 We were discussing about the chemical shifts and this is just a quick recap on what we did last time. So we have a nucleus which is here and you have an electron cloud around it which is indicated by the green dots here and we have an applied magnetic field which is B0 indicated here as B0 in the previous class I might have said it as H0 but it does not matter it is B0 here and this B0 field induces a current in this electronic cloud and which produces a magnetic field which opposes the externally applied magnetic field and therefore the field seen by the nucleus is Bi and that is given by B0 into 1 minus sigma i. So that is the sigma i is the so called a screening constant and this can vary depending upon the environment of your molecule and the processional frequency of a particular nucleus will then be new omega i or this is if you put it in terms of frequency in hertz that is minus gamma Bi divided by 2 pi and Bi is the field which is which is given here. Now this will then depend upon the magnetic field strength therefore since we want to characterize only the screening or the electronic environment around a nucleus we define an entity which is given here delta i is equal to nu i minus nu r divided by nu 0. Nu r is a kind of a reference compound and you measure the absorption frequencies with respect to a particular line of a reference compound and then this will be a very small number compared to nu 0, nu 0 is your spectrometer frequency this corresponds to B0. So divided by this nu 0 then of course this will be a number which will be very small and you multiply therefore by 10 to the power 6 although that is not I wrote it in the previous one and that is called as the PPM delta i is expressed in terms of PPM. Higher delta i means higher Bi implicating down field shift and therefore it is less screening and you will have a spectrum which is at a lower higher frequency of absorption and higher sigma i lowers delta implies that greater shielding that will means in a field shift and vice versa. So this is of course we already did this last time and let me explain that a little bit more here we take this example this is vinyl chloride. So vinyl chloride has 3 protons this has 3 protons here this is the electricity label them as A H, A HB and HC there is a double bond here and chlorine is an electron withdrawing group therefore it changes the electron density in these proton around the protons. So we said whatever is the electron density that determines how much will be the screening. So therefore if the electron density is lower then the screening will be less and therefore that will be at a higher frequency of absorption or the lower delta value I mean the higher delta value. So the frequency absorption is going like this. So the chemical shift here the HB which is the closest to the chloride which is the with the electron withdrawing group therefore this will have the lowest electron density here therefore the energy separation the Bi will be highest for this particular nucleus and this new B therefore will be the highest because the energy of absorption will be high the energy separation will be higher. This will be followed by new A which is indicated here this is a trans and therefore the effect of the chloride will be after this it will be more at this point which is trans to the chloride group and this will have the next frequency which is new A and the one which is assist to it that you look at the Van der Waals interactions and things like that the screening will be more here therefore the lowest delta will be for the new C therefore the frequency increases in this order and the chemical shifts numbers delta values will increase in this order. So this is how the chemical shift occurs because of the screening by the electron cloud around it and clearly this is an important parameter which will describe the electronic environment and therefore the structures of the molecules and there and a chemist used a large number of molecules to characterize this chemical shifts in the functions and what are the reference compounds I said there is with respect to a reference compound to define the chemical shift. So for the proton there are several reference compounds which are used TMS is tetramethylsilane and this is it has one line so that is taken as chemical shift that line is taken as 0.0 PPM and TMS is one which is used whenever your solvent is chloroform or DMSO and things like that and it is single line and which is the most upfield because the screening is the highest for that there are three methyl groups and therefore a silane SI CH3. So therefore this is the 0 PPM CH3 4. Now tetramethylsilane propionate this is the molecule which is used for soluble systems if you are working in water then typically you use this tetramethylsilane propionate and that is that also is one line and this other things are deuterated and you will have one line and that is you have 0.0. Another common compound which is used is DSS sodium 4-4 dimethyl 4-celepentane sulfonate this has many lines but if you deuterate all other protons accepting one then you will have only one line otherwise you have the most upfield line which is taken as 0.0. Other compounds are also used acetonitrile is used sometimes dioxane is used tertiary butanol is used and various different kinds of compounds are used for referencing and what numbers which are given here are with respect to the TSP or the TMS and you have this one as the reference chemical shift and others will be a reference according to that. For carbon 13 similarly you use TMS, TSP or DSS and these are all at zero frequency. For nitrogen 15 you use ammonium chloride or liquid ammonia and both are these their chemical shifts are taken as 0.0 and for phosphorus you 85% phosphoric acid and that also the line is taken as 0.0. So all your you either you add these things internally inside the sample itself or you put these in a capillary and insert that in your sample cube then it is called as an external reference if you put that inside the molecule itself then it is called as internal reference. So I can I can write that here. So you can have an external reference in that case the capillary containing the reference compound is introduced in the sample cube. If you directly put this molecule inside your sample then it is an internal reference. Sometimes you use if you are working with water you look at the water signal itself you look at the