 Hello everyone. I am sure you are having a good time. I welcome you once again to MSP lecture series on Interpretive Spectroscopy. In my previous lecture, I discussed it about first-order spin system taking several examples. One such example I consider was styrene. In styrene, although it was appeared in like a first-order spectrum, it's actually it's not first-order spectrum. So that means geminal coupling we came across. Although magnitude was about 1.4 or so. So how that happens and how to interpret let us look into now second-order spin system. Second-order coupling we call it as. So the examples of 1H NMR spectra we discussed in so far involved first-order coupling. So that means that chemical shift difference, we call it as delta-delta between the nucleus considerably greater than the coupling constant, usually five times greater. That means delta-delta is greater than J. Then it is called first-order coupling. When delta J is larger than nuclei involved in the spectrum are designated with letters from extremants of alphabet. That means for example, A and X, AMX, AX2, A2X2 depending upon how many such nuclei are there in a given molecule. That means when delta is larger, the nuclei involved in the spectrum are designated with letters extremants of alphabet that taken from extremants of the alphabet. AX we come across AMX, AX2, A2X2, etc. So when delta J is less than 5J, that means chemical shift difference is much smaller than coupling constant. Then for such a system, we are using letters close together in the alphabet. For example, A, B, A, B, C, A2, B, C, something like that. So in those cases, what happens? The Larmor frequencies of nuclei of different elements are always very large in those cases. In the first-order spectrum, if you consider or in first-order coupling, the Larmor frequencies of nuclei of different elements are always very large. Second-order coupling can occur only between the nuclei of the same isotope. One should remember that one. Nuclei of the same isotope only can exhibit second-order coupling. For example, as I said, protons on same carbon atom, when it is there and they are chemically equivalent and magnetically non-equivalent. In that case, we come across, but we similarly substituted fluorine and hydrogen. So that can never be a second-order coupling. There's always first-order spectrum. That means second-order coupling can occur only between the nuclei of the same isotope. For example, between 2, 1-H nuclei, 3, 19-F are 3 phosphorus. Since the chemical shift differences are very small for 1-H, the majority of second-order couplings are observed in 1-H NMR spectrum. Here, surrounding hydrogen, we have only 1-S electron and it is spherically symmetrical. Hence, chemical shifts are very small range. We have 1 to 10, only in exceptional cases where we have coordination compounds and metal to hydrogen bond is there. In that case, we may come across up to 60 hertz, 60 ppm. Otherwise, in most of the organic molecules, it doesn't go beyond 10 ppm. So that means as a result, what happens? Since the chemical shift differences are very small in 1-H, the majority of second-order couplings are observed only in 1-H NMR spectrum. We also come across in case of phosphorus NMR also. So let us consider a simple second-order spectrum such as AB. Since the chemical shifts of the nuclei in a second-order system are similar, the energy levels are much closer compared to the same in the first order. This is very, very important when we differentiate between first-order coupling and second-order coupling. The chemical shifts of the nuclei in a second-order system are very similar. That means they have chemical shifts much closer. The energy levels are much closer compared to the same in first-order spectrum, a first-order spectrum. So let us now, I have given a series of spectra here. They represent typically AB case. In AB case, the middle energy levels are similar to A2 here. If you see here, they are A2. As this middle level becomes closer, it becomes difficult to tell which nucleus has which MI value. That is the problem we come across in second-order splitting. As this mid-order levels becomes closer, it becomes difficult to tell which nucleus has which MI value, whether it has plus half or minus half. Similar to A2 case, consider the linear combination of two-way functions here. Since these new functions are mixtures, it is no longer possible to designate the transitions whether they are due to change in MI of A or B. As a result, by simply looking into spectrum, it is no longer possible to determine the chemical shifts of A and B. So general patterns of the AB spectrum depends on delta A of A and that of B and the corresponding J value. So to what extent delta J is comparable, delta delta is comparable with delta J. That would tell you the separation. The spectra shown here are for J equals 10 and a variety of delta delta values. So here in all of them, if you just look into it, coupling constant is kept 10 and then a different variety of delta delta values are given. When delta delta is very large relative to J, spectrum appears as a distorted doublet. The intensity of inner lines increases as J by delta increases here. So that means at delta delta equals 0, the spectrum is reduced to a single line as in case of A2 system. So this is how you can correlate and also all these cases are of different AB depending upon the difference in delta delta with respect to J value. Few points to remember in case of second order coupling is in a first order spin system, delta delta must be greater than delta