 Oxidative addition and reductive elimination are two key reactions in organometallic chemistry, especially when it comes to transition metal organometallic chemistry. This plays a very important role and so it is important that we spend some time understanding it and figuring out what are the conditions under which oxidative addition occurs and how we can have reductive elimination in a catalytic cycle. So, this is the second part of the description of oxidative addition and reductive elimination. So, we continue where we left off in the previous lecture. So, the important thing to remember is that in the oxidative addition, we have a change in the oxidation state of the metal and invariably you also have the formation of a new metal ligand bond and that results in an expansion of the coordination sphere of the metal. So, the name addition and the fact that you have a change in the increase in the oxidation number the formal oxidation number of the metal results in the oxidation part. So, this reaction as I just mentioned can be a reversible reaction and so you can have both oxidative addition and reductive elimination in the catalytic cycle and the thermodynamics of this process turns out to be extremely important. It is facile in the case of transition metal chemistry, but not in the case of main group chemistry where only oxidative addition happens. So, let us take a look at what else is special. The fact that we have written m as n plus in the oxidative addition reaction on the left side we have written as n plus and on the right side after the reaction we have changed it to n plus x plus x turns out to be a variable sometimes it is 1 very often it is 2. So, that is not a crucial aspect the ligand can accept one electron or it can accept two electrons or in some cases as we shall see in the first part of the stock it can accept one electron from two different ligands and so a new compound can result in the process. So, let us take a look at some of the possibilities in oxidative addition. So, we also classify the type of oxidative additions based on the addenda that you have Vasca's complex is a typical complex and it undergoes a variety of oxidative additions and we classify them into three groups. The first group was the group that could be polarized into plus and minus parts and so if you have methyl iodide we could mark it as m e plus i minus. If you have H g C l 2 then we can write as H g C l plus and C l minus. So, if you have an addenda that the molecule can be split into a positively charged and a negatively charged species then it is convenient to have oxidative addition in the transposition of the Vasca's complex. So, you end up with the three molecules which are marked in yellow in this projection. So, you have m e and i added in the transpositions of the Vasca's complex. So, this is the first group which is highly polarized and it will add in a transposition. Then we talked about a second group of molecules where multiple bonds are there in the part that is adding on to the metal that is being oxidized. So, one example would be oxygen and so during the oxidation O 2 which has got a double bond between the two oxygens adds on to the Vasca's complex and as a result it forms two new bonds to iridium and at the same time the bond between the two oxygen atoms changes from a double bond to a single bond. This change a double bond to a single bond requires that the two atoms that are added on to the iridium are now cis related to one another. So, at the end of the reaction in group 2 class of oxidative additions in the group 2 you end up with cis additions and we have two examples here which are again marked in yellow. These molecules are characterized by two key features. One of them is a fact that there is a multiple bond between the two atoms which are adding across the metal and the second is a fact that they are not highly polarized. So, either dimethyl acetylene di carboxylate or oxygen are not highly polarized as plus and minus species. At the end of the reaction you have the bond retain and so you have a cis addition. So, a brief comment on the oxidating on the stability of the oxidized species is an order. When you have Vasca's complex you have iridium in the oxidation state 1. At the end of the reaction when you have added methyl iodide for example, you end up with an oxidation state of 3. So, if you have more electronegative elements attached to the iridium then the plus 3 oxidation state is more stable. In both cases that I have listed here on the screen you have iridium attached to iodine and chlorine or iridium attached to two oxygen atoms and chlorine. Either way you tend to have a stable species because these electron withdrawing groups will stabilize the higher oxidation state. Electron withdrawing groups at the same time have got pi electrons which are pi donor atoms and so they stabilize the higher oxidation state. So, this oxidation or the oxidative addition proceeds from left to right in a very exothermic fashion. On the other hand there are species where the group that is adding on is not so electronegative. And so we have this example here of dimethyl acetylene which is oxidative added and at the end of the reaction you will notice that iridium is in a formal oxidation state of plus 3. But nevertheless you