 In this lecture, we will talk about a class of reactions which are called oxidative additions and reductive eliminations. These two go hand in hand because one is the reverse reaction of the other. For the major part, we will talk about oxidative additions in these reactions in this lecture first and then we will take a look at reductive eliminations. If you recollect, we have classified most of organometallic reactions and most of the organometallic chemistry into four or five major classes of reactions. In this grouping, what we have done is we have taken the first two to be those reactions where we do not have a change in the oxidation state of the metal atom. So, the metal atom does not undergo any change in the oxidation state in the case of A in A and in B. We do not have a change in the oxidation state of the metal. Whereas, in the class of reactions that we are going to discuss today, there is going to be a change in the oxidation state. It might either change by plus 1 or it could change by plus 2 units. In the reverse reaction, correspondingly, you would have a reduction in the oxidation state of 1 or 2. So, these are reactions which are very interesting and are unique to transition metal chemistry because it is only among the transition metals that you find a variable oxidation state. Different oxidation states have relatively similar energies. So, let us take a look at oxidative addition. The majority of the reactions that have been mechanistically studied are those reactions where you have a D 8 metal complex and many of the D 8 metal complexes have a square planar geometry. So, that is what we are going to take as an example. If you have x, y a molecule heteronuclear diatomic or a diatomic molecule x, y and if you add it to this D 8 metal complex, you can isolate a complex which is octahedrally coordinated and it is got a D 6 electron count. Now, the question is what has happened to the oxidation state of the metal? If we consider the metal to be less electronegative than x or y, then it turns out that you would have to write x as x minus in this complex and y as y minus. At least in the ionic method, it is easy to see that you now have 2 more negative ligands than what you started out with. You started out with 4 ligands. We do not know the character of those ligands, but you do not have a negatively charged species on the ligand. Then at the end of the reaction, you have 2 extra negatively charged species, which means the metal must have released 2 electrons to x and y. So, x became x minus and y became y minus. So, you end up with an oxidation state, which is plus 2 relative to your starting point. So, let us take a real life example now. You have a D 8 co-ordinatively saturated system. Then also, you can have the same situation. You can add this x, y group, but if you start with a pentacoordinated saturated co-ordinatively saturated metal complex, this is co-ordinatively saturated because you have if L is a 2 electron ligand, if L gives 2 electrons, then you have an 18 electron system right here. So, when you add x, y or when you reacted with x, y, then you end up with a complex, where you have the heterolytic splitting of x, y in such a fashion that you end up with M L 5 x plus. Now, notice the oxidation state of M L 5 x plus is plus 2 units compared to what you started out with because x is considered as x minus. And so, what you have is transfer of 2 electrons, 1 electron to y to form y minus and 1 electron to x to form x minus. So, you have oxidation state change of 2. Now, let us take a look at some of the reactions that this complex. It is actually a very popular complex. It is actually a very popular complex and it was one of the first ones that I discovered and very easy to understand and characterize. L is a phosphine or P R 3 group, L is P R 3 and very often it is P P H 3 and x is a halogen. X is a halogen and P P H 3 is a halogen. Then, you can have oxidative addition reactions that are plenty in the literature and you have species like H 2 adding to it and then you have this complex here and you have oxygen adding to it and you have this complex here. So, you can see that two very different species oxygen and hydrogen can add to this complex and this complex was discovered by Vasca in close to 1960s and so, it is called Vasca's complex and it is in fact a very popular system to study. Now, because you have a range of substrates, it is convenient to classify them according to the type of reactions they undergo. The steps involved are not always the same and so, they have been classified into three groups and we will consider these groups one by one. Now, if you take the first group, these are the molecules which are polar which I have for example, H X that is hydro halic acid H C L H B R H I and so on that belongs to one group and in the same group you can also add X 2 which is just a simple halogen molecule because they also behave in the same fashion and I will explain to you why that is the case. We can have very polar molecules very