J. That means what one should remember is this one is much greater than delta J, not delta J, J. All chemically equivalent nuclei must also be magnetically equivalent. This is very important. In first order spectrum, all chemically equivalent nuclei also magnetically equivalent. That means if you are considering two geminal hydrogen atoms, so they are chemically equivalent as well as magnetically equivalent. What is the meaning of chemically equivalent and magnetically equivalent? Two or more nuclei are chemically equivalent if they can be interchanged by the operation of some symmetry elements of the molecule. So two or more nuclei are chemically equivalent if they can be interchanged by the operation of some symmetry elements of the molecule. For example, let us say you take phenol. We have ortho-hydrogen atom survey and simply by doing a C2 rotation with respect to OH and para-hydrogen atom by rotating. So what happens? If we cannot distinguish them, then we can say both are chemically equivalent. In case of AB system, we are giving this delta AB equals delta 1 minus delta 4 into delta 2 minus delta 3 and this middle point will be given as delta usually. And then J can be calculated delta 1 minus delta 2 or that is nothing but equal to delta 3 minus delta 4. This is how the J is determined here and also chemical shift also determined here. Delta AB equals we have to take the middle of this one. Now let us consider some examples here. As I mentioned, if you consider here chlorobenzene and if you consider this as a rotational axis and if you do C2 rotation and 1 and 5 cannot be distinguished or indistinguishable. Similarly, H4 and H2 are indistinguishable. These two are called chemically equivalent and these two are called chemically equivalent. And again if you take in this one all these are chemically equivalent here. And then again if you consider here again they are chemically equivalent here because you cannot distinguish them. Similarly, if you consider this picture I have shown here H on C2 and C4. So this one and whatever the hydrogens we have on C2 and C4 I have not shown here or C2, CH3, CH3 and are equivalent so as H on C2 and C4 trans to CH3 plane. So plane extending through C1 and C3 makes a C2 and C4 equivalent. And similarly if you look into FE to CO9 here the axis through FE-FE bond if you make it terminal and bridging are equivalent these three are equivalent and these three are equivalent. So these are equivalent and these are equivalent. So this is how we can identify whether some symmetry operations can make them indistinguishable after that symmetry operation. So that means now we can have a better understanding of chemical and magnetical equivalence or non-equivalence. So if you say chemically and magnetically equivalent nuclei that means magnetically equivalent means each member of a chemically equivalent set of nuclei must be equally coupled to each member of any other chemically equivalent set in the spin system. So that means H5 are equivalent so as H2 and H4. So these two are identical these two are identical equivalent and the third set is H3 is H1 coupled to H2 ok H1 coupled to H2 same as H5 coupled to H2 answer is no because of the distance. That means if each member of one equivalent set is not equally coupled to each member of second set it cannot be first order. So this how you can determine whether a given molecule would give a first order coupling or second order coupling when you have situation like this. If H1 we are considering the H1 interaction of H1 in H2 is different from H5 and H2 similarly H2 interaction with H1 is different from H2 interaction with H5 or we can consider H4 interaction with H5 is different from H4 interaction with H1. So in this case what happens it cannot be first order. So chlorobenzene is an example of three spin systems if you designate this as A this is A prime and B B prime and C so we call it as this one and then A prime what we say is the one nuclei with prime says that they are not magnetically equivalent. A and A prime are chemically equivalent but they are not magnetically equivalent and similarly B and B prime are chemically equivalent but they are not magnetically. So the prime A here represents chemically equivalent but not magnetically. So letters in the same region or nearby alphabet indicate the similarity in their chemical shifts. That is the reason when we take A, A, B, B and C that means they have almost very similar chemical shift values. If the chemical shifts of H1 are significantly different from that of H2 and H5 then the spin system would be A, B, B prime and X, X prime. That can happen if the magnification is much larger as delta, delta would be larger. So this is where the significance of going from a low field NMR instruments to high field NMR happens. So when you go for higher and higher instrument what happens if you take chemical shift in hertz the separation will be much larger. As a result what would happen they all complicated secondary spectrum can be converted into very simple first order spectrum. This is where people always look for 400, 500, 600 megahertz instruments instead of using 200 or 300 or even 100 megahertz instruments. The moment we designate with prime A, A prime, B, B prime and C that means A and A prime are chemically equivalent but they are not magnetically that one should remember. And that one can understand by simple analogy whatever I shown here and ideal example is chlorobenzene. So remember that one. So now let's look into three spin order system. For calculating spin multiplicity as a number of