 Today, we will discuss reductive elimination, which is a simple reaction and it is the inverse of oxidative addition. So, during oxidative addition, we had an increase in the coordination number, the number of ligands coordinated to the metal. We also had a change in the oxidation state, which increased by 2 units or 1 unit. On the other hand, the reductive elimination involves the exact reverse of this particular reaction. So, it will involve a reduction in the oxidation number, the formal oxidation number of the metal. So, before we look at reductive elimination in detail, we should look at the thermodynamics of the oxidative addition. If you consider the three groups, where oxidative addition happened, you will realize that in the group 1, you had the addition of two groups, say CH 3 X and this is labeled in two different colors. Just to tell you that these are the two fragments, which are going to be added to the metal CH 3 and X and it invariably ends up, if one takes the example of iridium 1, it ends up in an increase in the coordination number by 2 and also in the oxidation state change of plus 2. So, the reverse reaction depends on the stability of the oxidase product. So, here I have pictured energy level diagram, where we have the reactants here, we have the reactant and the product on this side and you will notice that there is a energy of activation, but this reaction is exothermic by a certain amount and the thermodynamic stability of the product will determine how exothermic this reaction is. If the reaction is very exothermic, we will have an extremely stable product and the reverse reaction that is the oxidative addition will become more difficult. So, when will this be more difficult or when will the oxidase product be more stable? In this particular example that I have given, I have added CH 3 and X. Let us take another example, where you have two more electronegative groups, say halogen is added to the metal. Then the stability of the oxidized product is more. So, iridium 3 becomes more stable when you have two electronegative groups in the coordination sphere. So, the formal oxidation state of iridium as plus 3 becomes more stabilized when you have two electronegative groups. Then reductive elimination is lot more difficult and it is obvious from this energy level diagram that the energy of activation for this process is going to be greater. So, the energy of activation in this instance is going to be this large amount. So, this will become a more difficult process compared to the reductive elimination of CH 3 X. So, this is just to illustrate the principles involved in the reductive elimination process. It is primarily dependent on how stable the oxidized product is. If the oxidized product is extremely stable, then reductive elimination will be a high energy process. Let us now take the group 1 addenda and let us try to explain how reductive elimination might take place. The principle of microscopic reversibility suggests that if a reaction proceeds in a particular path, then the reverse reaction in this particular case the reductive elimination path should involve, should retrace the steps which it took when it formed the oxidized product. So, we started with a metal in plus 1 oxidation state and we ended up in a metal in plus 3 oxidation state. You will notice that the product which is given here, the product if it has to lose two groups can do so in a variety of ways. It can do so in a synchronous fashion in a concerted fashion, but what happened in the oxidative addition was a step wise reaction. It was an SN2 reaction between CH3 and X. So, in the reverse path one can imagine loss of X. So, that is X here is loss from the coordination sphere of the metal and you form an ionized product. So, X minus is formed and the metal complex positively charged metal complex is formed. So, this is basically an exit of this X. So, X has to be a good leaving group. You will immediately notice that there is a complication instead of X this other group Y can also leave. So, this group Y can also leave. So, let us take the next step. Now, X minus has to make an attack on CH3 in an SN2 type of fashion and if it does that CH3X will be formed and the original complex will be released. So, this is what is dictated by microscopic reversibility. So, you expect ionization of the product and in the second step you expect attack of X on the CH3 group. So, if you take the second group of molecules which carried out oxidative addition you notice that there are some addenda like oxygen where one bond between the two atoms which are adding on was retained. In fact in group 2 all of them will retain one of the bonds. In the case of acetylene there are three bonds and so it will end up with a product which have only two bonds. But invariably in the case of systems where you have an electronegative atom like oxygen it is highly unlikely that the reverse reaction is going to take place because the oxidized product will be extremely stable or it can undergo further reactions. As in this particular instance you have a dioxygen adduct which will react with some other species and carry out a redox reaction of say the ligand or a substrate. We also notice that in the case of acetylenes you have after oxidative addition which involved two acetylenes for example in this cyclic catalytic cycle that I have written here. You will have a second substrate coming in and so you can do an insertion reaction and then eliminate say an arene. So, elimination of arene in the final step which is indicated here elimination of arene turns out to be a thermodynamically very favorable process. So, you take acetylenes and convert them into arenes after performing an oxidative addition. What