 Hello everyone once again I welcome you all to MSP lecture series on interpretative spectroscopy. In my last lecture I started discussion on spin-spin splitting and also the term I introduced is coupling constant. Let us continue from where I had stopped in my last lecture. So coupling constant I had mentioned it is very vital in understanding and elucidating the structures using NMR and what one should remember is equivalent protons do not split each other. So that means usually protons present on the same carbon atom we are considering as equivalent and they do not split each other and especially methyl groups and it is likely that in some cases where we have methylene and if they are substituted with different on adjacent side then there is a possibility of splitting each other that arises when we have a second order situation. So as far as first order spectra are concerned equivalent protons do not split each other. Protons bonded to the same carbon atom will split each other if they are not equivalent due to unsymmetrical nature of the molecule. As I had mentioned the unsymmetry comes into the picture when they are differently substituted on both sides. Protons on adjacent carbon atoms normally will couple most of the time unless they are separated by atoms such as oxygen or sulfur or chlorine something like that then they may not couple but on other side in general protons on adjacent carbon atoms normally will couple and protons separated by four or more bonds will not couple or show very weak coupling. So as long as there are fewer bonds apart that means 1, 2, 3 or to an extent 4 they can show coupling but if they are beyond that one so they may or may not show any coupling even if they show coupling 5 bond 6 bond or something that is very very weak. So now let us consider one example of ethyl benzene and looking into the splitting pattern here and of course here while elucidating and understanding and interpreting the spectra the symmetry comes very handy. For example if you just look into the molecule if you want to focus your attention on aromatic protons here and you can see here you can rotate this way when you can rotate this way basically what happens you can rotate this through C2 axis in that case this one and this one are equivalent and these two are equivalent and this one is there that means it should show three signals in its aromatic region in a ratio of 1 is to 2 is to 2 but on the other hand this one would be very similar to what we come across in case of ethyl alcohol or chloroethane we see here this one will be appearing as a triplet and this one will be appearing as a quadrate and this will be overall it will be a multiplet so something like this it can show here and then to make you familiar with splitting pattern let me write again here how CH2 is getting quadrate here. So for example here now we have three protons are there and these three protons I am representing in this format something like this and then let us consider the situation one where all of them are aligned with the magnetic field this is one situation and the next situation is one two are and then one is down and now either possibilities one like this on down and one up so the middle one is down and the third case is this one is down and these two up. So these are the possibilities when we have two aligned and one opposing and next when we look into it we can have one like this and two down and then here I can write here so two down like this and one here or I can like one here one here and one here so these are the three possibilities and then the third case is all of them are opposing the applied magnetic field when they are applied magnetic field so here it is shielded and here it is deshielded that is in the same order the peaks will appear and this whatever the four peaks we are seeing here you can see this is one two three are there three one two three are there and one and one so we will see here one three three one so this is how it comes and same analogy holds good for identifying here one is to two is to one how that happens now CH2 as I mentioned here both of them can be something like this or one up one down or one down one up and then both of them are down so here one is to two is to one we will see one is to two is to one so this is how we can calculate and look into relative intensities simply by doing this one of course we cannot do for all of them that is reason we use Pascal triangle to identify the intensity of each line in a multiplied here I have shown the chemical shifts also all of them are here multiplied with very low coupling between them and you can see here this is 2.72 and this is 1.18 here so now let us look into the splitting pattern for isopropyl groups here and here we can anticipate three type of signals here this one is not coupled this may show a singlet and this would be showing a subtert 7 1 is there and of course I do not want to write here 7 you can calculate the intensity or look into the intensity by simply looking into the Pascal triangle and the spectrum would look something like this you can see here so this one if you see this one is a singlet it is appearing here and then this one will be split by these two n plus 1 this is 7 lines and 7 lines will be there here 1 2 3 4 5 6 7 will be there and then these two will be split by this one it shows a doublet here and if you just look into carefully this is spacing is same as this spacing this we call it as J and then if you calculate here 1 2 3 3 J H H coupling is what we are observing in these spaces I have shown the expansion here so you can see here this is the expanded one how nicely 1 2 3 4 5 6 7 lines are here so now distance between the peaks of multiplet measured in Hertz whatever the distance we I showed you that we call it as coupling