 Hello everyone. It is a pleasure to welcome you all to MSP lecture series on Transmittal Chemistry. This is the last but one lecture. So in this lecture let me give little bit information about few multinuclear NMR spectra and how to interpret the spectra and also brief information about IR spectroscopy. So in my next lecture I shall try to conclude and summarize whatever the 59 lectures I gave covering most of the aspects revolving around coordination compounds and to an extent argumentary compounds. So now I have given a few molecules here. Sketch is a 19 F NMR spectrum of PF5. If a question like this is asked one thing you should know if the information is not given about the dynamic process or it is not given information or it does not have any information whether the molecule is static or dynamic with fluxionality and of course PF5 is known for switching their equatorial and axial PF bonds via Berry pseudo rotation. But however if the question does not explicitly say about the molecule we should understand that fluxionality is there in it. In that case what happens all 5 will look almost identical with respect to NMR timescale. As a result what happens if you look into 19 F NMR spectrum it should show simply a doublet. All 5 will be equivalent they will be coupled with one phosphorus. In my previous lecture I was telling about remembering N plus 1 rule to identify number of lines in a splitting pattern. But I would say instead of this one it is better if you use 2 Ni plus 1 rule. I is the nuclear spin of the interacting nucleus and also N is number of equivalent nuclei. For example if CH3 CH2 OH is there and if you want to know of course you just leave this one. If you consider CH3 how we are splitting this one. So here 2 into 3 into half plus 1 it goes 4 lines. So this will give 4 lines for this one so it looks like this. And then for this one it is 2 into 2 into half plus 1 equals 3. So this will give 3. So I think this is better go with this one. This one would tell you the number of lines, the splitting pattern irrespective of number of equivalent nuclei and also the type of nucleus we are using whether i equals half i equals 3 by 2. It is applicable for all type of nuclei and also better to use this one rather than using this one. Do not use this one. Now let us look into so 19 F NMR would show a doublet. But on the other hand you should remember if they are assuming it is a static molecule in that case what happens to axial are different from 3 in the plane. So that means first axial will split by P F coupling to give a doublet and then doublet each line will be split into a quadrate because of 3 in the plane. On the other hand 3 in the plane would first split by phosphorus to a doublet and each line in the doublet will be split into a triplet because of this one. So if time permits I shall tell you about that one but you should be able to work out. So now we have this molecule, this molecule, this molecule is here and of course this is NMR active, this is NMR active. Now the question is asked about this one. So 195 platinum, 195 platinum when the question is asked you should not worry about abundance because we are looking into the NMR nuclei that itself. So in this case now we have to see this platinum is one bond apart from P 1 and two bond apart from P 2. So first it will be split into a doublet and then each line will be further split into a doublet. So this coupling would say this is for 195 P T platinum. So this would say 1 j P T P 1 and then this one or this one would say 2 j P T P 2. So that means spectrum will would look like something like this. For this one I have given in the next slide. So you can see here and of course this is how it looks you know 19, 195 platinum NMR and if you just look into this molecule here we have two phosphorus, one is trivalent, one is pentavalent. So they are magnetical and chemically non-equivalent so they couple each other with a large coupling constant and now we have selenium is there and this selenium will couple to this one as well this one but if the selenium NMR is asked here then you can see here these are the possible isotopes out of which only one is NMR active 77 selenium I equals half that means out of 100 molecules if you consider out of that one 77.65 let us say 7 molecules are NMR active remaining to 63 or NMR inactive that means basically but when we are looking to 77 NMR you have to consider very similar to here. So now first this one will be coupled with this one this is something like this and then each will be split by this second this due to 2 j selenium phosphorus coupling and this is how the spectrum would look like this coupling is this one and this coupling or this coupling is 1 j P S C. So this is how we should be able to write for other nucleus but only thing is when I am looking to 31 P NMR if I look into 31 P NMR of this one we have 2 signals in this case first let us say this one is coupled with this one gives a doublet this is due to 1 j P P coupling if I put P 1 and P 2 P 1 and P 2 and now P 1 will be coupled with this small coupling will be there okay so this is your 2 j P S C coupling 2 j P S C coupling this one also so it looks like and other one now this one also will show very similar to that one but it may show larger coupling. So now how it looks like the first this is coupled with so now it is not 100% as a result what would happen is first