 Today, we will consider some reaction mechanisms in organometallic chemistry, especially with respect to insertion of neutral ligands. We had already seen that carbon monoxide can be inserted between a metal carbon bond. We will continue with this discussion first and subsequently, we will look at some other neutral ligands, which can participate in this reaction. So, we have classified the type of reactions that can happen in organometallic chemistry broadly into these four categories. The first category comprises of rearrangement and simple isomerization reactions, where the position of the ligand or the type of ligand that is involved in these complexes undergoes a change. In the second group, we talk about ligand insertion reactions and that is what we are going to be concerned with today in this lecture. Basically, what we have is the insertion of a neutral ligand apparently into the m x bond. In this particular case, the way we have written it, m l is the bond that is broken and then x group is inserted in between. So, let us take a look at the type of reactions that can happen and what are the pathways by which these reactions happen. We first note that in the case of organometallic reactions involving transition metals, the reagents that are involved or the type of insertions that can take place are distinctly different from what can happen in main group organometallic chemistry. Main group organometallic chemistry is characterized by very polar substrates. In this particular instance, I have shown carbon dioxide here. Carbon dioxide is undergoing an insertion between an R-MG bond. In our convention, we know that carbon dioxide is polarized as C plus and O minus and R is also polarized as R minus and MG plus. So, it is very easy to understand an insertion reaction where you have insertion of carbon dioxide in such a fashion that the oxygen now bonds to the magnesium atom. However, we notice that in the case of organometallic, organotransition metal chemistry, it is possible to insert a neutral molecule like carbon monoxide which has a very small dipole moment. This neutral molecule is able to insert between a metal carbon bond. This is the basis, the insertion of carbon monoxide into methanol is actually the basis for formation of acetic acid. This particular process is used by the company Monsanto to produce a very large amount of acetic acid. So, it is an extremely useful process and it is also modified to generate a variety of small molecules which are useful in synthetic organic chemistry, especially in the pharmaceutical industry. So, this reaction is an extremely valuable reaction. Now, sometime back Calderozo figured out that there are two possibilities for having this reaction take place. In the first instance, we insert a molecule of carbon monoxide between the metal carbon bond. To do that, we will have to break. For example, if you want to insert carbon monoxide in this bond, we break this carbon bond and then the carbon monoxide can be inserted. I have color coded the carbon monoxide so that you can recognize what we are doing here. We are inserting carbon monoxide between the metal metal metal bond. In there is another possibility and in this possibility, what we are doing is inserting carbon monoxide into a vacant site in the coordination sphere of the metal. First the methyl migrates. First the methyl group actually migrates on to carbon monoxide and a vacant site is generated. On this vacant site, carbon monoxide is capping the vacant site and as a result we have octahedral complex again. So, Calderozo is trying to answer this question and through isotopic labeling experiments, he showed clearly that the methyl group is migrating to a carbon monoxide and generating an acetyl group. The incoming carbon monoxide is merely filling up a vacant coordination site. He found out that it is possible for the carbon monoxide to migrate or the methyl group to migrate. He showed very clearly that it is a methyl group which is migrating. He did this by utilizing an isotopically labeled compound which had carbon 13 labeled in the cis position of the acetyl group. Looking at the reverse reaction, most of these reactions are reversible. He heated this compound to slightly higher temperature till it lost the carbon monoxide. The reverse of this insertion reactions, I wish we could call it deinsertion reaction, but we have to call it an abstraction reaction or an extrusion. That is what is happening here. Carbon monoxide leaves and he showed that any one of these carbon monoxides which are cis to the acetyl group, any one of them can leave. So, you get a ratio of trans to cis, trans to cis to unlabeled as 1 is to 2 is to 1. This is exactly the ratio in which you have the four carbon monoxides which can leave the coordination sphere of the metal. So, very clearly he demonstrated that it is the anionic group or the formally anionic group, the methyl group here. That is the one which migrates to a vacant coordination site. So, he did this using isotopic labeling and through infrared spectroscopy. So, if the CO had migrated, a different ratio of compounds would have been formed and there were clearly it was indicated that the methyl group was migrating. So, we have answered some questions about the migration reaction. Now, in today's topic, we are going to talk about the stereochemistry of the migrating carbon. You remember it is the methyl group which is migrating. Now, a methyl group can migrate to a carbon monoxide as an anionic ligand. If it does so, it would migrate with a pair of electrons. In the neutral method of electron