constant this is always measured in Hertz and then they are independent of strength of the external field again when we measure the coupling constant in Hertz they are independent of strength of the external field the external field if you term it as be not they are independent of the field similarly chemical shifts when they are presented in ppm they are independent of strength of the external field so multiplets with the same coupling constants may come from adjacent groups of protons that split each other I repeat again multiplets with the same coupling constants may come from adjacent groups of protons that split each other so now let us look into coupling constant values for different types of molecules you can see here 1 2 3 bond coupling is there and then this is around 7 Hertz and then here 1 2 3 bond coupling is there and between we have a double bond so it is 10 Hertz 10 Hertz for sis and when we have trans that means magnitude of trans coupling is more than the magnitude of sis coupling you can see here sis coupling is 10 Hertz whereas trans coupling is 15 Hertz and the geminal if there is any non equivalence is there and the substitution in that case what happens you can observe very small coupling between the hydrogen atoms present on the same carbon this we call it as geminal coupling this geminal coupling is very low and then in case of aromatic groups this ortho proton coupling will be in the order of 8 Hertz and then meta will be 1 2 3 4 bonds here 1 2 3 bonds this is 3 j and then here 1 2 3 4 bond 4 j it is 2 Hertz and then in case of ally link what we have is 1 2 3 this is around 6 Hertz so some of these things one should remember or by repeatedly interpreting the spectra of several organic molecules you will be knowing exactly where the coupling constant value would come what is the magnitude of the coupling constant in absence of free rotation due to steric hindrance by a ring or bulky groups splitting and coupling constant values may vary again you cannot really set this as a standard value it can also vary due to free rotations or due to steric hindrance by ring or bulky groups because coupling constants are often conformation sensitive so now let us look into complex splitting let's consider styrene molecule if you look into styrene molecule okay these three protons are there they are designated as HA HB and HC if you look into molecule this is unsymmetrically substituted because HA is 6 to HC whereas trans to HB as a result what happens these two appear to be chemically equivalent no doubt but they are not magnetically equivalent so that means when two nuclei or two hydrogen atoms on same carbon atom are chemically equivalent but they are not magnetically equivalent then we come across complex splitting but let us consider assuming this is a first order spectrum in this case what happens you can see signals may be split by adjacent protons different from each other with the different coupling constants example HA of styrene is split by adjacent HB trans to it because first it should be split by trans because trans coupling is larger you know that it's 17 hertz and it is 11 hertz then it will be further the lines further split by a trans that means HC proton how that can be shown you can see here the spectra is given here and then values are also given here this comes around 5.25 and this comes around 5.76 and this comes around 6.72 and all aromatic protons are coming around 7.3 to 7.4 and then interestingly if you see here although it appears like a doublet it's actually a doublet of doublet same thing is true in case of this one this also doublet of doublet and this you can clearly see it's also a doublet of doublet that means four lines are there and here also four lines are there and here also four lines are there but spacings are a little smaller so how that's happening you can clearly see here that is first consider HA HA we have the chemical shift of 6.60 first it is HA is split by HP because the magnitude of trans coupling is more it first splits that into doublet the separation is 17 hertz here then each line in the doublet is further split into doublets because of coupling with cis coupling that is the magnitude of 11 hertz. Now at the end what we are getting is doublet of this is called as doublet of doublets the HA multiplied would be called as doublet of doublets and similarly when we look into HP HP again first it is split by HA because it's a trans coupling magnitude is more and now HP is further split by HC present on the same carbon atom because they are no longer equivalent so there is some non equivalence is there as a result they split each other as I mentioned this is called geminal coupling and geminal coupling values are of the order of 2 or less so here what we are getting is BC coupling is about 1.4 hertz that's the reason you can see in the spectrum you saw the spacing between two lines is very narrow so now let's consider into HC and in HC if we consider the larger coupling is one the cis coupling HA HC that is 11 hertz first splits into a doublet and again each line in HC in this one will be further split into doublet and the magnitude of this base is same what we came across here so this is 1.4 hertz that means this is also a doublet of doublets this also a doublet of doublets this also doublet of doublets and so as a result what we see is in this portion styrene the olefinic region what we are seeing is four signals each one is having four lines that four lines can be designated as a doublet of doublets doublet of doublets and doublet of doublets so this is what we saw in the spectrum here you can see here so this is all doublet of doublets here now let's look into another example here in this one chemical shift values are given here this is 1.63 a doublet is there because this