuclei. So we have to sum over the sum over all the nuclei how this term comes. So H0 the Giman interaction term can be given by summation of gamma i, H i and z i. Now H1 is a J coupling Hamiltonian this causes the mixing of two spins because two spins now talking to each other through bond. So that is how it is J i this is coming because of the coupling and here are two spins i, i, i and i, j. So i, i and i, j are operator of i and j spin and because of this we have a coupling constant j is a coupling constant i, j is the operator for the z component of the angular momentum from i spin. So these are the two Hamiltonian terms in liquid states. So now quantum mechanical term for spin half, spin half that mostly we are looking at at the moment. So spin half nuclei like proton or carbon 13 or phosphorous P31 or nitrogen nitrogen 15. Now in these states the half spin system can be either in alpha state or beta state alpha state or beta state and then we can take the summation over all states. So for a particular states say we have a n spins. So we can write a particular state by alpha 1 that means the spin 1 is in alpha state, spin 2 is in beta state, spin 3 again is in alpha state, spin 4 is in beta state and again spin 5 in alpha state. So multiplication of all those will be given by a particular function which is phi n. So phi n is our eigen function and Hamiltonian H0 that is Zeman Hamiltonian operated on phi n which gives the eigen value En. So that is what H Hamiltonian gives phi n is the eigen function and En is an eigen value. So now the another Hamiltonian which is H1 and that is j coupling Hamiltonian these as we discussed cause the mixing of two different states. Now for these two eigen functions there will be linear combination of the product. So what we mean by that, so if we have a spins like n spins so we can have a product of these spins and you have to sum over that and that gives you resultant eigen function. So eigen function will be determined by the solution of the secular equation you might have studied. So we can solve this equation where Hmn minus E delta mn will be 0 and you have to solve this. The Hmn is the matrix element of this Hamiltonian for solving this we have to define our operator which is Fg and the product function which comes from this j coupling Hamiltonian you can solve this by defining in Fz and if you take this if we take this product so Fj is eigen function and I have eigen value like total azimuthal quantum number. So one can solve it by doing simple quantum mechanical algebraic operation and Fz commutes with Hamiltonian H. So Fz has a degenerate eigen value and that have that the above communication does not imply that eigen function Fz will have a eigen function of H. So if we solve this equation you apply Fz on phi m and then you get Fm. So Fn and Fm are eigen value corresponding to two function phi n and phi m and you solve this commutation relation and one can find it out. So like this operation so Fn is not equal to Fm and then the matrix elements will vanish and that helps us actually solving these equations. So you can block diagonalize this representation taking the case of Fn not equal to Fm. So if you do that then secular determinants gets the block diagonalized and the problem of determining the eigen value will be reduced dimension and then each block can be treated separately for solving these functions. So that is how we can solve it the eigen function for each of these value can be solved and if you can substitute then one can find the by multiplying phi m and calculating the matrix element and one can do that then what we found the basic functions and finally we come to by solving this we come to such kind of coefficient. So now doing that we can calculate the Hamiltonian for two spins. So now the complex problem will be now simplified in two spin system and now next we move ahead and try to solve and look at how we can calculate the Hamiltonian for two spin system and from there we build up the concepts and we try we will try to get it to the solution for these spins. So today I would like to end up here just I will summarize that we looked at how the splitting happens and what is the basic rule for a splitting and where is the resultant chemical shift average comes what is what will be the rules for intensity of these terms. So for a detailed analysis of these splitting patterns one has to go to quantum mechanical formalism and we looked at basics of quantum mechanics how to solve this so we will continue from there to calculate the Hamiltonian for two spin system and then we build up from there. Thank you very much and looking forward to see you in the next class.