 I once again welcome you all to MSB lecture series on Interpretative Spectroscopy. So in my last lecture, I was discussing about chemical shifts and how a chemical signal we get in NMR for a molecule is denoted as chemical shift and what are the implications and why we call it as chemical shift and what are the information signals give. All those things we discussed, let us continue from where I had stopped. In my last lecture, I did mention about some of the chemical shift values for different type of protons in organic molecules. If you can see here in methyl groups in alkane resonate at 0.9 ppm whereas methylene groups resonate at 1.3 and methyl groups CH shows chemical shift at 1.4 and when methyl group is next to a carbonyl group it comes around 2.1 and next when acetylene carbon bond hydrogen resonates at 2.5 ppm and then when CH2 is next to halogens or oxygen the chemical shift comes around 3 to 4 ppm and then in alkenes the chemical shift due to H comes around 5 to 6 ppm and similarly methyl groups comes at 1.7 and aromatic region in phenol comes around 7.2 or even benzene for that matter and in case of toline methyl groups resonates at 2.3 ppm and aldehyde hydrogen comes around 9 to 10 and then acidic carboxylic acid H comes around 10 to 12 ppm and in alcohols alkyl alcohols OH groups comes at 2.5 and whereas in case of aromatic alcohols or phenols it comes around 4 to 7 ppm in case of amines H on nitrogen resonates in the region 1.5 to 4 or it can go further depending upon the type of substance we have and also the nature of the molecule. So now here I have given a different nuclei having spin nuclear spin I equals half and also I have considered hydrogen with instrument frequency 100 megahertz and you can see what is the corresponding frequency for different nuclei all having nuclear spin value of I equals half and also I have given natural abundance here H is 99.9 percent and it is 100 megahertz and the standard we are using is tetramethylsilane in case of 13 C also we use tetramethylsilane then if you will compare the frequency this is one fourth of it and similarly when you look into 19 F its abundance is 100 percent we do not have any other nuclei other than 19 F. So here the frequency corresponding to 100 megahertz in proton is 94 and similarly you can see here for 20 29 silicon 31 phosphorus 77 selenium 103 rhodium 117 tin as well as 1119 and 129 xenon 183 tungsten 195 platinum and 199 mercury so all these things corresponding frequencies are given here and this gives you approximate ratio from that one you can calculate if the frequency is known for hydrogen you can calculate the corresponding frequency for other nuclei and also I have given the standard references for some of these nuclei when we are using in NMR. So now before I proceed further let us try to make us familiar with the equation I showed you that equation is you just recall H nu equals 2 pi nu this is nothing but omega, omega i will be we have to add 2 pi nu so then it becomes even nu substitute for this one here this will be equal to gamma H into B naught. So now this H H will cancel so then what we left is nu equals gamma B naught over 2 pi so this is the equation we are going to use in NMR extensively. So let us try to understand the utility of this equation in calculating the precision frequency or normal frequency for different nuclei place in a magnetic field of different magnetic field strength. So now one such problem is here you can see here calculate the radio frequency necessary for the transition of 11 boron nucleus at a magnetic field of 10 tesla. So you use this equation of course here what is required is you need to know this gyromagnetic ratio that is given here and also 10 tesla is given simply use this equation here so nu equals gamma value let me write here 8.584 into 10 to the power of 7 this is 10 2 into 22 pi 7. So now if I simplify this one it will become 8.584 into 10 to the power of 8 into 7 by 44 on simplifying this one what you get is 136.5 megahertz. So this is the the radio frequency necessary for the transition of 11 boron when it is placed in a magnetic field of 10 tesla so you should be able to calculate this one of course here it 10.6 comes it is 136 hertz it comes if you remove this 10 to the power of 6 it becomes 136 megahertz ok. And let us look into one more example so that you will become more familiar. So the other two problems I have shown here is if the radio frequency of approximately 200 megahertz is required for the transition of 31 P nucleus and 400 megahertz required for the transition of 1 H nucleus calculate the applied magnetic field strength in Gauss as well as in tesla so you should remember 10,000 Gauss equals 1 tesla that also I have given here so conversion so let us do this one. Now again here use the same equation nu equals what we have is now gamma 4 2 pi into b0 so now we know this is this is known gamma is known and we have to find out a b0 in case of 200 megahertz phosphorous cinema so this is 200 or 200 megahertz can be converted into 200 hertz so this equals phosphorous gyromagnetic ratio is given here 10.840 into 10 raise to 7 into b0 and now again 2 into 22 by 7 so on simplifying this one it will become 200 into this 22 into 244 into 10 raise to 6 equals 10.840 into 7 into 10 raise to 7 b0 so then if you simplify this one what you get is 115,000 Gauss you get it so this can be nothing but 11.5 tesla so this is the magnetic field and this corresponds to 500 megahertz instrument ok this corresponds to 500 megahertz instrument so this corresponds to 500 megahertz instrument so you can I have given earlier this table of conversion of magnetic field strength and the corresponding frequency you can just check from that data it comes approximately 115,000 Gauss that is nothing but 11.5 tesla this corresponds to 500 megahertz instrument and if the 400 megahertz instrument 31p nucleotide request about 161.2 megahertz radio frequency now let us look into the second one the second one is 400 megahertz for 1h so it is the same you can do it again use the same equation gamma into b0 over 2 pi so this is 400 so simply I can write it here 44 this equals gamma is given here for proton this is 26.753 into 10 raise to 7 and then this 7 whatever was there it goes here so this is nothing but 2 into 22 by 7 so 7 comes here and 2 is going here so that is what I have written here into b0 so simply now b0 can be calculated from