 In this lecture, we will look at some reaction mechanisms in organometallic chemistry. So, far we have looked only at substitution reactions, which are simple replacement of one ligand by another ligand. Today, we will look at how some more complicated reactions take place in organometallic chemistry. Actually, organometallic chemistry has become an essential part of organic synthesis. I show this set of equations very often, because it illustrates how in a total synthesis at several points in the synthetic methodology, the researchers have used repeatedly organometallic and organic reactions interchangeably. It is indeed important for us to realize that organometallic chemistry cannot be divorced from organic chemistry or organic synthesis anymore. If you look at organic synthesis, it is a well developed field and the classification of organic reactions has been done long ago. Very often, it is easy to substitute one reagent with another reagent or one reaction by another reaction in organic chemistry. It would be useful if you could have a similar classification of organic reactions or organometallic reactions. Today, we will look at a plausible way of classifying organometallic reactions. First, let us just look at how organic reactions are classified. This is a broad, this is one way in which we can classify organic reactions. This is found in many textbooks. We have first of all substitution reactions where you can substitute one group in an organic molecule either by a nucleophilic substitution if it is an alkyl moiety, which where the substitution is happening. In which case also the reaction might happen either as an SN 1 reaction, either as an SN 1 reaction where the substitution is nucleophilic and it goes in a first order fashion or in a second order fashion and then we call it an SN 2 reaction. These are mostly nucleophilic reactions when it happens in an aliphatic chain, but if it happens in an aromatic chain, very often it is an electrophilic mechanism. Substitution reactions can either be a nucleophilic reaction or an electrophilic reaction and these are grouped as one class of reactions and then come addition reactions where an unsaturated molecule is converted to a saturated molecule by addition of a group. It could very often be in the case as in the case of halogenation, two x groups are added across a double bond. So, halogenation, hydro halogenation where H and a C L are added or an H and a B R are added and then they are called hydro halogenation. In all these groups of all these reactions we add a molecule which is split in two fashion in a heterolithic fashion very often and added to the unsaturated molecule so that it becomes a saturated molecule. In a few instances we do have addition of a neutral group like a carbon dioxide or a sulphur dioxide. These groups might be added to an unsaturated molecule in such a fashion that the whole molecule becomes saturated or more saturated. These reactions also proceed in electrophilic, nucleophilic or a radical fashion that depends on how the reaction is carried out and one can in fact tune the reaction so that the reaction proceeds in a particular fashion. Now, the third group of reactions are elimination reactions which are essentially a reverse of the previous group of reactions. So, instead of going from an unsaturated molecule to a saturated molecule in this case one is going from an saturated molecule or a more saturated molecule to a less saturated unsaturated molecule. When you extrude a neutral molecule in these instances you could extrude a carbon dioxide or sulphur dioxide and then you end up with a reaction which is the reverse of the insertion of a neutral group. The fourth group I would call it as organic redox reactions where you have addition of electrons. These are very rare but addition of hydrogen is also called addition of electrons because you are adding a group which is less electronegative than carbon to the unsaturated molecule. So, addition of hydrogen or addition of electrons is classified as an organic reduction reaction. Similarly, if you oxidize a species an organic molecule you could do it either by removal of electrons or removal of a more electronegative group. If you add a chlorine or an oxygen then it becomes an oxidation reaction of an organic moiety. So, in the final group we have what is called the rearrangement reactions. The rearrangement reactions could either involve purely sigma bonds or purely pi bonds or a combination of both. Many of these reactions are called pericyclic reactions. They are mostly orbital controlled and this is the basis or this is the topic of the famous rule the Woodward Hoffman rules which govern these reactions. Now, rearrangement reactions need not always be pericyclic. You can also have rearrangement of complicated skeletons like nobonyne and adamantine skeletons. These large ring systems can also rearrange depending on the ring size requirements and the strain that is involved. One can have rearrangement of a molecule where only sigma bonds are involved. So, either sigma bonds or sigma and pi bonds or pi bonds can be involved in this rearrangement reactions. Now, the reason for going through this classification of organic reactions is that I just want to draw parallel to the organometallic reactions and show how these can also be classified. Let us ask this