water signal itself and make as make that as the reference compound. Often you do not like to put additional molecules inside your sample especially you are working with biological systems with proteins or nucleic acids or any other biological system you do not want to introduce an external molecule external perturbation because it can change your conditions it may change the pH or things like that. So when you do not want to do that then what you do is reference your water molecule water signal with respect to that and used with the water signal itself as a reference compound that is that often is done. Now so having worked with a large number of organic molecules the chemist have come out with certain kinds of understanding as to what sort of a groups will have what kind of PPM in their values and this actually figure shows that sort of an information here. So 0 PPM is what we said earlier this is the TMS or TSP or whatever and if you have saturated compounds and those signals will appear between 1 and 2 PPM you have the allylic ones they appear here in this region and you have the CH2X which are the electron withdrawing groups here CH2O, CH2O1 or this ROH alcohol. Now what we are talking about is the signals from R not the OH and those ones will appear in this area then you have the vinylic compounds and the which I showed you the vinylic vinyl chloride earlier and those ones will appear in this region 5 to 6 PPM and if we have the aromatic compounds the aromatic compounds will appear between 6 to 8 PPM or 8.5 PPM all these benzene rings various kinds of rings which are there the protons in those ones will appear in this area and if you have the aldehyde group the RCHO then we are talking about this particular proton here this proton will appear at this place here and RCOOH this is the carboxylic acid group this proton will appear here and here of course the OH also will appear in this area it can vary from here this is a wide range here depending upon what the R is you will have a wide range of things and depending upon what the conditions are what is the pH of the solution you can have this OH proton appearing in a wide range of frequencies. So therefore it is very sensitive to the conditions of your sample and that proton chemical shifts can vary quite a lot depending upon the environment in your solution. Similar exercise now let me explain to you I explained last time about the aromatic ring effects the aromatic ring effects what is shown here is if I have a ring I showed here a ring and it has the electron cloud above and below and it produces a external induced magnetic field because of the currents inside this the field goes like this and like this and you can have a proton anyway present. So depending upon the proton you will have different kinds of chemical shifts what are shown here are the isochielding lines so let us say I have a ring here and then I draw a cylinder around this if I draw a cylinder the electronic current is here okay now let us say I consider a position I mean this I define the coordinate system here and this coordinate system here this is called the z coordinate and this is called as the these are the cylindrical coordinates so rho and z are the cylindrical coordinates so depending upon where your proton is suppose I say I have a proton here okay so it will have a certain coordinate z coordinate and the rho coordinate so this is the coordinate of these are the cylindrical coordinates for this particular proton what will be the chemical shift there so what is what has been done is the previous slide which I showed here so what is plotted is the z versus rho one particular one particular quarter of this so this is this entire circle instead of the entire circle you particular take one quarter of this so you have the cylinder lying on top here like this so I take this particular quarter here so if I plot this how the z axis is on this going up the z axis on this axis on the right orthogonal is the row axis and what are the chemical shifts in those ones so now you see here as you go up as you go up along the z axis so that is if you are on the top here if you are there then you have positive shifts here plus 4.00 3.0 2.0 1.0 in all these positions at all this position it will have the same 1.00 if you are here along this contour then you will have 0.00 if you are here 4.00 if you are here then it is 4.00 so this is all positive numbers here and this is the 0 axis this is the no change here and if you are on this side as the row is increasing here you start getting negative numbers which means you are on the horizontal plane of the plane of the ring if you are on the plane of the ring then you get negative shifts and that is to be expected because the field is opposing the field is opposing in the horizontal plane in the vertical axis the field is adding to the your main H0 field and therefore there it causes the positive shift the iso shielding curves and this is on the negative side so this is the calculation which has been done by theoretical procedures using the theory of chemical shifts in great detail and therefore this is extremely valuable to predict depending upon where you are with respect to the ring respect to the ring where you are you will have different kinds of chemical shifts this is important to know how far it can go and these are in certain units of 1.39 1.39 angstroms you see each one of these the z and the row are given in units of 1.39 angstroms for the some particular reason they have used this number 1 point is the chemical length of a carbon-carbon bond so particular bond which they have chosen and they even this in this is taken from this reference the general of chemical phase you see 1958 quite old and of course those units are valid all along so this is an important effect which is the ring current effect this is also called the anisotropy because of the ring there is an anisotropic in the chemical shifts depending upon the orientation of the ring depending upon the location of your proton with respect to the ring you will have different chemical shifts and that is drawn in the form of a contour okay so because there are two coordinates which are responsible one is the on the row and one is Z depending upon the combination of these you will have different kinds of shifts okay so this is what explained to you now similarly for the proton that is for the proton similarly you can also have the carbon chemical