 Hello everyone, once again I welcome you all to MSB lecture series on trans-metallic chemistry. We are almost we are in the last stages of this course, this is the 58th lecture and in my previous lecture I started discussion on NMR spectroscopy to make you familiar with interpretation to solve the structure of some simple molecules that you come across to make yourself familiar with NMR as I mentioned if you are more interested it is a wonderful topic and wonderful subject for you know studying and also doing research and if you are more interested always you can go to fully dedicated NMR spectroscopy courses and also the books I showed you in the beginning of my 57th lecture. So let me continue from where I had stopped. In my previous lecture I was telling about spin-spin splitting and how NMR is so powerful in interpreting data and structure determination for molecules containing NMR active nuclei. I was telling about Pascal triangle for i equals half so this is the one and I said it is unique for each nuclear spin this is for i equals half if you have 7 lines we will be having intensity ratio of 1 is to 6 is to 15 is to 20 is to 15 is to 6 is to 1. Of course you can also examine this one by considering 6 equivalent protons are 6 arrows and start arranging them 6 all 6 are first upward and then 5 are upward 1 is down and then 1 can be 5th 1, 6th 1 or first 1, second 1 like that you have 15 options are there and then when you have 2 upward spin low spin and 4 upward spin you have 20 options and again it repeats in the same sequence here and we end up with 7 signals 7 lines in a multiplet of septate having this intensity. As I mentioned I told you that I will show you in case of isopropyl that is coming I think I would show you about how 7 lines are coming the range of magnetic coupling we will see now. Equalant protons do not split each other for example if we look into ethanol CH3, CH2OH in CH3 protons they do not split each other same way methylene protons they do not split each other protons bonded to the same carbon will split each other only if they are non-equalant if the 2 hydrogen atoms present on the same carbon atom will split each other only if they are not equalant. So those cases are there I can show you and then protons and adjacent carbon normally will couple when you look into 2 groups alkyl say CH3 group CH2 group or CH they normally couple and show spin splitting accordingly. So protons separated by 4 or more bonds will not couple because when you move further the influence of the magnetic field generated by those will be less as a result what happens they do not couple that is the reason we do not see very long range coupling for example when CH3 is here after CH2, CH2 there is one more so really it does not have much influence or it is magnetic field generated in the same magnetic field when we keep it it does not have influence on the radio frequency of this one as a result they do not couple. So now splitting for isopropyl group I showed you so your structure is shown here now these 2 are equivalent and now this is different and then this is different here so that means in this one if you see this CH3 is not coupled it will show a single a singlet here and now this one is split by these 6 into 7 lines that is shown here and of course it is expanded you can see here 1, 2, 3, 4, 5, 6, 7 the ratio if you see it is 1 is to 6 is to 15 is to 20 is to 15 is to 6 is to 1 so that means 1 is to 6 is to 15 is to 20 is to 15 is to 6 is to 1 you can examine that one and then for these 2 will be under the influence of this one it splits into a doublet here so this is how it is splitting and as I mentioned in case of ethanol I should show you CH3 CH2 OH so let us not worry about this one now this one will be split by these 3 so first these all the 3 will be in this fashion and now what happens we will be having 1, 2 like this and 1 like this and now this one can have now 1 like this this one like this and this like this and now what would happen it can be here and these 2 will be here like this so this is one set 3 are there and now what happened 1 is like this 2 are down and then 1 down up like this or like this and like this you cannot have anything else now we have all the 3 down like this so now if you see 1 is to 3 is to 3 is to 1 so that is the reason so this one will show a signal something like this and the other hand for this one would be influence of under these 2 so this 2 will be 1 like this 1 is like this 1 is like this 1 is like this 1 is like this and 1 will be like this 1 is to 2 is to 1 so this is how you can also write for 6 also 6 if isopropyl is there we have to write like this 1 like this 3 4 5 6 now you consider 1 all the 5 and 1 like this and 1 keep on shifting here and then you write for 2 3 like that so then you will end up with this ratio here so this is how you can calculate the spin multiplicity and also their