we have achieved is oxidative addition with one substrate and subsequently through insertion reactions we have carried out the transformation of that substrate into a very different molecule in this case an arene. So, in the third group we have no multiple bonds nor do we have polar groups. In the first group we had polar groups, in the second group we had multiple bonds, but in the third group we have neither polar groups nor do we have multiple bonds. This is typified by hydrogen aromatic C H or an S I H on a S H bond. Here the polarity is typically very small. So, let us take a look at how these molecules will undergo redux reaction. If you remember what we said in the beginning of this lecture if you have two electron negative substrates oxidative addition becomes favorable. Here we have substrates which are not very electronective and so the redux the reductive elimination or the reverse reaction would be an extremely facile process. In fact, it is very difficult to carry out oxidative addition in these substrates and the reverse reaction is always favorable. For example, we have we took this case where you can photolyse a carbon monoxide complex and photolyses leads to elimination of one carbon monoxide from the coordination sphere of the metal. It leaves you a iridium one species which is co-ordinatively unsaturated. This co-ordinatively unsaturated species can add on methane. If it adds on methane such that hydrogen and methyl group are added to the iridium you end up with the iridium three species. So, this is an iridium three species and this oxidized product will readily undergo reductive elimination. In many cases it will reductively eliminate the two addenda that it added on or two different species and that is what leads to useful reactions. If it just eliminates methane in this case it just leads to a reverse of the forward reaction. It is also reductive elimination but it is not a productive reaction. Several examples were known and we had noticed this with in the beginning of 1980s. We started with variety of CH activation reactions and now this has become fairly common but it is still a particular challenge. Today we are going to talk about two or three different reductive elimination reactions. We will take some specific examples where these principles that we have dealt with can be discussed. The first example that I have taken is an example from the recent literature from 2011 where a cobalt three complex was synthesized. The synthesis itself was known for a long time. It is a old synthesis but then the reductive elimination was studied in detail and some very important principles regarding reductive elimination could be obtained by looking at these reactions. So, the reference the detailed reference is given here the DOI and the journal reference is given here for your information. You could read this paper in detail if you would like more information. The substrate in question is a cobalt three complex which was formed. The cobalt three complex could very simply be generated by taking a methyl cobalt complex and that is given here this PME three complex was synthesized and then treated with methyl iodide. Notice that methyl iodide now will do an oxidative addition but during this process PME three is generated and the PME three reacts with methyl iodide that is present. So, you need to keep two equivalents of methyl iodide so that the reaction can be pushed to one direction. In other words PME three is removed as it is formed and it forms this PME four plus I minus salt here which is indicated here and this cobalt three formally cobalt three species which we have two methyl groups on it. Notice that this molecule can now reductively eliminate in different ways. As I told you before the elimination of methyl iodide will lead to a non-productive reaction but this molecule has the option of removing two methyl groups in the coordination sphere of cobalt. One is this which is labeled in blue and the other which is labeled in black. So, this two molecules which are cis related not very polar so they can undergo reductive elimination in a synchronous fashion in a concerted fashion and eliminate ethane. In fact this is exactly what happens if you slightly warm the reaction and in many cases if the reaction can be carried out at slightly elevated temperatures it allows you to characterize the reactant very well. In this particular case the cobalt three complex was stable at room temperature but if you heat it it eliminated methane. The color coding is purely to tell you what are the groups which are being eliminated. Here there are two methyl groups which are going to be extruded. It is not methyl iodide which comes out but it is two methyl groups which come out. That makes it a favorable reaction. So, then you end up with a cobalt one complex which has got an hydro ligand and it is stabilized by these three PME3 groups and that is labeled as compound one. Now, if you think about this reaction it could be done in a concerted manner. Two methyl groups can come together a methyl methyl bond can be formed and the elimination can take place but because we talked about microscopic reversibility there are different possibilities and in fact these workers examined this reaction in detail and what they found was the first step required dissociation of PME3. So, if you have excess of PME3 in solution if you do try to do this reaction in the presence of PME3 this reductive elimination is hindered. In other words when you heat this reaction mixture the first step is loss of PME3. So, loss of PME3 is there and a vacant coordination site is generated. So, this is your vacant coordination site and that is essential for the reaction to happen. This is in