 This lecture deals with the catalytic processes where there is addition of a species like H A or H M, H X or X Y species where the second atom other than hydrogen is more electronegative than hydrogen or less electronegative. Now, there are several instances where this type of additions turn out to be important. We looked already at the addition of molecular hydrogen and we went through the argument as to why addition of hydrogen to olefins turns out to be unacceptable or it is not a process which can happen easily at room temperature and pressure and why a catalyst is necessary. The same problem in fact plagues the simultaneous addition of most species like H A to the carbon-carbon double bond. So, if you want to add a species H A to the double bond, you need a catalyst especially when the second atom is not very different in electronegativity from hydrogen. So, in other words if you look at the three atoms which are marked in black lithium, beryllium and fluorine, the electronegativity difference between hydrogen and these elements are quite large. So, you do not have a serious problem in terms of addition of these molecules. So, lithium hydride and hydrogen fluoride do not have a problem whereas when you take a boron hydrogen species addition of B H or addition of N H, then it looks like you need a catalyst because the polarity difference between these two species is not very large. C H is a special problem because many of the molecules that we are dealing with have got C H bonds. So, selective addition of C H or activation of C H is indeed a very difficult problem and we will deal with it in a separate class all together. Oxygen in fact is quite polar and so the difference between oxygen and hydrogen in terms of electronegativity is indeed large. So, it should be able to add on without difficulty, but as we will see even that addition of OH bond to the C double bond C turns out to be a reaction which needs a catalyst. First let us take hydroborulation, the addition of a B H bond. Here is a molecule which I have for you pictured for you here which is a in and a own. So, the two species are in fact separate. There is a double bond and there is a ketone moiety and the two are not conjugated. When the species reacts with the borane, a B H bond, the uncatalyzed reaction where you do not have any other species gives you a reduction of the ketone in such a way that the boron is now attached to the oxygen. The obvious reason or the obvious conclusion is that boron is oxophilic as indeed it is. So, this is the side of the molecule that reacts with the borane. So, you do not have a surprise in this uncatalyzed reaction. So, what would you do if you want to react the carbon-carbon double bond with the borane? Hydroboration is a reaction which happens quite readily in many cases. So, you want to force the reaction to go on the carbon double bond carbon rather than on the C double bond O. Here is a catalyst which is a familiar Wilkinson's catalyst which carries out the same reaction that we have just noticed that we have just discussed. It results in the addition of the B H bond on to the olefinic carbon rather than on to the ketone. So, here is a reaction which in which the chemo selectivity is important and the chemo selectivity is changed markedly with the rhodium catalyst which we are familiar with. Now, the mechanism of this reaction could be envisaged fairly easily because we know what happens with the Wilkinson's catalyst. I have a catalytic cycle drawn for you here which is very familiar, should be very familiar to you from what we have been looking at. Here is a resting state of the catalyst which on loss of P P H 3. So, it is minus P P H 3 loss of P P H 3 from the coordination sphere generates a vacant coordination sphere. At the same time you have oxidative addition of the B H bond. So, the B H bond is oxidatively added and if you remember the oxidative addition that we have discussed this falls in the class of molecules where the polarity difference between B and H is not very large. So, a cis addition is called for. So, we will have a cis addition of the boron and the hydrogen to the rhodium. Since, we have lost the P P H 3 molecule you have this vacant coordination sphere in which we can coordinate a molecule of the olefin. So, the olefin would be coordinated in such a way that would also be in a cis position with respect to the hydrogen. We now have migration of the hydrogen notice that this is delta minus and boron is less electronegative. So, it would have been B delta plus. So, if you look at the initial molecule also you have hydrogen which is slightly delta minus which now undergoes a migratory insertion which is essentially like a nucleophilic attack internal nucleophilic attack on the carbon carbon double bond. So, this is the carbon on which it would attack and as a result you would get a ethyl group coordinated to the rhodium if you started out with ethylene as the alkene. So, you have the oxidation state change the first step is in fact an oxidative addition process and then of course you have substitution or of the phosphorus by the olefin and then you have a migratory insertion. This is a migratory insertion that has happened where H minus or at least a pseudo H minus is migrating from the metal on to the olefin leading to the formation of another rhodium 3 species. The migratory insertion reaction does not change the oxidation state of rhodium. So, you now have rhodium 3 species from which the boron and the ethyl