relative intensities and this one simply you can do it from looking into Pascal triangle for this isopropanol spectrum I have shown here a septate is there and OH is coming around 4.8 and then this is coming around 1 4.02 and then this 2 are coming as a doublet around 1.21 what is coupling constant now another term is coming so distance between the peaks of a multiplied is called coupling constant if you go back to here if you see the separation between these 2 this is same as this separation this same as this separation is same and this separation is called j coupling constant and in this case what happens here if you consider this one 1 2 3 so if I write here like this this is 1 2 3 bond coupling 3 H H coupling and of course for this also it is same this also the 2 is there this also 3 J H H coupling this separation is same as the separation that indicates they are coupled to each other this is called coupling constant and distance between the peaks of multiplied then measured in Hertz not dependent on strength of the external field you should remember since it is measured in Hertz it is independent of magnetic field strength whether I record in 60 megahertz or 300 megahertz or 600 megahertz the coupling constant remains same and chemical shift in PPM remains same. Multiplates with the same coupling constant may come from adjacent group of protons that split each other so now values for coupling constant is given here for this one what happens when free rotation is there this is 7 Hertz and here the cis coupling is 10 Hertz and trans coupling is larger 15 Hertz on the same carbon due to some reason if it is molecule is unsymmetric and then the hydrogens on the same carbon will split then that is called geminal coupling that is 2 Hertz very small because they have very similar normal frequency 2 Hertz and then in case of ortho coupling in case of parametric group it is 8 Hertz meta it is 2 Hertz because further and normally you do not see para they are too far and allylic it is 6 Hertz the value of 7 Hertz in an alkyl group is averaged for rapid rotation about the carbon carbon bond rotation is hindered by a ring or bulky groups other splitting constant may be observed and complex splitting as I mentioned geminal coupling you can see here because this molecule is not symmetric and now the relationship between H B with H A is different from relation with H C with H A because these 2 are trans and these 2 are cis as a result they are not very similar they are chemically equivalent but magnetically they are not equivalent as a result what happens both of them will show different signals but margin is very less but they show coupling with this H A. So, signals may be split by adjacent protons different from each other with the different coupling constants once when you look into spectrum we should be able to assign the signal due to H A H B and H C very readily for example in the above example H A of styrene is split by an adjacent H trans to it by 17 Hertz this coupling is 17 Hertz whereas this coupling is 11 Hertz that means it shows a doublet of doublets one is 17 Hertz first you should write and this we called coupling or splitting or tree we call it splitting tree that we should be able to write it I will try to make you familiar in writing this one. So, H C is to it 11 Hertz that means trans is more 17 Hertz and the first it will be split by trans to give 2 signals and each signal will be further split into 2 each by H C with 11 Hertz you can see here 6.6 H A H A first you say 6.6 first it is coupled with this one splits into a doublet and then each line is further split it because the coupling constants are different it is stepwise you have to see the coupling and then this separation is 11 Hertz this separates 11 Hertz now it will be a doublet of doublets the separation this one is shown here and between this one and this one is shown here this is cis coupling and whereas this one is trans coupling and same thing this is for H A and then H B what happens of course H B first it splits into larger coupling and then the smaller coupling is between this one that is about 1.5 Hertz and now you can see the spacing this is due to B C coupling and this is due to whether this one you take or whether you take this one this is due to J A B coupling 17 Hertz so now with this one we should be able to analyze the coupling constant for the other one also this also shows the relationship between the mutually coupled protons so that writing their geometry or structure and correct conformation is also very easy so now spectrum of styrene is how it looks like this one will be showing a doublet of doublet this one will also showing a doublet of doublet and all three are showing doublet of doublet but wherever B C is there the coupling is very small 1.4 Hertz so with this one we should be able to make out which signal is due to which hydrogen atom very simple now