you get something like this a doublet and each doublet will be further split into small this one will be something like this and this will be something like this so this is your P S C coupling but when you take this one it will also show a doublet but here it will show a larger coupling so larger coupling how it shows something like this depending upon how much selenium selenium coupling it would be splitting like this. So you should see that one I will show you few more examples now let us look into this molecule here so here we have two phosphorus one is bonded directly to rhodium other one is two bond apart now let us look into first A here first this one will be coupled with rhodium this rhodium coupling is larger 103 rhodium 103 rhodium is i equals half and 100% abundant so first you get a line like this and then this is your 1 j rhodium phosphorus coupling rhodium phosphorus coupling and now this will be further split by this one to triplet this should triplet because P of coupling so each one this is your P of coupling triplet and now this each line will be split into a doublet because of 1 2 2 j P P coupling so now it looks like that so basically a doublet a triplet and a triplet each line will be doublet so that means doublet of triplets doublet of triplets of doublet so this is how it is doublet of triplets of doublet doublet of triplets of doublet this is how you you mentioned the spectrum in the same way if you go for Pb here first this would be coupled with say rhodium phosphorus 2j coupling will be there or if Pp coupling is more that would be there here Pp coupling would be more here so 2j Pp coupling this one will be little less and this is your 2j rhodium phosphorus coupling and then this will be further split by fluorine on this one so this smaller coupling then this is called doublet of doublets of triplets you should be able to distinguish and interpret properly so first phosphorus if you consider this one this is coupled with rhodium because the coupling is larger about 250 hertz and then each line of this one a doublet will be split by triplet because of Pf coupling Pf coupling will be less in magnitude compared to rhodium phosphorus 1j coupling and then each line will be further split into Pp coupling so this is how you can see the coupling tree you can write like this then one can write a spectrum like this so two are there doublet of triplets of doublets. So now when you consider this molecule here simply if somebody asks you to sketch 14 in NMR for this one we have three options okay three options are there what are those if you see this hydrogen is one bond apart these two phosphorus are one bond apart you do not know the magnitude of NP coupling and NH coupling so in this case we have to see whether Pn coupling can be larger than NH coupling one case or Pn coupling will be smaller than NH coupling or both of them can be identical in that case first let us look into Pn coupling is larger first in that case what happens it will first triplet into a triplet because Pn coupling is larger triplet and then each one will be split into a doublet like this so this is one second option is Pn coupling is smaller than NH coupling NH coupling is then it will first give a doublet due to NH coupling and then each one will split into triplet 1 is to 2 is to 1 so the spectrum will be something like this and in this case it will be in the third case both are identical the magnitude is so that means basically this is interacting with the three identical nuclei although they are of different elements so now how to say again use 2 ni plus 1 rule i equals half here so 2 into so 3 are there 3 into half 4 lines so what you get is 4 lines in third case so first it will split into something like this then it will be 1 is to 3 is to 3 is to 1 so these three conditions are there and you should see what condition matches which one you should know now let us look into the most common question we come across about boron hydride sodium borohydride if you take anion here BH4 anion so the spectrum is shown here this is one HNMR spectrum now we have to explain why these lines are 4 lines are equivalent and also we have some lines here now you should remember boron consists of 2 isotopes 10 boron with i equals 3 and then 11 boron i equals 3 by 2 so now you should see if 2 ni plus 1 rule you should apply for this one this is 80 percent that means out of 100 molecules 80 molecules have 11 boron and 20 molecules have 10 boron so first we should write for 80 now 2 into only one boron is there but spin is 3 by 2 plus 1 so you should give 4 lines of equal intensity something like this so this is for 10 11 boron H coupling 4 lines and this this separation is identical this is approximately 80 Hertz this is equals 80 Hertz now you should remember now we have 20 percent if this is 80 percent then 20 percent intensity will be for this 120 molecules they have 10 boron if 10 boron i equals 3 so 2 into only one boron is there 1 into 3 plus 1 so here it is 7 so there should be 7 lines the 7 lines you can see here 1 2 3 4 5 6 7 they will be essentially 1 2 3 4 5 6 7 something like this and then each separation is identical this separation is 10 boron H coupling so this is how you can interpret and you can also explain why the smaller lines are this it is overlapping of both because when you look into boron NMR it will account for 10 boron as well as 11 