 Good morning. So, welcome to today's lecture. We were discussing the scalar coupling or J coupling or spin-spin coupling and we will continue discussing today. So, in the last class we were discussing how these splitting happens because of two spins are in neighbor and these are connected through bond and then we went ahead in defining magnetically equivalent and chemically equivalent. We said that magnetically equivalent does not cross the splitting and chemically equivalent which groups have a same chemical shift and all those and then we moved ahead and looking at what actually is the reason for different coupling strength. So, we defined that spin-spin coupling over one bond is called J1 and then if it is two bond coupling then it will be called like 2J. So, here is the number of bond and this is J is J coupling and this is also called Geminal coupling. Now, three bonds coupling are called a Visinal coupling. So, that is denoted by J3 and long range coupling will be called like more than 4 bond they will call long range coupling and that is denoted by like as usual N is the superscript. Now, we ended it here and we defined the factors that affects the spin-spin coupling. So, what are those factors? So, some of those factors that we listed out are the hybridization of the atom involves in the coupling. So, this hybridization is one of the important factor and then that also is somehow connected with what kind of bond angle is there between the spins. So, that also affects the dihedral angle. So, dihedral angle is between two planes and generally this is very important in case of protein structure. So, dihedral angles also dictates the strength of the coupling and then the bond length like Cc single bond it is there or double bond there that also affects today we are going to understand all these and if there is a substituent like if it is a lone pair or there is a pi bond or some electronegative group are attached that also affects the strength of the spin-spin coupling or J coupling. So, that we will be discussing today giving some examples. So, let us start the one bond coupling. So, one bond coupling is like if we consider the coupling between proton and deuterium. So, suppose we have a molecule like here we H and D what is the one bond coupling between the one connected proton with deuterium. So, in this case the one bond coupling between D H is generally the 42.94 that is 43 hertz and as you know that proton. So, one bond H H coupling and D H coupling. So, D H coupling is 42. So, H H coupling can be quite high and if you look at this is dictated by the gyromagnetic ratio. So, the gyromagnetic ratio of proton is 6.5 times higher than the deuterium. So, in that case J 1 H H is 280 hertz. So, now we come to the heteronuclear J coupling. So, that as we discussed in the previous slide that depends upon the hybridization strength. So, let us take some of the example of J coupling between 13C and proton. So, if it is sp3 hybridized and you see this bond angle is also 109. So, here the J 1 bond coupling in this case is 125 hertz. Now, if this becomes sp2 hybridized the coupling strength actually increases and it becomes 165 hertz and when it is sp hybridized like then S contribution increasing. So, bond angle is also increasing and that gives us 250 hertz. So, you can look at the methane like here sp3 hybridized where 4 protons are attached in this case it is 125 hertz and as you increase the S contribution here actually the coupling constant strength increases. Now, other thing that affects the coupling strength is the substituent that can also change the spin-spin coupling and that value change can happen in tune of 20 to 30 hertz. So, just for an example now this was one bond coupling homonuclear and heteronuclear. Now, we will come to the two bond coupling these two bond coupling proton-proton two bond coupling are called geminal coupling and there is somewhere between now minus 23 to 42 hertz. As we discussed the splitting strength does not depend whether the coupling is negative or positive it is absolute value this is dictated by the orientation of spin. So, whether it is negative or positive. So, let us look at some of the two bond coupling geminal coupling and that is dictated by the bond angle between HCH here. So, if you look at here in this case two bond coupling is minus 12.5 hertz the same two bond coupling between these two decrease and that becomes minus 4.3. So, if you look at the bond angle is changing here here also two bond coupling because of bond angle change it is 2.5 hertz. So, as we move from here the sign of J coupling is changing and bond angle is changing and that is how it is also decreasing in the magnitude. So, some more example if you look at this molecule here the J bond coupling is minus 6 hertz and as we change so this becomes like now you can see it the substituent is also affecting the J bond coupling. So, the difference between this and this R2 oxygen are attached and that changes by like 