lines in a peak 2Ni plus 1 is not used for second order spectrum and it only for the first order spectrum. Again the 2Ni plus 1 rule where N is number of identical or equivalent nuclei and I is the nuclear spin this one holds good along with the corresponding Pascal triangle only in case of first order spectrum but in the second order spectrum we cannot use this rule at all. So now let us consider a 19 F NMR spectrum of this molecule here trifluoroethylene for example of AMX spin system consists of three sets of doublets of doublets here. It's a 19 F NMR we are considering and 19 F also is 100% abundant and I equals 19 F if you consider I equals half and 100% abundance is there. So it's as easy as 1H or it's very similar to 1H NMR or R31 P NMR. So now if you just look into FAA we have doublet of doublet FM we have doublet of doublet and FX we have doublet of doublet is there and corresponding coupling is given and MX and MX coupling is there and AM coupling is there and MX coupling is larger again trans coupling is there and then next AM coupling AM coupling is there and that is 56 Hertz and AX coupling is there that is So you can see here, this appears like a first order spectrum. So when two of the three nuclei have similar chemical shifts, I'm repeating again, when two of the three nuclei have similar chemical shifts, the spin system is designated as ABX. When two of the three nuclei have similar chemical shifts, we have three nuclei are there and two of them have very similar chemical shifts. In that case, the spin system should be called as or designated as ABX, in which the relative sign of the coupling constants of the appearance of the spectrum. So AB portion consists of eight lines that resemble two sets of quadrates, often they are overlapped. So AB portion itself will be consisting of eight lines that resemble two sets of quadrates and often they are overlapped and giving a different type of intensities for lines. And the X portion consists of six lines with two lines being very weak and almost there in the baseline. Transition from AB, AMX to ABX is shown in the next spectrum I'm going to show here. So you can see here how AMX spectrum, where it appears like a first order spectrum is converted into a second order spectrum because of the very little difference in the chemical shifts of A and M. So here and here. So now let's look into few points I have listed here. These series of spectra what I have shown from 1 to 4 represent transition of a first order spectrum, AMX to ABX and eventually to A to X. So now these are all simulated spectra. AMX with respective chemical shifts of 4, 2 and 1.1 ppm where JAM coupling is 12 and then JX coupling is 3 and JMX coupling is 10 hertz. Next in the second one, this is about the first one I told you in the second one. Again this ABX system with respective chemical shifts of same 4, 3.6 and 0.1 ppm and JAB value is brought down to 9 and then BX is 3 and AX is 7 here. So now spectrum 3 represents ABX system with AB chemical shifts of 4, X of 0.1 here only 1 and 0.1 ppm and the coupling constants are same as that we saw in case of 2, 9, 3 and 7 hertz. Now in the A to X system here what happens with A chemical shift of 4 here and then this is 0.1 and then JX is only showed that is about 9 hertz. So sometime you can see some of these variations are there and as I said these values depends on the chemical shift difference between A and M and as the chemical shifts of X nuclear moves closer to those of A and B it becomes ABC system. Eventually now we saw M moving very close towards A to become AB in the same fashion if the signal due to X also starts moving towards B it becomes a ABC system. Spectrum can have up to 15 lines here. You can see here this is a typical ABC spectrum here. ABC spin system I have given 2 here. 1 at 100 MHz the chemical shift values for ABC are 4, 3.7 and 3.4 for AB and C and the corresponding coupling constants of AB is 9 hertz, BC 3 hertz and then AC 7 hertz and the same coupling constants are kept and the spectrum recorded at 300 MHz would be something like this. Here you can see clearly as the field strength increases they are moving away from each other. When they are moving away from each other and for example if you go for maybe 400 or 500 MHz they will move even further apart and then it becomes simple AMX system. So this is the advantage of recording spectra at higher magnetic field strength. Let us consider another simulated spectrum with AB X spin system here. You can see here AB X spin system is there and delta AM is delta B equals 10 hertz here and then delta AB equals 10 hertz and since we do not know the sign always if the sign is not known always we represent the magnitude but we are not mentioning the sign and when we look into the sign here so when we are ignoring the sign it is always ideal to represent in modulus. So this one is minus 4 and this is 1 hertz. So that means basically if you consider this AB X spin system X resonance resembles a doublet expected for a first order spectrum here with X only coupled to B. Only X AB X only X is coupled to B. Close examination of expanded version actually shows six lines for X. This one shows six lines and it is not first order in such cases where resonance appears to be not affected by another nucleus HR X has said to be virtually coupled. So that means this a typical system where we have this kind of coupling values and chemical shifts we come across virtual coupling. Let me stop here and continue more discussion on second order system in my next lecture until then have an excellent time.