 Hello, welcome to today's lecture. So we will continue with high resolution NMR spectra of molecules. So previously we discussed about the important parameter such as J coupling and this J coupling causes the splitting of resonances. Then we ask this question how does this splitting happen? So what is the reason behind this splitting? And we looked at their two spins are coupled via bond that causes the splitting in the resonances. Then we discuss little bit in detail that what are the factors that influences this J coupling. And then we looked at two kinds of coupling, one was weak coupling and the strong coupling. This classification of weak coupling and a strong coupling was not on the basis of the strength of the J value but the J value compared to the difference in the chemical shift between two spins. So that was the origin of weak coupling and strong coupling. Then we went ahead and looked at the various spin systems such as ABS spin system, AX spin system, ABX spin system and AMX spin system. And the ABX spin system, AX spin system and all those spin system are based on the how far or how closely they are resonating. In that case AB means A and B their resonance frequency is quite close whereas AX, A resonance frequency is far from the X. So that is and in these cases actually these are weakly coupled system because the X in chemical shift differences this is quite far. So after that we defined something called magnetically equivalent and chemically equivalent spin system and we looked at that magnetically equivalent protons does not cause splitting and then we also said that we will look at in details when we discuss about the quantum mechanical aspects of this why magnetically equivalent protons does not cause splitting. However chemically equivalent proton spin system can cause this splitting. So today we are going to continue with this and then we are going to in detail we are going to discuss the first order analysis of the spectrum and how we start interpreting the first order spectrum. So interpretation of multiple plate structures we will discuss. So multiple patterns in different group of line that we are going to discuss today how we are going to interpret and what is the origin of relative intensity of this individual group and how we are going to measure this coupling constant between these lines. So splitting of individual groups and identification of these groups having common coupling constant that we are going to also look at how we are going to do. So these all points today we will try to cover up in today's lecture. So now we see that this center of the multiplates represents the chemical shift. So what I mean so if this was a peak before the J coupling consideration now if because of J coupling this is split into two so center of this will represent. So now this will be separated by JAB and the center of this chemical shift will match with the center of this the upper peak. So the center of multiplates structure represents the chemical shift. So the chemical shift for this spectra is essentially this center point. Now the magnetically equivalent spins like as we defined A2 or B2 or A3 the J coupling does not lead to splitting. So like for an example CH3 in methane for this actually the peak is going to be only one peak magnetically equivalent or in previously we had given an example of magnetically equivalent. So in those case if the spin system is like A2 or B2 or B3 there will be no splitting of line in these cases. Now next integral of any multiplate is proportional to total number of equivalent proton in that group. That means now what we are saying even if they are split say in methylene it is split into four for ethyl alcohol spectrum if you remember but this is coming because of the two protons so if you integrate this will the ratio will correspond to two proton. Similarly like CH3 in alcohol this will correspond this will be split into say 3 because of CH2 now this will corresponds to 3 proton. So integral of multiplate is proportional to the number of total number of equivalent proton in that group. Next the resonance line of a nucleus J couple to N equivalent of another type gets splitted into 2Ni plus 1 where i is the spin of the individual nucleus in the equivalent group. What I mean by that I will just explain each of these points one by one. So what I mean here is following that first thing we mentioned that the splitting of these two peaks the chemical shift will be centre. Second we mentioned that whatever is splitting is but this will match to the number of protons so that will be contributed by the number of proton that are contributing towards this signal then we mentioned that the number of how many numbers of it will be splitted that depends upon what is the i 2Ni plus 1. So if say it is because of proton which i the i of proton is or hydrogen is half so say there are two protons so two protons then it will split in 2 into 2 half plus 1 that is actually 3. So CH2 will split CH3 in ethyl alcohol into 3 so 1 is to 2 is to 1 that I am going to come how this ratio is coming so that is the idea that we discussed in the next slide. Now but what happens for a nucleus which is J couple to 2 groups of the equivalent nuclei with N and M nuclei having two spins like i1 and i2 what I mean by this suppose we are considering a case of CH where there are two groups like H and C group and here there are N and here there are M. So the splitting of these kind of things will be 2Ni 1 plus 1 and 2Mi 2 plus 1. So this again we are going to discuss in detail so how many equivalent that will be how are going to interpret this data. So next will be the spin of half nuclei say different groups like proton, carbon 13, F 19 and P 31 the multiplied for a particular nuclei would be N plus 1 into M plus 1. This