 Hello everyone, I once again welcome you all to MSP lecture series on interpretive spectroscopy. So this is the penultimate lecture in this series of 60 lectures. In the next lecture, the last one I am going to summarize and conclude about whatever we discussed in the last 59 lectures or so. So let me continue solving some more problems in this lecture. Here is a question. The iron III ruthenium II analog of prussian blue. So you are all familiar with prussian blue. It has iron III and iron II compound. And here we are talking about a hybrid of iron III and ruthenium II. So this is Fe 3 4 times Ru Cn 6 3 times 18 H2O. This is a dark purple complex unlike the homolyptic homologous analog of iron iron that is intense blue in color. So this is a dark purple complex which shows an IVCT, Intervalence Charge Transfer or Metal to Metal Charge Transfer, banded at 495 nanometer, compare its properties such as delocalization of electrons with prussian blue. So that means when we talk about delocalization it is nothing but the charge transfer. So the presence of an IVCT that is Intervalence Charge Transfer band that means metal to metal charge transfer band that indicates the delocalization of electrons in the mixed metal system. So one is a donor, one is acceptor. So here we to see the electrons are moving from the acceptor to the donor that we call it as delocalization. In fact all transitions, charge transfer transitions we will see delocalization to what extended delocalization happens would tell about the intensity of that band. The energy of the IVCT band is inversely proportional to the extent of electron delocalization between the metal centers. The energy of this charge transfer transition is inversely proportional to the extent of electron delocalization between the two metal centers. So that means more delocalization results in lower energy IVCT transitions. So more delocalization to occur what happen the energy should be very close to the matching or so. In that case what happens the movement of electrons is quite free as a result what happens delocalization would be more and when the delocalization is more electron transfer also would be very easy that is reason its energy is inversely proportional to the extent of electron delocalization and delocalization results in lower energy. So that is more delocalization is lower the energy of IVCT transitions. As a result of delocalization the shift in energy from 680 nanometer and if you calculate that one it will come frequency will be 14700 that means 1 over 680 will be wave number equals 1 by lambda. So you can do that it comes around 14700 centimeter minus one for prussian blue and then here it is 495 that is 20200 in case of this mixed iron ruthenium system. This indicates greater extent of electronic delocalization in the mixed iron ruthenium analog which is in fact dark purple. So that is the comparison here so what it says is compare its properties. So here delocalization is more extensive compared to iron and pure prussian blue as a result what happens here the energy required is much lower for transition all ruthenium analog if you consider all ruthenium analog means RU3 and RU2 analog it is a dark green complex shows an IVCT band at 1000 nanometer so indicating greater delocalization its even more delocalization system we come across in case of a pure ruthenium analog of prussian blue. So then here another question the allowed terms for D7R if you calculate some of them allowed transitions and find out the term symbols those term symbols are given here allowed ones for D7 system there is 2H2G4F2F2D2D4P2P among them identify the ground term well of course by just looking at you should be able to tell the one that has a higher spin multiplicity 2S plus 1 value and eventually 2S plus 1 is identical for both then we come across this one as well as this one in that can you have to consider the L value highest L value highest L value will be here if you take SEDF0123 so 3 is there here for this one whereas one is for this one so you can say without any hesitation this is the one this is the blindly but on the other hand you can also solve this one ground state ground state you can have electronic arrangement like this then you can find out 2S plus 1 value here 2S1 plus equals 2 into 3 by 2 plus 1 so this is equals 4 and then L is 3 F is there so F this is the one and then if you want to find out J value here J will be here L plus yes because this is more than half field so 3 plus 3 by 2 equals 9 by 2 so this will be 9 by 2 of course here this is the one for example if you want to know what is the microstate for d7 so you should be able to do it that is n factorial over r factorial into n minus r factorial and of course n is the total capacity of the sub shell here d orbital 10 factorial and then r is 7 electrons are there 7 factorial and this is 3 factorial so one can write here n factorial 10 factorial over 7 factorial into 7 10 minus 7 factorial so this is equal you can write 7 factorial into so this will be 120 so you can calculate this way so now let us look into another example here problem in its electronic spectrum has saw for vanadium 3 plus exhibits 2 absorption bands 1 at 17800 nu 1 and the second one is at 25700 the correct assignment of these bands respectively is so here already 4 options are given you have to choose