 So, this lecture is about carbon heteroatom bond formation and this is a reaction which is similar to what we have looked at when we studied the Negishi coupling where carbon carbon bonds were formed. So, in this lecture we will look at carbon heteroatom bond formation and we will also talk a little bit about high oxidation state organometallics because recently it has been found that some of these coupling reactions could actually proceed through high oxidation states. So, first let us take a look at what we had studied in the carbon carbon bond formation. The carbon carbon bond formation was mostly catalyzed by peridium and the carbon heteroatom bond formation was not as well studied as a carbon carbon bond formation. The key contributors in recent times have been two people Hartwig and Buchwald from two universities in the US and they have shown very clearly that there are features which can be controlled by the ligand and it can be useful to have specific catalysts generated for the reactions that we want to look at. Most of these reactions have been carried out with peridium and nickel complexes and first we look at the point why is it so rare, why is carbon oxygen bond formation so rare. If you look at the carbon oxygen bond strings, the carbon nitrogen bond strings and so on it is quite surprising that this reaction has not been studied to a significant extent. These bonds are quite strong they are 114 kilocalories per mole for the carbon oxygen bond and if you look at the peridium oxygen bond it is indeed much weaker although the single bond energy is not available the palladium oxygen bond would be roughly 90 kilocalories per mole based on the bond strength of the palladium oxide. So, one now has to answer this question why is it so difficult. Now, in 1983 Khosugi reported that there was a variant of the stilly reaction where the aryl bromide could be coupled with the tin amide. Now, this reaction which you have in front of you on the screen was catalyzed by palladium chloride and palladium chloride with a phosphorous ligand and the oxidation state of palladium was plus 2, but in the presence of phosphine and in a reaction medium very often you could have a reduction of the palladium 2 to palladium 0. So, it would be possible to have a reaction similar to the negishi reaction or the stilly reaction mechanism in this particular case. There were two serious limitations for this reaction one of them was a fact that no functional group on the aryl group was tolerated, which means if I have an aryl group with some other functionality other than the bromine the reaction would not proceed. Similarly, if you have a primary amine the reaction would not work. So, these are two serious limitations these two limitations were quite serious and the Khosugi reaction Khosugi reaction was not popularized significantly. Sometime in 1990s Hartwig and Buchwald started re-investigating this reaction and almost simultaneously in 1995 they published this procedure for making aryl amines. Hartwig's re-investigation showed that there could be significant improvements in the reaction if one could isolate the palladium complex that was formed on reaction of the palladium 0 intermediate. This is a palladium 0 intermediate that we are talking about and that is generated in situ. If that reacts with aryl bromide then an oxidative addition leads to a dimeric complex. In fact, Hartwig was able to characterize this complex thoroughly and then a reaction with the tin amide gave the aryl amine. Now, this tells you that if you have a palladium 0 intermediate palladium 0 catalytically active species it would oxidatively add the aryl bromide and generate a stable intermediate which can then react with the tin complex. The isolation of this intermediate also helped Hartwig to tune the reaction so that some improvements could be made. Specifically he found that if you have sterically bulky phosphines then the reaction was more successful better yields were obtained. But in spite of these limitations in spite of these improvements he found that only secondary amines could be used. This was a limitation which was there in the Kosogi reaction also. But what is interesting was that it is now possible to have substituents on the aryl group. So, a typical reaction in this particular case is shown here where you have a phosphorous stabilized palladium 0 as a resting state of the catalyst. In the presence of a base and the base is usually it could be K 2 C O 3 as it is indicated here it could also be other organic bases like sodium tertiary butoxide and the silal amide that I have mentioned here. So, these bases could also function equally well. In general the reaction proceeded well if you do if you carry out the reaction at about 100 degrees or so at a slightly elevated temperature and in the presence of a solvent ether solvent like dioxane. You need the ether solvent like dioxane in order to heat the compound to about 100 degrees without any difficulty. Now, if you have a bulky phosphine then the reaction would work even in the presence of substituents on the aryl group. So, you could have a variety of substituents and some of these substituents were quite bulky and close to the bromine and in spite of that the reaction worked. Some of them were highly coordinating like the amino group or the finacil group and in spite of that the reaction proceeded