6 hertz or so. Here again one oxygen is attached here replaced carbon is replaced by oxygen and you can look here it was minus 4.3 and now it will become 5.5. So, almost 10 hertz coupling has changed. Similarly, one can look at here the proton-proton coupling between these two change by almost 14 hertz and this is because now here this carbon was replaced by nitrogen and the substituent is also affecting and nature of substituent is affecting the coupling strength. So, in this case again if N was replaced by oxygen you can see this coupling has increased and now it has become 42.2 hertz. So, the bond angle the hybridization the substituents are changing the strength of the coupling that is what we understood here. Now three bond coupling between proton and proton-proton. So, this is generally is gives the dihedral dependence or phi dependence. So, this phi is very important and Martin Carplus actually provided an empirical relation which correlates the three bond HH coupling with the phi torsion angle in the peptide bond. So, this has become very instrumental tool to know this conformation of a protein and if one can calculate the three bond coupling between proton and proton essentially it is like this angle phi torsion angle between two proton. So, here is and H and N I will come to that in next slide. So, if one can determine the phi torsion angle between two planes one from one amino acid and another amino acid you know the dihedral angle and the coupling between this proton and this proton dictates the phi torsion angle by this empirical relation and if you know the three bond coupling you can determine the phi and that gives the torsion angle in the protein case. So, ABC are constant and they have empirical value for the hydrocarbons. So, like typically A is 7 hertz, B is minus 1 hertz and C is 5 hertz and you can calculate putting at the different phi value how the J varies between 180 degree and 0 degree and it is very small when it is 90 degree. So, I will come to the next slide to show the variation of the J couplings on the phi. So, here like if the phi torsion angle between these two planes, so this is a plane, if the phi torsion angle is around 60 degree, so you have a three bond proton-proton coupling is 2 to 5 hertz, if it is 180 degree then it is 10 to 16 hertz and if it is 60 degree it is 2 to 5 hertz. So, depending upon how they rotate here you can see the J value, the J value three bond coupling is changing. And so one if you can experimentally determine this J value we can calculate the phi torsion angle like similarly some more three bond coupling in this molecule the here 1, 2 and 3, these are three bonds. So, the coupling between this and this proton is typically of 4 hertz, if you take the same molecule but orientation is different now this becomes 9 hertz. So, if you compare this orientation with this orientation what I change is the angle and that angle change the J coupling value. Similarly, like if you look at here if J bond between cis protons and trans protons differs because there is a cis the angle between these phi angle is 0 degree in trans it is 180 degree. So, this is 8 to 12 hertz and here it is 14 to 18 hertz. So, now it is clear that even the torsion angle changes the three bond coupling in case of proton-proton. And therefore, this as we were discussing this is very important in case of getting the protein conformation. So, in protein suppose if we measure this coupling between this proton, this proton which is NH proton and this C alpha H proton. So, here is the torsion angle which is phi if one can measure experimentally we can determine the phi torsion angle. So, this is actually phi torsion angle. So, what we are measuring the coupling constant between these two protons and this is three bond one bond, two bond and three bond. So, by measuring this three bond coupling one can determine the phi torsion angle. And if you remember your basic Ramachandran plot it is between phi and psi. So, phi plot and psi plot minus 180 to 80 00 plus 180. So, one torsion angle just by experimentally determining from the three bond coupling one can know what is the phi and phi can change between plus 180 to 180. So, at least from phi you know that what is the phi angle and then one need to determine the phi angle then you know the conformation of protein or a secondary structure of a protein. So, therefore, the three bond coupling or the coupling in general is very important to get the molecular structure for a small molecule, small organic molecule using this coupling one can determine at least partially conformation of a protein because we can determine the directly phi angle from this three bond coupling. Now, what I will do today I will give some example to make more clear that how the spectrum looks like. So, I will give you some example of say let us take some of the example like benzyl acetate and we try to understand how the spectrum looks like and how you can interpret those depending upon the