highly polarized molecules like R S O 2 C L which are clearly R S O 2 plus and C L minus and we can also have R C O X which will polarize as R C O plus and X minus. Now, if you take all these molecules although they look very dissimilar they have one thing in common and that is a fact that if you take olefin if you take an ordinary organic molecule olefin and add this H X to it any one of these molecules which we have listed above all of them will add in this particular fashion. One part will add to one carbon and the other part will be attached to the other carbon. So, they will undergo simple addition reactions to olefins. So, if you take all these molecules which undergo simple addition reactions to olefins react them with Vasca's complex they will undergo reaction which will result in oxidatively addition which will undergo oxidative addition and form an oxidative addition product. So, let us take the second group in the second group of the class of molecules we have a unique feature that the two parts that the molecule has got two parts which are adding to the Vasca's complex representative of all the molecules which undergo oxidative addition and during the addition one bond between the two parts of the addenda are retained. In other words oxygen for example, has got a double bond between the two oxygens O double bond O and after the addition one bond which is indicated here one bond is in fact retained. So, although O double bond O is a molecule you started out with a single bond is retained at the end of the reaction. Now, what is the implication of this retention of the bond that means that this molecule can add only in a cis fashion to the molecule to the metal complex. So, in other words in the end of the reaction the product you always find the two parts of the molecule which are adding. So, this oxygen here and oxygen here both of them are this cis position in the final complex and secondly you do not you do not isolate intermediates during the course of this reaction. So, that separates class 1 from class 2. Secondly, you also find that class 2 is not necessarily polarized they do not always add two olefins to form addition compound with in the organic reactions which we know. So, let us proceed further the third group here you do have molecules which are extremely covalent and non-polar. They do not have any hint of polarity and not only that their electronegativity is almost similar to that of metal complexes. So, I have here listed several hydro compounds hydrogen compounds dihydrogen itself as a first molecule dihydrogen and we also have alkyl hydrogens. So, those are non aromatic C H bonds and we also have aromatic C H bonds and we have S i H bonds these are C i lanes and we also have R S H bonds. So, this should actually be written as R S dash H. So, the two parts that are being added are in fact R S and H. So, these add on to the metal complex and so the metal will now have R S and H added on to it. Now, many of these molecules are characteristic the characteristic feature of many of these molecules are as follows. They undergo an interaction with co-ordinatively unsaturated metal complexes in a weak fashion and this has been termed as an agostic interaction and we will discuss this in greater detail later after we finish talking about oxidative additions. But, nevertheless the third group what one needs to remember is a fact that they are not polar they are non-polar bonds which are cleaved by the metal complex in order to form the oxidatively added substrate substance. So, first let us take a look at the mechanistic studies with group one molecules. Group one molecules as I said can be very easily polarized into plus and minus and so we have several substrates which behave like this typically we can think about a molecule like methyl halide. Now, methyl halide is polarized as delta plus and delta minus and this is easy to see and as I told you halogen compounds are good any good leaving group on the metal is good enough as x, but x is in this particular example that we will be discussing x is actually a halogen and the metal if metal is iridium and it is in the plus 1 oxidation state it turns out to be similar to the Vasca's complex when l is p p h 3 it is in fact the Vasca's complex. So, what are some features of these molecules that we would like to consider that has been studied and the first thing that I would like to discuss is a kinetics. When you do a kinetic study of this reaction of methyl halide with Vasca's complex it turns out that they are first order in both iridium and first order in methyl halide. What that implicates is that during the course of this reaction at least the rate determining step involves most probably a collusion between these two molecules. These two molecules have to be collided in order for the reaction to happen and that is why the concentration of both iridium 1 and the concentration of methyl halide make a big difference in the rate of the reaction. So, that is the first thing that we note. The second