group can be eliminated in order to generate a catalytically active species. So, the elimination of this molecule the organic molecule the olefin which now has a B H moiety adder on to it generates or regenerates the catalytically active species which can add on another molecule of the reactant which is a borane which can carry on the catalytic cycle. So, once again you have the familiar oxidative addition migratory insertion or some rearrangement followed by the reductive elimination in the last step. So, let us now proceed to hydroamination as I mentioned C H activation turns out to be a special instance. So, you have it is carried out by several molecules and it is one of the most difficult challenging tasks because you make a molecule extremely reactive it will show up the organometallic catalyst. So, hydroamination is an instance where the hydrogen is slightly electro positive with respect to the other group. So, it is an instance where slight polarity difference is there just as we had in the B H bond, but now H is slightly positive. It turns out that these reactions are reactions where the mechanism is not necessarily the same as what we had proposed in the previous case. Which means it is not similar to the oxidative addition and migratory insertion followed by reductive elimination, but a variety of other mechanisms are possible. I have listed out one catalyst which is a titanium 4 species. Obviously, the titanium 4 species is not the one that will carry out an oxidative addition reaction, but the reaction mechanism is similar to what we had studied for the metathesis reaction itself. So, we will take a look at the mechanism, but to just convince you that this is only an addition of a mean. Let me just write out the equivalent of this molecule. So, this is the tautomeric form of that molecule and that just involves the addition of an amine to the alkyne. All we have done is add on a hydrogen and an R 3 NH group to the alkyne and this will readily isomerize to the species which we have listed. So, it turns out that you can also have a cyclic variant. The cyclic variant is what is listed in the lower half of the slide. This reaction turns out to be a useful reaction if you want to add an amine to an alkyne. You could do the same thing with an alkyne in which case you will get the saturated variety of this molecule. So, let us just take a look at the hydroamination reaction. Now, the hydroamination reaction because the nitrogen is more electronegative than hydrogen, you realize that C p 2 T i M e 2, this molecule will readily react with the methyl groups, the alkyl groups and release methane. So, elimination of methane generates a catalytically active species which could be written as if it is, I could write this molecule as if it is having N H R H R 3 N H R 3 and then another titanium molecule. So, here is the catalytically active species, but this species loses one molecule of R 3 N H 2 and generates an imido species which means there is a titanium double bond N R 3 group. Because we have to repeatedly write this framework, we are going to approximate this framework and I will do that by marking it in a different color. So, that you can see what we are talking about, we are going to mark out this species as the titanium species as the titanium species. So, the loss of an amine generates an imido species. Now, you will notice that T i double bond N R can add on to an alkyne in order to produce a metallocyclobutene. This is a species which we found in the is analogous to the metathesis reaction. So, this is the reaction which we had noticed earlier, only we had an all carbon variety. Now, we have both a nitrogen and the titanium on the cyclobutene metallocycle. So, if you form this cyclobutene, it can now ring open in a slightly different fashion and that is exactly what we are going to do. We are going to move the species in such a way that we will end up with a molecule which would have an imido compound. So, this again reacts with another molecule of R 3 N H and liberates the species that we are looking for, which means that this is the enol form, enol type, enamine type of a molecule which will isomerize to give you the final product. So, you will notice here that we end up, if this R 3 N H 2 reacts with this titanium, I can indicate this as follows. Then I would have hydrogen adding on here, one hydrogen from this amine adds on here and the nitrogen adds on to the titanium. So, you have both reactivity of this bond, which the titanium carbon bond undergoes, the titanium carbon bond undergoes an aminolysis or aminolysis in order to generate this intermediate, which is going to give you your product. So, this reaction is partly like the metathesis reaction. It has generated the catalytically active species, which can undergo a ring formation, a metallocyclobutene. Then the ring that is formed is kind of hydrogenalized by this amine in order to generate the final product. So, we will now proceed to the OH bond. This is just the amine was one example and the same thing can happen with a water molecule or any OH bond for that matter. We will see the celebrated Wacker process, which has been the mainstay of the industry, especially the Wacker Chemie, which is in Germany, where it was originally discovered. The reaction that we are talking about is a reaction of ethylene with water molecule in order to generate acetaldehyde. Once again you will realize that this is in fact the enol form. If you write the enol form of the molecule, you