structure. So, if you look at benzyl acetate is like this. So, there is a benzene ring then CH2 is attached to this and then COO and then CH3. So, how many types of proton we have essentially we have a three types of proton. One type is coming from like if you are recording on a relatively smaller magnet we have a like one proton coming from the benzene ring here we have five proton 1, 2, 3, 4, 5 then second kind of protons we have here two kind of proton coming from methylene group here and then third kind of from methyl group here. So, three kinds of proton we have. So, therefore we should have at least three peaks in the spectrum and their intensity will vary depending upon how many protons are contributing. So, here we are having contribution from three protons here we are having contribution from two proton and here we are having contribution from five proton. So, if this is 0 ppm is our TMS then the first one will be like from the methyl proton that will be around two ppm and this will correspond to three proton. Then will be from methylene this is this will be little bit downfield shifted because it is connected to ring as well as the COO group COO is a electronegative group. So, that will be downfield shifted or higher ppm value shifted and that will correspond to two proton. Now, here for benzyl ring we have a five proton so that will correspond to five proton. So, that will be the spectrum of benzyl acetate. Now, let us look at some of the some other example and we take the example of suppose like say benzyl acetate the carbon spectrum if we look at. So, how many types of carbon now we have? So, let us look at the carbon that we have same molecule here let us go back let us look at the carbon molecule. So, we have the carbon here we have a six carbon here one carbon here seven and eight and nine carbon. Now, how many types of carbon we have? So, we go back here and we have like here if you look at these three carbons are chemically equivalent these two carbons are chemically equivalent this is another one now CH2 is another one and COO is another one. So, and yeah and here is the methyl is another one. So, we have six type of carbon therefore, we should get six peaks and where are those? So, the first one we will look at the methyl peak because that will be most upfield shifted or lower ppm value so that comes around 20 ppm. So, this is for CH3. The next one here the next one is from the methylene group and that will be here. Now, after that what we have is three aromatic peak. So, three aromatic peak like two here then so, three aromatic peaks are coming here. So, that are quite close here if you look at like these three three and four are quite close and two is because two is connected to CH2 and all those. So, this will be ortho so that is two and one is carbon ill peak that will be most downfield shifted. So, this is for CO this is from the aromatic carbon two and these are three and four they are quite close and then we have from CH2 and this is CH3. So, that is benzyl acetate carbon spectrum. Then we move ahead and take some another example to understand how this spectrum looks like. I will give you and now we have to analyze. Our molecule is suppose CH2 H4Ca and suppose we have a spectrum which is like this here is our 0 ppm and this is our signal coming from the standard reference and just suppose we get only one peak. So, what will be this molecule? Let us define. So, one peak are coming, four protons are there this is for proton spectrum and this comes suppose around 4 ppm. So, here one two three four five six something like this. So, what this molecule is? So, now it says that it looks like four all protons are in the same environment. So, there are two carbons here four protons. So, we will write four protons here and all four are in the same environment. So, it is this molecule right. One two dichloroethane. Let us take some other example. Suppose our molecule is now C9 and H12 and we got this spectrum which is 0 ppm standard and we got here and here and this ppm value is suppose to 0.5 ppm and this one is around say 6.5 ppm. So, if you integrate here we get ratio of 3 to 1. Now, what this molecule can we solve this? So, two things are clear. First thing is if something coming around 6.5 that must be coming from the benzene ring. So, let us write a benzene ring and something coming around 2.5 most likely this is coming from the CH3 or CH2. So, since now there are 12 protons. So, you look at where we can adjust and proton and all three like here. So, 6 carbons are adjusted here we have 3 carbons and then 6 carbons are gone here. So, we have actually 6 proton left here. So, now that means we can if we have molecule like this because the methyl are in the same environment. So, therefore, it should be something like this. So, now we have a 3 protons left here and these 9. So, we have two kinds of proton one coming from methyl group. So, 9 methyl here and 3 here from the benzene ring. So, therefore, 9 ratio 