question. Is there a way in which we can classify or is there a way in which we can use a functional group approach in which we can substitute one reagent by another reagent. Now, in organic chemistry if you want to convert one functional group to another functional group, we can have a group of reagents which will do this. So, if I want to convert for example, here an unsaturated molecule I have shown here an unsaturated molecule which is unsaturated in two places. One is a carbon-carbon double bond, another is a carbon carbonyl group. I only want to reduce one of them or I want to reduce both of them. In this case, I have reduced both of them. I can use a particular reagent in the same way. We can take another more complicated molecule and ask the question, can I carry out the same reaction? Can I carry out one of these reductions or can I carry out both reductions simultaneously which reagent would I use? If one has the knowledge of converting this molecule, this unsaturated molecule to the second unsaturated molecule that we have shown here, a similar transformation can be carried out with this more complicated molecule using the same reagent. This type of transferability gives organic chemistry an excellent way of extending the chemistry and also helping the student understand many complicated reactions. So, it is important that we also have some kind of transferability in organometallic reactions also. Is there a way to do this? Let us take a look at a few examples now. Here are two reagents which you might be familiar with. One is a rhodium complex which is the Wilkinson's catalyst and another is a cobalt analogue. Here are two other examples where we have the same cobalt complex, but now I have substituted the two ligands. The answer to the second question may be yes. If you substitute the ligand on the metal complex, one can very often transfer the reaction capabilities. So, the reaction that is carried out by this particular complex might be carried out by this complex. It might be possible, but on the other hand the answer to the first question is actually no. The reaction that is carried out by rhodium is not carried out by the cobalt complex. This turns out to be a reasonably well established fact because the chemistry of the 3D and the 4D are very different and the 4D and 5D are reasonably similar. So, very often transferability might be possible between 4D elements and a 4D complex and a 5D complex, but it is very difficult to transfer the reaction which is happening on a 4D element to a 3D element. So, let me give this example where we have multiple insertion reactions. I want to talk about this insertion reactions in more detail and I have chosen this example where we can use acetylene and convert them to arenes. That means you are taking 3 C triple bond C molecules, 3 acetylene molecules and converting them to a 3 acetylene molecules. So, you can see that we have stitched together 3 acetylenes. Now, metal complexes can be chosen in such a way that one can get exclusively arenes or one can get only cyclobutadiene molecules or even poly acetylenes. The best catalyst for carrying out this reaction so that you can get mostly the areane. So, you want to synthesize only aromatic molecules from alkynes and the best molecule for this particular reaction turns out to be the cobalt complex. So, one can ask this question is the same analog capable of carrying out the synthesis of areanes with cobalt is replaced by rhodium. If the cobalt complex carries out the synthesis of areanes can we also use it for can we also use the rhodium complex for carrying out the same reaction. It turns out that for synthesizing areanes it is more easy to use a palladium complex and not the rhodium or the aridium analog. So, in some ways it is very difficult to have transferability in organometallic chemistry. We will classify the reactions on the basis of transformations organic transformations they carry out and we will show this in a few slides later. Now, it is important for us to when we classify these reactions it is important for us to understand the fundamental of the key steps involved in these reactions. When we change an acetylene when we stitched together acetylene to form an areane is it going through the cyclobutadiene or is it directly converting from an acetylene to an areane. So, are we going directly from the acetylene to the areane or are we going through an intermediate. How does one figure out which of these steps are really happening are more importantly we have to ask the question which steps are not involved. This will help us to design better catalyst it will help us to understand these reactions better and probably even promote the same type of transferability that we have organic chemistry to organometallic chemistry. So, I have classified the organometallic reactions into four broad classes and probably we will have to extend it to five, but we can for this initial lecture use only four broad classification of reactions. First of all I would like to talk about rearrangements or isomerizations. These are situations where the ligand rearranges from one position on the metal to another position. So, you can have trans-isomerizations and so on. These are rearrangement reactions and these can probably happen in both main group and organometallic reactions. Then come ligand insertion