will realize that ethylene has undergone a partial oxidation and you have instead of adding H and OH across the double bond, you have added H and OH and then you have removed the molecule of hydrogen. So here in this case, you have both an oxidation and an addition that is taken place. So, that oxidation is carried out in an indirect fashion. Initially, it is a palladium 2 plus, which gets reduced to palladium 0. Then in a second step, because you have cupric chloride in the reaction, the cupric chloride oxidizes the palladium 0 to palladium 2 plus. In turn, it gets converted to cuprous chloride and cuprous chloride is oxidized in a very fast step in aqueous solution with oxygen to cupric chloride. So, this is a combination of two catalytic cycles one in which oxygen oxidizes the copper 1 to copper 2 and the copper 2 in turn oxidizes the palladium 0 to palladium 2. In a second reaction in which only the organic species reacts with palladium 2, there is an oxidation of the ethylene to the enol form of a stanlihyde. Now, why are we classifying or why are we using this reaction, talking about this reaction in the chapter, where we are discussing a nucleophilic attack on the ethylene. We will see this in a particular in a few minutes, but I just want to point out that this reaction is a fantastic reaction in terms of atom economy. There is absolutely no wastage of atoms when you convert the ethylene to acetaldehyde and the reaction medium is also water. So, there is a very green chemistry that we are discussing right here. In fact, although wacker process is not used by many companies, wacker company itself uses this methodology to generate acetaldehyde and recently the discoverer of the wacker process Reinhard Jira gave a small account of the discovery in a journal, which is referenced here. It makes very interesting reading, but let me just tell you a few points from that article, which will allow you to appreciate the whole discovery. At the point where Ira started his work, it was known that silver oxidized ethylene to ethylene oxide and he was trying to discover what would happen if you oxidized ethylene with other metal systems. He chose palladium on charcoal as a catalyst system and he added both oxygen and hydrogen. So, both oxygen and hydrogen were in the reaction mixture and on passing ethylene, he found that the reaction had proceeded by but there was no way for him to analyze the reaction mixture in a very modern way and he smelt acetaldehyde in the reaction mixture. That is all that he had as a hint about what had happened to the reaction, because at that time as he in his own words, in his own words he says the consortium of the company did not have even a simple gas chromatograph. So, based on the smell, he decided that he would look up the reactions where acetaldehyde is generated and lo and behold, he found another report of chat, which indicated that zaisalt, which is an olefin platinum complex, which had a molecule of water. So, here is a molecule of water coordinated to a ethylene complex of platinum, same group as palladium and that also generated acetaldehyde. So, he felt that if you had only oxygen and oxygen, platinum and palladium, probably you would generate more acetaldehyde and hence the rest is history. He patented this method for generation of acetaldehyde, which was very important at that time for making a variety of chemicals, including acetic acid and butyraldehyde. Turns out that in recent times at least acetic acid synthesis has moved over to Monsanto's process, which is something that we had discussed when we discussed carbon monoxide insertion chemistry, but Wacker company still uses this particular procedure in order to make acetaldehyde. If you want to read more on the mechanism of the Wacker reaction, you could refer to this article, which is there at Angukem and which also makes very interesting reading apart from Ira's article that I mentioned earlier. So, let us take a look at some of the modifications of the Wacker process. As I told you, if you use acetic acid instead of water, you would be able to generate the instead of OH on the enol. You now have vinyl acetate, which is also extremely useful. This reaction proceeds at best when you have sodium salt of tetrachloropiladium and also it is promoted by gold. So, these reactions are still used in an industrial scale to generate vinyl acetate. You can also generate vinyl ether if you do the reaction in the presence of ROH. So, essentially there is the addition of a nucleophile, RO minus or OAC minus to this double bond. It is a double bond. It is instructive to see the mechanism of this reaction, which has been studied in great detail. Here I have for you the first few steps of the reaction, which involves the addition of ethylene to palladium chloride in the presence of water in order to generate a olefin palladium 2 complex. Now, olefin complexes of palladium are plenty. So, there is plenty of precedence for this such a reaction. If you remove a proton from this water molecule, which is very often easily done. If you have a metal in a plus 2 or a plus 3 oxidation state, the p k a of the water molecule changes in such a way that it becomes lot more acidic. You would lose a proton fairly readily. You can generate a molecule, which has got an OH group and an olefin coordinated to the palladium. If this molecule undergoes a migratory insertion. So, notice here that in this reaction we have not carried out an oxidation state change. We have only