fact strange in the previous instance we said the two of the two groups which are reductively eliminating one group comes out first and leaves a vacant coordination sphere. That was what we learnt from microscopic reversibility but in these cases it is a third group the one which is not added methyl added was added but a PME3 which is a spectator ligand has to leave the coordination sphere of the metal and generate a five coordinated intermediate. If you remember the reductive elimination involves reduction in the number of ligands coordinated to the metal. Surprisingly for the reduction in the number of ligands you need to generate a vacant coordination site. This turns out to be a paradox which is there in organometallic chemistry in specially in reductive elimination. Why do we need a vacant coordination sphere before you can generate more vacancies on the metal. So, this is a mystery which needs to be solved. Now, this is the first step of the reaction there are more interesting things to follow if you take the second possibility after generating a vacant coordination site by loss of PME3 that is indicated here in the first step on your left side top you have this five coordinated intermediate that is what is generated in the first step. If you remember your coordination chemistry five coordinate complexes are extremely flexional which means they are labile they will change the coordination site according to the need and that leads to very interesting reactions. Now, there are two possibilities which the authors considered one is to dissociate a methyl group generate a methyl radical. So, this is a methyl radical which is generated and a cobalt two species a formally cobalt two species is generated. If you remember we said generation of methyl anion is difficult I do anion would have been possible, but that leads to a non productive reaction. Here we are generating a methyl radical which means a cobalt two complexes a formally cobalt two complex is generated, but this methyl radical could be in a cage and it could react with the methyl species which is still attached to the methyl combination of these two could lead to elimination of the dimethyl group or the ethane. So, this is the group which is eliminated from the coordination sphere of the methyl and so you have ethane elimination from this intermediate. Surprisingly this is not the path that is this is not the path which is favored it is also possible that you have a concerted elimination after you form the five coordinate intermediate. Now, that will give you a species where the two methyl groups have a cis coordinated and as I told you because of trigonal bipyramidal species being fluxional you can have rearrangements which will lead to cis arrangement of the two methyl groups and elimination concerted elimination can give you ethane. A third possibility was also considered by these workers and that involves both reductive elimination and also insertion reaction considering the fact that you require a vacant coordination sphere on the metal. Let us generate it with this five coordinate intermediate having generated a five coordinated intermediate you could in principle generate from this methyl group. You have three hydrogens on this methyl group three hydrogens are there in this methyl group you could generate a cobalt hydride species that would correspond to a propane. Abstraction of hydrogen from the methyl alpha hydrogen abstraction this is the alpha group alpha carbon and if you abstract one hydrogen from there then you would end up with a methylene unit along with a hydrate. Now, this is not an unreasonable reaction to hydrate considering the fact that this is just a migratory insertion of reverse of a migratory insertion of a hydride unit. This is the hydride unit if this migrates on to the CH 2 group you would end up with the reverse reaction that is the methyl group formation. So, this is basically just a hydrogen abstraction from the alpha position you can think of this as a nucleophilic attack of a hydride on to a neutral methylene group. So, this is a migratory insertion and it is reverse what we are talking about in the black in the forward reaction is actually the forward reaction is actually the hydrogen abstraction. This is the hydrogen abstraction reaction. So, now having formed this methylene group you could now have a second step in which the methyl group which is against this to the methylene group. This can now carry out a nucleophilic attack on the methylene group. So, that is now a migratory insertion of the methyl if that happens this is a migratory insertion of the methyl on to the methylene you would end up with a ethyl moiety on the cobalt. You will notice that these are color coded the migrating methyl group the anionic group is the one which migrates to the neutral group. So, that is what we have written it is a migratory insertion. So, the methyl group attacks the methylene unit and an ethyl moiety is formed is generated on the cobalt. If you now eliminate ethane you can very easily generate C H 3 C H 2 H by eliminating these two units together. So, you can just eliminate these two units and both of them are cis related the cis related and it can have a reductive elimination to have C H 3 C H 2 H. So, this is a another alternative mechanism and what the researchers carried out was to do isotopic labeling in order to figure out whether this reaction was happening or not. In fact they showed that this was not happening in the reaction doing the course of the reaction. So, they considered three reactions and based on their results they concluded that path B where you have a concerted elimination. This is path B concerted elimination is the most favorable pathway and alpha