will notice that the only electron withdrawing group that is there on the atom on the iridium atom is the chlorine mole chlorine atom or chlorine ligand. So, as a result of this particular factor the reaction proceeds from left to right rather reluctantly and the oxidized species is not as stable as in the case of Vasca's complex undergoing oxidation with electronegative species. So, this is apparently a disadvantage for oxidative addition, but nevertheless it allows for many catalytic cycles to take place and that is what we are going to take a look at this in this lecture. So, oxidative addition with carbon-carbon coupling is a special case where you have the oxidation taking place with concomitant formation of a carbon-carbon bond. Let us take a look at exactly how this happens. Imagine a cobalt 1, this is a cobalt 1 atom which is coordinated to a C p ring. A C p ring is very simply a cyclopentadienyl ring, a cyclopentadienyl moiety which is coordinated to the cobalt 1. So, this is your C p coordinated to cobalt and this particular moiety is now interacting with two acetylene molecules. Now, when you have oxidative addition in such a fashion that these triple bonds become double bonds and the bond between cobalt and oxygen is formed. The bond between cobalt and oxygen is formed. So, exactly what is happening is you have cobalt giving one electron. So, we indicated with a half arrow and it gives one electron to the acetylene and acetylene gives one electron. So, forms a covalent bond. This covalent bond now results in the formation of a radical species which will have two electrons sitting on the two carbon atoms here. The covalent bond has been formed between cobalt and carbon and a radical species, di radical species is formed. If we now form a bond between these two C H units as indicated by this arrow, if a covalent bond is formed between these two C H bonds, then what you end up with is the formation of a metallocyclopentadiene. The new bond that is formed is marked in red here. So, metallocyclopentadiene can be formed by oxidative addition which is indicated by these two half arrows on the cobalt coming out and forming a covalent bond with acetylene resulting in a vinyl radical. The two vinyl radicals combine to form a metallocyclopentadiene. So, now this metallocyclopentadiene is involved in a variety of reactions which resulted in a plethora of catalytic cycles with cobalt as a key player and acetylenes as a initial reactants. Now, after the formation of the metallocyclopentadiene which is indicated here and this fast reaction results in the formation of a species which can interact with a sector molecule of acetylene. Now, let us come back to the slide, but first let us take a look at what are the ways in which these two species can interact. Let us say you have a carbon-carbon coupling followed that is what we said is the formation of the metallocyclopentadiene followed by an insertion reaction. Remember insertion is a simple addition of an anionic species from the metal interacting with a neutral species on the metal. So, here the neutral species turns out to be acetylene. So, acetylene is a neutral ligand which is attached to the cobalt and the anionic species or the negatively charged ligand is the one which we have marked in red. So, if this bond migrates to the neutral species and the neutral species in turn forms a bond to the cobalt exactly what is going to happen is that we are going to have a reaction such that two new bonds are formed one between the vinylic carbon which is here vinylic carbon which is here and the acetylene carbon which is here both of them are marked in red. If you mark the other acetylene carbon and the cobalt center if a new bond is formed here that will lead to the formation of this blue bond here. So, let us mark the two colors that we have the two new bonds in the two colors. So, the red colored carbon atoms are joined together to form this new bond and the blue bond is formed between the cobalt and the acetylene carbon. In the next step we can in fact do the opposite of oxidative addition. In other words if we remove the cobalt carbon bonds in such a way that they get back the cobalt gets back its two electrons in this particular unit in this molecule cobalt is in the plus 3 oxidation state. So, if we remove half an electron from here and half an electron from here then cobalt will become plus 1 again and then these two vinylic carbons will be left with. So, let us just mark the vinylic carbons. So, they will be left with one electron each and so in the subsequent step you will be able to form in the subsequent step you will be able to form a carbon-carbon bond and that carbon-carbon bond results in the formation of an acetylene. So, basically you have 1, 2, 3, 4, 5 and 6 carbon atoms attached to the cobalt and if atom numbers 1 and 6 combine together if 1 and 6 combine together. So, that is exactly these two carbon atoms 6 here and 1 here then you will end up with the formation of this new bond between 6 and 1 and that results in the formation of a benzene ring and the stability of the aromatic benzene ring results in a very fast reaction to form the 6 membered ring and regeneration of the cobalt 1 species with the cyclopentadienyl group. The same reaction could be described as a Diels Alder reaction