reactions. In these cases if you have a ligand that is pictured here as X it can be a neutral molecule. If it can add between the metal and the ligand then one can have such a transformation that is you can have the insertion of this X group in between the metal ligand bond. So, it looks as if this metal ligand bond is broken and you have insertion of the X group. So, we have the same group here X group which is present in this particular place. So, M X is left unchanged and M X is there and you have this transformation of M L to an M X L. Now, these reactions are very often possible in both main group chemistry, main group chemistry and transition metal chemistry. So, I would write it as main groups and transition metal chemistry both main group and transition metals are capable of carrying out these reactions. Now, there is another group of reactions here where organum transition metal reactions exclusively. These are situations where you have oxidative addition and reductive elimination. I have pictured here the oxidative addition where the metal undergoes an oxidation state change. This change can be anywhere between one oxidation state change or two oxidation state change and we will look at these situations as we go along. The second set is a reactivity change on the ligand. This is again a reaction which is exclusively or mostly in the domain of transition metal compounds. So, transition metal organometallic compounds have this capability of carrying out oxidative addition and reductive elimination because transition metals have got the capability of changing the oxidation state. Main group elements are very often species which will not change the oxidation state. So, if you look at all the reactions together organotransition metal chemistry comprises of these four major groups. In these four major groups, we have the last two which is oxidative addition and reductive elimination and reactivity changes on the ligand mostly restricted to organotransition metal chemistry. Now, today we are going to look primarily at ligand insertion reactions. These are reactions where the X group that is added on is added on to a ML group in such a fashion that we have a MXL complex being formed. It turns out that the X group can come from outside as we have pictured here. It can be added externally or it can be added internally. So, if it is added internally, you have the formation of MXL. If it is added externally, then we can have this X group intact. So, we will look at these reactions in fairly great detail today as we go through this lecture. Let us take a look now at the main group insertion reaction. If you have a simple Grinard reagent where you have RMGX, you have a simple Grinard reagent. It is highly polarized as R minus and M G plus. This is the polarization of R minus and the Grinard reagent. So, if a molecule like carbon dioxide is allowed to react with it, carbon dioxide is also polarized. The polarization is quite obvious. This is a delta minus and this is a delta plus. Because of this polarization, a very easy addition of the R minus group very easy addition of the R minus group occurs on the carbon. So, you have the formation of an R COO M GX. The magnesium, because it is positively charged or positively polarized, then it adds on to the oxygen group. So, we have an R COO M GX being formed. Now, this is a very favorable reaction and this happens very readily. On the other hand, let us take a look at an insertion reaction that is common in organometallic chemistry. That is the insertion of carbon monoxide. Carbon monoxide has got a very small dipole moment and carbon surprisingly is the negative end. So, even though it is polarized, but because the dipole moment is very small, if you mix methanol and carbon monoxide, even at very high pressures of carbon monoxide or very high temperatures, this reaction is unlikely to happen. You would not have the formation of acetic acid starting with methanol and carbon monoxide. On the other hand, if we add a small amount of this catalyst, which is an organometallic species, this organometallic species has the capability of transforming methanol to acetic acid in the presence of hydro adic acid at 80 degrees. Just a mild heating of 80 degrees and 3 atmospheres pressure of carbon monoxide is sufficient to bring about this remarkable transformation, where we have added a neutral carbon monoxide molecule. We have added a neutral carbon monoxide molecule to methanol to form acetic acid. So, this is almost looks like magic as far as the transformation is concerned. We have added a molecule of carbon monoxide between the methyl and the carbon. The C H 3 O bond has been broken and a carbon monoxide molecule has been inserted. Now, this is the basis for the famous Monsanto's acetic acid process. More than a million tons of acetic acid are made annually every year. You make more than a million tons of acetic acid using this particular process. This is very surprising considering the fact that hydro adic acid is a very corrosive molecule. You have to pump hydro adic acid into the presence of carbon monoxide, which is a toxic gas of course. But you add hydro adic acid and carbon monoxide. You pump it in using heavy duty pumps, which will get easily corroded because of this hydro adic acid. Still, this process turns out to be a viable process and it turns out to be