counting, we consider that as a one electron donor, but if you consider that as a two electron donor, it is an anionic ligand. That seems to be more appropriate method to count electrons in this particular talk. So, we will use the anionic method. If it migrates to a cis vacant site, then we can ask this question, what will happen to the stereochemistry? If it migrates as a free methyl radical, then what would happen to the stereochemistry? Now, it is also known that the methyl group can migrate as an alkyl cation, especially when it is substituted. So, we can think of it as if there has been a nucleophilic substitution carried out on the methyl group, in which case it will have to undergo inversion during the process. So, let us proceed further. There was this experiment carried out by white sites, which was an extremely useful experiment and a very creative experiment, where he carried out an insertion reaction by taking a molecule, which had stereochemistry marked out by hydrogen and deuterium. So, if you have an R group, a hydrogen and a deuterium, you can see that this carbon center is actually chiral. So, let us mark this chiral carbon center. This is the carbon center, which is chiral. This is the chiral carbon center and this is the one, which is undergoing migration. It is migrating to a carbon monoxide, which is on the ion. C p is in fact a cyclopentadiene group, but for the purposes of today's discussion, we can just keep it as one of the three groups, which are present on the ion atom. Now, after migration, we can look at the chirality of the center. We can look at the chirality of the center and try to understand what is the geometry of the carbon after migration. So, as I mentioned to you before, there are two possibilities. One is to undergo migration with retention of stereochemistry. Now, that is what is pictured on the top half of the slide. If the carbon just migrates, let us say this is the bond that is migrating. If that migrates with retention of stereochemistry, the product that we would get is this particular product, which I have on the top part of the slide. This can be, it should be mentioned that this is now chirality is retained. So, here chirality is retained in doing the process of this migration. So, an acyl group is formed by migration of the alkyl group on to the carbon monoxide. The same thing could have happened with an inversion of stereochemistry, in which case you would have this carbon undergoing an inversion. So, let us try to understand what will happen in during a Walden inversion. What you have is inversion of this particular carbon, which we have put a star on. So, the deuterium atom, which was pointing upward, this is the deuterium atom, which is pointing upward. That will have to go down and that is what has happened. It has come down and so has the hydrogen. So, the hydrogen and the deuterium atoms, which we are pointing up on this, in this molecular structure, the hydrogen and the deuterium were pointing upwards. Now, they are pointing downwards and the R group has come up on the other side as well. So, here inversion has taken place in this case. Now, because we want to compare it in a similar fashion, I have rotated the, rotated this chiral center with respect to this carbon CO bond. When I do that, I get this molecule, which is pictured here. You will notice the chirality of the carbon, which has undergone migration. Here it is undergone inversion. So, it is very clear that in one case, you will retain the chirality. In the other case, if you have inversion, if you undergo Walden inversion, the chirality would be exactly the opposite chirality of what we started out with. Now, how will you figure out whether the molecule has undergone inversion or not? That is where the ingenuity of the experimenter, in this case, wide size comes into play. What he did was to attach another chiral center next to the carbon undergoing migration. So, the carbon undergoing migration was here and he put a second chiral center next to the carbon undergoing migration. So, what he did was to compare the chirality of one carbon center with respect to the other carbon center before and after migration. In this experiment, he attached in a fairly large group. This is a tertiary butyl group here, a very large group on the carbon. So, that rotation about the carbon-carbon bond would not be facile. If the carbon-carbon bond cannot be rotated, then the relationship between the two deuterium and the two hydrogens is forced to be the same before and after migration. So, let us see what happens in this particular reaction. He took the erythro compound, which means the hydrogens were on the same side. So, he took the erythro compound and you can see that they are on the same side because we have put the large groups trans to each other. If you rotate around this carbon-carbon bond, you can see that the two hydrogens are on the same side. So, this is the erythro compound and after migration you can see that the chirality was still retained and the two hydrogens are still on the same side with respect to each other. So, he started with the erythro compound and ended with the erythro compound showing that the stereochemistry of the migrating carbon is retained. So, this is so retain the stereochemistry after migration. This seems to indicate that the migrating group is actually migrating like an anionic group, which would have a lesser propensity to undergo inversion. As opposed to the methyl radical, which will undergo inversion fairly rapidly or as a methyl cation, where also a planar structure would be involved. So, how