and when you have very electron withdrawing groups, when R is very electron withdrawing then you end up with a Diels Alder type of reaction and in this case also it is important to understand the 4 carbon atoms which form the diene. These are the 4 carbon atoms which form the diene and these are the 2 carbon atoms from the acetylene which form the dienophile. So, the diene and the dienophile react together such that a new bond is formed between atom numbers 4 and 5 and atom numbers 1 and 6. So, that results in a bicyclic intermediate which is pictured here. This bicyclic intermediate can again do a cyclo reversion reaction or a reductive elimination and so in this particular instance you have to transfer half an electron, one electron from here and one electron from here resulting in the formation of 2 new pi bonds and the formation of the stable aromatic ring. Such that you end up with a very nice aromatic system at the end of the reaction. So, these are 2 ways by which you can have the formation of an aromatic ring using the metallocyclopentadiene which was generated from cobalt 1 and 2 acetylenes. So, the same reaction could have happened with an acetylene. It could have also happened with a nitrile species. So, if you want to use a nitrile then you could carry out the reaction in such a way that you end up with a parody. So, instead of acetylene if you use a nitrile then the same reaction could undergo a Diels Alder type reaction and result in the formation of a substituted parody. This was exploited by several workers notably by K. P. C. Ohlhart who used this reaction to synthesize a variety of interesting molecules. Now, the first step which involve the formation of a metallocyclopentadiene was a very fast reaction. In fact, that is the reason why you can in fact do this cyclo cyclic ring formation reaction with 2 acetylenes and 1 acetonitrile or 1 nitrile molecule. So, you can vary the molecule in such a way that the 2 steps can be distinguished. The very fast formation of the metallocyclopentadiene and the subsequent Diels Alder addition can be distinguished. They appear to be 2 separate catalytic cycles. So, let us now take a look at a slightly different reaction. In this case instead of adding 2 acetylenes to the cobalt we slow down the reaction by adding a molecule of triphenylphosphine. The rate at which the 2 acetylenes combined together is now slowed down because the dissociation of the triphenylphosphine is a rate limiting step. So, by adding an acetylene with an alkene we end up with a metallocyclopentene. So, this is the metallocyclopentene earlier we had a metallocyclopentadiene. So, in this metallocyclopentene we can have a choice of adding the ethylene in this case. We can either have an ethylene adding which is what is pictured here or we can have an acetylene which is pictured here and then we can form 2 different types of molecules. Let us first take the reaction cycle which is pictured on the left. So, if an ethylene molecule is added we end up with if the molecule does an insertion reaction. So, this is an example of an insertion reaction and the insertion is carried out by this vinylic group forming a bond to the neutral molecule which is the ethylene molecule which is pictured here. So, the ethylene is a neutral molecule and the vinylic group is an anionic group which migrates to the ethylene molecule and as a result the ring expands just as we described in the previous set of slides. We have explained how this insertion can increase in a ring expansion reaction. The ring expansion leads to the formation of a 7 membered ring a metallocycloheptene. So, here you have one double bond which is in the third position in the third position with respect to the cobalt and this particular molecule can now have an option of doing reductive elimination either directly to give you a metallocy sorry this intermediate now has the ability to do a cyclo reversion reaction such that it forms cyclohexene. On the other hand if you realize that there are two hydrogens on this carbon there are two hydrogens on this carbon and one of them is very close to this alkyl group which is attached to the cobalt. So, if you have this CH2 group abstracting one hydrogen from this carbon center you can end up with a CH3 unit such that you now have the formation of butadiene which is substituted in the one position. So, you now have butadiene which is coordinated to the cobalt but again you have had a reaction in such a fashion that cobalt has undergone a reductive elimination and the vinyl group the alkyl group the CH2 unit has abstracted a hydrogen so that it forms a CH3 and the CH2 which we will mark with a different color. So, that it is easy to follow so this CH2 group is what we have here this is the CH2 group and the CH which we had marked in green is right here. So, the CH2 which was on the right side has become a CH3 and that is what is pictured here and the vinyl group is generated by the abstraction reaction that was on the ring. So, now we have another possibility you can also do an addition of a acetylene to this metallocyclopane team. If you do an acetylene addition then you can end up with a cyclohexadiene generated from this ring system and to do that one has to do the migration of this vinyl group on to acetylene now. So, this is