thing that we note is that if this is in fact occurring through a collusion of these two molecules we can talk about the reaction as if it is a reaction of the iridium 1, iridium 1 as a nucleophile displacing the X group on the methyl group. The X on the methyl let me just complete this. So, if X on the methyl group is in fact a halogen this is leaving the methyl halide as X minus and the iridium 1 is coming in and reacting with it. So, that metal methyl bond is being formed and that is what is pictured here. Now, we might be wondering how can you have a nucleophilic substitution by a metal complex. If you look at the electronic structure of the iridium atom which is there in iridium 1 you find that it is a square planar geometry and it has got 8 electrons in the D shell. So, if you remember the splitting diagram for a square planar complex is in fact in such a fashion that the D x square minus y square is the highest lying orbital and the highest lying orbital among the D manifold and the highest occupied molecular orbital turns out to be the highest occupied molecular orbital turns out to be the D z square orbital. The D z square orbital has two electrons and that is the one which is the homo. So, if you turn the square planar complex in such a way that you have the z axis in along here then you can see that the D z square will be pointed towards the methyl group and this will in fact result in a pentacoordinated structure at the iridium and if you consider the X minus also a tbp geometry at the methyl halide. So, X minus will be going away and you have a simple SN 2 type reaction on the methyl halide and that is carried out by the nucleophilic addition of the iridium to the methyl group and X minus is the leaving group. So, you can always ask what is the stereochemistry of the carbon a question that we will come to later on. So, let us take a look at some of the features that support this type of reaction. First of all it has been possible at least in some instances to find this pentacoordinated intermediate that itself is stable and it can be isolated. But in most instances the X minus in the case of Vasca's complex the Vasca's complex the X minus attacks the iridium atom again in the transposition to the methyl group. So, if it attacks here you end up with the transcoordinated methyl halide and this is the Vasca's complex after oxidative addition. So, that is a octahedral complex. So, if you think about this reaction sometimes it has been shown that you have a small isomerization of the X and Y. Remember Y is the species which was attached it is also a halogen many times it is a halogen and it is attached to the iridium atom and X is also a halogen. So, sometimes it has been possible to isolate an isomeric species and that gives you the indication that the intermediate must have been this pentacoordinate pseudo square pyramidal geometry that is that was indicated earlier. So, if you have this as an intermediate X minus can now come in and react with this pentacoordinated complex which I have indicated here and result in the species which is drawn below. Now, if it is thermodynamically more favorable to have X minus attack in this fashion then you would have the formation of the product here. On the other hand if X minus it is thermodynamically more favorable to attack here and we will indicate this with a different ink color. So, if X minus attacks here we will have the product which is listed here. So, depending on where the X minus attacks in the intermediate you are able to isolate two different compounds. One where the X is in fact trans to the methyl group and that is what is characteristic of these reactions in the first class and sometimes if it is more stable to have X in the Y in the transposition then it has a small rearrangement or the attack takes place in a different place and you can have the isomeric product isolated. Now, what are the various factors which support such a hypothesis? First of all you had this bimolecular reaction between the methyl halide and the iridium complex. Secondly it was found that the X group which is CH 3 X here CH 3 X the methyl halide that you have here undergoes reaction at different rates depending on the nature of X. So, if X is iodine then the rate turns out to be fastest compared to bromine and chlorine iodine reacts fastest. On the other hand chlorine is the slowest and this is fitting the leaving group ability of X. So, this turns out to be correlating with a leaving group ability. So, the leaving group ability of X correlates with the rate of the reaction. Now, there is another indication that in fact it is a nucleophilic attack by the iridium complex. How do we know this? Depending on the nature of L, L is ligand which is attached to the iridium. If L is electron rich then the reaction happens at a faster rate. So, here we have varied the second line. We have varied the nature of L and we can see that if it is triethyl phosphine, the ethyl groups donate more electron density to the phosphorus which in turn