let us look into stereochemical non equivalence what is the meaning of stereochemical non equivalence usually two protons on the same carbon are equivalent and do not split each other I mentioned you and then if the replacement of each of the protons of a CH2 group with an imaginary Z gives a stereo isomers then the protons are non equivalent and will split each other now time dependence molecules are tumbling relative to the magnetic field so NMR is an average spectrum of all the orientations axial and equatorial protons on cyclohexane interconvert so rapidly that they give a single signal for example if you take NMR spectrum of cyclohexane at room temperature what happens they are rapidly converting axial into equatorial equivalent axial because of this dynamic process what happens NMR time scale doesn't identify separately axial and equatorial and it they appear as single signal rapid tumbling is there what would happen if you go for low temperature yes if you go for low temperature NMR in that case you can arrest this dynamic process so that you can clearly distinguish between axial and equatorial protons and you will see separate signals so proton transfer for OH and H may occur so quickly that the proton is not split by adjacent protons in the molecule what happens they are very acidic as a result what happens they will be exchanging this hydrogen atoms work quickly as a result what happens at a given time the neighboring groups will fail to get influenced by the magnetic field of that one as a result what happens you don't see normally coupling due to OH or NH peaks because what happens if they flipping the flipping is so fast the if it is much faster than NMR time scale of 10 to the power of minus 8 as a result what happens so you cannot see the influence of those protons unless otherwise you make it static or stop the dynamic process for example hydroxyl proton if you see here ethanol we have taken here in ethanol you can see this CH3 will show a triplet because of N plus 1 here and then this one should show a strictly speaking a quadrate because of this one but each quadrate is split into a doublet that means we are seeing each quadrate is split into a doublet the doublet of quadrates so that means that is because of OH proton here that means in very pure ethanol if you take 100% pure ethanol probably you can see this kind of things but rarely we will see it on the other hand what happens normally you see something like this here you will see a signal here and this is not influenced by this one is a quadrate and a triplet here strictly speaking one should get a signal like this of course between OH and CH3 no coupling they are very far but certainly CH2 will be first coupled with the quadrate here or it can be a doublet if this coupling is more and then each line will be split into four lines because of this one so a doublet of quadrates or a quadrate of doublets so you can see here whereas this one is split by this one into triplet rarely you see this kind of spectrum so ultra pure sample of ethanol can show splitting very pure 99.99 will not be enough it should be 99 100% 99.9999 you can keep on telling so that can show this kind of thing and ethanol with a small amount of acidic or basic impurities will not show splitting at all here and NH proton again NH proton would be very broad here for the same reason because of rapid exchange and it will not show any coupling to any other protons present on carbon but these two will split each other so this will be giving a quadrate and this will be giving a triplet here so moderate range of exchange and peak may be very broad so identifying OH and NH peaks so chemical shift will depend on concentration and solvent it is very very important and to verify that a particular peak is due to OH or NH shake the sample with D2O keep it for 24 hours and I record and then the signal due to this one will be missing in the NMR that indicates you have OH or NH group as I mentioned earlier Dutero will exchange with OH and NH to form OD and ND on a second NMR spectrum the peak will be absent or much less intense and now we look into carbon 13 so 12 has no magnetic spring 12 carbon and then abundance of 13 C is only 1% that means if I take 100 molecules out of 100 molecules containing carbon one molecule have 13 C and remaining 99 would have 12 C their NMR inactive so whatever you see signal that is because of one so that means signals are very weak and you need large quantity of sample to observe 13 C NMR and gyromagnetic ratio of 13 C also one fourth of that of almost approximately one fourth of that of 1 H that means 60 megahertz NMR for proton would be about 15 megahertz for 13 C or if is 25 megahertz in case of 100 or 100 megahertz in case of 400 megahertz so signals are very weak get in glass in noise and if the signal to nice ratio is very high you may not even see a signal due to carbon at all the sample