the correct one before that you have to identify what system it is it is a d2 system and you should know that d2 system comes under the second category of d2 d3 d7 d8 system they show 3 transitions so here it is a special case I would tell you later why it shows only 2 absorption it shows 2 absorption so it is d2 system and in d2 system from simple Argel diagram you should be able to find out which is the ground term here and then here the ground term is this one so here it is there here it is there and then among them which one is correct you have to choose let us say you are choosing this one so high energy transition will be this one and this one will be the high energy one and now just go back to this Argel diagram written for all this d2 d3 d7 d8 octahedral as best tetrahedral complexes of high spin complexes only Argel diagrams holds good only for high spin complexes if you want to use for both high spin as well as low spin you have to use Tanube-Sugunov-Tanube diagram so that is you have every system you have to have a Sugunov-Tanube diagram so here you can see here these are the two transitions are there one 3t1g2 3t1gp higher in energy and 3t1gf2 3t2gf is the second one so this is shown here these two and then the question is why only two and of course you also know why this is deflected out we also showed you about Taka parameters and all those things we can go back and check and now here the ligand feed strength of water results in transitions occurring close to the crossover point between 3t1gp and 3t2gf here this so as a result what happens here of course this not a problem this is more pronounced when you come for this side that means a d7 system or d8 system here this is opposite of d2 so d8 system it comes more here as a result what happens the energy difference is very small and it merges and you get only two transitions not three transitions so they are not resolved so vanadium 3 plus iron with three different ligands to consider instead of hexa aqua let us have ammonia chloro and aqua something like that in that case we can see three distinct peaks or absorptions here so now let us come back to again this mass NMR related problem this compound has the molecular formula C9 H11 NO2 included in this problem or the infrared spectrum infrared spectrum is given here and then 1H NMR spectrum is also given and some 13C NMR spectral data is also given here so you have to depict the structure all you have to find out the right compound molecular formula is there you have to identify and write the structural formula or structure of the molecule so first when you look into this one you should know that we have carbon hydrogen nitrogen and oxygen is there and then we can also find out hydrogen deficiency index here for that one it is very simple 9 plus 1 minus 11 by 2 plus half so now if you take it here 10 and it will be 11 by 2 means 10 10 and half it will become 5 so we have 5 here so here hydrogen deficiency is 5 here so we have to find out that means at least there will be a ring there will be a ring here plus 4 double bonds will be there here so this information comes from hydrogen deficiency index and also you should see here there is a CO is there you should remember carbonyl group is there and also you can see here very intense bands are there 1200 something is there you will check those things now and also some information is there just look into how many signals are there 1 2 3 4 5 6 7 8 and 9 signals are there all are different and then some splitting further splitting is also shown for G F and Ed G F and Ed and G F E and D so this is in the aromatic region a further splitting is also shown here and whereas here we have a quadrate and we have a triplet and also we have a small one it represents maybe OH or NH 2 or something like that so we should remember these things in mind and we shall try to work out the structure now the solution means first calculated hydrogen deficiency index this we have already calculated it is 5 and all of the spectra shown in this problem suggest an aromatic ring aromatic ring is there and plus 4 double bonds are there the remaining index 1 CO group found in 17 O A okay that is also there and this value for the carbon group is too high for an amide so it is not an amide it is in a reasonable place for a conjugated ester so that means it is a conjugated ester the 17 O 5 gives that information while N O 2 present in the formulas suggest a possible nitro group this cannot be the case because we need 2 oxygen for the ester functional group so since we need 2 for 1 carbonyl and ester means CO O R then you do not have any other oxygen to think of N O 2 so N O 2 is absent by a simple reason you can rule out the possibilities of N O 2 if N O 2 is not there the other possibility is a primary amine or secondary amine type so the doublet about 3400 in the infrared spectrum is perfect for a primary amine this value stretching frequency suggests its NH that suggests that it is a primary amine so that means we found out CO and there is a O and NH 2 is there then 13C NMR spectrum has 9 peaks which corresponds to 9 carbon atom that means all are unique we have 9 different type of carbon atoms in the molecule the ester group carbon appears at 167 so CO is appears at 167 the remaining downfield carbons are attributed to the 6 unicaromatic ring carbons