carried out a deprotonation. We have carried out a deprotonation. We have carried out a substitution. We have carried out a substitution, a deprotonation and a migratory insertion in this reaction. That would give us the species, which I have indicated for you here, where an OH group has migrated from the palladium on to the double bond. So, you have this migration of this OH to this double bond and you have the formation of a single bond between the palladium and the terminal carbon or olefin. So, you would get a species like so. Notice that because we deprotonated that neutral molecule, we end up with a negative charge on the whole system. The migratory insertion does not change that process. So, you have a negative charge on this molecule, which I have indicated for you at the bottom of the slide. Now, let us do a couple of things. This molecule can carry out a couple of reactions or undergo a couple of reactions. One of them is the movement of this proton of this hydrogen from the organic moiety on to the palladium. That will give you that is a reverse of the migratory insertion. So, this is a migratory insertion in the reverse. It is a migratory extrusion, if you might call it, which leaves you with a palladium 2 complex. But the palladium 2 complex now has got a enol coordinated to the palladium. It has got a hydrogen and a chlorine. It assist to each other on the palladium coordination sphere. Loss of HCl, which would what would happen if this underwent reductive elimination. So, reductive elimination gives you the enol and palladium 0 and the enol can be converted to acetaldehyde. Now, if you look at this reaction, you started with palladium 2 and ended up with the reductive elimination. So, the oxidation state of palladium has been reduced through the catalytic cycle to palladium 0. That is why you needed the copper in order to oxidize it. There is also an alternative mechanism for this whole process and that is the reverse of the migratory insertion. If the migratory insertion were to proceed from the hydrogen on to the CH2, then you would end up with if this hydrogen adds on to the CH2, you would end up with this molecule where you have a CHOH attached to the palladium directly. The advantage of this particular mechanism is that it is easy to see how the reduction of the palladium could happen because the transfer of this hydrogen as a H plus and transfer of this pair of electrons into the palladium would completely reduce the palladium to palladium 0 instead of doing a reductive elimination of HCl. So, these two parts are difficult to distinguish, but nevertheless both mechanisms are common in the literature. So, the two mechanisms are different only because of the way in which we have carried out the transfer of the hydrogen. In one mechanism, we have transferred this hydrogen directly from the organic moiety on to the palladium. In the other case, we have transferred it back to the organic moiety. We have isomerized the alcohol. This alcohol has been isomerized in order to generate a new alcohol which in which the alpha carbon has got the OH group and that led to a decomposition or a reaction where the acetaldehyde is generated directly. Now, let us put all these things together and look at the full catalytic cycle and that is there in front of you. You have as I told you the palladium 0 is oxidized to palladium 2 plus, palladium 0 is oxidized to palladium 2 plus and the coordination of an olefin gives you an intermediate which can undergo a migratory insertion of a water molecule or an OH group depending on the pH of the medium. That gives you a hydroxy ethyl group which is the key molecule in this whole catalytic cycle which can either undergo a hydrogen abstraction or a hydrogen migration back and forth in order to give you the final product. So, the palladium 0 being oxidized to palladium 2 in an independent step with oxygen from air is also a very important step because if that was not present all the palladium 2 plus would be reduced to palladium 0 and this would render the catalytic cycle completely inactive. The two cycles are independent of one another and that is also a very important point to notice. If you avoid the copper you could do several reactions and we have for you a few which are of mechanistic relevance. For example, if you do not add copper and instead you add lithium chloride and cupric chloride this is the reaction cycle. We have done this or we have shown this reaction with isotopically labeled species so that you realize that the hydroxide which we showed in the previous slide we showed it as an internal attack from the palladium it can also happen from the outside. So, here is a reaction where you add OH minus and add lithium chloride and cupric chloride and the concentration of these two species is quite high in solution. Then you have a strong probability of forming the chloro ethanol and that is the product that is formed because you have hydroxide hydroxide ion also in the reaction medium the alcohol is deprotonated and it carries out a internal attack an internal attack on the second carbon atom ejecting the Cl as Cl minus. So, let us take a look at this stereochemistry in detail what you have is a OH minus which is attacking this carbon and when it attacks the carbon you end up with the palladium on the opposite side where opposite side of the hydroxide ion which was incoming. So, I have exaggerated the length of bond so that you can see that the hydroxide is coming from outside from towards