should be very pure and it should be in larger quantity at least 12% to 20% should be there in a carbon NMR active if not at least it should be little bit more or probably had to run the NMR for a longer time hundreds of spectra are taken and average so now let's look into hydrogen and carbon chemical shifts for various groups all the day I showed you in previous ones this gives some idea about the chemical shift range in both 13 C and 1 H for various groups here for CH of aldehyde or keto it will be around carboxylic group it is here and when you have a halogen here it will be 7 to 8 and when you have a alkene bound one and then when you have a a halo it will be around 4 to 3 also something like that similarly one can also see for carbon also corresponding where they appear so now combined 13 C and 1 H spectra have shown here for this molecule here and this one is the 13 C spectrum 13 C spectrum if you see 1, 2, 3, 4, 5 carbon atoms are there all 5 carbon atoms are very different as a result you will see 1, 2, 3, 4, 5 carbon atoms are very different you will see 1, 2, 3, 4, 1 more should be here I think probably it is here so 2 are there here in case of 1 H NMR you can see this one is here much D shielded and then NH proton is there and then what we have is these 3 protons we have these 3 signals here and this NH is somewhere here it is small NH very weak signal because of exchange so differences in 13 C technique so resonance frequency as I mentioned is 1 4th or 15.1 megahertz instead of 60 megahertz and then peak areas are not proportional to number of carbons and carbon atoms with more hydrogen atoms observe more strongly that means it is very difficult to see signal due to quaternary carbon if carbon has hydrogen the possibility of seeing strong signal is more like in looking to spin-spin splitting here so it is unlikely that a 13 C would be adjacent to another 13 C so splitting by carbon is negligible that means carbon carbon splitting normally we do not come across unless it is enriched sample if it is enriched sample means you know for example if I take methanol all carbon atoms are 13 C in such molecules like ethanol if I take CH3 CH2 OH all CH3 CH2 have 13 C 13 C then probably you can see coupling otherwise you do not see it because what happens if I take 100 here and 100 here and each one has NMR active one if you mix together the possibility of seeing those two coming together to establish coupling is very rare and it is negligible so that is the reason we do not see carbon splitting at all 13 C will magnetically coupled with attached to protons and adjacent however it can couple with other magnetically active nucleus such as hydrogen or phosphorus or fluorine or any other NMR active nuclei these complex splitting patterns are difficult to interpret now protons spin decoupling what is this one decoupling means to simplify the spectrum protons are continuously irradiated with noise so they are rapidly flipping so that means when I am recording 13 C to make it simple and interpretation is easy all hydrogen present will be decoupled from the carbon atom so that you will see signals only due to carbon since carbon carbon coupling is not there what happens you can just by looking into the local symmetry and you should be able to identify how many signals and if you see that many signals in NMR yes you can say you got your compound so the carbon nucleus see an average of all the possible proton spin states and then thus each different kind of carbon gives a single unsplit peak all show unique one and in around 72.1 you may see a trip rate of 1 is to 1 is to intensity that is due to CDCl3 sample we are using CDCl3 normally we use as a solvent so that one the carbon is split with deuterium deuterium is i equals 1 if you use N plus 1 rule deuterium is i equals 1 so what basically you will see is 2 Ni plus 1 the splitting is so you can see 2 into 1 into 1 plus 3 3 and now you get a triplet here N plus 1 peaks will be there but splitting will be 4 so basically what happens 4 lines will be there so here in case of deuterium CDCl3 if you see 13 C CDCl3 signal what happens it appears like this one here so this is because this this signal is coupled by this one D okay so this is D is i equals 1 i equals 1 so you get a triplet 2 Ni plus 1 number of lines will be there of resonance decoupling so what is that one so 13 C nuclear split only by the protons attached directly to them so the N plus 1 rule applies a carbon with N number of protons gives a signal with N plus 1 peaks so interpreting will be little bit complicated as a result what happens we go for decoupled one you can see here for example this is a coupled NMR and this is a decoupled decoupled is very easy we have 1 2 3 4 carbons are there you can see 1 2 3 4 carbons plus TMS standard whereas here also we have 4 but you can see because of hydrogen coupling this one is also showing a quadrate