from this we know that ring is not symmetrically substituted since we have so many it indicates that certainly we do not have a very symmetrical structure at the aromatic carbon atoms that DEPT results confirm the presence of two carbon atoms with no attached protons so that indicates we have two carbons without having any hydrogen atoms that is 131 and 147 by DEPT measurement experiment and four carbon atoms with one attached proton and four of them have one attached proton here that means we have CH unit here from this information we know that the ring is disubstituted so that means these four in the aromatic region are there that means out of C6H6 to have gone with its two substituted one that means if amine group is there and if ester group is there it has to be on the aromatic ring the ring must be disubstituted because four protons appear in the aromatic ring it should be disubstituted and also it is not symmetrically substituted that means there are no groups in the relative para position one four position now this information is given so here one so this one no peaks are there that indicates they have no hydrogen atoms are there in this one so now this is the structure for this molecule this is we have CH2 and CH3 is there and then this is the ester group and then this is substituted something like this the reason why it is substituted like this this one would show a triplet and then this one would show a singlet or it can be coupled with this one and this will show a doublet and this will also show a doublet here so based on this the structure is given this is the correct structure so this pattern suggests one three substituted pattern rather than one four or one two one four or one two have a little bit more symmetry they are ruled out the key observation is that the proton F is narrowly spaced triplet or a doublet of doublet suggesting 4j coupling but with no 3j couplings in other words the proton must not have an any adjacent protons it is sandwiched between two non-proton groups amino and carbonyl protons G and F appear downfield relative to protons E and D because of the D shielding effects of the anisotropy of the CO group this further information you can draw from the positions of these peaks in the NMR spectrum the aromatic out-of-plane bending bands in the infrared spectrum suggest meta substitution 680 760 and 880 this comes by practice if you solve more and more problems you will be familiar with the stretching frequencies under that corresponds to the groups and their positions the 1H NMR spectrum shows an ethyl group because of the quartet and a triplet quartet and triplet is a typical of ethyl group present in a molecule 4.3 and 1.4 respectively for the CH2 and CH3 so CH3 comes at 1.4 as a triplet and 4.3 a quartet for methylene protons so finally a broad NH2 peak integrating for two protons appears in the 1H NMR at 3.8 ppm 3.8 ppm we have saw very tiny peak the compound is ethyl 3 amino benzoate you can see this is for NH2 so this is the molecule here so now with this information you go back and try to analyze once again the positions and how they look like expansion is also given for this aromatic region that should tell you the interaction of some of these nuclei together to give different type of multiplicity for each signal. Now let us look into one more example this compound has a molecular formula C5H7 NO2 following are the infrared 1H NMR 13C NMR spectra so C5H7 NO2 is there again we can go for hydrogen index deficiency here it is 6 minus 7 by 2 is 3.5 and plus 0.5 here so this indicates this is 3 this is going to be 3 here so hydrogen deficiency index is 3 here and also you should remember 1 2 3 4 5 signals are there and this is for TMS 5 13C signals are there and then we have a quadrate and triplet is there again probably a ethyl group is there that should come automatically to your mind and then we have one this is coming in the region where CH2 and CH3 aliphatic are expected but it is not coupled means it is isolated that we should remember then we have here again a CO group is there and again it indicates here probably we have a ester group here so with this information let us try to analyze the spectra and illustrate the structure. Now we calculate an index of hydrogen deficiency of 3 we already calculated so the infrared spectrum reveals just go back to infrared spectrum one more I forgot so this is characteristic of C triple point N nitrile group here around 2300 something so now we have CO and we have ester group and also we have C triple point N so this information is there now so now the infrared spectrum reveals the source of unsaturation implied by an index of 3 a nitrile group at 2260 so it unsaturation is 2 for this one and then we have a carbonyl group 1 so that is taken care of hydrogen index deficiency so one triple bond is there and one double bond is there the frequency of the carbonyl absorption indicates an unconjugated ester so this is an unconjugated ester it is coming around it will higher 1747 it should be a little bit lower 1710 17 up to 1710 you can see conjugation the appearance of several strong CO bands near 1200 confirm the presence of an ester group so CO group the 13C NMR spectrum shows five peaks and this is consistent with the molecular formula which contains five carbon atoms that means all five carbon atoms are