I am going to again explain you in a moment and a single group of N equivalent so N equivalent nuclei with spin half coupled to a particular nuclei say A the different line in the A multiplied will have the intensity of proportional to binomial coefficient of expansion AB plus N. So let us explain one by one for a nucleus J coupled to 2 groups of equivalent proton N and M nuclei of i1 and i2 spin. So like there are two kind of spin so let us say i1 is half and i2 is say 3 by 2 or something like this. So then total splitting will be 2 plus suppose i1 is 2 half plus 1 and for because of this so multiplied with because of this 2 into 3 by 2 plus M suppose 2 plus 1 so that will be total how many so 3 multiplied with 3 plus 2 plus 1 that will be 6 so total 18 kind of so such this will give a such complicated spectrum whereas like in a heteronuclear coupling so say proton is half spin carbon 13 is half spin so 1 proton coupled to 1 carbon 13 so for carbon 13 and proton both are half so this will be number of nuclei 1 plus 1 so that will be total of 3 convert so like carbon 13 and proton will cause two lines separately. So now let us look at some of the example here so just to remind you we are now discussing the interpretation of multiple structure and for first order rule so only we are discussing the first order rule. First order rule means the comparison you have to do with a difference in their chemical shift compared to the J value so here suppose resonance A and resonance B there is a difference between them in terms of chemical shift is mu and their coupling constant is J AB so for first order rule to be valid the difference in mu that is here divided by the J should be much much larger than the 1 that means the difference in the chemical shift should be larger than the J. Now for half nuclei the multiplicity of splitting that will be n plus 1 where n is the number of nuclei from the neighboring group and for spin more than half it will be 2 ni plus 1 so suppose for deuterium which spin is 1 so deuterium is like this spin is 1 so this will cause the splitting 2 into suppose 1 deuterium is contributing 1 plus 1 so this will be 3. So deuterium will split into 3 whereas a proton will split into 1 proton will split into 2. So now for spin half as we said that will be splitting according to the coefficient of the binomial expansion of A plus B to the power n that is given by something called Pascal triangle. So suppose there is a zero spin that will be only one line so one line if there is no other spin but suppose there is one proton near to another proton so that n is 1 here if you look at here so that will split into 2 and intensity will be dictated by the coefficient of this expansion 1 1 so that means one proton will split the neighboring signal of equal intensity 1 1. Now there are suppose there are two protons so they will split their neighboring proton in the intensity ratio of 1 to 1. Now if there are three protons that will split into 1 3 3 1 and that is all coming from the expansion of this and this is called Pascal triangle. Similarly if there are n equal to 4 they will split 1 4 6 4 1 and 5 will be 1 5 10 10 5 1. So I will just give you little bit of understanding of this so what we are saying suppose there is only one proton near another proton so one one this is our observation proton or nuclei and the other one is say X here so AX system. So now X will split A into 2 of equal intensity because X can either be like up or down and that will be split into 2 of ratio 1 to 1. Now suppose there are two X which are two X here two molecule of X so that means they will be oriented both up up down down up and down down and that will split A into 3 of ratio 1 is to 2 is to 1. So 1 2 1 that is that is what we are saying suppose X is now 3 so how they are going to split so their orientation is going to be like up up up 2 up 1 down like this and then 2 down 1 up 3 down so that is 1 3 3 1. So 3 spins A sorry the X3 will split A into now 1 to 3 to 3 to 1. That is what we are saying that the intensity of the line will be according to the Pascal triangle so 2 spins 1 2 1 3 spin 1 3 3 1 4 spin 1 4 6 4 1 and so on and so forth. So that is the intensity of the lines how they are coming from. Now similarly again I will go little more detail about the energy diagram that is what we have said so our let us talk about our famous molecule ethyl alcohol and at the moment I just consider these two CH3 and CH2. So CH2 will be in up up state that is alpha alpha state alpha beta beta alpha state and beta beta state and that will cause the splitting of CH3 in 1 to 1 ratio. So this is the multiplicity of this now CH3 will be in alpha alpha alpha state alpha alpha beta state so that is what I mean by alpha alpha state and alpha alpha beta state something like this. So that will be again 3 type 3 type 1 type and that is why the CH3 will split CH2 in ratio of 1 3 3 1. So now that let us look at little bit of energy level diagram what is happening for 2 and 3 non-equivalent spins. So 2 spins here is the case energy level diagram suppose for alpha state of A and alpha state of X here. So now in this case alpha is flipping and that is giving origin to this mu A2 and here also alpha is flipping so this is giving rise to mu A1. So therefore because of this coupling with X now two lines are coming one is from here this one and another is from here to here and the splitting between them is actually that dictated by this what is the coupling constant between them Jax. So and the without splitting this would have been the original frequency here but now because it is coupled to X we have two lines and they separated by that Jx. So now A because of coupling with X splitted into 2 and of equal intensity. Suppose A is coupled to non-equivalent spin so here will