 Hello everyone. I once again welcome you all to MSP lecture series on interpretive spectroscopy. I am sure you are all enjoying some of those interesting problems. Let me discuss more problems today. Once again, let me begin with inorganic problem related to phosphorous cinema. I just look into the problem here. Consider all possible isomers that could be obtained for the eight membered ring compound P4N4Cl4PH4 and indicate the ideal 31 PNMR spectrum expected for each. I have already given the spectra here, but let me explain. So, this is cyclotetra phosphazine because we have alternate double bonds. This is cyclotetra phosphazine. Usually the parent compound has 8 chlorine atoms, 2 chlorine atoms on each phosphorous and then 4 chlorine atoms have been replaced by phenyl groups. So, we have this composition P4N4Cl4PH4 and one can start writing all possible isomers. You can also make an attempt. I think only 5 isomers are possible here and then by using different arrangement of 4 chlorine atoms and 4 phenyl groups here. And the first one is here like this and if you just look into it, the equivalent phosphorous atoms I have given the same color here. So, 2 blue are there in the first one and then 2 red are there. They are chemically and magnetically equivalent and the second one also same thing, but their positions are little different. They are close to each other and they are opposite to each other and one bond apart whereas, here these 2 red phosphorous are identical, these 2 phosphorous are identical and then in the third one we have 1 phosphorous has 2 phenyl groups and one has 2 chlorine and other remaining have 1 each of chlorine as well as phenyl and in this fashion if you just see these 2 are identical and then P red and P blue are again unique. And then if you look into this one all of the phosphorous atoms have 1 each of chlorine as well as phenyl and this one is very symmetrical molecule and the chemical and magnetically equivalence of all phosphorous atoms is same as a result it is expected to show a single resonance here. And now the last isomer possible is we have something like this we have a diphenyl phosphorous here and opposite to that one we have phenylchlorine and then here of course, 8-membered ring you should remember it is not planar it is little bit of puckered one only 6-membered ring cyclo triphosphazene is planar and in this case we have all 4 phosphorous atoms are unique all different all are chemical and magnetically very different. So, first let us look into the first one here first one these 2 are identical they couple equally with red ones to show a triplet and then the blue ones will also show a triplet here first one yes 2 triplets and the second one if you look into it here this one and this one looks identical, but they are different because one is here p 1 is here and these 2 couple with these 2 to show a triplet and then these 2 would also couple with this one should to show a triplet here and then if you look into 3 here these 2 are identical, but these 2 are different first this one would couple to form a triplet and then each triplet will be split into doublet. So, we will see this here 2 doublets doublet of triplets here and then this will couple with first a doublet and then another doublet. So, this if you look into here is doublet here and again it is a doublet here. So, these 3 signals are here this the ratio if you look into it 1 is to 1 is to 2 integration and then all are different and the long range coupling is ruled out. So, 2 bond coupling is possible here first it couples with this one a doublet and then it couples with this one doublet of doublet and similar red window couple with this one doublet and then this is doublet red one and pink one here this couples with the first blue and then the green one doublet of doublet and similarly green would couple with first with this one and then this one it is. So, it gives 4 sets of doublet of doublets first one will give 2 types of triplets this one would give again two sets of triplets very similar to one whereas, this one will show triplet and then this one will show a triplet whereas, middle ones will show doublet of doublet here pink one and then this is all are equivalent it shows a singlet and then we get. So, this is how you can draw the splitting pattern for all 5 isomers and understand. For example, when we take this octachlorocompon P4N4 say Cl8 and if you add phenyl ethium for equivalents we do not know what kind of isomer we are going to get and probably we may get a mixture of isomers and if you look into the mixture by simply looking into the coupling constants and the splitting pattern we should be able to identify how many isomers are formed and also we can also do quantification in what ratio they have formed. So, that