did he measure this inversion or retention in chirality? Here, we have to use what is called the Caplus equation. The Caplus equation merely connects the hydrogen hydrogen coupling constant. The coupling constant of the two hydrogens, which were connected to the two carbons and as we said it was in erythro position. So, the coupling constant was, the cis has a smaller coupling constant. The trans orientation gives a larger coupling constant and as I showed you because of the two large groups that are present on the two carbons, the hydrogens will rotate in such a fashion that they will maintain a 180 degree angle. So, this angle was close to 9 to 12 hertz and that clearly indicated that the coupling constant between the two hydrogens was in fact trans both in the starting material and in the product. So, in order to visualize this, we have shown here a Newman projection of the two molecules before and after migration. So, before migration, what you notice is that this is a compound before migration and the migration happens on to a carbon monoxide group and there is a vacant coordination site, which is generated. So, you would have the same molecule, you would have the same molecule with hydrogen, deuterium and iron and with one CO. So, you have, this is the carbon monoxide which is undergone migration. So, you would have a vacant coordination site on. So, here is a vacant site on the iron, which has been formed because the alkyl group has migrated on to the CO group. So, this is the CO on to which the alkyl group has migrated we can see that, we can picture that here by moving this bond here and this is what you would get and a vacant site would be there on the iron, which is now occupied with p p h 3. If you react that with p p h 3, this is the product that you would get. So, this is a two step reaction, where you have the migration, you have a migration and subsequently you have the reaction of, subsequently you have the reaction or the addition of a p p h 3 molecule on to the vacant site. You get the final product, but you notice that the migration has happened without affecting the trans relationship between the two. You can see that we have not affected the trans relationship here between the two carbons. So, this relationship is maintained before the reaction, it was in a transposition and after the reaction it is still in a transposition. So, this allows you to only measure the coupling constant and because the coupling constant you have is very different at 60 degrees. When you come close to 60 degrees, the coupling constant becomes close to 0 to 5 hertz and if you have a cis geometry, where you have 0 degrees or 180 degrees, the coupling constant is close to 9 to 12 hertz. In this particular instance, this was close to 11 hertz and so you can without measuring the stereochemistry using optical activity or any other method, X-ray method, you can just conclude that the stereochemistry has been retained in the migrating carbon. This brings out a very interesting possibility, that is the migrating group can jump from one center, that is from the ion center on to the carbon monoxide molecule, carbon monoxide group without losing the stereochemical identity. Complete retention was observed in this particular case. So, no resolution was necessary and white science performed this excellent experiment in order to show the stereochemistry of the migrating carbon. Let us now take a look at what is happening at the metal center. The metal center before it undergoes an insertion reaction, it has got two coordination sites occupied. I have indicated this by the before the reaction, you have this fragment here. There are other groups on the metal, but we would not concern ourselves. At the reaction site, we have an R group, which will in the neutral method it will donate one electron and a carbon monoxide and that will give you two electrons. Two coordination sites are occupied and metal, the formal oxidation state of the metal in this particular fragment is m is 0 in the neutral method. In the ionic method on the other hand, we have an R group, which is an R minus and that will give you two electrons. That is what we have pictured here, two electrons from carbon monoxide, because we have an R minus the metal has been oxidized to plus one oxidation state. So, for this fragment alone, one can see for this fragment before insertion, you have the electron counts as indicated here. Let us take a look at what happens after the insertion. You will notice that after the insertion, one coordination site is occupied and one coordination site is free and that is the site here, that site is in fact free. So, you have a free coordination site on the metal and as far as electron count goes, you have one electron from RCO. So, you have a co-ordinatively unsaturated system when you undergo insertion reaction. In the ionic method, if you had two electrons from R minus, initially you now have two electrons from RCO and the oxidation state is plus one. So, basically what you are doing is that you are generating vacant coordination site on the metal. You are generating a vacant coordination site on the metal, but you are not changing the oxidation state on the metal atom itself. So, this is fairly important. We have to remember that during an insertion reaction, whether you use the neutral method, if you use the neutral method, M remains as M 0. If you use the ionic method, M remains as M plus, whichever way you count, you are not changing the oxidation state of the metal. Whereas, the electron count on the metal undergoes two electron loss. So, if