donates more electron density to the iridium. So, iridium makes is much more suited to do this nucleophilic attack. So, more electron density at the metal these two factors suggest that more electron density at the metal is favorable favours the reaction. So, this is clearly supporting the fact that we have in fact a nucleophilic attack where electron density on the metal does a nucleophilic attack on the methyl group and the X leaves as X minus. So, there was one result which was slightly difficult to understand based on this what we have just seen here and that is the nature of Y. If Y is a halogen, but if Y is electron rich it should have promoted the reaction, but on the other hand turns out that fluorine is much faster than chlorine, then bromine and iodine. This is contradicting what we just said regarding the electron density of the metal and there was another slight difference. If it was possible to show that p p h m e 2 is much faster than p e d 3. These two results can in fact be understood if we remember that the nucleophilic substitution at the methyl halide has two factors that will favour it. One is electron density on the metal, another is the steric factor which indicates that the incoming group has to be small in order to carry out the nucleophilic attack. If you have a very large group coming and attacking the methyl, obviously that would lead to an unfavourable situation because the intermediate has to be a 5 coordinate intermediate at the carbon and that is very difficult considering the fact that carbon is a first row element. It has got only 1 s and 2 s orbitals, 1 s, 2 s and 2 p orbitals and so you do not have too much space around the carbon in order to accommodate all this electron density. So, you would like to have a small group coming in and if you have larger groups then the rate turns out to be smaller. So, this is how it was understood and in fact this is consistent with most of the data that we now have which suggests that it is a nucleophilic attack and there are two components. One is an electronic component, electronic component is indicating that if you have greater electron density on the metal you will do a faster nucleophilic substitution. The second factor is that the smaller the incoming group the faster the reaction. So, here I have indicated to you the net result of this nucleophilic displacement of nucleophilic displacement of a x by the iridium one. You have x on the metal group which is being replaced by the iridium and so what you end up with is this pair of electrons which is almost like a lone pair of electrons on the iridium. You do have a lobe on the other side and we can see this we do have a lobe on the other side and we have a torus here. That is mostly the d z squared orbital that is how the d z squared orbital will look like, but it is primarily this lobe which we are concerned with because that is the part that is carrying out the attack on the methyl halide. In the transition state you would have the famous umbrella inversion or the Walden inversion where the x minus would be leaving and the iridium would be coming in. So, the stereo chemistry of the carbon would be inverted. At the end of the reaction what you would have is a d 6 complex. Now, we counted as a d 6 complex because we would have a iridium 3 species why are we calling it iridium 3 because you now have 3 anionic groups. We have 3 anionic groups one is x another is y another is a methyl methyl because it is relatively more electronegative with respect to the carbon is more electronegative with respect to the metal. We consider that also as a negative group. So, you end up with a tri positive metal and as a result if iridium 1 becomes iridium 3 this is iridium 1 here and this becomes iridium 3 here. You would end up with 6 electrons on the metal and these 6 electrons are now in octahedral geometry. If you ignore the small symmetric perturbations it will be caused by different ligands in an octahedral field they will be split in this fashion and 6 electrons can be conveniently accommodated in the t 2 g set of orbitals and you will have a stable situation. So, that is why d 8 metal complexes undergo oxidative addition to form octahedral complexes which become d 6 species and they are also 18 electron in nature. So, let us move on now to determine the stereochemistry a series of experiments were carried out. It is possible to have a single substituent on the methyl halide which is undergoing substitution. So, if you make that R let us assume that that is a alkyl group. Then if x is a halogen then you can have a Walden inversion at x and let us say the Vasca's complex undergoes a reaction with the species you would end up with at the end of the reaction a Walden inversion and the product should have been in this case in this instance. Let us write this out this group and so you would have this carbon undergoing Walden inversion or inversion in stereochemistry. So, several reactions were done with Vasca's complex and surprisingly they observed complete