and this is showing a triplet and this is showing a quadrate again so this how it is but we do not want that information whatever the information proton comes from 1 H NMR so we do not want to complicate 13 C NMR that is the reason what we do is we nullify the coupling due to hydrogen and we call it as a decoupled 13 C we write in this one decoupled one will be written like this 13 C in flower bracket 1 H means 1 H decoupled similarly in case of 31 P we see 1 H something like this now MRI it is magnetic resonance imaging it should have been nuclear magnetic resonance imaging but since the nuclear coming people are worried that it may be radioactive that is the reason the term nuclear is removed and it simply it is called as magnetic resonance imaging in this case what happens only protons in one plane can be in resonance at one time and computer puts together slices to get a 3D information 3D structure tumors readily detected and this is very important what is the FTR F T NMR Fourier transfer NMR this is nothing but nuclear in a magnetic field or given a radio frequency pulse close to the resonance frequency and then the nuclei absorb energy and precess spin like little tops like precess and then they flip and then a complex signal is produced then decays as the nuclei lose energy that means when it comes back to the ground state free induction decays convert into spectrum we store FID and then later you can convert that into spectrum so now let us look into few problems here an organic molecule shows two absorption peaks at 870 and 975 Hertz in a magnetic field of 3 Tesla what are the corresponding chemical shifts in PPM this is a question and gyromagnetic ratio for 13 C 6.7 to 6 3 into 10 to power 7 radians per Tesla per second for 13 C it is I equals half it is given and also it is organic molecule we are saying it is a 13 C so now we have to see what we have to do is we have to calculate now new corresponding to this one because 3 Tesla is there corresponding to Tesla what is the magnetic field strength that we have to calculate once if we calculate magnetic field strength then if you divide that one the value 870 by that one you get the corresponding value in PPM so let us do that one so now you should remember only one equation here it is very simple delta equals h nu equals gamma into h over 2 pi into B naught is the applied magnetic field and this is Planck's constant so now if these two goes then 2 pi nu equals gamma into B naught and then nu equals gamma B naught over 2 pi this is the thing so now B naught is given B naught equals this gamma is also given let us apply here 6.7263 into 10 to the power of 7 into 3 over 2 into 3.14 approximately this comes around 32.1 into 10 to the power of 6 CPS that equals to 32.1 megahertz this is the field strength now this is the field strength so calculate the chemical shift in PPM in PPM so what you have to do is 870 is given here first 870 divided by 32.1 and another one is 975 divided by 32.1 will give you corresponding in PPM that one is 27.1 PPM and then this is 30.4 PPM so this is how one can do the calculation now let us look into another example in a magnetic field of strength 2.349 Tesla the resonance frequency of 15 nuclei is 10.13 megahertz what is the resonance frequency of 15 n in a magnet of 11.745 Tesla so now we have to see is a 15 n frequency at 11.745 equals 15 n frequency at 2.349 is 5 times so that means basically 10.13 is nuclei 10.13 into 5 would give you 50.65 megahertz that is it if you simply if you see here it is almost 5 times is there so you should be able to do that one magnetic field strength this one and this one is there this is 5 times that we know by simple calculation so 5 times for this one you should multiply and you get the answer so this is the answer and then in NMR spectrometer commonly used in medicine the resonance frequency for the protons in water is 60 megahertz if such an instrument was to be used to observe 31 P what frequency of radiofix radiation would be required this one so 31 frequency we should know from gyromagnetic ratio frequency equals 1 H frequency into 0.405 so that means if it is 60 megahertz we are using 60 into 0.405 would give you approximately 24.3 megahertz for 31 P so this is how you can write so now I have given one NMR I have given the molecule you have to interpret 195 NMR is shown in the figure 1 this is the figure 1 for the platinum complex as I in the coupling constant now what you can see is the structure is given and also this is a 15 n and of course it is not labeled means it is understood is 14 now we have to see why we got 2 triplets and each triplet has intensity of 1 is to 1 is to 1 and usually a triplet with 1 is to 1 is to 1 has to do something with I value of 1 not half 15 n means it has I equals half I equals half that means it has a 1 J there should be 1 J P T so that means basically it first fits and this is due to 1 