you started with an 18 electron complex, you will end up with a 16 electron complex. Similarly, one coordination site is vacant at the end of the reaction. So, let us take a look at what are the species which can undergo insertion reactions because of the explanation that we just gave regarding the migration, which keeps the stereochemistry of the migrating carbon. We will look at them as anionic ligands. So, there are two species which are undergoing a reaction. The anionic ligand migrates to the neutral ligand. In the case of H RCO, R is anionic ligand and it migrated to a carbon monoxide. That is a neutral ligand. Similarly, we can have various other combinations of anionic ligands migrating to neutral ligands. Here on my right, I have a group of ligands which are anionic ligands which can migrate on to neutral ligands. So, the neutral ligands that are most often studied are species like carbon monoxide, which we have just considered. Today, we will continue with some reactions where you can have migration of an anionic alkyl or a hydrogen on to olefin as well and acetylene as well. So, these are the anionic ligands and these are the neutral ligands which can undergo insertion reactions. In general, species like phosphine are not capable of undergoing insertion reactions. The reasons for these will be obvious as we proceed further. So, let us take a look at the insertion reactions further. Now, based on the discussion that we had, we can in fact think of this whole insertion reaction as if a nucleophilic attack has taken place on the neutral ligand. So, in the previous instance, we had carbon monoxide as a neutral ligand and an anionic group was attacking it. So, if you have an olefin, can we have an attack of a nucleophile? Isn't it true that the olefin is electron rich and it will prevent a nucleophilic attack? That is in general true, but if you remember after coordination to a metal, the electron density on the olefin has been reduced. Electron density has been moved from the olefin to the metal center and it undergoes typically photochemical type of reactions because you are populating the lumo, which is the lowest unoccupied pi star orbital. The lumo is in fact a pi star orbital and the homo is in fact a pi orbital. So, you move electron density from the pi to the pi star just as you would do when you photo excite an olefin. So, these reactions are cases where you are pumping in electron density into the lumo, where nucleophile attacks the lumo, pumps in electron density into the unoccupied orbital. So, turns out that when a olefin is coordinated to a metal center, you can still have attack of a nucleophile because it is the one, which is less populated with electron density and the homo has been depleted. So, it is ready to accept electrons as well. So, the migratory insertion can in fact be thought about as if it is a nucleophilic attack on a coordinated olefin. I have shown here a coordination of an ethylene to a metal hydride species. So, this is the anionic group that we are talking about. This is an anionic group. This is an anionic group and that is attacking an olefin, which is coordinated to the metal. When it does so, this hydrogen has migrated here. So, this CH 2 group becomes a CH 3 and a CH 2, this double bond is now instead of a coordinated bond, you now have a sigma bond formed between the metal and the CH 2 group. So, let us take a look at a simple example of a hydro zirconation reaction. Now, this reaction simply involves the insertion of a hydride on the zirconium and the addition of this hydrogen to this olefin, which is pictured here. A mixture of olefins is shown, but any one of them is sufficient. But let us take a look at what happens when you have a mixture of olefins. If you add the hydrogen and the zirconium across the nucleophile, if you add the hydrogen and the zirconium across the nucleophile, if you add the hydrogen and the metal across the olefin, what you end up with is addition of a hydrogen on the one side of the olefin and the metal on the other side of the olefin. So, we have added m h into the unsaturated olefin. Now, this is exactly what we are going to do here. We are going to add this hydrogen and we are going to add zirconium to either side of the double bond. So, what we have pictured here as a product is addition of zirconium to the terminal position, zirconium here and hydrogen here. So, this is where zirconium as added and hydrogen as added and the product would be this. Notice that even if you use a mixture of olefins, there was only one product isolated at the end of the reaction. So, this was a puzzle which people solved and it was understood that it was a sequence of addition of hydrogen and abstraction of hydrogen, which means it is a insertion of hydrogen and it is a de insertion or rather abstraction of hydrogen from the olefin, from the alkyl group. So, let us take a look at what we expect, what we would have expected from a simple hydro zirconation reaction. That means addition of hydrogen and zirconium on to the three olefins. Now, I have color coded the hydrogen on the zirconium so that we can see where it is ending up at the end of the reaction. So, let us assume that the pink hydrogen is coming from the zirconium. If you added to the three olefins, you would end up with three different products as pictured here. So, notice that the addition reaction is performed in such a way that zirconium is adding to the least hindered position on the olefin. This is understandable because zirconium seems to be bearing three large groups, two cyclopentadienyl units, two cyclopentadienyl