loss in the chirality at the carbon center undergoing inversion undergoing oxidative addition. So, this is a big surprise to the organ metallic chemistry community and this was true whether one started out with a single substituent which had a chiral center generated with hydrogen and deuterium or if you have R 1 and R 2 either way you ended up with loss of chirality at the carbon. So, what now is left at a loss to explain how all the experiments which were carried out earlier indicated an ascent 2 type of substitution, but now we are observing loss of chirality. So, the mystery was in fact solved when people started doing reactions very carefully by deoxygenation of the solutions in which the reactions were carried out. Now oxygen is paramagnetic and has the capacity to induce radical nature to a reaction and so one has to deoxygenate the solution carefully. When one did that this loss in chirality was sometimes not completely observed and so people started doing reactions with azoizobutronitrile or in the presence of duroquinone, duroquinone is the molecule that I have listed here. Now azoizobutronitrile generates radicals, this generates radicals. So, this inhibits radicals. So, it was shown that it is possible to promote this reaction in the presence of azoizobutronitrile and it is possible to inhibit the reaction in the presence of duroquinone. So, this clearly indicated that there was a radical nature to this whole reaction and that would easily explain why chiral substrates were racemized. So, this whole mystery regarding loss of chirality at carbon was explained very simply by reactions which were carried out in the absence of oxygen and by then carrying out the reaction in the presence of radical promoters and radical inhibitors. It was possible to show that the nature of this reaction was in fact affected by radical substrates as a radical intermediates in the reaction and that also explained the chirality problem. So, let us take a look at what exactly is happening. Let us assume that there is a radical initiator in the reaction. So, if you have a radical initiator then you could have the formation of an iridium 2 intermediate. This you should remember is only a catalytic initiator. So, q dot is in fact an initiator. It could be any r dot or any radical species and if you use isobutronitrile the isobutron radical would then be r dot. So, if r dot reacts with iridium 1 it generates a small amount of an iridium 2 species because once again because if you use an r dot then we would consider irr as r minus an ir plus. So, this would lead to an oxidation state change of plus 1 in on the iridium. So, oxidation state of plus 1 a change of plus 1. So, that becomes iridium 2. Now, this iridium 2 is again you remember this is a d 8 species. So, if it becomes iridium 2 it will be a d 7 species it will be an odd electron species. So, that is why we indicated by adding this that is why we indicated by adding this dot. So, it is a radical. So, this whole thing is a radical species. Now, if that reacts with r x. So, if that reacts with r x remember r x is the group that is adding on to the iridium. So, this is purely the first step is purely an initiation step where it led to the formation of an iridium 2 radical that iridium 2 radical reacts with r x grabs x and forms i r q x which again is an iridium 3 species and generates r dot. Now, this becomes our key intermediate we need very little of the initiator that could be anything from oxygen to a radical species in in the flask which will lead to starting of this reaction. So, once you form r dot then you can write a cycle a catalytic cycle where you would react r dot with iridium 1 to generate the iridium 2 species r i r 2 which again would be the net species would be a radical species because you have reacted a radical r dot with even electron species. Now, this iridium 2 species which is radical which is got a radical nature will react with r x. This is identical to what we saw here these 2 reactions are similar these 2 reactions are similar we just have r x reacting with this radical species to generate free radical again. Now, because you have a radical intermediate in this whole reaction what we will end up with is loss of chirality at the carbon which is undergoing oxidative addition. Now, if you add up this 2 reactions if you add up these 2 reactions you see you can see that iridium 1 plus r x gives the iridium 3 iridium 3 i r r i r x which is the oxidatively added product. So, this is the oxidatively added product and that is what is isolated at the end of the reaction. So, you can see that radicals can induce this type of reaction and whatever was studied with methyl was in fact coincide was fortuitous because methyl is a substrate which is got very little steric interaction and iridium can do a nucleophilic substitution reaction on methyl halide very easily and carry out this nucleophilic attack. All the factors that we had looked at with methyl halide will point to the fact that you have a SN 2 type of