boron 11 boron out of 100 molecules molecules with 11 boron are 80 and 10 boron are 20 so if you look into intensity of this one if it is 80 say each one is 20 intensity and then here 10 means 10 by 6 you can see how much it is about 1.5 intensity or something at the end both of them should be a match 100 percent so this is how one can interpret. Now one more question the element lead has several naturally occurring isotopes including all these things of these isotopes only 207 has a non-zero nuclear spin i equals half what would you expect to observe for 1HMR spectrum of tetramethyl lead okay given the 2j coupling is 60 hertz so then how do you explain this one the question is very simple we have given here so now it is 1HMR spectrum so that means we just look into all isotopes except for 22 percent 22 percent is NMR active i equals half other you can consider 78 percent NMR i equals you can ignore so that means basically we have 100 molecules 22 molecules have i equals half 207 lead others are we should not worry that means those 78 percent will first show a singlet and then this 22 percent what happened this will be split into doublet and then like you could distance from this one these are called satellite peaks these are and now it is this is 11 percent this is 11 percent this will be 78 percent if you look into the intensity and now this operation is called 1j Pb H coupling this is equal 60 hertz so this is what given something like this this is how the spectrum looks like so now let us look into this one this is very interesting first let me write 31 PNMR and let me write the structure of the molecule quickly N and remember this is 15 N is 100 percent it is 15 N so now Si we H this is something like this all are in NMR active so something like this this one is strongly coupled with two fluorine atoms first it should give a triplet so let me write little bigger because lot of coupling is going to come here so triplet this is your PF coupling PF coupling same triplet and now this is coupled with N a doublet so doublet each one is a doublet this is done this is your PN coupling now this H coupling so H coupling will come another doublet doublet this is your NH coupling and now these three are there these three will also couple each one will be split into a quadrate and intensity will be 1 is to 3 is to 3 is to 1 so you should be able to sketch this spectrum beautifully so 1 2 3 4 5 6 7 8 9 10 11 12 12 into to 48 lines will be there totally so then you can say triplet of doublets of doublets of quadrates this is how the NMR would look like you can see here this is how it looks like and this is how you can interpret and do it try for 15 N also one H NMR also you should try this is been interesting just look into this molecule here this PPR far away and then if you see here it will show single peak for both the phosphorus atom one is a nitrogen one is an oxygen they do not have any coupling because they are far away from each other but in the complex now they are two bond apart because of platinum chelation now they can show P P coupling now this small separation is P P coupling and now platinum again if you consider platinum 34 percent of platinum is NMR active equals half that is 195 platinum rest is 66 is now we should not worry and 0 196 platinum so that means basically if I have 100 molecules so each one would show a doublet P P coupling this also shows a doublet now they will show satellite peak due to this one each intensity will be 17 percent so this one you can see this is one and this is one so here this separation this separation or this separation is identical this is for this P T coupling and this one and this one for this P T coupling so this is satellite peaks are there and if you just look into this one this is 17 percent and then this is 17 percent and then this is 66 percent so this is how you can write satellite peaks equidistance from the central one that is for i equals 0 species isotope so this is how you can see and the magnitude I have given the one J P T P coupling is very large when they are 6 4000 to it can go even up to 6 to 7000 and usually when two phosphorus are trans to each other P T P coupling will be less than 2 hertz 2000 hertz or it may be 2400 that means if any isomeration is there or if you have cis compound or trans compound you should be able to tell immediately simply by looking into phosphorus platinum coupling. So now here for example this compound is taken bis triphenyl phosphine ruthenium CPCL this compound was prepared using I showed you the method and when it is reacted with a bisphosphine all possible so we got all the products as a mixture and how to assign this one where simply 31 P NMR can tell you what are the products you have in the mixture for example this will show a singlet whereas this one show a doublet for these two and a triplet for this one and then this would show a doublet for this one and these two will couple with this one it shows a triplet and then this would show a triplet and each triplet line will be split into a doublet and then if you look into the intensity you can also get the information in what ratios these compounds were made and also if you know the reaction condition whether in case if there is a problem in separating these three we should think of separate reaction method for isolation of these compounds by looking into the stoichiometry and reaction