would also give you some idea about the overall yield and then probably we can alter the reaction condition to make one or two in their purest form. Another problem is there now this is with respect to cyclo triphosphazine N3P3Cl6 in that one two chlorine atoms have been replaced by two fluorine atoms as a result we have N3P3Cl4F2 here and then again one can draw 31 PNMR for this compound after writing all possible isomers here one the two isomers are possible one is both the fluorine atoms are geminal and the same phosphorus atom or they can be on two different phosphorus atoms two fluorines can be on the same phosphorus or on the different phosphorus. So, these two isomers are there and in this one this we would be coupled with these two. So, that means we should know the fact that PF coupling is larger and also it is one bond it first fits into a doublet here and then each doublet will be split into a triplet because of this one. So, we should get triplet of triplets here. So, each one will be against triplet because two are there. So, triplet of triplet here this is for this isomer for say A and then this is for B this is for B here these two will be coupling first with phosphorus to a doublet and then each line will be further split into 1 2 3 J coupling is there. So, 3 J coupling is there each one will be split into a triplet. So, what we get is a doublet of triplets here and then if you look into this one here again this will be coupled with these two to first form a triplet and then with fluorine atoms it will split into a triplet here. So, we will get triplet of triplet here the same thing and this one these two will first couple into fluorine to give a doublet and then each line will be split into doublet. So, doublet of doublet. So, we can interpret the data in this fashion for two isomers of N3P3 Cl4F2. Now, let us look into another example this is about some reaction that is carried out we have to identify the product all the details are given here an air sensitive tungsten complex WOCO3 tris-astronitrile this compound can be prepared by taking WOCO6 reflecting in astronitrile for 24 hours in absence of air under argon it gives the maximum number of carbon monoxide we can get rid of using nitrogen donor synthesis only three. So, we cannot get anything else other than this one, but if we stop the reaction in the beginning or for two hours or something like that probably we may get two substituted one substituted also, but if we drive the reaction for completion for reflecting for more than 24 hours we can knock off three carbon monoxide to have a composition something like this three. So, this can be prepared in situ and then it was treated with a bisphosphine two equivalents of bisphosphine let us not worry what is the bridging group here two bisphosphine are there then we are getting a compound A here the IR spectrum of A showed three new CO stretching frequencies at 2040, 2000 and 1950. So, all of them are in the terminal region. So, there is no bridging that indicates we have three terminal carbon monoxide metal to carbon monoxide bonds then 31 PNMR spectrum shown below relative intensities are one is to two is to one. So, we are getting this pattern here one doublet one is a doublet another one is triplet of doublets. So, that means how to arrive at this now we have to deduce the structure and then interpret the data given here it contains octahedral geometry possibly because it is a bidently and it has a tendency to form chelation thermodynamically favored one and another phosphorus is also bound. So, here unless the coordination is expanded to 7 it cannot bind let us assume it is not expanded to coordination number 7. So, in that case what happens one should be uncoordinated this is coordinated this is coordinated this is coordinated. So, now if we just look into this compound here this one would only couple with this one to show a doublet and also this is uncoordinated uncoordinated show a chemical shift very close to the that of the free ligand and free ligand value is also given 62 PPM. So, possibly this one for P uncoordinated. So, now this is chelated CHI would represent now we have to see this one. So, this one can coordinate it is coordinated it can couple with this one as well as this one because both are two bond apart. So, in this case first it couples with this one a triplet and then it shows a doublet. So, that means it should be something like this. So, now other option is it can show first a doublet and it can show a triplet here then it should be now we have to compare which pattern is given this pattern corresponds to this one. So, this is correct first it couples with this coordinated one chelated ones and then these two chelated ones are chemically and mentally equivalent they couple only