a profitable process. They have to change the pumps often, but in spite of that this is the best way of making acetic acid. That is why we have a million tons of acetic acid produced using this method. Here is another method where 7 million tons of aldehydes are generated. This is the conversion of propene to butaryldehyde. What we have added in this reaction is a molecule of hydrogen, a molecule of hydrogen and a molecule of carbon monoxide. We have added these two species to a molecule of propene. Here once again we see that all the three species that are involved in this reaction are almost neutral or hardly polarized. They do not have strong polarization of the bonds. In spite of that you have managed to insert carbon monoxide, which will indicate in different colors so that it is easy for us to follow. So, here is the carbon monoxide and we have inserted it here. This is the carbon monoxide which has been inserted. Here is the hydrogen. We will indicate the hydrogen using a dark red color. So, this is the hydrogen which has been added. This hydrogen one of them has been added here and the other hydrogen has in fact ended up in this position. So, this is the original the CH2. That we had here the CH2 that we had here has remained intact in this particular position. So, the two hydrogens are in this position and they are unchanged. So, you can see that you have transformation of one unsaturated molecule to another molecule which has added hydrogen as well as added a carbon monoxide molecule. The essential catalyst or the catalyst that is involved in this process is this simple carbonyl species, this hydro carbonyl species. We have not talked about hydro carbonyl species, but I would like to give this example so that you will see the value of such a transformation. So, this insertion reaction is involved in several of the industrial processes and here I have pictured for you another transformation which is extremely useful which has been carried out using cobalt and rhodium catalyst. You transform allyl alcohol. This is allyl alcohol right here. Allyl alcohol is converted once again in addition of a hydrogen and a carbon monoxide has been carried out and we can see that it is essentially the same reaction that we talked about in the previous case. Here also we have added a hydrogen and carbon monoxide and that is exactly what we have done here, but in the industrial process the product that is formed the aldehyde is converted to an alcohol by a reduction reaction using hydrogen and rennickel which is a heterogeneous process. But in the course of this transformation allyl alcohol is converted to 1 4 butanol and this is an extremely useful molecule because it can be converted to tetrahedrofuran which is T H F a very popular and a common solvent and a very useful molecule. So, this is a simple way of converting allyl alcohol using hydrogen carbon monoxide and elimination of a water molecule after hydrogenation to produce T H F. This is again carried out in very large quantities because 1 4 butanol is an extremely useful molecule and is an intermediate in the formation of several organic molecules as well other than T H F. So, here is one last example where an insertion reaction is involved in this particular case we add a simple addition of carbon monoxide this involves methyl acetate. So, these are methyl groups. So, you have a methyl group here and you have a methyl group here this is methyl acetate and you convert it into acetic and hydrate. This again sounds like magic you have just added a molecule of carbon monoxide. We will indicate this by showing the molecule of carbon monoxide that is coming in and here is the molecule of carbon monoxide that you have added. So, it is the same molecule of methyl acetate except that you have one CO added in between and this is the best way of making acetic and hydrate starting with rhodium and no organic catalyst will actually bring about this transformation. The chemistry is completely unique to organotransition metal chemistry. So, it is very important for us to understand such insertion reactions where there is hardly any polarity change that is bringing about the transformation. It is primarily the ability of the metal which is able to bring about the insertion of a neutral organic moiety. So, main group insertion reactions the polarity is very important both fragments have to be charged or polarized charged or polarized and this is very important. Whereas, in transition metal chemistry very often you have a neutral ligand that is undergoing insertion reaction. So, neutral species can be added using organotransition metal chemistry whereas, polar compounds can be added using main group organometallic chemistry. So, let us take a look at what are the steps that are involved in this insertion process. Because this is a well studied reaction and many examples are known and we will take a look at one example that has been extremely useful because of some kinetic reasons which I will explain as we go along. If we take Mn2C010 or this dinuclear manganese carbonyl complex and if you reduce it with sodium and mercury if you reduce it with sodium and mercury and then you end up with a sodium salt of manganese pentacarbonylate anion MnCO5 minus and this sodium salt can undergo further reaction with a methyl iodide and this will give us a CH3 MnCO5 