the viewer from the side of the viewer towards the olefin and the palladium is behind the olefin away from the viewer. So, that leads to a molecule in which the deuterium the two deuterium atoms if the OH and Cl are present on the same side the two deuterium atoms are trans in this molecule and you would end up with a rotation in order to do an internal attack of the OH minus or the O minus on the second carbon you would have to rotate the hydrogen and around this carbon and the hydrogen and deuterium would change places and that is what we have shown here the hydrogen and deuterium have changed places and as a result the epoxide has got cis deuterium atoms. So, this tells us two things the hydroxide can come from the opposite side of the molecule A and B it also tells us that it is possible to modify this reaction in such a way that other products can be obtained. Here is another reaction and we will go through this rather quickly this time because you have a similar situation hydroxide attacks from the outside because OH minus was added and it turns out that adding hydroxide from outside can be combined with insertion of carbon monoxide into this palladium carbon bond which as I showed you in the last trans last projection the palladium carbon bond length has been exaggerated. So, that you can follow what is going on if the insertion happens with retention of configuration at the carbon this is the carbon which is undergoing migration and if that happens with retention the product that you obtain would have this following configuration so where the two hydrogens are now still trans to each other and rotation of this oxygen in such a way in order to generate a lactone in this case a lactone can be achieved only if you bring the two atoms on the same side the two deteriorated atoms on the same side. So, you will notice that the migration has to happen with retention of configuration so the OH minus must have come from the opposite side of the palladium. So, these are two mechanistically relevant and important reactions that have been studied. Now, the Wacker reaction as I mentioned earlier has been modified in a variety of ways and it has also been modified in such a way that can be industrially useful. Here is a modification where you carry out acetoxylation into a diene and that results in OAC minus being added on to do two terminal positions of the butadiene. This gives us a very useful molecule because reduction with ranne nickel this is ranne nickel and hydrogen leads to a diol formation and this diol is a useful molecule because it can be used in the polymer industry in a fairly large scale. Here is another reaction where if you want to carry out two addition of two different nucleophiles that also is possible specially when you have palladium acetate and benzoquinone, benzoquinone to reoxidize the palladium to palladium 2 plus and in the second step you can convert the chloroacetoxylated species to an amine and this happens in the terminal positions only the nucleophiles have come in from the terminal positions and as a result you can generate this very interesting and novel molecule fairly easily. Now although we are not strictly adding bi-molecular species in this reaction here is a molecule which has got a cyanogroup instead of H A we have a pseudo halogen cyanide which is attached to hydrogen. Turns out that this is a reaction which can be carried out in two different ways. The nickel catalyzed fashion is the one which is used in the industry significantly and we look at the catalytic cycle in a minute. But the interesting fact is that if you use dichobalt octacarbonyl you end up with a different regeochemistry. We will take a look at this reaction which can be carried out with alkenes, alkenes and even carbonyls. So, you can even add H C N to carbonyls to get a cyanohydrin. So, this is a very useful reaction you can carry out the addition with a variety of species and here is a catalytic cycle. The catalytic cycle is shown with addition of H C N and we now move over to what we had seen earlier. An oxidative addition followed by oxidative addition of H C N followed by addition of nolefin in the vacant coordination sphere of the nickel 2 plus. Now, nickel has become 2 plus at this particular stage and this oxidized nickel 2 plus would prefer a 6 coordinate species or at least a 5 coordinate species. That is what happens when you in the second step you have oxidative addition followed by coordination of nolefin leading to a 5 coordinate intermediate. Now, you have migration of a hydrogen and this migration of a hydrogen could have taken place in 2 different fashions. If it added on to the CH 2 group it would have ended up with a hindered alkyl group. Instead it seems to add on to the hydrogen to the carbon which has got less number of hydrogens. This is like the anti-marconic of addition of H C N. So, you end up with an alkyl group where the bulky group is usually away from the metal atom and the hydrogen that you have added is also added in this position. This is the new position of the hydrogen and this alkyl nickel species can reductively eliminate let us indicate that with a different color here. This can eliminate a C N group and the alkyl group and then you would get the product. So, this results in the catalytically active species because reductive elimination brings back the nickel to nickel 0 and it is now capable of doing the oxidative addition back to nickel 2 plus. So, the interesting fact is that the regio