condition for example here RUCL bond is broken that means the the yields of this compound can be enhanced by using a polar solvent whereas if you use non polar solvent you can retain RUCL bond and simply you can replace two triphenyl phosphine with the chelate ligand here what happens entropy is the driving force for this one so that you can analyze using spectrum and also you can understand the reaction condition and accordingly you can change the reaction condition to make the desired compounds I will show you the spectrum here this is how the mixture looks like the one compound where we have simple chelation or identical one you are getting as I mentioned here you have very nicely you can interpret the data for three type of compounds we have this one will show a singlet and this will show a complicated three lines are there one and then this will show a doublet and a triplet these two and whereas this one is a doublet for this one and then this is a doublet of doublet here and then this one show a triplet and each triplet line is split into doublet so you can very nicely interpret and understand what kind of products you got in your own reaction and now if you just look into n-butyl lithium it has a cubane structure for example two spectra are given at 195 k okay it is static structure here three lithiums are connected to each carbon the cubane structure is there just recall my teaching in my you know main group chemistry lecture if you have looked into or you can just go to my lectures and look into it the structure so three lithiums are connected to tertiary carbon here so this one will show seven lines here because lithium has i equals 1 so if you use 3 ni plus 1 3 into 2 plus 1 7 lines it shows but on the other hand when the fluxional process is there we have four equivalent lithiums are there in cubane four alternate corners are occupied by lithium and methyl group so we have totally four lithium atoms are there all are interacting with carbon because of fluxional process in that case we get nine lines so very nicely you can interpret the data so this is about similarly you can look into this one and you can just see I will stop now I shall proceed for IR spectroscopy for few minutes so infrared spectroscopy is a vital tool to chemists wishing to identify what functional groups exist in unknown samples the light our eyes to see is but a small part of a broad spectrum of electromagnetic radiation in the visible region on the immediate high energy side of the visible spectrum lies the ultraviolet and on the low energy side is the infrared one so you can see here this is the range so with a wavelength range of 2500 to 16000 nanometer the infrared region is capable of revealing information not easily uncovered through basic means and what are those things so proton energies associated with the infrared region are not large enough to excite electrons but they are still strong enough to induce vibrational excitation of covalently bonded atoms and groups so when you look into a bond so covalent bonds are not rigid sticks but rods but are like spring that can be rotated if the single bond is there or they can be stretched like this or they can be bent this is how with by simply absorbing energy in the infrared region this will be the stretching is there bending is there all rotation is there so this wide variety of vibrational motions is characteristic PM molecules component atoms all organic and inorganic compounds will absorb infrared radiation that corresponds in energy to these vibrations and infrared spectrometers permit commits to obtain absorption spectra of compounds that are a unique reflections of their molecular structure so now let us look into a typical spectrum I have given here and this is for formaldehyde so IR readings are inverted compared with UV visible spectra UV spectra you can see like this is coming but it looks like inverted one the sharp peaks while that one that did absorb a significant amount would be towards the bottom if they are observing they will be falling down those do not observe will go flat like that each peak dips on the graph represents various bond characteristics and because most functional groups have specific bonds they can be readily identified high given the groups and also corresponding stretching frequency here for maldehyde what is important is ketone carbonyl group is there and shows around 1750 centimeter minus so this how this is the inverse one we are not considering this one so this how it looks like so now how it helps it is very very important in carbonyl complexes infrared and it is very vital in structure determination also by simply looking into the number of bands in the CO stretching region if the number is consistent with that provided by the selection rule of a particular point group may be assigned to that molecule and also for CO around 2000 centimeter inverse is very very important relationship between the molecular structure of a compound and the activity of stretching modes can also be understood and analyzed to arrive at the structure of a molecule for substituted carbonyl complexes we can think of assigning point groups considering the local symmetry of the metal and the carbonyl groups provided ligands of spherical symmetry or by considering the symmetry of the molecule as a whole so we have two options before we assign look into stretching frequencies for