units which are pictured here and one chlorine. So, all of them have to be attached to the methyl group. So, invariably an anti-Marconikov's addition happens and the product is as shown here from the one substituted olefin. You would end up with the product here which is the product which we had shown earlier. If you take the two substituted product, you would end up with the product shown here and with the three substituted olefin, you would end up with this hydro zirconation product. But surprisingly at the end of the reaction what they found was that there was only one hydro zirconation product. So, that is what we are trying to explain. This was expected and due to a series of insertion and abstraction reactions in this leads to isomerization. So, let us take a look at what happens. Let us mark the hydrogens which are undergoing reactions with different colors. So, that we will be able to follow where the abstractions taking place. So, in the case of the hydro zirconation which has happened in such a way that the middle carbon is now bearing the zirconium. You can easily see that the zirconium is quite crowded and so it abstracts a hydrogen and it keeps shuttling between these two sides. So, what is happening is that this hydrogen is abstracted and an olefin bond is forming in this case. So, what you end up with is a hydrogen abstraction or beta hydrogen abstraction reaction which is very simply reverse of the migratory insertion. So, if this hydrogen goes back here you would have ended up with this product. But instead of going back into this position what happens is the hydrogen goes into this position which we will mark in a different color. Let us mark in a blue color. So, the preferred attack of hydrogen now is in this position. So, what you end up with is if hydrogen attacks the internal part then you end up with the zirconium migrating from the carbon which we can label as 1, 2, 3 and 4. It is migrated to this position here. From the 4 position we have added the hydrogen to the 5 position here. So, let us take a look at what will happen here. If you keep doing this migration you can see very easily that the zirconium can migrate from one carbon center to the other carbon center. So, in the second step we can do the same thing again once more. But now we will have a terminal olefin generated as a result of hydrogen abstraction. This time the hydrogen that is abstracted is from the carbon atom number 1 and if you do that then you end up with a hydridor zirconium species which can now transfer the hydrogen to the second place. That is the second position in the carbon. So, you end up with the compound which was the only product which was isolated at the end of this reaction. So, in other words what we are doing is that we are allowing the metal to walk along the alkyl chain by a series of hydrogen insertion and hydrogen abstraction reactions. So, you can in fact use a mixture of olefins, but all of them will walk around in such a way that only this one product the preferred product is this. This is the preferred product and that is the only product that is formed at the end of the reaction. So, this end result is a formation of a single product whichever product is taken whichever hydro zirconium happens the walking of the metal along the alkyl chain results in the formation of a single product. So, this gives us a very clear example of how the metal can add an abstract hydrogen from abstract a hydrogen from an alkyl group add a hydrogen to a olefin. So, lead to isomerization it is very common to observe this kind of an insertion and abstraction reaction with H COCO 4. So, a similar situation was encountered when deuterium in the presence of deuterium it was found it was found that H COCO 4 will in fact isomerize a series of olefins even if you take a terminal olefin you end up with a terminal olefin, but now you have deuterium in all the positions at this olefin. So, in other words the amount of olefin that you that you add an abstract keeps increasing as time progresses. If you have deuterium gas you can in fact populate all of these sides with deuterium because you are adding an abstracting hydrogen with H COCO 4. So, this can be a useful exercise that you can attempt in the presence of deuterium how you can add an abstract hydrogen atom from a olefin. So, this is also the insertion reaction is also the basis for the famous Ziegler Nutter process. In fact, it is possible to make large amount of polyolefins using simple olefins and this is a basis for the generation of a large amount of easily usable poly plastics. It is estimated that more than 55 percent of products plastic products are generated using this particular insertion process. And it is very interesting that polyethylene you can take ethylene and polymerize it using this particular process which involves a simple insertion reaction and as much as 100 tons of polyethylene are generated using just one gram of the catalyst. So, that means the insertion is happening very efficiently and the insertion time for this particular process has been estimated to be 10 power minus 5 seconds. And that is close to how fast enzymes carry out reactions and so this has been labeled as enzymatic activity of metal complexes for olefin polymerization. So, olefin polymerization also involves the simple insertion reaction and it has been extensively studied. The basic process just involves the migration of an alkyl group into the olefin. Let us take a look at the two steps, the key steps that are involved without concerning ourselves with what else is present on the metal. Very