unique and then the carbon atom in the C triple bond N group has a characteristic value of 113 ppm in addition the carbon atom in the ester appears at 163 ppm one of the remaining carbon atoms probably lies next to an electronegative oxygen atom the remaining two carbon atoms which absorb at 25 and 14 are attributed to the remaining methylene and methyl carbon atoms which are coupled in one HNMR to show a triplet as well as a quadrate so now the following are the in the same thing I am coming back here to write the structure so this is the structure here the structure is we have CH3 CH2 CH2 CN so you can see here this will be a quadrate and this will be a triplet for this one this one and then this one should show a unique a singlet here and then of course see these are the three peaks we come across and then C triple bond N is there since we got a peak B singlet that is for methylene which is not coupled to anything in adjacent carbon atoms we do not have any CH and also this is little bit of the D shielded region because next to CO group here so you can see here now so this one is for this here P and then quadrate you can see here and then triplet is here and then CO you can identify here and this is there and then C is here so this is a structure so I mean how nicely we can interpret the data with little bit of knowledge about the positions of absorption bands in various spectral patterns so now let us look into few examples here an organic compound Dx is composite of carbon hydrogen and nitrogen with carbon constituting over 60 percent of the mass mass spectrum of A shows a molecular ion at M by Z equals 1 1 2 answer the following questions so this is again it's already some information is there 60 percent is carbon so 60 percent means 100 it is if you take 60 percent of that one there should be at least seven carbon atoms should be there so you can do like this and then yeah 70 to 60 percent so it is 70 to six carbon atoms will be there and then corresponding H 12 N 2 you can find out from rule 13 easily you can do that one from rule 13 and that should tell you I have shown quite a number of molecules in my earlier slides so that shows you 1 1 2 once with that one you can also calculate hydrogen index that will give you two so this is the answer so plausible molecular form for X is this one C 6 H 12 N 2 and then hydrogen deficiency index is two here similarly one can also do the same structure but what happens here we don't have nitrogen we have carbon hydrogen and oxygen is there and then again if you find out it comes around C 6 H 8 O 2 oxygen atoms are there and hydrogen index is three here next you can do the same thing here here M by Z is 180 and carbon is 60 so 6 into 12 so basically what happens 108 108 and 8 had hydrogen and 64 4 are there and it comes around 180 and here in this one if you calculate hydrogen index that is 6 here so here one example is there a liquid compound gave a mass spectrum in which the molecular ion appears as a pair of equal intensity peaks at M by equals 122 and M by Z equals 124 small fragment ion peaks are seen at 107 and 109 and here again you should notice here M by Z equals 79 80 81 82 this is very important and this is also very important to see what kind of isomer is there here so what happens these things are characteristic of isomers if you have the same structural isomers are there with three or more possible structures in that case some of this will give you a precise value and then the large fragment ions are at M by Z equals 43 so now it is 122 and 122 is there means probably M M plus 2 if you say then let us take 123 as the mass if you take 123 as a mass we use 13 rule so here so 9 will be there 9 and then if you take so it is 9 plus 6 so that means basically C 9 H 15 this will be because that is 15 123 so from this one the moment we see here 79 80 bromine is there so bromine let us take 80 so 80 will come by taking C 6 H 8 so that means basically we have to take out this one now it becomes C 3 H 7 B R when C 3 H 7 B R is there then the when we look into the fragment we can easily write this one the formula can be C H 3 C H 2 C H 2 B R is one possibility so what is the other possibilities so now this is one bromo and it is two bromo propane so now we have to distinguish between these two at this stage with this information we cannot really tell unless we look into those things or some of these things are characteristic of two bromo propane I let me show you now the spectra here you know mass spectra of both isomers I have here we can see this is for one bromo propane and this is for two bromo propane so here we have these two and here we have these two and here it is missing here in this one and also further what it says 79 to 82 we are seeing peaks yes we are seeing 79 to 82 peaks are there and then we are seeing 41 and 39 so 39 is there 41 is there and here it is missing so from this by comparing and looking into the some of these fragments we should be able to tell that this is not one bromo propane but two bromo propane and we should focus some of these things very minute information that looks like very trivial but when it comes to identifying the right isomer this is very very important so let me stop here continue in my next lecture about summarizing and concluding whatever the lectures we discussed in the last 15 minutes