selectivity in the reaction is very much dependent on the ligand that you use. If you use a very bulky ligand the regio selectivity is very good and you get only one product and that product has got the R group away from the cyanide or in other words the addition of hydrogen will happen in such a way that the hydrogen is added to the bulkier side of the olefin and you get a terminal carbon where you can add on the cyanide. So, this happens only if you have a bulky ligand and a phosphite as opposed to a phosphine. So, the presence of a phosphite is important and the presence of a bulky group is also important. So, if we do the same reaction cycle with a butadiene then you can do the reaction in two steps. In the first step one cyanide is added and the second step another cyanide is added and this leads to the molecule which has got which is now ready for if you add on hydrogen in this step you add hydrogen to this step to this molecule you end up with a six carbon molecule and six carbon diamine and this is exactly the molecule that is used to prepare nylon 6 6 and nylon 6 6 can be prepared by oxidizing this to the di acid to the di acid and reducing it to the amine and that gives you a very easy way of converting butadiene to an extremely useful molecule which is nylon 6 6. So, here is a catalytic cycle which will go through fairly fast because we have already discussed it with a single in suffice it to say that each in behaves independent of the other in. So, much so that you end up with the the C N group added to each end of this diene. So, if it that if the two double bonds are cooperated we would have obtained a different result. So, the importance of the bulky phosphate is not to be minimized. Now, let us just ask the question what about elements in the second row. So, far we have talked about the three elements boron nitrogen and oxygen and what about elements in the second row can the aluminum or the silane or the phosphane add on to the olefin is that possible. In fact, hydrogen add on attached to an aluminum is known to add on olefins without a catalyst and this was in fact studied by Ziegler and we will talk about this in a moment. But the moment you move on to other species it is interesting that they can either add on without a catalyst or with in the presence of Lewis acids they do not really need a transition metal catalyst. Let us just take a brief look at the hydroalumination reaction because that is fairly interesting reaction and it is a challenging reaction to understand as well. Ziegler's discovery of polyethylene is in fact quite celebrated, but he did extensive work with aluminum alkyls. What he showed was that aluminum in the presence of hydrogen and ethylene can lead to triethyl aluminum. So, triethyl aluminum is a species which is generated in the presence of just ethylene and pure aluminum. Now, if aluminum and alkene are in the ratio 1 is to 2 you end up making aluminum which has got an aluminum hydrogen bond and this turns out to be like a stoichiometric reaction. So, you can do this in a batch process and you will see in a moment how this can be extremely useful. So, it looks as if the direct reaction of hydrogen aluminum, hydrogen and aluminum is quite feasible and the catalyst for this reaction it is auto catalyzed process because you aluminum triethyl aluminum itself functions as a catalyst. Another interesting fact is that this reaction is completely reversible and it can generate the reaction can go in the backward direction as well. The last important point is that a small amount of titanium promotes this reaction. So, you will notice that the Ziegler Nutter reaction is catalyzed by or promoted by aluminum and the hydroalumination that was discovered by Ziegler was promoted by titanium. So, here is a net reaction for this whole process. If you treat aluminum with hydrogen in the presence of triethyl aluminum which is a catalyst then you get a growth process where a growth step where you end up making aluminum and this aluminum can add on ethylene. So, each of these aluminum hydrogen bonds adds on a molecule of ethylene and it leads to the triethyl aluminum. So, if you have a small amount of triethyl aluminum it can catalyze this whole reaction. The net reaction of course is the formation of triethyl alumine. So, the use of hydroalumination is extremely valuable because you can use this growth step to keep adding ethylene groups in order to generate multiple or very long alkyl chains attached to the aluminum. This can be done in a bulk process. So, in a batch process triethyl aluminum can add on aluminum in multiple steps in order to generate linear species. As I told you the bulky alkyl group on the aluminum atom is unstable. So, it would exchange with a linear group. So, if it keeps doing it then a very long polymer not a polymer, but very long oligomeric chain can be attached to the aluminum as much as 10 to 20 ethylene groups can be attached. After carrying out this reaction with a high pressure of ethylene high pressure of ethylene you would end up with a long chain attached to the aluminum which can now add on oxygen. If you add oxygen to it it generates in an alkoxy group attached to the aluminum and this would be in only one step is shown, but multiple after multiple steps you would end up with a long chain alcohol where this the end that is attached to the aluminum is the one that is having the oxygen. So, this would end up with aloh thrice and alcohol. This long chain alcohols are extremely useful