example let us look into example shown here C P M N CO 3 it has C 3 V symmetry and C P V CO 4 it has C 4 V symmetry and if you just look into this one Arene M CO 3 it has again C 3 V symmetry so all of them show two stretching frequencies for carbon monoxide and both C P and C 6 H 6 Arene are axially symmetrical with respect to these point group that means you can rotate through that one passing through the middle of your C P group or benzene you can do a rotation and in case if we replace Arene group with say Aniline or something or thiophene what happens symmetry will be lowered then the symmetry of whole molecule has to be considered so you should not consider local symmetry of metal you have to consider the whole symmetry in that case what happens we will see three stretching frequencies electron distribution okay of course electron distribution I have discussed in length how a pair of electrons and carbon will go to metal through sigma bonding and metal T 2 G electrons will be transferred to pi star through back bonding so depending upon to what extent electrons are moving from metal to ligand stretching frequency will vary accordingly so this is the typical bond explanation so this is sigma bonding and this is pi bonding these two modes of bonding are mutually reinforcing and strengthening metal to carbon bond this is called synergy effect or synergistic effect charge removal through pi bonding leads to more extensive sigma bonding while charge donated donated through sigma bonding thus specializes further back bonding so that means it is give and take because of sigma donation see carbon will become electron deficient and metal become rich it gives through back bonding with back bonding what happens metal become poor and carbon monoxide become rich so sigma bonding will be stronger so both of these opposite direction bonding strengthens each other as a result what happens we call it as synergistic effect so eventually metal to carbon bond is strengthened and carbon to oxygen bond is weakened so to what extent this is weakened or strengthened depending upon how much electrons are coming from metal to pi star of carbon monoxide so that is a measure of the overall electron density at the metal center and if you are looking into the combined or mixed ligand complexes what is the share of other back bonding ligands and how they compare in terms of their back bonding capabilities for example while it is not necessary to remember all of the peaks for each functional group learning a few key characteristics will be enough to answer 99 percent of all your questions in the reading of formaldehyde spectrum we have lot of peaks here we should not focus on those things we have to focus on only few things like sea stretching co stretching where it is and also whether we have ch3 or stretching or any other so these are very important while characterizing this we call it as fingerprint region so the complexity of infrared spectra in in the 1450 and to 600 centimeter region makes it very difficult we get lot of peaks in this region it difficult to assign all the absorption bands and because of unique pattern found there it is often called the fingerprint region it is not necessary to remember any peaks in this region as they will not help you determine any significant functional groups and not any significant functional groups come in that region either so absorption bands in the 4002 1450 are important are usually due to stretching vibrations of diatomic units and this is sometimes called the group frequency region it is this region which is most practical to identifying groups some notable frequencies that should be remembered I have shown here you can see alcohols and acids OH group is there in this range it is there for amines NH this will be this range and for alkanes CH okay they come here narrow peak and ketones acids CO group will be here and for triple bonds CCRCN it will be important first neutral binding is there or acetylene binding is there these are very important so one thing to keep in mind about IR spectra is that it can only tell you whether a group is present or not it will not tell you how many groups are how lot the molecule is once you remember it is just qualitative not quantitative for example if you see a sharp peak at 1750 centimeter minus 1 it will tell us that there is some sort of a carbonyl group it can be ketone ester or part of an acid etc or sometime it can be metal carbon monoxide with extensive back bonding or it may be bridging two metal centers but it will not tell us perhaps there are two or three ketone groups present so it will not tell you whether how many such groups are present as I mentioned it is qualitative not quantitative with this in mind and using only the selected few important peaks we have to remember so we can say easily or we can guess what functional groups are present and thus identify the molecules from a few selections. In the following spectrum if you see we should focus our attention on 3000 and 1750 one can most easily notice a sharp peak around 1700 easily so that means you can conclude that there is a CO group that it is most likely a ketone or aldehyde this range also tell you secondly there is no broad group or a sharp group around 3000 so that means there is no OH or NH so there is however a sharp peak just below 3000 which most likely corresponds to all KNCHCH2 and CH3 bonds it