often in the industrial setup and industrially relevant complexes a large amount of methyl alumoxane is used and it is not clear what is the group that is present on the metal. So, let us just mark it as M. There are two groups, one is an alkyl group and the other is an olefin. The group that migrates is an alkyl group and the alkyl group migrates to the olefin in such a way that it generates a new alkyl group. And this is marked here for simplicity we are starting with a methyl group, but it could be a butyl or any group that can be added. Because we have color coded, we can now see that the entering carbon had a double bond and now you have all single bonds and the vacant site that was generated on the metal. So, that metal is now capped with an olefin. So, the initial olefin is shown in black and the second olefin is shown in blue. Now, what happens is that you have a migration of this alkyl group on to this olefin. Remember the anionic group is the one which migrates on to the neutral moiety. And if it does that then once again you have the formation of a metal carbon covalent bond and you have the formation of a longer alkyl chain. Now, this process is repeated several times. You have now the coordination of an olefin in this position and subsequent migration, you will have the formation of a very large alkyl chain. And this is the process that is essentially at the back of the Zeig-Lenata process. So, earlier we had discussed the reaction with hydrozirconation and zirconium and titanium are probably some of the most common elements which are used for the Zeig-Lenata polymerization reaction. Let us take a small look at. So, here I have the olefin coming in and then it inserts and the second olefin which is pictured in red again and then again it inserts and gives you the polyolefin. So, we can see that each insertion leads to lengthening of the chain and the second group is added sequentially. So, apart from olefins one can also introduce acetylene and the last part of this lecture, we will look at how acetylene can undergo insertion reactions. So, I have given here a simple reaction where an acetylene molecule is polymerized in the presence of palladium chloride. Surprisingly, this involves an insertion of a chlorine and surprisingly it involves addition of a pdCl bond across the acetylene molecule. Although the final product is a completely organic moiety in many instances, it could be the simple formation of hexamethylbenzene if you use dimethylbenzene. So, if dimethylbenzene, dimethylacetylene is a starting material then hexamethylbenzene is a product. However, when you use a bulky group as a phenyl group on the acetylene. So, if you use diphenylacetylene then the product is different. It is usually the cyclopyridine complex of palladium chloride. So, this particular reaction was studied extensively and it was shown by experimentation that in fact it is going through a sequential addition reaction. Let us take a look at this reaction and see how the addition reaction is capable of carrying out the cyclization. So, first step we have the coordination of palladium chloride to acetylene. What you have pictured here is a complex which is a dimer and very often organometallic chemists or coordination chemists will only draw half of the molecule. So, half of the molecule is drawn and 2 is indicated to show that it is in fact part of the molecule. The other half is a mirror image of what is pictured here. So, this dimer could be undergoing the reaction in a sequential addition fashion addition of another molecule of acetylene. So, from this point onwards we will draw only part of the molecule. So, let us take a look at this species. You have a chlorine which is undergoing insertion into the C triple bond C and a carbon palladium bond is formed as a result. A carbon palladium bond is formed and this is a product that you have. Now this product which could be a dimer as I told you earlier could now react with another molecule of an acetylene. Notice that what you have ended up creating is another anionic group, a vinyl group attached to the palladium. Now if an acetylene is coordinated to it, what you would end up with is a neutral molecule coordinated to the palladium and this vinyl group migrating on to the R group. So, the product would be once again it would be a group where you have this chlorine which originally came from palladium is now sitting at the end of this vinyl moiety. The vinyl moiety is now attached to another alkyne which has been inserted between the carbon and the palladium group. So, this carbon palladium bond is broken and another carbon bond is formed as a result of which you end up with a molecule where there are two vinyl moieties stitched together. Now notice that in this molecule you have this R group which is sitting here and it is quite close to the palladium and if this R group is very large it would impose some steric interaction and as a result it rotates around this carbon-carbon single bond. So, rotation of this carbon-carbon single bond leads to the formation of this molecule, this molecule which I have pictured here and here the chlorine which was originally on the palladium is brought close to the pdcl bond again. So, these two groups which were originally together are found close together. Now if you can do an abstraction reaction or a de insertion reaction such a way that you move the chlorine back from the carbon-chlorine bond if you can move it back on to the palladium. So, let us change the color of the ink here. So, if I move this chlorine back on to the palladium then what I would end up