in the industry from making detergents. So, these are extremely useful molecules for generating detergents. So, you can also make simple terminal alkenes with long chains n equals 10 to 20 as I mentioned to you before by treating it with a high pressure of ethylene. So, that the whole chain growth is stopped and the alkene is ejected from the coordination site. So, this turns out to be an extremely useful hydro elimination turns out to be an extremely useful reaction. If you take hydro silalation hydro silalation is also useful because you can convert the silicon that is attached to the carbon to other groups in a subsequent step. So, alkenes and alkenes add on R 3 S I H and the interesting aspect of this addition addition of hydrogen and the hydrogen and R 3 S I group hydrogen adds on here. So, you can see that the hydrogen adds on to the sterically more encumbered encumbered carbon and you can do this reaction with alkenes alkenes carbonals and shift bases and so on even nitriles. And for all these cases it turns out that platinum is the best catalyst. Now, the catalyst of choice is usually in this as shown in this slide it could be in any oxidation state of platinum because the silane reduces the platinum from whatever oxidation state you start with. In this case it is shown as a plus 4 oxidation state it reduces the catalyst to 0 oxidation state platinum 0. Platinum 0 is the active form of the catalyst here as two systems this compound has been crystallographically characterized by x ray diffraction and this car set catalyst system involves the use of a vinyl silane which stabilizes the platinum 0 resting state of the catalyst. So, H 2 P T C L 6 is reduced by silanes to platinum 0 and the platinum is found coordinated to the olefins and resting in the stable state. You can also reduce it isopropanol and that is called the spear catalyst. So, these are two catalyst systems which are popular for hydro silation reactions. The mechanism of hydro silation has been invariably it is been discussed in great detail and it turns out that you can add the silicon first or the hydrogen first conceptually and you can either way you can end up with the product, but the reaction mechanism where you have platinum platinum in the 0 oxidation state undergoes first of all oxidative addition. So, an oxidative addition is essential after the oxidative addition you do have a step in which there is a migratory insertion and the migratory insertion happens in such a way that the alkyl group now has got a silicon attached to the beta carbon of the carbon attached to the platinum and now it can reductively eliminate to give you the product. So, this can now reductively eliminate this is the reductive elimination step. This is the migratory insertion step that we have and that will give you the final silane which can be manipulated either with treatment with oxygen or hydroxylating species. So, that you can form a variety of interesting molecules. Now, the chak harrowed mechanism for hydro silation suggests that it is in fact the hydrogen which is added first and this is the more popular of the mechanisms which are used in the literature to explain the way in which the stereochemistry goes and you have the addition of H and S i R 3. Only difference between these two mechanisms is that the hydrogen is added on first the hydrogen migration happens first to the olefin and that leads to an alkyl species which reductively eliminates the silane. So, in conclusion one might say that as long as there is a significant polarization in the two groups that means hydrogen on one side and either it is an aluminum as in one as in some examples that we have seen or a fluorine then direct addition of an olefin to these molecules is possible. So, there is no activation necessary there is no catalyst that is necessary for this whole reaction, but if you have a non polar molecule or less polar molecule where the polarization between hydrogen and the molecule A is significantly less than the then you end up with the need for a catalyst and the catalyst is usually a molecule that can do oxidative addition of the species. Oxidative addition followed by a migratory insertion and a reductive elimination makes up the catalytic cycle. CH is a special case where activation is even more difficult we will discuss this later. Usually it is the negative end of this species A H that carries out the attack on the alkene. Although C N seems to be an exception when you have H C N we notice that there was the cyanide is obviously the negative end of this molecule, but still it was hydrogen which carried out the attack on the alkene. So, in general the nucleophile is the one which attacks the alkene and this could be followed by hydrogen abstraction. So, in the case of Wacker the Wacker process hydrogen abstraction happened, but in other cases hydrogen abstraction is not happening and you have just an addition of the two molecules the H and the A to the double bond and then you have the activation of these molecules or then we say that we have the activation of molecules to add on to C double bond C or any unsaturated molecule C triple bond C or the C double bond O as well. So, in future lectures we will discuss the CH activation which is even more difficult because it has to be a rather specific chemo specific oxidation chemo specific activation is necessary because the ligands on the organometallic compound are also capable of undergoing a reaction if you do not have chemo specific reactivity.