species. So, basically it is as if you have a Mn minus species and a sodium plus species and the Mn minus species as made a nucleophilic attack on the methyl that is exactly what is happening and the iodine goes away as I minus and as a result we will have in the net reaction we will have sodium iodide being left out and the methyl compound of MnCO5 being formed. Now, it turns out that if you add a molecule of carbon monoxide, if you add a molecule of carbon monoxide then you form an acetyl complex of MnCO5. So, this is a classic example where you have this is a classic example where you have added a neutral carbon monoxide molecule and carried out an insertion reaction. Now, this reaction which was studied by Calderazzo in great detail turns out that the carbon monoxide that you add and the carbon monoxide that you have in the molecule cannot be distinguished. So, he wondered whether it is indeed an external carbon monoxide which is inserted itself between the methyl group and the manganese. So, this is one possible scenario. The other possible scenario is that one of these carbon monoxide groups has in fact, one of these carbon monoxide groups has in fact, been utilized and it is this carbon monoxide. It is this carbon monoxide which has now transferred to this second position and the entering carbon monoxide is occupying a vacant site on the molecule on the manganese. So, there are two possibilities and the he realized that if he used labeled carbon monoxide which means you cannot color the carbon monoxides in real life using a color code, but you can add an isotope. You can use an isotope which is different from the naturally occurring isotope. So, usually you have 12 C most of the carbon monoxides will be isotopically will be having the isotope carbon 12 and instead if you have isotopically enriched carbon monoxide, then we can use it as if it is color coded and then we can look at where the isotopically labeled carbon monoxide has ended up. This is exactly what he did. So, if you add carbon monoxide and here he is added labeled carbon monoxide and that is what we mentioned in the previous slide. So, we have 13 C labeled carbon monoxide coming in and the expected product if it was a direct insertion of carbon monoxide, we will have the 13 C labeled in this position. So, this would be labeled with carbon 13 and you can easily monitor this using carbon 13 NMR one can distinguish the position at which the carbon monoxide has come in. The other option is that you initially have a conversion of the existing carbon monoxide molecule on the manganese to an acetyl group. So, this could have happened in two ways. We could have had the migration of a methyl group. You could have had the migration of the methyl group to a carbon monoxide or you could have had the carbon monoxide in the molecule going in between the methyl and the manganese. So, in both instances you would have got an acetyl manganese complex which has now a vacant coordination site. So, this coordination site has now become vacant and the carbon monoxide which is labeled would come in and occupy this position. So, this is the other alternative and this is the product if the entry carbon monoxide is merely filling in a vacancy on the manganese. So, when the reaction was carried out what he found was the direct insertion product which is pictured here below that is not happening. It is not the direct insertion product and what he had was the formation of carbon monoxide in this position where it was not found as acetyl. So, entering carbon monoxide was merely filling in a vacant coordination site on the manganese. Now, he used infrared spectroscopy extensively to figure out where the carbon monoxide was placed in the final molecule. So, one other interesting aspect of this molecule is that if indeed the carbon monoxide was entering a vacant coordination site if the reaction was under equilibrium then you would have lost the positional identity of this molecule of this entering carbon monoxide. If the carbon monoxide was going out and coming in then the position would have been lost because once you form this molecule you will notice that all four carbon monoxides are identical with respect to this acetyl group. So, any one of them if the reverse reaction was possible any one of them could have been lost and the entering carbon monoxide would have mixed with the available carbon monoxide on the molecule. So, the positional identity would have been lost in, but on the other hand the positional identity was not lost and this reaction was under what is called kinetic control. Once the carbon monoxide was inserted the product remained as an acetyl moiety and no reverse reaction was happening. So, two things were discovered in the process in this course of this study. One is the fact that carbon monoxide that is coming in is not inserting itself in between the manganese and the methyl group. Secondly, the fact is that it is under kinetic control and the reaction was not going in was not under equilibrium with a non inserted product. So, the reaction was not going back and forth. So, this is a reaction which was not happening that is it was not a direct insertion. Secondly, the reverse reaction was also not happening in this reaction. So, let us take a look now how he confirmed whatever he had studied in this first case. What he did was a very clever