with is a pdcl2 moiety which will again be a dimer. So, I have indicated that as a dimer, but then the carbon-carbon there is a carbon-carbon bond being formed between these two centers which I am indicating with two dots. So, these two centers will form a carbon-carbon bond and a palladium chloride is extruded from this reaction. So, an abstraction of the anionic cl minus back to the pdcl results in a de in an abstraction reaction which gives you a cyclobutadiene now coordinated to a palladium. So, notice how this reaction can either stop here because you have a large group this steric repulsion resulted in turning around of this moiety in such a fashion the vinyl group in such a fashion that the two chlorines were brought together. Now the alternative is especially when the R group is small as in the case of R equals methyl R equals methyl then the reaction proceeds further and undergoes a second insertion third insertion. So, let us take a look at this is the case where the R group is only a methyl group. So, I have indicated it in this fashion in this slide first the moiety that we start out with is the species which we said which we showed in the previous transparency. This is the species that we showed if R group is small R is small then we can end up with a second insertion a second insertion of this vinyl moiety. This is the vinyl group and this vinyl group can migrate here and the carbon palladium bond can be formed. So, what you have here is three alkyne stitched together the three alkyne stitching can happen only if you have a reasonably small alkyne group otherwise the reaction will not proceed in this direction it will not proceed in this direction, but it will in fact proceed towards the cyclization of the initial group. So, if you have three alkyne groups alkyne moiety stitched together you now have once again you have the rotation of the carbon carbon single bond here you have the carbon carbon single bond rotating. So, you end up with palladium and chlorine close together an extrusion reaction or an abstraction reaction follows these two groups are reacted with each other in such a fashion that the chlorine migrates back to the palladium it is exactly the reverse of the migratory insertion. So, the chlorine migrates to the palladium and you have the formation of a hexamethyl benzene moiety and this hexamethyl benzene moiety is more has got more number of electrons and what the palladium would like. So, the molecule falls off the coordination sphere and it prefers to coordinate to another molecule of the acetylene and so you end up with after two insertions you end up with this moiety which I have colored in green color and that is your key intermediate which keeps in inserting one more molecule of an acetylene to give you a catalytic cycle. So, a simple insertion reaction the simple insertion reaction that we have pictured here is capable of generating two different molecules one is a cycloputrodiene if the R group is large and the other is an aromatic ring benzene moiety if the R group is small surprisingly the previous reaction was carried out with palladium the same reaction is carried out with nickel and there are two nickel species that I have pictured here. The use of nickel a cac results in the formation of cyclo octetetreine what that means is that four molecules of acetylene has just together in order to make this acetylene tetramerize to give you cyclo octetetreine. On the other hand surprisingly if you use nickel zero as a catalyst you also end up with a benzene moiety. So, this is an interesting variation in fact this reaction was not understood for a long time and the formation of cyclo octetetreine was solved only in nineteen was solved only in nineteen eighties although the reaction was known for a very long time for more than a hundred years and still the reaction was studied only recently. What they showed was through some isotopic labelling experiments they showed that it was only the zipper reaction the zipper reaction which was happening and four acetylene molecules are stitched together sequentially if they showed through the isotopic labelling that cyclobutadiene is not involved as an intermediate nor is an aromatic ring involved as an intermediate. So, these two are not happening and it is a zipper mechanism which is happening as the possible is best possible mechanism to explain the formation of cyclo octetetreine. So, I will conclude by saying that the migratory insertion is an extremely important reaction, but one must now ask the question is it really necessary to call it a migratory insertion. It seems to be a simple nucleophilic attack of an anionic group attached to the metal atom on to a coordinated ligand the coordinated ligand as I mentioned earlier could be any neutral molecule like carbon monoxide or it could be an acetylene molecule and the anionic group of course could be a hydride or an alkyl group or it could be even some other groups which are quite easily obtained in an organometallic molecules usually the migration happens only in the cis position. So, the cis geometry is universally accepted as a place where the migration would take place or the insertion would take place and the anionic ligand is the one which migrates on to the neutral species. This is again something that has been shown repeatedly in several instances and finally, we should remember that the stereochemistry of these molecules are maintained especially when the migrating carbon is stereochemically labeled then one can see that it is my retention of stereochemistry rather than inversion of stereochemistry.