J P T 15 n so this coupling let us not worry about that when there is a coupling and now this will be split by 14 and 14 Y N so N equals 1 so if you write 2 Ni plus 1 so 2 into 1 is there nucleus and into spin 1 plus 1 it will be 3 so that means 3 of equal intensity why 3 equal intensity I equals 1 so 1 will be like this 1 will be like this so now intensity will be equal so now we will see like this so intensity 1 is to 1 is to 1 1 is to 1 is to 1 this is what we see here yes we can interpret so it has coupling of both 14 N that is a 2 J coupling this value is 2 J P T 14 N so that is it so this is how you can interpret the spectrum very easy we have to think little bit you have to give some time try to understand how the splitting look likes what is the intensity you should be able to do it. One more question I have given here so I have given 4 spectra here write the appropriate structures and met the corresponding 31 P N M R spectrum for the following inorganic cages and the explained so this is basically the spectrum was given observed for a mixture of samples when white phosphorus was reflected with certain quantity of sulfur. Sulfur what happens is very similar to white phosphorus reacting with oxygen to form phosphorus pentoxide or phosphorus trioxide P 2 O 3 and P 2 O 5 here also we can get different compounds and now based on what kind of compounds we get by just looking to the spectrum we should be able to identify so if you see here and of course here chemical shifts I have not given if you are familiar with chemical shift range then you should be able to tell in the absence of this one still one should be able to speculate that there are two samples having one type of phosphorus atoms assume in white phosphorus all the four phosphorus atoms are intact so P P bonds are not broken it is not fragmented maybe insertion may be there of sulfur in that case two cases where we have four equivalent phosphorus atoms in this one what happens we have out of four two of one side two of one side because they split each other into triplet and here what we have is P and 3 P so something like this we have so that means one equivalent or 3 P and one is P so this one will split this into doublet and this will split into now we have to write the structures so probably by just by looking into the 31 P spectrum of pure white phosphorus and comparing yes maybe a small quantity of white phosphorus is left unreacted you can see here yes this is there this can be one of them and now if you just look into this one this is another product in this one all phosphorus are equivalent and very similar to this one so this also shows a single signal so this is for this one that is assume and this is for this one and now here these are of one type and this is one type so this will show a quadrate and this will show a doublet so that means basically so this one is for this one and now the other one is left if you see here two this is of one type and this is of one type these two will split into triplet and these two will split this into triplet and this is for this one so this is how you can very nicely interpret and it is very very interesting when you make a molecule or when you get a mixture of compounds and when you get a good NMR spectrum you can really enjoy interpreting the data is wonderful so if you want to label them this AX3 spin system it is A2X2 spin system of course what is YX, Y0, AB or something that probably you should learn from a fully dedicated NMR courses as I mentioned here it is only for interpreting the sake of making you how to interpret without learning much basics I am giving you only brief introduction to NMR spectroscopy and of course if you are more you can also look into that one what would happen to the chemical shift difference what happened to the chemical constant coupling constant differences and all those things and here I have listed several compounds containing about 17 compounds containing phosphorus and also I have given the chemical shifts for each one and also the splitting pattern I have mentioned deset for example here deset means 10 lines 10 lines are there if you see here it is trimethylphosphite it is a trimethylphosphite 9 protons are splitting phosphorus into 10 deset line and pH coupling is 9.5 2J pH coupling if you interpret and write the structure in each case I am sure I can conclude that you know little more than 50% of 31 PNMR spectroscopy you should be able to interpret any spectra that comes on to you with respect to 31 PNMR nucleus just try all those things I am sure you should be able to get the correct structure in each case. So let me stop here and few more examples I would continue in my next lecture before I move on to interpretative IR spectroscopy to understand spectral IR spectra of coordination compounds with more emphasis for carbonyl groups until then have a great time and thank you for your kind attention.