experiment between the sodium manganese salt that I talked to you about earlier. He treated that with labeled acetyl chloride, labeled acetyl chloride is what is pictured here. So, you take acetyl chloride which has got the carbon 13 label at the acetyl group. Now, if you treated with the manganese salt you end up with a product that you had prepared by insertion of carbon monoxide into the methyl complex. So, once this labeled complex was made he was able to study the reverse reaction. Now, I told you that under the conditions of adding carbon monoxide you did not have the reverse reaction, but if you heat it if you supply sufficient heat it was possible to convert the labeled acetyl group into the methyl complex by extrusion of carbon monoxide. Now, it turns out that this is the exact reverse of the reaction and usually in chemistry we know that there is a principle of microscopic reversibility, which means that the forward reaction is mirrored in the backward reaction. Backward reaction just retraces the way by which the complex or the product was formed. So, when he heated this molecule of acetyl group and then you know that the carbon monoxide on the metal should one of them should have left. And here in this particular instance you have carbon monoxide which is trans to the acetyl group this could have left and then you could have had movement of the carbon monoxide which is in this acetyl group. You could have moved this carbon group back on to the metal. So, it could have been one of these carbon monoxide it could have been either one which is cis to the acetyl group which I put in which I am marking as a square or it could be one which I have marked as a circle. And exclusively the product that was formed turned out to be the one where you had loss of a carbon monoxide which is on the cis position to the acetyl group. So, one of them which I have labeled as a square is one of the carbon monoxide should have left and the metal group must have migrated to the manganese. So, you have migration of let us just write this out. So, you have an intermediate where you have a vacant coordination site. So, you have this acetyl group and then you have a vacant site generated because one carbon monoxide has left the coordination sphere. And now the metal group migrates to that vacant coordination site. So, because we had this carbon monoxide which is labeled this carbon monoxide was labeled. So, we have the label in that position the label is still in that position. And the metal group the metal group has in fact migrated to this vacant site and that is what we have here. So, we have reaction which suggests that the reacting carbon monoxide acetyl group and the vacant site should be in cis position with respect to each other. In other words the CO and the metal group should be cis to each other and that is exactly what we have here they are cis related. So, that is the only way in which you can have a migration. So, this cis migration was proved by Kandler also using this isotopically labeled substrate. Now, the third reaction that Kandler also did was to check if you can use this product where which was formed in the first reaction. In the first reaction he added labeled carbon monoxide to a methyl manganese complex and he heated this to product which was formed in the first reaction. He heated this product and he got both trans and cis labeled molecules. How is that possible? Because you have 4 carbon monoxides in the cis position. Now, to the acetyl group you have 4 carbon monoxides which are equivalent. One of them of course is labeled because that is the carbon monoxide which came in from the external addition of carbon monoxide. Now, if you heat it, it is possible that this 13 C labeled carbon monoxide leaves. If that labeled carbon monoxide leaves then you would have migration of the methyl group to that position and then you would get an unlabeled product. You would get an unlabeled product if you had loss of this carbon monoxide when you heat it. If you have loss of one of these carbon monoxides which is cis to the labeled carbon monoxide then you would end up with this product. So, I am labeling this as a cis product because there are 2 of them. 2 of the carbon monoxides are there. It is possible that one of them leaves and both of them are cis to the acetyl group. Any one of them can leave and then you would get a cis product where the methyl group is cis to the methyl group is cis to the labeled isomer labeled carbon monoxide. Now, it is also possible that this carbon monoxide which we will mark with a different color. So, it is also possible that this carbon monoxide leaves. If this carbon monoxide leaves then you would get a trans isomer and labeling this as trans with respect to the I am identifying this as a trans isomer because the labeled carbon monoxide and the methyl group are trans related. Similarly, the methyl group and the 13 C are cis related here. So, depending on which carbon monoxide you lose in this reaction depending on which carbon monoxide you lose you will end up with one of these 3 isomers. If the labeled carbon monoxide leaves you get the unlabeled isomer. The probability with which these carbon monoxides leave are about one fourth in the case of the unlabeled isomer because there are 4 carbon monoxides and each one of them can equal have a equal probability of