elimination to give you a series of migratory insertions and alpha abstractions is not happening. So, in other words this is not happening this path is not happening and the radical elimination was also not happening and it is in fact the formation of a co ordinatively unsaturated intermediate from which a concerted elimination of two methyl groups happen and you have the reductive elimination. So, the various paths that the molecule can take because it is a five coordinate intermediate are listed here. We can start with the five coordinate intermediate which is here and then we can carry out very pseudo rotation which is basically a reaction in which the two equatorial groups become axial one equatorial group remains stationary and that is called the pivot atom. So, if the black methyl is the pivot which means this is the pivot let me color it for you in a different color. So, if the pivot atom is this if this is the pivot atom then we will move the equatorial groups the two green equatorial PME 3 units and make them axial you would end up with an intermediate like this. This intermediate which will be a slightly higher energy intermediate can undergo another very pseudo rotation and that can be carried out this time you take the three equatorial groups are the methyl the methyl the Ido group and the another methyl group. So, you can note you will notice that the two methyl groups can be interchanged through this very pseudo rotation process and you can end up with a second intermediate which is listed here on which is the intermediate here through another very pseudo rotation this time the blue methyl is a pivot. So, that means we keep this constant and we carry out we carry out a very pseudo rotation in such a fashion that the methyl and the Ido units which are labeled in black become axial groups. So, this is the axial group now. So, this is the axial group. So, this angle contracts this angle between the methyl cobalt and the Ido group contracts and the PME 3 units come together to form a very pseudo rotation product which is listed here. Now, you can see how the two methyl groups can interchange very easily during the course of this reaction and you can have elimination of methane resulting in the formation of a product which is a cobalt one process cobalt one compound. So, reductive elimination is a fairly complex process and recently more detailed investigations have been carried out and in this particular study which brings out some very important principles it has been possible to figure out how palladium 3 compounds and palladium 4 compounds can undergo reductive elimination. In the example that I am going to first discuss it is primarily a palladium 4 compound which is again a mononuclear intermediate which is formed and it undergoes reductive elimination to give you a C O coupling. As I mentioned to you if there is a possibility for C C coupling and C O coupling reductive elimination may very often favor a C C coupling. But in this particular instance because of considerations that are steric in nature and also C C electronic in nature basically you end up with elimination of two cis groups which are C and O and you have only a C C elimination product. C C elimination product is not favored C O elimination is favored and the research has been has been focused on how we can make C C coupling products. So, let us take a look at this reaction in detail which will help us to understand some very important principles. If you take a species which is palladium 2 you can have a palladium 4 intermediate through an oxidative addition of a C H bond. In this particular instance the C H bond there is oxidatively added is given here. This is the C H bond which is going to be oxidatively added and it forms a bond with palladium and one of the acetate groups leaves with the hydrogen which is present on the arene. So, much so that you have an intermediate which is a palladium carbon product palladium carbon bonded product and you will notice that from palladium 2 you have changed the oxidation state to palladium 4. So, in this palladium 4 intermediate which had a two acetates to start with you end up with only one acetate and H O A C has been removed H O A C has been removed and so this proposed intermediate can undergo reductive elimination. This time it undergoes reductive elimination again the elimination happens in a cis fashion. So, these two bonds are broken the two bonds that I have marked in green are broken and an acetate aromatic ring bond is formed and the product is the acetate functionalized aromatic phenyl pyridine. So, this product is formed exclusively and the intermediate is supposed to be a palladium 4 intermediate. What makes this study very interesting is a fact that you can generate the palladium 4 intermediate you can make it in a stoichiometric fashion and study the chemistry of it. The palladium 4 intermediate that could be formed in this particular instance is listed here and it has got a very stable palladium 4 coordination geometry octahedral geometry which is very stable for palladium 4 and because of the cyclometallation it leads to a very rigid geometry and that is illustrated in this particular picture which I have given here. The carboxylate ligands the two acetate ligands which are present in the coordination sphere of the metal are the ones which allow for reductive elimination and that leads it can be introduced by oxidation with phenyl Idozoacetate. So, let us just take a look at this reaction. If you take the bis cyclometallated product which is again palladium 2 formally because these are anionic ligands. So, if you look at this ligand. So, you have a lone pair on the nitrogen which is coordinated to the metal and then you