reaction. But the moment you have any other alkyl group on the methyl on the carbon undergoing substitution so in other words if you have r or 2 r groups then the chances are that you will end up with the radical reaction and not the SN 2 clean SN 2 type of reactivity. So, nevertheless this has helped us to understand although we have this complication it has helped us to understand how one can carry out nucleophilic substitution or an oxidative addition on a metal atom. Now, there are more reactions that were studied which led to a beautiful understanding of how you can have radical intermediates in organometallic reactions. Let me tell you something fascinating about a molecule which is called a radical clock. This radical clock is a cyclopropyl carbonyl radical. Now, if you have cyclopropyl carbonyl radical it can ring open to generate the homo allylic radical. So, if this radical is formed in solution it ring opens at a very fast rate which is known to be 1.3 into 10 power 8 second minus 1. That means every second 10 power 8 molecules will convert from this ring open state to this ring an acyclic form of this radical. So, you can understand how fast this reaction happens. So, if it is involved if the radical is involved as an intermediate one can easily detect the formation of products from this species. If the reaction in question happens at a faster rate than the ring opening of the cyclopropyl carbonyl radical then you would not have this ring opening reaction. So, let us take a look at what was reaction that was carried out in order to understand this ring opening this radical nature of this reaction. So, the reaction in question is cyclopentadienyl ion di carbonyl species. This species as we are not discussed the cyclopentadienyl group itself we will we can just ignore it just let us do the electron counting. If you have C p as C p minus that means this is the aromatic C p minus anion ion is ion plus, but nevertheless there is a negative charge which is present on this whole species because sodium is a counter ion. So, the net charge on ion is actually 0. So, this is again a d 8 species this d 8 species can react to a nucleophilic substitution. In fact, the ion if you write the ion in this fashion to indicate F e C p C o 2 then this can indicate this can carry out a nucleophilic substitution on this carbon and have a product which is indicated here. Now, this is a simple nucleophilic substitution that can happen in this reaction and that is what people expected. But if radicals are involved if a radical intermediate is generated during the course of this reaction as I have indicated here if the cyclopropyl carbonyl radical is involved as an intermediate then it can ring open to give you the homo allylic radical which is again on the right hand side of the screen. So, people carried out the reaction this nucleophilic substitution reaction using the C p F e C o 2 minus anion. So, this anion was reacted with the cyclopropyl carbonyl halide and when x the halide was iodide they found that 70 percent of the reaction was in this channel which means the ring unopened product was formed in 70 percent and the ring open product was formed in 30 percent. That tells you that almost 30 percent of the reaction went through a radical pathway. So, a radical reaction has happened and the ring opened very fast and so that gave you the homo allylic product which is indicated with the formation of this radical as an intermediate. On the other hand if one used the bromocompound surprisingly the bromocompound resulted in greater than 97 percent of the nucleophilic substitution reaction product which is a simple nucleophilic substitution on the cyclopropyl carbonyl halide and exactly what we have pictured here has happened exactly what we have pictured here has happened and this is with with we are as a leaving group we have only this channel with a iodine as a leaving group we have the radical channel. So, with iodine as a leaving group we have x equals iodine and x equals b r gives you the ring unopened product. So, this was a beautiful chemistry which illustrated that radicals are in need involved and the surprising feature of this whole study was that ESR was not able to detect radicals in the reaction mixture in spite of that they were able to show because of the radical clock reaction they were able to show that radicals were indeed formed during the course of this reaction. This brings me to the fact that absence of evidence is not evidence of absence it is a very important principle that one needs to know in science in general and in reaction mechanism studies in particular if you do not have a particular evidence it does not mean that it is not there it is just that you are not been able to detect the radicals. In this case ESR was not able to detect radicals, but a different technique a chemical technique was able to show that radicals were enrolled. So, now this brings me to the second group of reactions. The second group of reactions involves oxygen, sulfur or azides and