with this one it shows a doublet here two intensity. So, this is one this is one this is one. So, now it is done so this one for PC chelated ones and then this is for PC from the marantate ligand dangling one and this is for uncoordinated one. So, this so interpretation is done here. So, now I should tell you something about rule of 13 and to understand the mass only if molecular weight is given is it possible to arrive at the molecular formula yes it is possible. We have to have little bit of hint from one or the other spectroscopic, but nevertheless we can distinguish and possibly we can write several structural formula and from that one we can compare once we have some data obtained from one of these spectral methods are more of them. Let us look into what is this rule of 13 is very very important to arrive at the molecular formula for a given mass identified from mass spectrometry. So, what we have to do is once when we get m plus or m divide by 13 it may or may not be completely divided we get the quotient and also the reminder. In that case what happens let us take say we divide by n and then n plus or we should put here n, n is the quotient here and n is quotient plus whatever the reminder. So, this would make the first basic formula and from that one we can check here for example let us say 78 is there to begin with before we add some hetero atoms I would come back to that one later say 78, 78 let us divide by 13 we get 6 that means it is completely divided. So, if you see this is equal to n. So, C6 has 6 this is 72 plus 678. So, we can say without anything this is a benzene molecule. Now, let us look into 92 here 92 92 by 3. So, 7 reminder is 1. So, C7 H7 plus 1 equals C7 H8 yes, we have C7 H8 here cyclohaptor triene. Now, we have 161 what we get is. So, we get 5 extra. So, we get this is 17 here. So, this is the formula for a molecule. Now, we have to work out again using another formula that is hydrogen deficiency index. So, that also we can use it. So, I will come back to those things in next couple of problems. Now, the question is this is fine it is oliphatic or aromatic hydrocarbons what would happen if we have hetero atoms first derived formula as a both. So, first we have to find out this one is Hn n plus R R can be 0 or any number. Once after identifying that one we have to see whether any other hetero atoms are there. If oxygen is there oxygen atom equate is 16 112 carbon is 12. So, you eliminate 1 CH4 to add 1 oxygen atom to add 1 oxygen atom eliminate 16 that is CH4. So, next if you have nitrogen and nitrogen is atomic weight 14 you have to eliminate CH2 to add nitrogen or subtract 14 to add nitrogen and if we have both O and N then we have to go for C2H6 C2H6 we have to add here and then if you have 19 F 19 F is 12 plus 7 19 we have to subtract or take away CH7 to replace 19 F and then 28 silicon we have to take out C2H4 is 24 plus 4 28 and then 31 P we have to take out C2H7 and then 32 S we have to take out C2H8 we have to take out from the actual formula we get after rule 13 this is 24 plus 8 and then if we have OS we have to take out 4 carbon atoms and then if you have chlorine we have to take out C2H11 and if you do not have sufficient hydrogen atoms for we can we can also do this one take out 3 carbon atoms and add additional hydrogen atom for example if one more one chlorine you can either take out C2H11 or add one H and take out 3 carbon that means CLH you add and take out C3 CLH you add and take C3 and then bromine is there you can take C6H7 we will come around 79 or if you consider 80 then you have to take C6H8 and 127 iodine you can take C10H7 so how do we know those things by simply looking into the parent ion fragmentation we can see for example whether it is M M plus 2 M plus 4 immediately that can tell you about the presence of halides this already we discussed while dealing with mass spectrometry so remember from there and you use this simple formula and then after arriving at the molecular formula what we should do is we can go for identifying whether saturated or unsaturated how many rings are there how many double bonds are there and then whether in the CO group is there that information comes from either IR spectroscopy so we should proceed like that so again I am telling you so for example we take 108 here 108 if we take it 108 divided by 13 so it will be 4 will be left so 8 and H12 if you do it 104 plus 4 will be left so it will be 12 so this is it so that means possible candidates with heteroatoms so here take this one remove one CH4 add it can be C7H8O or take out two CH4 and add O2 it can be C6H4O2 or you can add two CH2 remove CH4 and CH2 O and N it can be something like this so we can work out all possible ways to arrive at the right kind of structural formula by having help from