have you have anionic center which is coordinated to the metal. So, let us take a look at. So, these are the two species. So, because this is anionic now and you have two of them you have a palladium 2 center. You can oxidize this species with phenyl Idozoacetate which is an oxidizing agent which has got iodine in the plus 3 oxidation state. If you treat this molecule with phenyl Idozoacetate you will end up with removal of P H I. This is actually minus P H I and O A C is added on to the palladium. So, palladium 4 complex is formed and this is formed in good yield and this now allows you to characterize this intermediate very well and then study the reductive elimination. Here is a crystal structure of the palladium 4 complex and it is interesting that this type of a geometry coordination geometry leads to 16 different C H signals. Each one of these hydrogens is unique and gives you a different signal in the NMR spectrum because the environment around them is different. So, you have 16 of these hydrogens on both rings and all of them have a different environment. Since this is not particularly a spectroscopy class, we will not discuss the spectroscopy of it, but that allows them to characterize the structure of these molecules very well. If you heat this molecule you now have the possibility of forming both carbon-carbon coupled product. You can have a carbon-carbon coupling happening and that will give you this particular molecule where you have the two phenyl pyridines coupled together or you can have the acetoxy compound. As I mentioned earlier, although two products are possible, during the catalytic cycle you ended up with only the acetoxy product. Only the acetoxy product 3 is formed when you heat it. So, we would like to understand how exactly this whole reaction is happening. The reaction is carried out by heating at 80 degrees and in astronitrile this leads to formation of 3 and the other metal containing product is the acetate bridged palladium 2 dimer. So, now the question is how exactly is the reductive elimination happening? Because we know from several other examples that reductive elimination although it is paradoxical it requires a vacant coordination site. The first thing that they examined was what would happen if you have added acetate ions? Suppose you have an acetate or an aromatic acid. If you add the anion then it should suppress this equilibrium which is given here. If this equilibrium is suppressed then you would have only the pentacoordinate intermediate and this pentacoordinate intermediate is presumably the place from which reductive elimination happens. So, you could have a concerted mechanism in which the two groups which need to be eliminated which is what I am going to mark out in green here. These are the two groups which need to be eliminated. These can be eliminated in a concerted fashion and could give you a palladium 2 compound with organic material which is listed here. So, these are the two possibilities and initial studies in fact suggested that there is no effect of added acetate or the carboxylate anion which means whether you suppress this formation of the intermediate or not the reaction was happening. This was in fact very surprising because one normally has the formation of a vacant coordination sphere. Elimination of one of these bonds would be higher than highly unlikely because you now would have to generate a chelated compound breaking one of its bonds and forming a mono coordinated species which is highly unlikely. But nevertheless that has also been considered for this mechanism. The third possibility was another mechanism where you would have some rearrangements and we will come to that. Before that let us just see how these compounds can be detected. For this particular example they used electro spray mass spectrometry electro spray mass spectrometry to study the reductive elimination. They found that if you equilibrated the system with labeled acetate which means it is C D 3 C O O minus this is the labeled acetate where deuterium labeled acetate was used. They could exchange one of the acetate groups. There are two acetate groups here but only one of them underwent exchange and they identified this using mass spectrometry and a combination of crystallography of a differently labeled carboxylate. In the mass spectrometry experiment they could show that loss of acetate from the palladium 4 species was readily happening from this position. That means opposite the carbon because carbon was having a large trans effect this group was very labile and even if you have the labeled acetate you would end up with mass of 473 m by z value of 473. If you took the unlabeled acetate then also you ended up with a mass of 473. If you took the labeled acetate generated using acetate which was added externally then also you ended up with 473. On the other hand if you independently synthesized if you independently synthesized doubly labeled acetate which could of course be carried out by taking the adosovacetate which is labeled then you can generate this fragment which gave you a mass of 476. This 476 is coming from a species which would be generated by loss of this one acetate group but because the second acetate is labeled you have a increase of 3 AMU. Even if you take an acetate which is only a di acetate molecule palladium 4 di acetate molecule as 2 B D 3 if you take this molecule which can be generated by taking 2 D 6 and treating it with NaOAC. In excess if you treated with sodium acetate in excess which is unlabeled you would generate 2 B D 3. This molecule also gave 476 as the mass. So, using mass spectrometry they were able to show that in fact the only species which was being formed in solution was 2 A D 3. So, this