effectively. So, this is a significant improvement on the Kosugi reaction and this was brought about by the fact that the palladium 0 intermediate that Hartwig was able to characterize and show what were the factors that were stabilizing the intermediate. So, now we can ask ourselves the question what is the mechanism of this particular process. The mechanism of this process could be very simply modeled after the reactions that we have studied with the negishi coupling. So, if you take a palladium 0 intermediate and do an oxidative addition you would end up with a species which would be 3 coordinate or 4 coordinate depending on the number of phosphorous atoms that are attached to the palladium. In most cases you would expect the palladium to have a square planar intermediate which would be 2 phosphorous ligands. He also showed that if you have cis related phosphorous ligands stabilizing the palladium then the reaction worked much better in some cases. So, after you make this oxidative added product where you have aryl and x groups the x group can be replaced with amine and when you do that when you have an amine coordinating to the palladium and displacing the x as x minus x minus comes out this group comes out and the amine is the one which is going on to the palladium and during this process the base would mop up the H x that is generated in the reaction medium. This is very important because if you have this H x a the reaction can go back in the reverse direction. That means you could have the reverse direction reverse reaction happening you could also have other unnecessary side reactions as we shall see in a moment. Once you form the palladium compound which has got the aryl group and the n r 2 group simultaneously coordinated to the palladium and now this is a palladium 2 species it can reductively eliminate palladium 0 and the product can be formed. So, this is the product which is formed as a result of reductive elimination. So, as we found in the previous reactions with negishi coupling and Suzuki coupling you have an oxidative addition of the sigma bonded intermediate and reductive elimination in the final step which gives you the product and regenerates the catalyst. In between you could have insertions you could have trans metallations you could also have an exchange reaction a sigma bond metathesis type of reaction as is shown here. So, if you look at the reaction it is also easy to understand why primary amines cannot be used or amines where there is a beta hydrogen present in the amine. So, you cannot use in this reactions r C H 2 and r. So, if you have an r group in which there is a beta hydrogen then also the reaction does not work and this helps you to understand why this restriction is present. After the oxidative addition and the sigma bond metathesis or the substitution by the amine you would end up with a species where if I have an N H group. Whereas, if it is a primary amine I would end up with a r N H group and this hydrogen would be capable of the amine of doing a reductive elimination with the a r group. So, the a r group and the hydrogen this hydrogen can be eliminated together as a r H and that is a serious side reaction that is present in the reaction which converts the important aryl halide to a simple arene. So, you cannot carry out the reaction when you have the possibility of beta hydrogen elimination as I have shown here. You can have a beta hydrogen elimination which will give you a palladium hydride and that palladium hydride will eliminate we can write that out. We can have a species like this you can have a species like this which would eliminate the a r H that is also feasible if I have amine which has got a beta hydrogen or as I have shown here if I have an r N H primary amine and I carry out the reaction then the primary the alpha nitrogen has got a hydrogen which can be reductively eliminated. So, these two are complications which still plagued this Kosugi reaction variant which was promoted by Hartwig and Buchwald. But the way in which they were able to isolate the intermediate and also improve the reaction is remarkable and that led to the recent resurgence of C X bond formation. If this is the way in which the amine reacts one can also have an alcohol reacting in the same fashion. So, if an amine can be a nucleophile and r O H or an alcohol can also be a nucleophile and it can also carry out a similar reaction. So, just to remind you this that this is identical to the or very similar to the negishi reaction. I have shown for you here the negishi reaction in this slide where the only difference is this trans metallation instead of an amine you have a metal which is doing carrying out the trans metallation and everything else in this catalytic cycle is exactly the same. So, here is a here are the two catalytic cycles together and you can see that the parallel is remarkable. It is only in this one trans metallation step or the reaction of the amine that the two reactions differ. Now, during this time it was both organic chemists and organometallic chemists developed this reaction significantly and interestingly if you have an intramolecular attack of an amine as it is shown here the reaction is much more tolerant to the groups which are present and the reaction can be carried out at much milder condition. So, here you have the book walls group has shown for example, during