the vinyl group that is attached to the cobalt which will migrate to the acetylene which is the neutral molecule. So, this is again an example of an insertion reaction. So, in this type of a reaction you first have an oxidative addition you have an oxidative addition oxidative addition followed by an insertion reaction and then subsequently you have the reductive elimination. So, you now have the reductive elimination and to understand the reductive elimination you can think of the cobalt taking back one electron from this from this bond in such a fashion in such a fashion that cobalt three becomes cobalt one again and you have the formation of a cyclohexadiene at the end of the reaction. So, this helps us to understand how the carbon carbon bond formation coupled with oxidative addition can result in the formation of a cyclic metallocyclopentene or a metallocyclopentadiene and these groups can subsequently add on a variety of molecules through an insertion reaction. And if you have very electron withdrawing groups they can also undergo deal solid type reaction subsequently. So, during the course of studies with quadricycline it was noticed that quadricycline is a molecule which will ring open to give nobona diene. So, this is nobona diene and this is quadricycline. Now, it was interesting that nobona diene could be converted to quadricycline by photolysis and because this is energy rich molecule quadricycline is an energy rich molecule it was thought that if one can store energy from by absorbing sunlight converting nobona diene to quadricycline. Then it would be possible to harvest sunlight using this particular reaction because the forward reaction of converting quadricycline to nobona diene was exothermic. So, here if one could generate heat and so you could harvest light energy and converted to some chemical energy. Now, it turns out that the reaction which leads to conversion of quadricycline to nobona diene has to be catalyzed. It does not happen at room temperature without any catalyst and that is because this is an orbital controlled reaction and the cyclo reversion turns out to be difficult. Rodium one complexes so this is an example of a rodium one complex was a good catalyst for this process. Initially it was thought that this was reaction which was case of orbital controlled reaction where the symmetry rules were circumvented by addition of the rodium orbitals into the equation. So, this theory however was proven to be wrong by Jack Halpern who showed that in the presence of carbon monoxide and new compound could be isolated. So, what was really happening was that the strained cyclopropane ring which can be marked here this strained cyclopropane ring which I will mark with a single line. So, this cyclopropane ring could ring open and you could have an oxidative addition of rodium. So, that a rodium three species can be formed. Now, this rodium three species turned out to be capable of doing a cyclo reversion reaction in such a fashion that no bone or diene can be generated. On the other hand if you have carbon monoxide in the reaction medium here we have added carbon monoxide then you could have an insertion reaction of carbon monoxide in such a way that a new product could be isolated. So, because this product was stable this is stable and could be isolated the intermediacy of rodium three compound was sufficient to explain both the formation of no bone or diene and the formation of this insertion product. So, this was theory which was short lived it was thought that orbital symmetry control with metal atoms is useful for converting quadricecline to no bone or diene. On the other hand what was really happening was an oxidative addition and a cyclo reversion reaction leading to the ring opening. So, now we come to the third group of molecules which can undergo cyclo. We come to the third group of molecules group three where you have oxidative addition of a set of molecules which are not polarized, but at the same time many of them are found to be adding in a cis fashion. These molecules because they are not very ionic in nature they do not ionize as plus and minus nor are they multiply bonded, but nevertheless after addition reaction they are very often found in the cis geometry. There are two examples which are listed with Vasca's complex here and both cases you have an iridium three complex formally an iridium three complex being formed as a result of oxidative addition of an S i H bond. Here the S i H bond has been broken and you have the addition of S i X here this should have been H and an S i R three group. So, on the other hand you have a hydrogen molecule that also undergoes an oxidative addition and iridium now has a formal oxidation state of iridium three and the two groups which have added on are found in the cis position. So, group three type of molecules also add in a cis position they are not polar and they are not multiply bonded. So, these molecules are strange because many of them undergo they have a common feature and that is the fact that they have very little ionization of this group S plus and minus they do not ionize like this nor do they ionize as minus and plus, but still they are able to undergo weak interaction with the metal in what is called an agostic way. This agostic interaction is something which