shifts now the carbon chemical shifts also you see this carbon has a much wider range of chemical shifts while proton went from 0 to 10 or 12 ppm carbon goes from 0 to 200 and 230 ppm this is a quite a wide range of chemical shifts you have here and once again depending upon the nature of the carbon your different kinds of chemical shifts the highlighted carbon is indicated in red colour here so you have RCOC that appears in this area C double bound CC that appears here in this range and this range and C aromatic you have this one here CSR NR2 CH saturated alkanes CBR and CI so all of these these are the halides here iodine chlorine fluorine and those ones will appear in this area R2C CH2 is if you look at this CA carbon it will appear here and if you have the aromatics the aromatic carbons will appear from 100 to 150 ppm hetero aromatics means if you have this hetero atoms substituents over there they will appear here and RC triple bond N that will appear here sulfoxide sulfones see this is the result of enormous amount of data collection there is a huge database such thing databases are available in the literature and there are also many many volumes of database of the chemical shifts reported and from this you draw this sort of a picture so amides appear here RCO NR2 RCO2 R dash so these are the acids esters here and these are the carboxylic acids and they appear in this region 150 to 200 ppm typically the carbonyl carbons so you see the typically the carbonyl carbons appear in this area aromatic carbons appear in this area aliphatic carbons appear in this area and mostly we will be dealing with these kind of chemical shifts when we do with proteins or nucleic acids because there you have the aliphatic carbons they will all appear in this area and the aromatic carbons the carbonyl groups in peptide bonds and things like that they will all appear in this area aromatic carbons in this area and so on so forth phenylalanines and all of these aromatic ring rings in amino acids they will appear in this area so this sort of a classification is extremely useful for understanding the structures of molecules first of all you have to assign the individual resonances to particular nuclear both proton and carbon and once you have done that then you can go forward to see what sort of a structural indicators they are and now there is another important parameter which is called as the spin-spin coupling now the protons while they are surrounded by the electrons and that is of course important to screen the magnetic field but the protons themselves they also interact with each other suppose I take this CHCl2 and CHO this proton and this proton they are separated by 3 bonds okay so let me write here C let us say CH okay this is this molecule right so now you have this proton and this proton and here you have CLCl these two protons interact with each other because they are all magnetic they are magnetic moments they are magnetic dipoles therefore they interact with each other either through directly directly through space interaction so there are two kinds of interactions possible dipolar interactions dipolar interaction or there is other what is called as through the electrons through the electrons in the intervening bonds now dipolar interaction is a separate story this is actually most applicable in the case of solid state NMR and relaxation phenomena and we will also talk about it a little bit later and mostly for what we see in organic molecules or molecules these the interaction through the electrons and this is also called as J coupling spin spin coupling is also this coupling is called as J coupling okay what is the result of this I am just showing you the spectrum here you have the delta B this is the chemical shift you have the chemical shift of the particular nucleus let us say we call this one of them as A other one as B so let me say this is A and this is B and you have the chemical shift of B here and the chemical shift of A here but this line is now not appearing at delta B it is actually getting split into two lines this is split into two lines too and the separation between them is the so called spin spin coupling constant and that is J AB why does this happen how does this happen so we can represent this as in this manner let us say I have a proton here I write a proton here and this is my H C C H so this is the orientation of the spin here if I call this as let us say the alpha state is the alpha oriented parallel to the field so there are also electrons here so there are also electrons here right in the bonds there are electrons so these electrons what happens is the proton nuclear magnetic moment polarizes the electron spin electron magnetic moment and makes it orient in a direction opposite to the alpha state out to the proton nuclear orientation now since these two are in the same bond so this electron which has to this bond has two electrons the second one will be oriented in this manner because they have to be anti-parallel that contributes to the lowest energy state and therefore it will be anti-parallel once this one is there then suppose this were carbon 13 suppose the carbon this is also has a nucleus and now it will orient this nuclear I should do a different color here again so this is the nuclear magnetic nuclear spin it oriented like this and this in turn orient the electron here it will put in this direction and the pairing electron it puts in this direction and now once again let us let us put the carbon will again be up so the carbon here will be up if it has to be open now then they put the proton once more and the electron I put it here like this and then like this and what will happen to this so therefore this will become like that so these two protons are anti-parallel to each other okay so therefore they are oriented in opposite direction now in terms of the interaction between these two how do we describe this we describe this interaction as J I A dot I B now if they are opposite to each other if they are opposite to each other this interaction will be it will produce a negative number right I A dot I B is I A I B cosine 180 I Z A I Z B cosine pi so if they are exactly one day opposite to each other this will be negative therefore if they are opposite to each other they will have a different energy level okay so because of this now if J is positive this interact this state will be the lower in energy compared to the other state where