this time that you can have an intramolecular variant where you have orthobromo group here which is missing in this slide and we will put that in. So, this orthobromo group can be eliminated in order to generate the cyclic compound which is pictured here which is a very familiar group or which is a recurrent moiety in many drug molecules. So, it is very important to have this type of intramolecular attack of the amine in order to form an indole complex indole compound. So, this was developed by Stephen book walls group and surprisingly you can have the ring size can be in fact varied and you have a variety of rings that can be generated not just the indole that I have shown for you as an example where n equals 1. So, after the generation the isolation of the catalyst and the first generation catalyst as it was called where you had only bulky substitutions on the phosphorous atom it was shown by Hartwig that it would be better to have if you have a chelating phosphine then it is possible to even do carry out the reaction with the primary amine. So, in some ways the primary amine reaction side reaction that was described for you a few minutes ago is not possible when you have a chelating amine chelating phosphorous on the palladium. So, apparently the palladium goes to palladium 0 and it is stabilized by this phosphorous ligand and the same catalytic cycle can be present except that now you do not have the possibility for the A R H elimination and. So, you get very good yields of the desired product and the same type of conditions have to be followed it is approximately 100 degrees and the base and potassium carbonate or cesium carbonate are good are good bases that can be inorganic bases that can be used these are called solid bases and they are very useful because they are very efficient in mopping up the acid that is generated in the reaction medium without causing any side reactions. So, here is an example of a natural product that was synthesized using this reaction in a very efficient way one of the key steps involved in this reaction was this cyclization reaction. So, in this particular compound it was possible to use the simple palladium phosphorous tetrachys triphenyl phosphene ligand complex of palladium. So, this is an extremely simple step that has to be carried out it does not even require the first generation catalyst the bulky phosphorous ligand on the palladium in order to carry out this reaction very efficiently. So, you can see that this can be an extremely useful reaction extremely useful and valuable reaction for fine chemical synthesis. Now, we can give several examples, but what we will do is to summarize the key findings that are available for us due to the extensive work done by Buchwald and Hartwig. The key variants or the modifiers that they have used are mentioned here primarily it is the ancillary ligand that was changed significantly and some of the ancillary ligands are shown here for you on the on the slide. They are extremely electron rich if they are more electron rich the reaction proceeds to a better degree and that in an easier fashion and it has been possible to activate even chlorides. That is understandable because you can have easy oxidative addition when you have electron rich phosphorous atoms which are ligands on the palladium. Secondly, if you have a bulky substituent on the phosphorous then there is a weak interaction there is a stabilizing interaction where the bulky group prevents other moieties from attacking the palladium to coordinated palladium complex and which is susceptible to oxidative additions. So, it is important to have bulky substituents and that is also mentioned here as sterically demanding and in some cases in the second generation catalyst it is been possible to have its advantages to have a secondary amine in the ligand. So, that it would stabilize the palladium intermediate that is formed. So, there is a weak interaction between the palladium and this nitrogen which is indicated here. So, that weak interaction stabilizes the intermediate step or the intermediate that is formed after the reductive elimination and it only carries out oxidative addition and this reductive elimination without carrying out beta hydride elimination. That has even allowed ammonia to be used as incoming species. So, that you can make primary amines as a result of this reaction. Of course, palladium precursor it can be either palladium 0 or palladium 2 and I mentioned here D B A which is dibensalidine acetone which is capable of stabilizing palladium and palladium 0 state. So, this is used with palladium 0 and acetate and the chloride are used with palladium 2. So, these two are used with palladium 2. Now, the base as I had shown you it is either potassium or sodium tetra butoxide. Butoxide again is a bulky ligand which even if it substitutes on the palladium is readily replaced by the less hindered amine. But, cesium carbonate is an extremely useful effective and good base that can be used for generating this reaction in very high yields. As I had mentioned earlier dioxin as a solvent is the most convenient solvent for the simple reason that one can heat it to higher temperatures without loss of the solvent or having request to a high pressure reactor. So, having discussed the amine it is not very difficult to push this one step further and have C O bond formation. So, this