leaving when you heat this molecule. There is only one carbon monoxide which is trans to the labeled carbon monoxide. So, this will be formed in the same ratio as the ratio in which the unlabeled molecule is formed. However, you have 2 carbon monoxides which are cis related to the labeled carbon monoxide and they are also cis related to the astell group. Remember the carbon monoxides which have to leave in order for the reverse reaction to take place are the ones which are cis to the astell group. The astell group is present here and there are 2 cis related carbon monoxides which are also cis related to the labeled carbon monoxide. So, these 2 have twice the probability of leaving the molecule. So, you will end up with 2 equivalents of the cis product with respect to the trans product and with respect to the unlabeled product. This is exactly what he found. He found that there was no formation of the there was no direct elimination possible. It was only these molecules which were formed and they were formed in 1 is to 2 is to 1 ratio. Now, if this is the ratio in which the 3 products are formed, it is very clear indication that the cis elimination pathway or the cis insertion pathway is the one that is being followed by this molecule. Methyl migrates to a vacant site and to a vacant cis site. If carbon monoxide had migrated inside and it was not a methyl migration, we would have obtained very different result. So, let us just take a look at that. If carbon monoxide migrates, what would happen? In the previous instance, we assume that the carbon monoxide leaves the coordination sphere and then the methyl migrates to a vacant site. So, here we are going to check what would have happened if indeed the carbon monoxide migrates. If carbon monoxide migrates, then of course, you would have got the probability of forming this molecule. Let us take the first one, the carbon monoxide. If the CO leaves and first we are assuming that this carbon monoxide leaves. So, you would have a vacant site here and then the carbon monoxide migrates to the cis position. So, let us just mark that here in this fashion. So, the carbon monoxide migrates here. Then in that case, you would form exclusively this product. This product here, which I have marked as 3x and we will explain why 3x is possible, because there are 3 carbon monoxides like that and each one of them would have and all of them would end up with the formation of this product. On the other hand, if one loses this carbon monoxide, let us mark that with a different color. If this carbon monoxide leaves, you would end up with a vacancy in the cis position also, but you would have if CO migrates, then it is this CO that is migrating and then you would end up with this product. So, the possibility of the species being formed from one of these carbon monoxides migrating and one of these carbon, the label carbon monoxide migrating, the ratio would have been 1 is to 3 and there would have been only 2 products. Because we have the formation of a set of 3 products, the labeled unlabeled one, the trans one and the cis one. We suggest that the reaction is happening only by methyl migration and not by CO migration. So, here we have assumed that it is the CO migration is happening and we have shown clearly that the ratio of these 2 products would have been 3 is to 1. We would have got 3 times the labeled product and 1 we would have got the unlabeled product and that is not happening. So, in fact, we say that it is the methyl migration to a vacant site which is happening. So, let us just proceed further now. Let us take a look at what happens when you move from this methyl migration. Let us take a look at what happens when you have methyl migration. I have used different complex here. What we are doing is that we are generating a co-ordinatively unsaturated complex. Because you can have both 16 valence electron species and 18 valence electron species, one should think about the fact that what will happen if you have a migration first and then coordination sites being filled. In the previous example, we had 18 electron complexes, but we have pictured here in this example a 16 valence electron complex. If you have 16 valence electron complex, one has to go through the intermediate where you would end up with a 14 valence electron species. If this species is if this reaction is reversible, then you would end up with rate dependence which is not dependent on the added triphenyl arsene. As we have pictured here, the second step the triphenyl arsene is coming in and blocking the vacant coordination site on the platinum. First it is migration of the R group. First it is the migration of the R group. Then in the second step the vacant site on the platinum is blocked by the triphenyl arsene. So, here is what is happening. We have conversion of a 16 valence electron complex to a 14 valence electron complex. Then the 14 valence electron complex is now converted back to a 16 valence electron complex by the addition of the ligand. So, one can check these reactions using kinetics depending on whether the first step is rate determining or the second step is rate determining. We would have a first order dependence on first order reaction or a second order reaction. If the second step is rate determining clearly, it would depend on the concentrations of both of these species. So, you would