when you react them with molecules like the Vasca's complex they undergo a simple oxidative addition reaction. As I mentioned one bond is retained always in the second class of molecules which do oxidative addition. Now Vasca's complex itself was the first one which was studied it was treated with dioxygen a way back in 1961 and that reaction was in fact studied by Vasca himself and what was shown was that the product has got a cis addition and the product is in fact having one oxygen one oxygen oxygen bond retained by the oxygen at the end of the reaction. If you recollect it is important that we homolithically break one bond in all the oxidative additions we homolithically break one bond and remove two electrons from iridium added to the two add under and form metal metal add under bonds. So, in this particular case it is iridium oxygen iridium oxygen bonds that are being formed and that leads to these are the two new bonds that are formed the O O bond first you have O double bond O and you have break in we sorry. So, we have broken this oxygen oxygen bond. So, the double bond has been converted into a single bond and the final product has got a bond which is intact between the two oxygens. A similar situation happens with acetylene here you have a C triple bond C the acetylene adds on to the iridium and you can see easily that the two carbons are attached to the iridium in a in a fashion that indicates that you have these two new bonds and if you assume that the oxidative addition involves carbon as carbon minus because it is more electronegative than iridium then you end up with an iridium three species. So, oxidation state is in fact a formal a formal oxidation state it is not the charge on the metal atom this is something which one has to bear in mind when one is studying oxidative addition. So, the characteristic feature of this oxidative addition in this case is the fact that you have a change of plus 2 in the metal atom in the oxidation state and the fact that it is retained in the cis position and the fact that it is such it is added in such a way that one bond is broken and if you start with the molecule which has three bonds as in this case you have three bonds then you end up with retention of two bonds. If you start with oxygen as in the previous case we had two bonds and we ended up with retention of one bond. So, let us proceed now with a slight variation in the oxidative addition reaction until now we have been talking about oxidative addition with carbon with no change in the carbon center. So, we will take a slight deviation and talk about oxidative additions which involve a change in the organic substrate also. So, let us take a look at a molecule which is very well studied and this is a cyclopentadienyl cobalt complex and one can in fact draw this molecule in a different fashion if. So, this is a cobalt center it is bonded to all five carbon atoms and you have two neutral ligands. If you remember cobalt is in the plus one oxidation state in this molecule because C p is C p minus and this molecule is neutral this is a neutral species. So, because this is a neutral species this should be plus one if this is minus one. So, if you react this cobalt molecule with acetylins you have at the end of the reaction you have the formation of aromatic rings. What is interesting is that this reaction is of course, catalytic what I am describing, but it can be done in a stoichiometric fashion also, but this turns out to be very interesting because it is a convenient way of generating an aromatic molecule starting with simple acetylins. So, let us just take a look at this molecule this particular reaction. First of all you can have an oxidative reaction in which you have two acetylene molecules reacting with this C p C o 2 which has lost the two ligands. So, you can replace the two carbon monoxide that we had on the cobalt with two acetylins. So, the reaction that we are talking about is C p C o C o 2 plus two acetylene molecules will lead to this intermediate and this intermediate now can carry out an oxidative addition reaction. Remember when you do an oxidative addition reaction you break one of the bonds in the acetylene molecule and make two bonds to the cobalt. So, in this particular instance if we broke two bonds on the acetylene, two acetylene molecules, two different acetylene molecules then you would end up with a species which will formally look like this, which will formally look like this. So, you can see that there are two there are two unpaired electrons on the two acetylene molecules which have oxidatively added. Instead of taking one acetylene and carrying out an oxidative addition you can take two acetylene molecules and add two electrons to them in such a way that you form cobalt carbon bonds and then you will end up with two electrons sitting on the two acetylene molecules which can now form a bond very readily. Let us indicate this with a different color. So, that we remember that this bond is a newly formed carbon carbon bond and that is what