be the case for that. So now two non-equivalent spins so that means each of them will split into 2. So M is 1 which is coupled with A so now M will be splitted because of coupling with M, A will be splitted into 2 and then further it will be splitted into 2 because of coupling with X. So now it will give us four lines with equivalent intensity which is here M1, M2, M3 and M4. So here it is shown here. So in this case M is flipping and that is why M alpha is going to beta that is giving line to M4 here then here actually now if you look at this also M is going to beta but that has different energy and that will be M2. So these two are coming from here. Now it is coupled with X as well therefore this giving line to M3 and M1. So one line now splitted into 4 because they are coupled to M and X and that is how we are saying two non-equivalent spins will cause splitting into 4 lines. Now let us look at some heteronuclear splitting pattern. So here suppose we take a molecule where we are detecting carbon 13 and it is coupled with proton. So carbon 13 has its spin half and proton also has carbon 13 the spin is half and for proton so i is half for this for proton also spin is half. So now how this coupling is happening. So for a CH3 group that C is now coupled with three protons of half spin and these three are equivalent therefore this carbon spectrum will be splitted into 4 with a ratio 1331. So this methyl carbon spectrum will be like this. Now coming back to methylene here we have two equivalent proton that will split carbon spectrum into 3 and ratio of this will be 1 to 1. Now coming back to CH, CH here there is only one proton which is splitting this carbon resonances and that will be 1 to 1 and quaternary carbon now this quaternary carbon is not attached to any of the proton so there this will not be splitted and this will give single line. So now but now if you integrate all of them since all of coming from one single carbon so their integral of this will be 1, 1, 1, 1. So that means each of these carbon contributing to equal intensity however splitting pattern depends upon how many equivalent protons are attached to different carbon. So we can conclude that splitting pattern is independent of sign of the coupling so it does not matter the coupling is negative or positive it just how many protons are attached to it and magnitude of J coupling between nuclear spin decreases as the number of bond increases. So that we have seen that 2 bond coupling is quite a strong compared to the 3 bond, 4 bond coupling so as we move far the effect of coupling constant decreases. So let us look at some of the more interesting examples. So like we have defined AMX system so for AMX system the separation in the resonance frequency between AM and X is quite a far and we assume that the coupling between J AM and J AX is different so J AM is more than J X so how they will split. So now A will be firstly splitted because of the coupling with AM because this coupling constant is higher then eventually it will split because of the coupling with X. So here for an example now this splitting is happening so there are say 3 resonances separate for A, separate for M and separate for X. Here what we are saying that A is coupled with M and X as well so first splitting will happen into 2 so here are that splitting that mu A will be splitted into 2 one is this line another is this line. Now each of them will be splitted into 2 so that is why we have 4 line but the center of that will be the original frequency of A. So now A will split and give 4 lines. Here if you now M is also coupled with A therefore now the splitting first happens because of A here into 2 and MX is smaller so that means again it will be splitted into 2 and here it will be splitted into 2. Similarly X will be is coupled with M so it will also be splitted into 2 and further it is coupled with A so that will be splitted into 2. So a system like this here A, X and M we have spectrum like this and that will be each of them if you are integrate here this will be intensity 1, intensity 1 and intensity 1 because each of them each contributed by only one proton but the difference in the line coming because they have different J coupling value. So here that is why we have 12 line first it is splitting because of J AM now further it is splitting because of J AX so each of them gives 4 line and that is the separation between these lines. So now that is what here for A 4 line, 4 line for M, 4 line for X total of 12 line. So what are the factors that affects this spin-spin coupling? We have looked that single bond coupling is called one bond coupling is denoted as like J1. The geminal coupling is 2 bond coupling, the visceral coupling is 3 and long range coupling we can call it JN. So the factors that affect this spin-spin coupling is basically hybridization of the atom what is the contribution of each of the like SP, SP2, SP3 what is the contribution and that dictates actually the what will be the coupling constant then it is it also depends upon what is the bond angle because that bond angle also plays an important role and dihedral angle and what is the length of like single bond or double bond or triple bond so bond length changes according to the bonding pattern and if there are substituent attached like electronegative group or neighboring group or different kind of bond, pi bond and lone pair and all those affects the coupling constant. So in next class we are going to in detail discuss how these each of these parameters affects the coupling constant and how using these concepts we can start interpreting some of the data. So we are looking forward for you to attend the next class and if you have any question don't hesitate to write or ask us we will try to respond those questions. Thank you very much.