is coupled with this one these two coupling is there and next 3.52 3.52 here here it is there and this is coupled with this one and then 3.78 again 3.78 here and then what we have is 4.10 we have here so now you can see the number of lines you should be able to tell what are the couplings involved in those things okay you try to work out if it's not possible we will solve that one again now let's look into another aspect while determining the coupling constant is stereochemical non- equivalence so what is this stereochemical non-equivalence in general two protons on the same carbon atom are equivalent and do not split each other so it means I gave emphasis about that one that no geminal coupling as long as the protons are hydrogen atoms on the same carbon atoms are equivalent but that is not the case in case of styrene because styrene if you just look into it I showed you HA is coupled is having transrelationship with one HB whereas it has cis relationship with HC as a result what happens the entire molecule is not symmetrical as a result what happens you observe geminal coupling so if the replacement of each of the protons of a CH2 group with an imaginary Z gives a stereo isomers then the geminal protons becomes non-equivalent and will split each other that's what exactly happened in case of styrene molecule so molecules having non-equivalent geminal protons I am showing you here few examples in this one you can see HA and HB are on the same carbon atoms but that non-equivalent and you can anticipate coupling between HA and HB they are geminal protons and same thing is to in case of styrene also so HB and HCR geminal and they are non-equivalent and you can anticipate coupling between them but however the magnitude of this coupling is very small same thing is to in case of this one also you can see you can see here D and C so geminal coupling is observed here so now the time dependence so what is the time dependence molecules are tumbling relative to the magnetic field so NMR is an average spectrum of the all orientations if you recall when we placed H2 and F2 molecules perpendicular to the magnetic field and parallel to the magnetic field these are the average orientations considered taking average of all possible orientations that means the molecules will be tumbling relative to the magnetic field so NMR is an average spectrum of all the orientations that's the reason we do spinning axial and equatorial protons on cyclohexane interconvert so rapidly that they give a single signal at room temperature if you record NMR spectrum for cyclohexane they interconvert axial and equatorial although we can differentiate them but the exchange or interconversion of axial to equatorial is much faster than NMR times scale of 10 to the minus nine seconds as a result what happens we get a single signal so proton transfers for OH and NH may occur so quickly that the proton is not split by adjacent protons in the molecule again same thing happens the exchange happens when we have groups such as OH and NH in a molecule and they the exchange is so rapid that basically what happens we don't see any splitting due to the presence of OH or NH on adjacent protons now let's look into hydroxyl proton here I have considered ethyl alcohol here and if you consider ultra pure samples of ethanol it shows splitting ultra pure sample of ethanol if you record quickly NMR it can show splitting of methylene due to OH as well for example you can clearly see here this is a triplet for methyl because of split by two methylene protons here and then this will be split into quadrate and each line in the quadrate is further split into a doublet because of HO and HO again split by this one into a triplet this is a very interesting one not often you come across very clearly you can see here so basically first it splits into quadrate this is 1 is to 3 is to 3 is to 1 and each one is split into doublet then it would appear something like this so this is what we are seeing in case of CH2 here and of course here in this case what happened this will split this one into a triplet here but ethanol with a small amount of acidic or basic impurities will not show this splitting this splitting due to OH vanishes with a small amount of acidic or basic impurities so now if you see here in this one it is not pure as a result that splitting whatever we saw is disappearing here so OH will show only one signal here and CH2 shows as usual quadrate and CH3C shows a triplet here so how about NH protons so moderate rate of exchange is observed when we have NH protons peak may be broad you can see here this compound we have here CH3 is there here and CH2 is a quadrate here and then this NH2 is showing you very small broad peak here so this is how most of the time we observe for OH and NH are NH2 peaks in the NMR spectrum so identifying the OH or NH peaks how to do that work chemical shifts will depend on concentration and solvent to verify that a particular peak is due to OH or NH shake the sample with the D2O for example because of difference in concentration we may not are exchange we may not see signals due to OH and NH or in the beginning we see it and then to confirm that one what one should do is the keep the solution in D2O for 24 hours and then record and then if the peaks due to OH or NH disappears that confirms that OH or NH is there deuterium will be exchanged with OH or NH protons and a second NMR spectrum the peak will be absent or much less intense so you take NMR for the freshly prepared solution and then keep it in D2O overnight and take it and if the peak disappears that confirms that they are due to either OH or NH let me stop here and continue more discussion on coupling constant and type of spectra we come across in my next lecture until then have an excellent time thank you