have a second order rate constant. Now, here the generation of a 14 valence electron intermediate is postulated. In general, the reactions go through either a 16 or an 18 electron intermediate. Now, how can one go through a 16 electron intermediate in the case of complexes which start with a 16 electron complex? So, here is an example. In this case also we have the migration of a carbon monoxide or a migration of an R group onto a carbon monoxide. If we add carbon monoxide here in this reaction, let us add plus CO and you have minus CO. Then, we can have the equilibrium with an 18 electron species. So, 16 valence electron species adds on a molecule of carbon monoxide becomes an 18 valence electron species. Now, we have the migration of an R group onto the carbon monoxide in the cis position. Remember, the two groups have to be cis related. So, the CO group that is cis is the one on to which the R group migrates. So, if you have the migration of this R group now on to the cis position, you end up with 16 electron species again. So, you go from a 16 to an 18 to a 16. It is generally agreed that most catalytic cycles or most reactions would like to go through a 16 electron, 18 electron cycle rather than a 16 electron, 14 electron cycle. So, it is generally understood that this is the reaction mechanism. Lastly, one can also look at migratory aptitudes. By migratory aptitudes, we want to look at either reaction rates, mostly in mostly reaction rates. If you have a set of complexes where you have alkyl, benzyl, allyl and vinyl species, then these groups will migrate much faster than vinyl or proparagel. Proparagel group does not migrate very easily compared to the vinyl. On the other hand, the alkyl group has the greatest tendency to migrate and it is the one which will undergo migration fastest. If you have a molecule in which two different alkyl groups are there, say an alkyl and a benzyl, it is an alkyl group which migrates and not the benzyl group. Similarly, benzyl migrates faster than allyl, allyl faster than aryl. Now, these could be because of the electronic nature of the R group which is attached to the metal and this is something which has to be studied further. Now, some groups have very little propensity to migrate and these are groups like hydrogen, trifluoromethal. So, this is a methyl group where all the hydrogens have been replaced with fluorine or an acyl group. So, if an acyl group migrates, you would end up with double carbonylation. So, you already have an RCO group. If it migrates on to another carbonyl group, then you end up with a dicarbonyl species. Now, this usually does not happen. So, a single carbonyl migration is what you observe in many cases and you have no double carbonylation reactions. Very few double carbonylation reactions are known. Finally, I would like to look at the stereochemistry of the migrating carbon. The migrating carbon could be migrating as an anionic species or it could be migrating as a radical species. If it migrates as a radical, then the radical species would lose the stereochemical information on the R group. If it undergoes inversion as in the case of an SN2 process, then the stereochemistry of the migrating group would be changed. Nevertheless, the stereochemical information is retained. In other words, if you start with an optically active product, you would end up with an optically active substrate. We will end up with an optically active product. So, if we use the neutral method, then we neutral method would be changed. So, after electron counting, we would think of it as a radical species. I have mentioned it here as a methyl radical, but it could also be an alkyl radical that is migrating. Now, if you use it as an anionic species, then we would imagine that it is undergoing an SN2 type of reaction. Let us look at some of the features of this migration reaction. If you have a neutral ligand, a 2 electron donor like carbon monoxide, you have an anionic ligand or if you use the neutral method, then you talk about it as if it is a 1 electron donor. Then the principle is that the anionic ligand migrates from the metal on to the neutral species. So, carbon monoxide is a neutral species and the anionic ligand is migrating on to the carbon monoxide. Initially, we have a total of 3 electrons on the ligand which is being donated to the metal. At the end of the reaction on the neutral ligand counting, neutral electron counting procedure, we end up with a 1 electron ligand that is the RCO group. So, the RCO group is a 1 electron donor and the R group is a 1 electron donor. Then the CO group is a 2 electron donor. So, total of 3 electrons are present on the metal and finally, you end up with a 1 electron donor on the metal. So, this turns out to be a loss in the number of electrons around the metal and this is something that we have to bear in mind. If we look at the net effect on the metal, we realize that after the insertion reaction, we also lose 1 coordination site. It becomes 1 coordination site on the metal becomes free. So, based on whatever electron count procedure you might use, you end up with a loss of 2 electrons on the metal and you also end up with a loss of 1 coordination site on the metal being free at the end of the reaction. So, these are the effects of migration and we will discuss the reaction in greater detail in a future lecture.