this can be positive also it is not the other thing is not possible it is also possible that this fellow will become negative down here this goes up the other configuration is also possible the other configuration will be you have this here the other one is let us say it is also like this is another configuration okay this is configuration one this is the configuration two this is also possible this will have a different energy compared to this and this energy will be lower compared to this okay so in other words the proton alpha state will be split into two levels so now I draw here the alpha state of the proton this will have two energy levels because it will be split into two because of the orientation of the other spin now this is let us say I call it as nu A and this is alpha state of nu A and here the I will write here the spin state of the X nucleus let me write the X nucleus and if this is beta this will be down the energy because these are opposite orientations and this is alpha here this is the orientation of the other spin this is the beta spin the B spin orientation so therefore these two have different energies obviously instead of one line coming here there will be two lines coming there if I call this as the beta state so here again it is split into two okay it is split into two why in the beta with the alpha goes down so this one this is the alpha and this is the beta because I 1 dot I 2 if the two are in the same direction if the two are in the same direction it will be positive energy if they were in the opposite direction then it will be negative energy so therefore the beta alpha goes down beta beta goes up alpha alpha goes up alpha beta goes down so therefore they will have this orientation like this and what is this energy separation here this will be of the order of few hertz and that is J which is indicated similarly here now what are the transitions possible here for the transitions will be dependent on delta M is equal to plus minus 1 so what instead of this transition I will have two transitions now which are the two transitions possible I can have a transition M is equal to from M is equal to 0 to 1 so what is the what is the M value for the beta beta the M value here let me write the M value this is minus 1 the beta value M value is 0 here because of the beta alpha one is down other one is up the sum of the two M values is equal to 0 here similarly this one is 1 and this one is 0 the total M value the summation of the M values I am writing here of the individual spins this is now I should have a transition where delta M is equal to plus minus 1 this is summation M i of the two now which transitions are possible I will have a I will have a transition possible from here to here 1 to 0 and 0 to minus 1 I will have another transition possible here there are two transitions possible okay so therefore this leads to a separation okay how much is this separation here this is J by you can we can calculate what is the value of that this will be J by 4 this will be J by 4 the separation will be J by 4 this will also be J by energy which is going down will be J by 4 okay that is J by 2 total so now if you look at this therefore you will have two lines for each one of those so we will have two green two sign lines so one is between these two states this is from here to here which I indicated and the other one is from here to here 0 to minus 1 up and down both are possible okay therefore delta M is equal to plus minus 1 will have these two states and the separation between them will be J okay this is how the splitting occurs between the two energy levels so we can do the same thing for the other spin as well and that is beta spin I wrote it for the nu A notice this is in megahertz okay this nu A is in megahertz and this is in hertz therefore this is a very small number this is a very small number so similarly one can write for the nu B as well so I can write for nu B as well I will also have two lines corresponding here and the separation between them will be this is nu B and this separation will be J AB I showed you for the three bonds you can have coupling from one bond you can have one bond coupling that is let us say proton to C 13 or you can have proton to a nitrogen 15 you can have two bond couplings this is one we can also have two bond couplings that is like CH CH here so that is a coupling between these two this is a two bond coupling I showed you the three bond couplings that is H C CH these are three bond couplings the same mechanism applies in all of those cases and notice here this how far does it go how far this relay goes it doesn't go beyond four bonds okay because if you go beyond four bonds then the effect will be relatively small then you don't see the coupling as much so typically so you will see from one to four bonds coupling and they have different ranges so if I want to write the these are typically of the order of one bond couplings of the order of carbon couplings are of the order of 140 to 200 hertz that is a and then you have 90 to 100 hertz for the proton this one and this is for this one and the two bond couplings are of the order of 14 to 18 hertz and three bond couplings will also be of the order of 3 to 15 hertz okay and incidentally one more thing one has to notice is the two bond couplings can have a negative value now go back to this previous slide here now I showed you the value of J interaction is J I1 dot I2 right so if J is positive I1 dot I2 if it is negative then there is a decrease in energy if J is negative then it will be I1 dot I2 negative will be positive right so the total will become positive in the case of two just if you take two bond two bond let us say HCH and I draw the spins here like this and let us say this one is like this this one is like this electron spin and I have here this one let us say like this this one like this and this is the most stable state let us assume this is the most stable state now you see the two proton spins are parallel to each other and this is if the stable state this actually has to be energy has to be negative so this parallel state has lower energy so this implies that J is less than 0 because it decreases the energy value and therefore the geminal couplings the so-called geminal couplings are negative whereas the visceral couplings and the one bond couplings they are all positive okay so I think I think I will stop here again we go to the next one in the next class