3 is 1 ratio 3 and their chemical shift position is 2.5 and 6.5 ppm. This is the molecule that that is a spectrum is like this. Next move to next molecule we have a molecule which is C4 sorry, C4 H10 and O2 and this gives us a spectra here say 0 ppm and this gives us spectra of 2 quietly close and the ppm value is between 3 to 4 ppm and the ratio is 6 to 4 what this molecule is. So, how can we solve it? Now, it looks like that only two kind of protons are there. So, this is reference and then we have to write it like that. So, 4 carbons are there. So, 2 carbons here and since they look quite downfield shifted. So, probably there is two oxygens here. So, now we have to fill the protons. So, 6 protons are of same chemical environment. So, that can be CH3 and CH3 and then 4 protons are of same kind. So, here will be 4 protons. So, this is the spectrum of this molecule. So, at the moment we are looked at the chemical shift only and how to interpret the spectrum and generate the spectrum from a given compound. So, that we looked at. Now, I will give you some examples of the splitting. So, we will try to determine the chemical structure of the compounds. So, suppose now we consider the spin-spin coupling to determine what molecule is this. So, let us take some example like C3H7Cl and we have a spectrum where our reference is at 0 ppm, TMS and we get here 2 peaks and here we get a multiplied like this. So, here the ratio is 1 and this ratio is 6. So, what this molecule could be? So, let us try to understand this. So, one thing is clear that 6 protons are here, one proton is of this type. So, only two kind of chemically different protons are there and then there is a chlorine group attached to it. So, let us attach a chlorine group, Cl group. So, then two kinds of protons. So, it looks like that and since the value is here, this is around like 1 ppm and this is 2 ppm and this one is 4 ppm and 5 ppm. So, it is very clear that one is methyl proton. So, here is CH3 and another CH3. So, now 6 protons are gone, one proton which is left is this one. Now, this we can explain it. Now, 6 proton will split this into how many 7 and that ratio one can know that what will be the ratio of this splitting. So, one can identify these molecules very easily. For 6 proton it should be 1 is to 6 is to 15 to 20 to 15 to 6 to 1 and these proton will split these chemically equivalent 6 protons into this. So, that is why it is doublet. So, these two are equivalent therefore, there should have been one peak if there was no coupling, but because of coupling we have a splitting into 2 and the 6 proton split this into 1, 2, 3, 4, 5, 6, 7. So, this is accepted. So, next I will give one more example before we try to close up. So, let us have a molecule which is C12 is 14 and O4 and the spectrum for this is coming something like this. Here we have a reference compound 0 PPM, then we have something like this and a quartet like this and then we have a singlet like this and this corresponds to 4 peaks, this corresponds to also 4 peaks and this corresponds to 6 peaks and the value here comes around say 7 to 8 PPM sorry 7 to 8 PPM and this is somewhere around 4 PPM and this comes around say 1.5 PPM. So, 1.5 PPM. So, what this compound could be? So, 1.5 is clearly methyl group is there and there are 6 carbons probably 2 methyl group are there. Here there is a benzene ring. So, let us try to write it a benzene ring first and 2 methyl are attached to there. Now, we see the 4 peaks are also there. So, that means now these methyl and there is here also that there is a something called methylene group. So, probably it looks like like there is a structure which will be something like this because now methylene groups will come here that are 4. So, if you look at 2 to 4 this will be 6 and this corresponds to 4 proton and then you can also explain this splitting because each of these equivalent proton split this into 4 and this will split into 3, this CH2 will split CH3 into 3. So, 6 proton will be splitted into triplet. Now, these 4 protons will be splitted into quartet and this will be mostly singlet because this there is no splitting here. So, these are some of the example, rest examples you can do in the tutorials. So, we will continue now we have a good exposure of the chemical shift how to interpret using chemical shift. We also introduced the coupling constant and we looked at how the coupling constant gives the final structure. I mean we learned looking at the splitting pattern and the chemical shift value how we can identify the chemical structure of a molecule. Now, we are at a stage going little deeper and understand the quantum mechanical formalism of the coupling constant and then we looked at interpretation of some of those. So, I hope these things are clear to you and rest the concepts will be cleared in the tutorial session. So, I am hoping you to see in the next class where we will start with the quantum mechanical description of the coupling. Thank you very much.