is the bond that is formed as a result of this reaction. We can see that this intramolecular ether formation is in fact exact analog of the C N bond formation. Everything else would be identical in terms of the mechanism or in terms of the ligands that have to be used. So, you need a phosphorous stabilizing ligand which is usually a P R 3 P R 3 as a ligand is used in this reaction also. Now, it is possible to replace palladium with nickel in the case of the ether formation. It has been shown that nickel 0 as nickel cod twice. So, this is nickel 0 where cod is cyclo octadiene and you can have cyclo octadiene coordinating to the palladium. That is very readily displaced ligand and this falls away from the nickel. So, this falls away from the nickel generating nickel 0 species which is extremely reactive. So, this is the bis cod nickel species that is used very often as a good starting material. What was called by Wilke as naked nickel and it is convenient to use this in this reactions. So, everything else appears to be the same. So, nickel 0 also must be going through the nickel 2 intermediate. However, these two species nickel 0 and even copper copper 1 which has been shown to be a useful alternative. They are not as effective as a palladium. Now before we proceed further, I would like to point out that there are two small variations of this ether and amine formation reactions. These are interesting variants which allow us to start our discussion on high oxidation state chemistry. So, here is a reaction that has been studied. Now, you first make a nickel complex which has got two phosphorus atoms coordinated to it. If you treat it with phenyl azide which is P H N double bond N double bond N, then it is possible to insert a phenyl nitrogen which is this group here directly into the carbon nitrogen bond. So, this is the carbon nickel bond. If you introduce the P H N 3 here, you tend to form this five-membered indole moiety or indole derivative which I have drawn for you here. During the course of this reaction, this turns out to be a stoichiometric reaction because the phosphorus atom gets oxidized also to the species which is called the H is given here. So, this is the variant of the Staudinger reaction which is during which also you have oxidation of the phosphorus. But in this reaction, you also have insertion of a P H N group between a nickel carbon bond which is indicated here. So, we can have oxygen insertion in a similar fashion and this turns out to be conveniently carried out with N 2 O nitrous oxide. Here is an example where oxygen is inserted between the nickel carbon bond. So, the nickel carbon bond is the one which I have converted to a red bond and I have inserted N 2 O. It is not very clear what the intermediates are is possible that you have a coordinated N 2 O to nickel which then inserts the oxygen to the carbon nitrogen bond. Then you have of course, the ether formation after oxidation with iodine. So, these are stoichiometric variants. There are no catalytic variants for these because invariably the N 2 O and the phenyl azide which we discussed in the previous slide, they destroy the complex completely. So, these are stoichiometric reactions. But that seems to suggest that if you have a source of oxygen, you can add it between the metal carbon bonds. There are instances where you want to do that. Suppose you want to form an epoxide or if you want to insert oxygen into a ketone, a cyclic ketone in order to generate an ester, a cyclic ester. These are examples where it would be nice if you can insert an oxygen or for that matter an amine group. So, we will discuss those later. Now, I want to talk about the ether formation which was known for a long time. A slightly more drastic condition was used in the Ullman reaction and copper was a catalyst that was used. In these reactions, you normally used dimethyl formamide as a solvent and it also needed a base. This was discovered as early as 1903. But no mechanism for this reaction was available. The mechanism was not known. But Ullman showed that you can very efficiently make a variety of ethers using this particular procedure. This has been recently reviewed and you can have Ullman type coupling reactions with copper in a fairly facile fashion. Now, Buchwald showed that much milder conditions could be used and cesium carbonate can be used as a base and the reaction can be carried out in the presence of palladium and in the presence of carbonate. Copper, if you use copper if it was shown that if you have non-coordinating bases as anions then they are the best for this whole reaction. So, usually cesium carbonate is the best base and if you have a copper triflate as it is shown here or for example, perchlorate it is also possible to carry out this reaction in reasonably mild conditions. Now, we can ask this question. I want to move over to this question of oxidation state. In all the reactions that we have shown here, we have carried out the reaction with palladium 0. We have carried out the reaction with palladium 0 or with palladium 2, but with the possibility of converting palladium 2 to palladium 0. So, the catalytic cycle basically involved oxidative addition on palladium 0 to convert it to palladium 2 and then a reductive elimination step regenerated the catalyst. But is it possible that palladium 2 gets converted to palladium 4 through oxidative