we will have to deal with little later, but suffice it to say at this point that all these molecules are sometimes found to have weak interactions with the metal and in spite of that weak interaction they are able to break a very strong bond the metal is able to break a very strong bond between the carbon and hydrogen which is worth about 100 kilocalories per mole and similarly H two which is also quite a strong bond or S i H and S H are of course weaker bonds these are weaker, but nevertheless the interaction of the metal with these bonds is able to break the bond and form two new bonds one with a metal and a hydrogen and the other is with a carbon silicon or another hydrogen atom. So, let us take a look at how exactly these reactions turn out the first thing that we need to notice is that this reaction is in fact a formal oxidation it is not indicative of the charge on the metal atom. Now, M N 2 C O 10 that is dimanganese deca carbonyl can in fact react directly with hydrogen under pressure. So, if you treat M N 2 C O 10 with H 2 then you get two molecules of H M N C O 5 and this can be understood as a very simple breaking of the manganese manganese bond. So, in M N 2 C O 10 you have manganese manganese bond and you have in the other reactant H H bond and. So, if you now break this in such a way that you form a new bond between the manganese and the hydrogen then you end up with H M N C O 5. Because hydrogen is slightly more electronegative than manganese it is written as a hydride complex, but as we have mentioned in the lecture in metal hydrides H M N C O 5 will actually behave like a weak acid. It is in fact possible to ionize it as H plus and M N C O 5 minus. So, this should be regarded as a formal oxidation reaction not a real removal of electrons, but nevertheless we can write the whole reaction to indicate the transfer of electron from manganese on to the M N H bond. Now, the situation is even more amazing when you consider the reaction between M N 2 C O 10 and sodium which results in the formation of M N C O 5 minus. M N C O 5 minus has got an oxidation state of minus 1. So, this is in A plus and M N C O 5 minus M N is in minus 1 oxidation state that negative charge is in fact nicely stabilized by the carbon monoxide which are surrounding the manganese. So, this stable molecule now reacts with a proton and that proton is coming from acetic acid. So, the proton is transferred to manganese and H M N C O 5 results the same H M N C O 5 which we encountered in the reaction with hydrogen and in the course of this reaction sodium acetate will of course be extruded. Now, H M N C O 5 again has got a formal oxidation state of plus 1 and so you have changed oxidation state in this process by 2 units. So, it looks as if there are 2 electrons which have been removed from manganese. So, this is a interesting case of oxidative addition reaction which has in fact not changed in reality the number of electrons around the manganese, but nevertheless formally we have to call it an oxidative addition. Now, Vasca's complex was in fact the first complex where many oxidative additions were studied. So, this is another example where direct addition of hydrogen can be carried out with Vasca's complex and interestingly you can add hydrogen under pressure and you can form a compound in which 2 hydrogen atoms are added to the iridium and again the formal oxidation state is plus 3. So, notice that the 2 hydrogens are present in a cis position. So, this cis position is characteristic of group 3 add under. So, this is a cis addition of 2 groups. So, this is discovered as early as 1961 and subsequently a variety of reactions have been studied where oxidative addition has been observed. Now, the more interesting and challenging reactions are those in which you have oxidative addition of very non-polar bonds. In this particular instance I have shown you CH bond so you can see that this is a CH bond that has been oxidatively added in such a fashion that ruthenium now goes from 0 oxidation state. This is 0 oxidation state to plus 2 oxidation state, but notice once again in the final compound the coordination number has changed from 4 to 6. The coordination number has changed from coordination number 4 to coordination number 6 and the oxidation state has increased by plus 2 units and 2 non-polar groups have been added to the ruthenium. In this particular case hydrogen and an aftile group. Now, exactly how the ruthenium is able to carry out this type of an addition reaction is in fact the key feature which separates transition metal chemistry from mean group chemistry. Here is another example which was again a very interesting example because here we are going to take not a aromatic ring hydrogen, but an alkyl hydrogen. The challenge in carrying out alkyl group hydrogen atom activation is quite remarkable. Here we have taken a iridium complex which again has got cyclopentadienyl moiety and this cyclopentadienyl moiety now stabilizes the iridium one. If you is iridium three in this particular instance this is still iridium three. If you excited with light it extrudes a molecule of hydrogen and results in the formation of an iridium one species. That iridium one species because you do not have the hydrogen any longer in solution H 2 is liberated H 2 is liberated from the