tells you that one acetate is labile in the reaction in the doing the course of the reaction. The palladium 4 complex has got one acetate which can be removed and replaced but surprisingly that was not helping the reaction. So, they also showed that the acetate which was left behind which was not exchanging was the one which was doing the carbon oxygen coupling. So, here is a crossover experiment they carried out. They took 2 aromatic carboxylates and these aromatic carboxylates were added on to the palladium 4 and now after exchanging with acetate they carried out the reductive elimination. If reductive elimination is carried out from this species because you have 5 equivalents of the acetate this carboxylate would now be replaced by an acetate. In spite of that you have only one product being formed and that is being generated by a combination of the aromatic carboxylate and the phenyl pridine. So, this is the only product which is formed during the course of this reaction and that is because of the way in which the species which is exchanging is not the one which is undergoing reductive elimination. So, in the first step they showed that only one acetate is exchanging and that is a group which is trans to the carbon and that is the one which is present here which I have built in green. This is the one which is undergoing exchange and in the second step they showed that even if you exchange that acetate it is the carboxylate which is left behind intact on the palladium 4 that is the one which undergoes reductive elimination. So, 94 percent of this product was formed in spite of removing the carboxylate which is trans to the carbon. So, clear indication that it is not this carboxylate which is going out which is carrying out a nucleophilic attack on the ligand and carrying out reductive elimination as we had seen in the group one chemistry group one oxidative additions. So, one more experiment they showed this is very clear that even if you carry out this exchange without carrying out the reductive elimination you could end up with acetate labeled in position 8. So, now we have two sets of experiments here which are which are listed here. So, with all with labeled compounds which I told you could be independently synthesized they confirmed very clearly that it is only the labeled acetate which is left behind on the palladium which is reductively eliminated. This was very clear if you treat this with unlabeled acetate you would end up with only this product 2 B D 3. So, if you treat this with sodium acetate which is not labeled you would end up with 2 B D 3 this molecule would only give you this as a compound because only this acetate is exchangeable this can be exchanged. This other acetate is not exchanged and if you heat this compound only one product is formed and that compound has got the label. So, two things have been established very clearly. The acetate which leaves the coordination sphere is not the one which is attached to the carbon at the end of the reaction and only one acetate is exchangeable. So, the third possibility which I told you which will require this chelate ring breakage was also considered, but this requires that you have the chelate ring which is not coordinated and it is highly unlikely and through a series of experiments where they tied up the two chelate rings they showed that this is highly unlikely to happen. So, the concerted mechanism path B is probably the one which is happening from the co-ordinatively unsaturated intermediate that is formed. So, rational design for carbon-carbon coupling is one thing that we talked about. Now, how would we make this carbon-carbon coupling happen in this case? If you use a large amount of a carboxylate the carboxylate that is coming out if you use a large amount of the carboxylate then what would happen is the ionization of this intermediate is suppressed and when you heat the reaction mixture this stops the unsaturated intermediate being generated. Now, if you have reductive elimination it turns out that carbon-carbon coupling can happen these two carbon bonds are cis related. So, in principle it is possible to have reductive elimination by forming a bond between these two carbons. So, these are the two carbons which need to be bonded together in order to form the carbon-carbon coupled product which is marked as 4 here and you will know you will remember that if you do not have this added carboxylate then you have 96 percent of the CO coupling. CO coupling is formed in 96 percent and only 4 percent of the carbon-carbon coupled product is formed. But by adding a large amount of the carboxylate you suppress the reaction the ionization and then a concerted reaction happens and this time you will have only a carbon-carbon coupling. So, if you go back to this general scheme you notice that the first step is in fact generation of a co-ordinate will unsaturated intermediate then you have CO coupling then you have CO coupling. If you suppress the ionization then you end up with C C coupling. Suppress ionization then you end up with if you suppress the ionization you end up with C C coupling. So, this is the first time where both C C and CO coupling in a reaction have been studied and also a rational way by which you can bias the reaction towards one product has been achieved and this has been done through the formation of a palladium 4 mononuclear intermediate. In the same year surprisingly there was another reaction which was also studied and we will come to this reaction after summarizing what I have just told you. So, what I have told you just now is that suppressed ionization of the acid leads to C C bond formation and the C C coupled product and if you have ionization of the acid it leads