addition and then a subsequent reductive elimination regenerates the palladium 2. Now, palladium 4, copper 3 and other such high oxidation state species have always been proposed in some instances, but it has been very difficult to isolate and characterize the intermediates. Recently, Melanie Sanford from the University of Michigan has shown that this is indeed possible and they carried out some excellent studies with palladium 4 complexes, isolated them and characterized them and then shown that reductive elimination can happen from these palladium 4 intermediates. So, in fact it is possible to cycle between palladium 4 and palladium 2 and if you did that you can generate both CO and CC bond forming reactions. Both of them were possible in those cases. Now, interestingly very close to the discovery of this palladium 4 it was Tobias Ritter from MIT who showed that it is possible to have even palladium 3 and this he showed starting with palladium 2 dimer. So, the two palladium units are involved and both palladiums were oxidized to palladium 3. So, reductive elimination can then happen from this palladium 3 intermediate and generate palladium 2. In other words, you can have all the oxidation states that are listed here palladium 0, palladium 1, palladium 2, palladium 3 and palladium 4 all of them participate in the reaction. It is not an impossibility. So, it should be possible to see much more chemistry coming out from these interesting species that have been discovered by Ritter and Sanford. So, let me just show you what has been discovered. This is from a paper by Melanie Sanford in recent times and the 2009 paper which is the 2009 paper that has been shown here. So, here what she has shown is that copper 3 has an intermediate. There is a plus charge here. So, you can have copper 3 which is capable of carrying out in the presence of oxygen the oxidative acid toxylation of an amine group. But this happens in the ortho position or in the adjacent the ortho position of the adjacent ring to a pyridine moiety. So, the same thing is capable or possible with palladium 4 and that of course happens through the intermediacy of this complex of the six coordinated palladium 4 intermediate that I have pictured for you here. So, this is high oxidation state organometallics where there are very clear indications for a palladium carbon bond being involved in the reaction medium. Now, high oxidation state organometallics although we have not discussed it in great detail, we do find that occasionally in the literature. The reason for its absence from the literature to a great degree stems from the fact that we have concentrated on carbon monoxide based chemistry. Carbon monoxide requires that we have low oxidation states if you want to stabilize the metal carbon monoxide bond. So, the metal carbon monoxide bond is significantly weakened if you oxidize the metal center. So, here is a reaction where if you take a molybdenum species which is molybdenum 1, singly oxidize to the presence of cyclopentadienyl group. If you have a molybdenum 1 species and if you pass oxygen into the reaction mixture, you end up with a dimeric species which has got oxo ligand. This oxo ligand is in fact a strange species because it is not expected to be found in combination with a cyclopentadienyl group. One would expect this group to be oxidized further. In fact, in the case of rhenium, here you have a rhenium carbonyl compound which is rhenium 1 and it is having 3 carbon monoxides. So, this is C P star. It has got these or these methyl groups attached to the cyclopentadienyl unit and you have 5 of them. So, this is C P star. We write this as C P star ligand and the C P star ligand stabilizes the rhenium and it is possible to oxidize it with oxygen and generate rhenium tri oxo ligand. So, this is all oxygens that are attached to the rhenium and we normally write it as oxo groups because of the bond distance is indicative of multiple bonding between oxygen and rhenium. Now, interestingly if you use a high pressure of carbon monoxide, you can convert it back to the rhenium carbonyl complex. So, you go from the carbonyl complex. You can oxidize it with oxygen and then carbon dioxide is liberated on reaction with a high pressure of carbon monoxide and you regenerate the carbon monoxide complex. So, it is possible to have high oxidation states and it is possible to carry out these reactions very easily by adding oxygen. But, very often you would end up with horrible mixtures. The compounds are completely oxidized. The carbon containing ligands are also oxidized, but if you have a C P star ligand, it seems to stabilize it and you can have some chemistry which is being actively pursued. So, we will discuss a few cases where oxygen can be introduced into organic molecules with the help of these oxo complexes. Now, rhenium complex that I just introduced to you in the previous slide turns out to be a very interesting molecule because you have rhenium in the plus 7 oxidation state, the highest oxidation state that you can think of. In this highest oxidation state in which the rhenium is present, can be converted into an organometallic species. This is of course an alkyl organometallic species. You have a very clear metal rhenium, metal which is rhenium and methyl group. So, you have a carbon rhenium bond and this is a extremely stable molecule. This stable molecule, this