this one 400 into 44 divided by 26.753 into 7 so this will give you into 10 to the power of minus 7 so then when you convert into megahertz what you get is you get approximately 9.39 tesla that is equal to 9.4 tesla so that means basically 94,000 Gauss so this is the magnetic field strength for this one so this is how you should be able to calculate. Let us move on to further discussion okay now let us look into the aromatic protons for example if you take a benzene molecule and subjected to NMR what would happen to its chemical shift we all know that the chemical shift range is 7 to 8 and why that happens you can see clearly here and of course here also we should look for all possible orientations of benzene molecule in a magnetic field strength of b0 and then if you take average of all of them you can consider only 2 possible orientations one is perpendicular one is parallel and here I have considered perpendicular one in this one the circulating electron density would generate a magnetic field or induced magnetic field generated because of circulation of electron density in the benzene molecule under the influence of b0 generates a magnetic field that is aligned with the magnetic field b0 you can see the direction shown here okay this is aligned with the magnetic field as a result what happens the net magnetic field experienced by the protons in aromatic molecules would be going to high frequency shift so that is the reason we call it as D shielding and then how about the parallel one so parallel one you can consider here so you consider like this molecule here and here the little circulation of induced magnetic field can be seen little circulation of electron density and hence whatever the magnetic field generated can be totally ignored or neglected so that means not much contribution is coming from the parallel orientation the bulk of the influence of electron density that generates say induced magnetic field comes in the perpendicular orientation that is responsible for pushing the protons into D shielded region of chemical shift range 7 to 8 ppm so now let us look into vinyl protons again vinyl protons I have considered the perpendicular one and in the parallel orientation same thing is observed what I had described in case of aromatic molecules so in this perpendicular one what would happen again induced magnetic field is acting in this direction that is it is reinforcing the applied magnetic field as a result net magnetic field experienced by vinyl protons is much more compared to the protons in the absence of induced magnetic field as a result the net effect is D shielding so they appear in the range of 5 to 6 ppm and again in this case also perpendicular orientation has little circulation of electron density and hence there is not significant magnetic field developed because of this effect and hence that can be ignored so now let us consider aldehyde proton again in this case also we can ignore the parallel orientation where the induced field magnitude is negligible so we have to focus on again the perpendicular orientation in case of perpendicular orientation again this group is attached to electro negative oxygen atom again in this case also the induced field generated because of this ring current that is produced is aligned with the applied magnetic field here you can see applied magnetic field as a result the net effect is D shielding or shifting the signal to high frequency and it appears in the range of 9 to 10 ppm and same energy holds good in case of oscillating protons also here the parallel one is what dominates here and in this case because of the alignment of hydrogen protons in this way the induced field whatever that we are seeing is opposing the applied field is opposing the applied field as a result what happens we see the net shielding in case of oscillating proton and hence the chemical shift appears in the region of 2.5 ppm and then if you consider the perpendicular orientation again here in case of perpendicular orientation there is no significant contribution from bi developed here so it can be ignored ok only the parallel one is what matters here so once if you know these facts it is very easy to understand why a given molecule nuclei are shielded or D shielded or appear in high frequency region or low frequency region what would happen to the OH and NH protons chemical shifts of OH and NH depends on concentration very very important hydrogen bonding in concentrated solutions hydrogen bonding in concentrated solutions D shield the protons so signal is around 3.5 for NH and 4.5 for OH so that means if considerable hydrogen bonding is there that results in D shielding of the protons involved in hydrogen bonding and hence the chemical shift due to NH protons or OH protons would shift to respectively for 3.5 ppm and 4.5 ppm so that means hydrogen bonding is there in molecules in solution and one should be able to diagnose and tell approximately the to what extent the hydrogen bonding is there by just simply looking into the chemical shifts of such molecules. So the another reason for broadening of this OH and NH peaks in NMR is because of proton exchange between the molecules so now let us look into carboxylic acid so acetic acid is here I have considered and delta 11 for OH comes here ok this is for OH and whereas for CH3 it comes around 2.1 here and you can see here considerable D shielding is observed for carboxylic proton and appears at 11 ppm. Now look into another molecule here we have 3 different type of protons are there so one is next to carbonyl group another one is next to oxygen and another methylene group is in between two carbonyl groups and the corresponding chemical shifts I have shown this one comes around 2.25 so this is here and then this is 3.41 so this is here and then this one is here and again what you should notice is equivalent hydrogens have the same chemical shift that means all hydrogens present on this carbon show single resonance and similarly all hydrogens present on this carbon show single resonance and all hydrogens present on this methyl group also show single resonance ok you should remember we will come back to looking into why they are equivalent and whether there can be some non-equivalence and hence we can see some interaction of the hydrogen atoms present on the same carbon atom. So now let us look into the importance of intensity of signals you can see here in this molecule we