reaction mixture. So, you have a vacant coordination sphere on the iridium and the electron poor iridium grabs a C H bond on the tetra methyl methane and is resulting in the formation of alkyl iridium one species and a iridium H bond. Notice in these cases a strong C H bond is replaced by a iridium H bond and iridium carbon bond. Many of these reactions can be carried out with 4 D and 5 D elements, but they are not favorable or not favorable the reaction is not possible with 3 D elements. The reason why 3 D elements do not work out well is because the 3 D metal carbon bond is much weaker than the 3 D then the 4 D metal carbon bond or the 4 D metal hydrogen bond. The 4 D and 5 D elements form very strong hydrogen bonds to hydrogen and bonds to carbon. So, this reaction was discovered only in 1982 by Bergman which spearheaded lot of effort on C H activation. C H activation is important because you can now use that to functionalize what comes out of the petroleum feedstocks and what cannot be used for as a fuel can now be functionalized and used for making useful chemicals. Here is another example which was discovered very close to the discovery of Bergman and that was done by Graham in 1983. Here a iridium one complex was again photolyzed and during the photolysis carbon monoxide was liberated CO was liberated minus CO and because of the vacant coordination site and rhodium on iridium and the presence of methane you can have an oxidative addition of a C H bond to the iridium one species resulting in the formation of a methyl iridium hydride which is formally the iridium is in plus 3 species is in plus 3 oxidation state. So, you have several cases where you can have oxidative addition of bonds which are not polarized and bonds which are quite strong. They are broken up by the metal because you have a vacant coordination site and in this particular instance the intermediate would be as follows. This is a iridium one carbonyl complex now it is highly co-ordinatively unsaturated and so it will take any molecule that it can react with that is why you have to use C 6 F 12 as a solvent. Other solvents where you have a C H bond would become too reactive and they would react with the iridium one center. So, you have to use a non-reactive solvent which is a perfluorinated hexane cyclohexane. So, subsequent to the discovery of Graham and Bergman several people several more people studied the C H activation reaction purely because C H activation turns out to be a key reaction if one has to use petroleum feedstocks more effectively. So, what we have seen in this class of reactions is that C H activation or neutral bonds like aromatic C H bonds or S I H bonds can be activated and made to form M S I M H bonds or M C M H bonds. During the course of this reaction the oxidated state of the metal changes by plus 2 units. The important criteria whether this reaction will work or not work or will proceed from right to left or not is dependent on the fact that you have strong 4 D and 5 D metal hydrogen bonds and strong metal carbon bonds. If the metal is 3 D, then the carbon bonds 3 D element, then the metal carbon and metal hydrogen bond energies are not sufficient to compensate for the bond that is broken which is a C H or an S I H bond. So, very often they turn out to be an endothermic reaction and it is only possible to push it to one side by the addition of a third molecule which might in fact compensate for this loss in energy of the C H bond. So, 4 D and 5 D transition elements undergo oxidative addition with the third class of molecules more readily than 3 D elements. If indeed there are examples where you have C H activation or H H bond breakage and formation of 2 M H bonds, then with 3 D elements the reverse reaction is a favored reaction. So, the reverse reaction happens much more readily and you have reductive elimination. Now, coordination number increases by 2 during oxidative addition and so, with the same count you can have reductive elimination, but reductive elimination will be accompanied by a reduction in the coordination number and in the oxidation state. So, reductive elimination will lead to the opposite behavior that is they will lead to decrease in the oxidation state and decrease in the coordination number. So, it is also possible to have oxidation state changes where the oxidation number of the metal increases by 1 and manganese is one example where you have manganese 0 going to manganese plus 1 in the case of a manganese hydride. And it is also possible in other cases as in the case of lithium for example, you can have an oxidation state of 1 and you have an oxidation oxidative addition reaction. So, you can have a coordination number increase that is also 1 or 2 oxidation number can also change by 1 or 2. Remember that it is a formal oxidation state that is going on during the course of this reaction it is not a real oxidative addition. So, with this we conclude our discussion on the nature of oxidative additions and reductive eliminations. During the course of this talk we have seen how some reductive eliminations can happen especially when you have the third class of molecules or in the second class of the molecules where we can form carbon carbon bonds and readily eliminate neutral groups so that catalytic cycles can be accompanied.