could be part of the aldehyde CH bond as well so that means if the choices are there like 2-meter cyclobutanol 3-meter 2-butanol and 2 chloropentonic acid if the options are this it should be very clear that we will recognize this as 3-meter 2-butanol which has both necessary CO band and they and it lacks a OH bond as well so immediately we can pick the right one among these three to say that this spectrum belongs to this molecule here it is that it is easy. So now let us look into metal carbonyls metal carbonyls with nitrogen donors you can make a maximum you can replace a maximum of 3 carbonyls I mentioned so now if you see here a free carbon monoxide is around 2133 and metal hexa carbonyls will show around 2000 and increase in negative charge on metal is observed by new CO changes for example you can see here so it is a positive charge is there and indeed here neutral and its negative charge is there so when your negative charge is there more electron density is there as the result stretching frequency will drop more electron density goes to the pi star and here it is a midway 2000 whereas here is a positively charged so less electron density goes from the metal as the result stretching frequency is more so we should be able to analyze the electron density that resides on metal very easily by looking into the stretching frequencies and of course the two extreme cases of metal carbonyls is shown here very strengthened and very weak and very strong here CO group and here of course metal to carbon strength increases in this order this information very nicely comes from IR spectroscopy so pi acceptor ability of CO in metal hexa carbonyls will tell you that it can take anywhere between each carbon monoxide can take anywhere between 0.1 to 1.2 electron paste to its pi star orbitals so new CO decreases as more and more groups are substituted that means more and more groups are substituted with non back bonding ligates in that case what happens the their responsibility falls on few carbon monoxide left on the metal as a result the stretching frequency decreases so complete substitution of CO from MCO 6 has been achieved only with strong pi acceptor ligand such as PF 3 the best competitors for CO or phosphines especially having strong electron withdrawing groups or electronegative groups on phosphorus so advantage with phosphines is coordination properties can be readily altered but you cannot do with carbon monoxide so among no pi acceptor ligands example diagram ammonia and all those things if you want to know what is a diagram it is an ether something like this and how to spell it you can say diethylene glycol dimethyl ether many reactions you come across a diagram you should be able to write the structure and you can tell it is diethylene glycol dimethyl ether so mixed metal carbonyls with one or more diagram like ligands can also show some trends in the stretching frequencies for example if the diglime is there that shows higher stretching frequency than having simple diene here and also when we have these things sigma orbits for backburning decreases and then relative sigma donor ability of about donor atoms may be estimated from the stability of their addition complexes especially with aluminum trichlorides and Lewis acid the order of donating ability is also follows this order and this order it is about only backburning this has to do nothing with stretching frequencies so now I have given depending upon the type of point group possess it by these molecules and how many IR active modes are there for example MCO5L is there you can see three CO bands and if you have something like this sis you can see four and if you have trans you can see one and it can just go through this one I have given for octahedral geometry and I have given for trigonal bipyramidal geometry as well and also I have mentioned position of carbon borne oxide and also the point group here just you can look into it and now for example you take hexacormyl and if you replace one three and you get a facial in that case we get only two provided all ligands are identical and then in case what happens if you have one MCO5L you get three stretching frequencies and if you have trans we get only one stretching frequency similarly if you have sis here we can get four stretching frequencies and if we have something like this here we can get three stretching frequencies meridional so something in the same way when we go for pentacormy and if you start replacing carbon monoxide with L the number of stretching frequencies would be shown in this figure here so three active here four active are there and in this case CS so all of them are unique you can see three different and same thing here and same thing here with this so let me conclude about IR spectroscopy and MR spectroscopy and overall whatever I wanted to tell as a part of this course probably in my last lecture I said try to consolidate and summer up and as if anything is left little bit I will highlight about that one and conclude my lecture so that would be the 60th lecture so thank you for your kind attention and I have to go through quickly in case of IR and NMR because of lack of time of course but everything is there in the slides and also I have spoken already please listen to what I say and also try to look into all the slides in detail and try to understand the chemistry behind these things thank you so much.