have two type of hydrogen atoms are there I have designated with blue color and H color here this is one methyl group is there and here on this carbon we have three methyl groups are there and then if you just look into the signals by simply looking into the intensity you should be able to tell which signal is for which group so this one shows chemical shift here and this one shows chemical shift here and if you just measure the intensity it is almost three times if this is say this quantity is there and this is about three times here you can see here three times one two three so that means basically the intensity what we see can also tell you about how many such groups are there and how many protons are there so that understanding the structure and elucidation of the structure becomes rather easy. Now I have simplified here in the graph you can measure and you can see the intensity here and then you can show here and up to here if you go it can clearly tell you about here nine protons are there and here three protons are there that means it is one is to three ratio is there that could be clearly seen from this one so that means intensity looking into the intensity is very important in identifying similar type of groups in the molecule. So the area under each peak is proportional to the number of protons so this is shown by integral trace this is what we call it as integral trace usually you can get this information directly from NMR spectrum. So now let us consider this molecule here and now let us see here we have three protons are there here two protons are there and here we have one is there and then we have here six are there ok and that means basically one two three four these two are one and this is two this is three and this is four four type of groups are here and we should have four signals of course one is here one is here one is here one is here and just by looking into the intensity we should be able to tell these two will be corresponding to this one and now CH2 CH3 is there CH3 will be coming to this one and then CH2 will be coming to this one so simply by intensity you should be able to tell and then values are there so you can clearly see now where exactly these signals are located and this is the integration. So far whatever we saw we saw only single signal do we see always one signal and no matter how many protons are there how many nuclei are there that is yes now let us look into this term what is spin spin splitting so non-equivalent protons and hydrogen non-equivalent protons and adjacent carbons have magnetic fields that may align with or oppose the external field you should remember when we are subjecting a molecule to the magnetic field under NMR conditions if you are focusing on let us consider a ethanol molecule here so when I am focusing on this one this methyl group and of course here methyl group what happens a circulation of electron density also induces a magnetic field and that can be aligned or oppose the magnetic field based on that one we get a chemical shift at some reason so but on the other hand when these protons are persisting you know the neighboring would not keep it they will also be persisting as a result what happens they whatever the induced magnetic field they generate can also influence on the chemical shift of this one and to what extent they influence depends on whether they are aligned with the magnetic field or opposing the magnetic field so that means the neighboring protons can split in the adjacent protons and adjacent carbon atom depending upon how many such protons are there and how they are aligned or how they are behaving in the magnetic field so this would leads to the term called spin-spin splitting as a result signals would be seen as multiplets so that means this magnetic coupling causes the protons to absorb slightly downfield when the external field is reinforced and slightly upfield when the external field is opposable so that means whatever the magnetic field they are generated can be something like this or some can be something like this if there are something like this you can see signals are more deshielded and if they have something like this shielded so this is how they can influence and that means all possibilities should be considered so signal is split. So now another useful property that means now I am giving some more details about spin-spin splitting so another useful property that allows NMR spectra to give structural information is called spin-spin coupling so which is caused by spin coupling between NMR active nuclei that are not chemically identical if they are chemically identical what would happen you will see only one signal for example if we take ethane molecule 2 CH3 are there they are chemically equivalent and you will see only one signal but if you take CH3 CH2 OH or something CH3 CH2 CL then the neighboring ones are different so then you can see interaction between them that what we call it as spin-spin coupling so different spin states interact through chemical bonds in a molecule to give rise to this coupling which occurs when a nuclei being examined is disturbed or influenced by a nearby nuclear spin so in NMR spectra this effect is shown through peak splitting that can give direct information concerning the connectivity of atoms in a molecule. Nuclear which shares the same chemical shift do not form splitting peaks in NMR spectra so this is very very important in general neighboring NMR active nuclei 3 or fewer bonds away leads to this splitting this splitting is described by the relationship where in neighboring nuclei results in N plus 1 peaks and the area distribution can also be seen from the Pascal triangle so however being adjacent to a strongly electronegative group such as oxygen can prevent spin-spin coupling so that means in between two methyl groups if you have oxygen that can prevent the interaction of these two groups for example a doublet would have two peaks with intensity ratio of 1 is to 1 while a quadrate would have a 4 peaks of relative intensities 1 is to 3 is to 3 is to 1 this information comes from the triangle the magnitude of the observed spin splitting depends on many factors and is given by coupling constant J which is in units of hertz. So let us discuss more facts and look into more examples to understand in a better way spin-spin coupling in my next lecture until then have an excellent time reading about spectroscopy. Thank you.