r group that is introduced by this triane brutal tin here or it could be any other r group. It can be transferred on to the rhenium turns out to be a stable species. There are some variations which are shown here. Here, I have oxidized it further which suppose you have hydrogen peroxide, this trioxo rhenium can be converted to this peroxo complex and that has also been characterized. Here is another species where you have rhenium in the plus 7 oxidation state and you have 3 methyl groups instead of the single methyl group which I showed you earlier. So, these are species where the rhenium is supporting an alkyl group and forming a metal carbon bond and at the same time having oxo groups and these two seem to be compatible. So, it must be possible to have significantly significant chemistry with these species. In fact, MTO or methyl trioxo rhenium, MTO is a little deceptive because it does not talk about the rhenium or the r is missing, but MTO is abbreviation that is used in the literature methyl trioxo rhenium. This species right here where r is methyl was prepared as early as 1979, but its value was not realized until Ulfgang Hermann who took over as a professor in Munich after Fisher who discovered the arene complexes started popularizing it. He showed that it is very easy to make these alkyl rhenium oxo complexes and these alkyl groups can be anything it can be transferred from a zinc alkyl and an inorganic oxide r e 2 o 7 is a inorganic oxide and that can be these two can be reacted in order to generate a inorganic alkane. So, here was r e o 3 which is an inorganic alkane and that was soluble in water. So, it had some very interesting properties and he looked at the catalysis of that is capable with this metal center and he showed that olefin oxidation, olefination of aldehydes and even metathesis could be carried out by these interesting inorganic alkanes. So, here is the catalytic cycle which is in the presence of which is olefination of an aldehyde and this is a reaction which requires a conversion of removal of an oxo group using triphenyl phosphine. So, triphenyl phosphine because of its oxophilic nature removes one of the oxo groups from the rhenium and. So, you now have a rhenium 5 this is in the plus 5 oxidation state and this coordinates to carbene source which is a diazoalkane. The diazoalkane generates a carbene complex and this carbene complex now can react with a aldehyde. So, you need r c o h here and you can see that this is your r c h o. So, this is the r c h o which is added on to the to the rhenium center and now you can see that olefination can take place very easily because this rhenium center can now become r e o 3 and you can have for sake of clarity we will just use a different color so this group can be eliminated as the product which is given here. So, it is possible to have olefination of aldehyde very catalytically you can carry out this olefination that is another advantage only thing that you are doing is that you have indirectly carried out a vitic type of a reaction. So, where you have used a phosphorous phosphine to remove an oxygen exactly the way vitic does, but you have used an azide in order to generate the methylene fragment. So, these two fragments have been reacted on a metal center and you can do this catalysis. So, here is another example where you have bare williger oxidation being carried out, but this example is restrictive it has been done with cyclobutane butanone and the cyclobutanone moiety reacts with the peroxo complex of rhenium that we just showed you little earlier. This peroxo compound can be generated by reacting it with H 2 O 2 hydrogen peroxide and two molecules of hydrogen peroxide are required to carry out this reaction where you have rhenium peroxo bond. This rhenium peroxo bond can react with the cyclobutanone in such a way that it adds on to this double bond here. So, it adds on to this double bond the R E double bond O and you end up with the trioxo system and that can do a rearrangement and that rearrangement is easy to envisage. You just have to move this carbon carbon bond to a carbon oxygen bond and you would end up with a reaction in which you have a cyclo acyclic ester being eliminated and which comes from these groups. These groups if they are eliminated you would end up with this organic molecule and you have one more oxo peroxo group on the rhenium which can carry out the same reaction once more and regenerate the C H 3 R E O 3 molecule. So, this reaction can this hydrogen peroxide generated intermediate which is a di peroxo molecule can carry out two oxidations of two molecules of cyclobutanone. So, you have one more possibility you can just transfer the oxygen to an olefin that is an epoxidation reaction. If you do that you will get an epoxide which is pictured here. So, here again you end up with a rhenium peroxo complex. The rhenium peroxo complex has got a ligand usually a pyridine which stabilizes this system and once you have this pyridine stabilizing the rhenium it turns out that it adds on to olefins or this oxygen tends to attack olefins directly and this is the intermediate that is formed where you have again a pyrocyclic type of compound where oxygen is at the bridge head and this can eliminate epoxide and generate one more oxygen rich intermediate which can carry out one more oxidation. So, you can see that the methyl trioxo