the data we have obtained from various spectroscopic means for an unknown sample nitrogen rule is also there a molecule with even numbered molecular weight must contain either no nitrogen or even number of nitrogen so that means if the molecular weight is even in that case it should have nitrogen only in the even number or it should not have any nitrogen and if the odd numbered molecular weight is there it must have odd number of nitrogen only not even number of nitrogen this also information also quite helpful in for example when you go for proteins and other amino acids and other things where possibility of more are there we can use this rule so molecular weight is even only even number of nitrogen atoms will be there or there will be no nitrogen atoms when odd number is there there has to be odd number of nitrogen atoms so with this again showing you consolidating here divide the molecular weight by 13 and reminder less than 13 has to be there is termed as R and the quotient is N so that the molecular formula is Cn Hn plus R here and then element soft unsaturation may found out from the molecular formula if O is there add O and subtract CH4 if N is there add N and subtract CH2 again remember this odd and even molecular weight and if CL is there add CL and subtract C to H11 or subtract C3 H minus 1 that means add one more H and take out 3 carbon atoms after obtaining correct molecular formula find out hydrogen deficiency index using this formula here C plus 1 minus half into all H plus half X X is halogen and plus half N with this one we can arrive at molecular formula and also about saturation or unsaturation and the presence of aromatic group or so ring I would not say aromatic group there can be aliphatic ring can also be there so now let us look into problem here a compound composite of carbon, hydrogen and oxygen has a molecular weight molecular ion at m by z equals 1 1 2 in its mass spectrum the base peak is at m by z equals 28 atomic mass units the infrared spectrum shows strong absorption in the region 28 50 to 29 80 very strong absorption at 17 17 very strong absorption at 17 17 centimeter minus 1 the 1 H NMR spectrum shows a single sharp signal at 2.7 ppm and 13 C NMR also has 2 signals at 37 and 208 ppm. So, this one we should take here first 1 1 2 is there 1 1 2 divided by 13 equals what we get is it is 1 0 4 1 0 4 and then another 8 will be left. So, it becomes C 8 H 16 and now since it says 17 17 there is strong absorption probably 1 or more CO groups can be there. Let us try both options first let us in add 1 oxygen for adding we have to remove C H 4. So, for this what we do C 7 H 12 O and now we can try to remove 1 more because the strong it says C 6 H 8 O 2. So, now with this one let us identify let us assume this is the correct molecular formula now let us look into hydrogen index differences. So, 6 are there 6 plus 1 minus 4. So, 3 are there. So, 3 means basically ring plus 2 double bonds 1 ring and 2 double bonds are there this indicates the ring may not have inside double bonds. So, in this case and then we are seeing around 2.7 2.7 does not fit into aromatic hydrogen 1 H NMR signal. So, that means possibly it is not an aromatic ring 6 are there let us try to write something like this C H 2 C H 2 let us assume because if we write something like this say 8 are there 8 it is satisfied and 6 are there and then O 2 then possibly this one. So, now this is symmetric and they show only 1 sharp signal at 2.7 any other possibility is there something like this. So, then also it is symmetric, but what happens here 1 is there and this 2 on this. So, they are all different 1 2 3 signals we get it in 1 H NMR and another possibility is another possibility is this one in the. So, if you do like this this one and this one are identical and we get 2 type of signals here. So, that means these 2 are not possible this is possible here and probably the structure is something like this. Now, I have taken for this one here you can see we are seeing only 1 signal here at 2.73 we are very close to so, yes this is this compound it is confirmed from 1 H NMR let us look into now 13 C 13 C shows only 2 signals 1 is this one 1 is this one here 208 and 36.5 the molecule what we predicted is correct by using 13 rule number 13 and also hydrogen deficiency index with these 2 and also getting some vital information from IR spectroscopy and also NMR we now we without any problem we identified the right product. So, this is how we can combine all information and also use empirical information empirical formula that are given by people who have worked thoroughly with this kind of spectroscopic methods we can arrive at the structures easily. So, let me come back with more problems in my next lecture until then have an excellent time. Thank you.