to C O coupling of the product. The nice aspects about this paper is that they have studied a very well behaved system which could be thoroughly explored and they also characterize the intermediates that are involved using mass spectrometry NMR spectrometry spectroscopy. They have also used a variety of other tools like X-ray crystallography to show the intermediates that are involved. It is important to note that multiple options have been explored to prove exactly what was going on in the reaction. Now, before we go to the next step we have to briefly review some literature that was prior to the 2010 where it was known that reductive elimination could involve this concerted reactions and that is labeled as one here in this projection. You have one which is a concerted elimination. You could also have the formation of a dinuclear intermediate where you could eliminate X and Y in a bimolecular after undergoing a bimolecular reaction. In other words these two react together form may be a bond or bridged intermediate and in this bridged intermediate X and Y are reductively eliminated. Microscopic reversibility favors the first possibility but as I told you there are several surprises in reductive elimination. Several paradoxes one of them is a fact that you need low coordinated complexes and they undergo reductive elimination much better than the co-ordinatively saturated intermediates. Second surprise is a fact that dinuclear complexes seem to be involved and this is contrary to what is expected from microscopic reversibility. So, one particular example that needs to be talked about is the case of osmium. Osmium 2 intermediates undergo reductive elimination of methane and they can also undergo carbonyl insertion. What is shown here is a carbonyl insertion and during the course of this insertion you can also have one of the products is elimination of methane. It could have in principle come from a concerted elimination of M E and H from a single intermediate but using double labeling they were able to show double labeling basically means that you make another molecule where you have C D 3 and deuterium instead of methyl and hydrogen. So, if you mix the two then you can see whether it is only methane which is coming out or if you are getting C D 3 H or M E D. The observance of M E D and C D 3 H clearly tells you that you have binuclear intermediates and possible that elimination is happening from a binuclear intermediate. So, these two were considered both C D 3 D and C H 3 H species were mixed together. If indeed only the mononuclear intermediate is formed you would only have the possibility of C D 4 but because you have both C D 3 H, C H 3 D, C D 3 D and C H 3 H when you start with a mixture and we will write this down here now. This is a osmium compound we have C O, C O, C H 3 H. If you take a mixture of these two and then pyrolyse the reaction mixture you would get only C H 3 H and C H D 4 if a mononuclear intermediate was involved. If a dinuclear intermediate was involved then you would end up with all four possibilities. In fact this was observed and this tells you very clearly that the elimination is happening from a dinuclear intermediate and it is possible that species like this are involved and that is how the elimination was happening. One more possibility is to look at species which are heterogeneously functionalized and this was carried out. This was a study which was carried out by Hegdes who showed that you can localize the molecules on a silica gel surface and then you can prevent dinuclear elimination if they are sufficiently well separated. You could have elimination if they are close by and then if the concentration of the catalyst is very high loading level is very high then the probability of dinuclear elimination will keep increasing and they showed by carrying out this functionalization to different extents that the loading level was correlated with the number of non isolated sites. So, if you load the catalyst to a very large extent then the percentage of non isolated sites will be increasing. That means the percentage of dinuclear species is increasing as you increase the loading level and based on some spectroscopic features they were able to distinguish the two systems and subsequently they showed that elimination was in fact happening from a dinuclear intermediate. So, the turnovers of dinuclear elimination decreased with amount of site isolation. If you increase the amount of site isolation only mononuclear species are formed. If mononuclear species are formed then reductive elimination is unfavorable and in fact that is what they showed here by plotting percentage of site isolation with turnover number. So, there was very clear indication that the elimination was happening in a dinuclear intermediate. Now, we will conclude today's lecture by just briefly looking at what we have studied and what we have seen is that reductive elimination what we have seen is that reductive elimination happens in a variety of ways. The mechanism of the reaction is fairly complex. We have also seen that dissociation is essential. This turns out to be a very important and paradoxical factor, but dissociation is essential before reductive elimination can be carried out. We have also seen that the different compounds whether it is cobalt or osmium or palladium the reaction mechanism has to be examined in detail in each case before one can conclude that the reaction is happening by a particular path. But the only thing that can be told for sure is that dissociation of one ligand is needed and in all probability the cis groups will be reductively eliminated in a concerted fashion. So, we will consider some other complicated examples in a future lecture.