rhenium is regenerated at the end of this reaction and pyridine is also just present as a catalyst. So, all you are carrying out is hydrogen peroxide and ethylene this is very green chemistry and the catalyst itself is soluble in water. So, you can methyl trioxo rhenium is soluble in water and so you can carry out this reaction extremely efficiently. Before we proceed or close this discussion on high oxidation chemistry we should mention surplus epoxidation where you can have epoxidation using oxo groups and this is a reaction which is been extremely useful and we will discuss the chiral variant of this reaction little more detail, but suffice it to say that the oxo groups that are generated by various metals molybdenum, manganese, ruthenium and osmium are capable of carrying out oxidation of different substrates. So, it is not always the epoxidation or the bare williger oxidation that we have seen here a variety of substrates can be used and here are some reactions of the oxo containing metals complexes and these complexes are sometimes the intermediates have been characterized, but sometimes the intermediates have not been characterized and so it has led to some debate about whether these reactions are really organometallic species. Let me give you the surplus epoxidation catalytic cycle and tell you why this is the case. So, the surplus epoxidation the osmium tetroxide can in fact react with the olefin. So, here is the olefin coming in and it can in fact react with the olefin to form a osmium olefin bond which can react in an add on to the oxo group and form a metallacycle oxo osmium metallacycle and that is pictured here and if you now expand this ring, if you expand this ring if this carbon moves over to this oxygen, if it moves over to this oxygen you would end up with another pyrocyclic compound with osmium at the bridge head which can easily regenerate the catalyst in the presence of water to give you a diol and osmium dioxide which can be used to oxidize further with hydrogen peroxide to osmium tetroxide. So, this surplus epoxidation was postulated to go through this osmium olefin complex, but many people believe that it is not necessary to form this complex and here I have for you the reasons why we need not do this with per magnet which is again a high oxidation state species. We know that it carries out a dihydroxylation reaction. The dihydroxylation reaction first forms a complex a manganese olefin complex or at least it is possible to form the manganese olefin complex which can now add on to the oxo group. It is can add on to this oxo group in this particular slide you should have three oxo bonds and so you have three oxo bonds formation of a oxo metallacycle which can again ring expand as I showed you in the previous transparency to give you this five membered manganese cycle which can eliminate MnO2 minus and generate two aldehydes. So, technically this is this key species here is a 16 electron species and it can coordinate an olefin, but in the reaction the people doubted the value of this organometallic complex being formed purely because the reaction is much better carried out in the presence of a ligand such as L pyridine or an organic base which coordinates to the manganese. If that is indeed the case then there would be no 16 electron intermediate and olefin complex need not be formed in the first place before generation of this key intermediate. So, the same reaction could be carried out without coordination of the olefin to the metal. Now, epoxidation by coordination compounds is quite well known variety of biologically relevant systems, porphyrin systems are capable of carrying out this epoxidation reaction. So, epoxidation chemistry is known and in many of these cases there is no possibility for the metal to interact with the olefin because the oxygen if it is present on one side and the olefin would be present on the other side. So, a direct attack of the oxygen on the double bond is what is envisaged. So, here are cases where you have epoxidation of an olefin, but there is no organometallic chemistry that is really involved. Let me summarize what we have seen in this lecture about carbon oxygen bond formation. In the first part we saw that carbon oxygen bonds can be formed in a catalytic fashion just like carbon carbon bonds are formed by using palladium catalyst. However, one had to use bulky electron donating phosphines to stabilize the palladium 0. Interestingly, recent literature also shows that palladium 3 and palladium 4 complexes can stoichiometrically make carbon oxygen bonds. These complexes do not have phosphine ligands to stabilize the palladium 0 state. So, one might wonder are high oxidation states and high oxidation state complexes better for CO bond forming reactions. One is reminded of the classic shapeless epoxidation reaction and the permanganate promoted dihydroxylation reaction. These are good examples where carbon oxygen bonds are formed. However, we notice that they do not involve metal carbon bonds and so are not organometallic complexes. Lastly, we reviewed a few examples of rhenium catalysts which catalyze the formation of CO bonds. These complexes are interesting because rhenium belongs to the same group as manganese and they are truly high oxidation state complexes and they are truly organometallic systems which promote the formation of CO bonds effectively.