is indicated here which is now a cobalt three species, which is now a cobalt three species the cobalt three species because you have C p as C p minus and two vinyl groups attached to the cobalt atom. So, this species can now react with another third acetylene molecule a third acetylene molecule and you can do this in two different ways. Remember what we looked at when we talked about insertion reactions. This is a anionic group which is this is an anionic group and this anionic group can now add on to the neutral acetylene molecule that we have here. So, you can have an insertion reaction and that will give you a six carbon chain which is attached to the cobalt which undergoes the reductive elimination and gives you the aromatic species. It turns out that this reaction has been well studied. We will look at this in a little bit of detail, but the main factors are that kinetic analysis first showed that you need a loss of the ligand which is attached to the cobalt. Before the reaction can happen and it will also shown by isolating some of these intermediates that you can have intermediates like this where you have oxidative addition on the cobalt and you have metallocyclopentadiene. I am calling this a metallocyclopentadiene because it is a cyclopentadiene ring in which there is a metal. So, this is a metallocyclopentadiene which has also been isolated and characterized and we can replace if you replace p p h 3 with p m e 3. If you stop this reaction you can stop this reaction by adding p p p m e 3 then you can isolate and characterize an intermediate where you have this metallocyclopentadiene on the cobalt. So, the second step which involve the reaction of the metallocyclopentadiene with a third acetylene molecule was replaced with an acetonitrile or an alkyl nitrile and then it was shown that you can in fact make pyridines. So, if you do the reaction with R C triple bond n then you can form pyridines that I have indicated here. You can generate pyridines if you treat it with R triple bond C n and if you have a simple R C R then you can form acetylenes. Now, this can happen as if it is a Deals alder reaction. If you have a Deals alder reaction then between this metallocyclopentadiene and this alkyne you form a Deals alder reaction this is the product that you would end up with. And that Deals alder adduct the Deals alder adduct can now lose a molecule of C P C O which is the catalytic active species. This is the catalytic active species which can react with other acetylenes and go back into the catalytic cycle. But the product that you have got is the aromatic species. Now, turns out that if you have a mixture of acetylenes you can have a variety of arene molecules generated. And if you have nitrile present you can have pyridines generated. It is convenient to make pyridines substituted pyridines using this method because it is often found that the first step undergo the first oxidative addition and coupling happen with cobalt. And then the second step happens with the Deals alder reaction with nitriles or acetylene substituted acetylenes. So, that you get aromatic compounds as I have indicated here or substituted pyridines as I have indicated here. Although there are two possibilities and these two possibilities have can be distinguished one can bias it. And the way one can study this is by treating it with a molecule which will undergo Deals alder reaction very readily. So, if you have dimethyl acetylene dicarboxylate which is a very good dinofil then you can in fact isolate an intermediate like this where you have PME 3 coordinated to the cobalt and the compound can be generated which will on heating generate dimethyl thalate. So, these are some reactions which have been carried out where you have oxidative addition and the coupling reaction and an insertion reaction or it is a Deals alder reaction. So, you have seen a variety of reactions which are possible today. Let me now conclude by going to the conclusion slide. Usually in all these reactions the coordination number increases by 2 during oxidative addition. The coordination number increases by 2 and the oxidation state increases by 2 as well. It is possible that one can carry out oxidation state increase of 1 that is also possible. It is possible to increase the oxidation state by 1 as well. We will look at some of these reactions where oxidation state can be changed by plus 1 units in the following lectures. But oxidation number and coordination number usually increase and this is a reaction that is characterized by a change in the oxidation state of the metal. So, it is called its characteristic of transition metals and it is called an oxidative addition reaction. Invariably you have a coordination number increase and that can also change by 1 or it can change by 2. Remember it is a formal oxidation state change and it is not a real change in the charge on the metal. So, with this we conclude the first part of oxidative addition reactions. We will take up oxidative additions in a future lecture as well.