 So, we start today's lecture with a small three dimensional model and the demonstration of how allyl species can be fluxional. The allyl species has got three carbons and the carbons are 1, 2 and 3. These three carbons are oriented in such a way that the central carbon has a unique hydrogen, which is marked with yellow, which is shown in yellow color. And the two hydrogens, which are on the terminal CH2, on the terminal carbons, there are two hydrogens or two alkyl groups, which are on the terminal carbons. And these are marked in different colors depending on whether they are sin with respect to the central hydrogen. So, for example, if the yellow hydrogen is pointing towards me, that is it is pointing towards me and the blue hydrogens are also pointing towards me. So, these are the two hydrogens, which are called the sin hydrogens. The anti hydrogens are the ones, which are pointing towards the camera. So, these are the two anti hydrogens and they are pointing towards the camera. So, there are two types of hydrogens on the terminal carbon or two types of positions. One is called the sin position, which is closer to me. The observer is closer to the anti hydrogens, which are marked in light green. Now, what happens when you have a dynamical behavior of this allyl groups? The central metal atom, the metal atom is usually in the central position. So, it would be like my fist, which is where I am holding this allyl group. That is where the metal is situated usually. And it is almost equidistant from all three carbon atoms. Now, during the time that the allyl group becomes fluxional, it moves from the central carbon atom or the central position to the terminal position. So, here I am showing you the hand, which has a ring is the one, which is holding the central carbon. And this hand I am holding the terminal carbon. Now, if the metal is closer to the terminal carbon, then the two carbons, which are held here have the double bond have the pi bond. So, this carbon is the one, which is free to rotate. And I explained how it is possible to move this carbon in a simple rotation, which would take the metal from the position, which is below the plane of this allyl group to the plane, which is above this allyl group. So, the metal can go from one plane to the opposite plane. During this rotation, what happens is that the position of these two hydrogens, what was sin with respect to the central carbon hydrogen is now becoming anti to the hydrogen. And so, the sin and anti hydrogens become equivalents when you have a simple rotation of this metal from one plane to one side of the plane to the other side. Now, one can imagine the same thing to happen either through a flip, which is happening by the metal by changing the position of the metal or by changing the position of the allyl group. So, I can rotate the allyl group in such a way that the allyl group now goes from this position, from the position where it was earlier to the other position. So, rotation around this axis, rotation around this axis, which would be rotation of the vinyl group or rotation of the metal, both of them are equivalent. Now, we will see this with two dimensional pictures now. This is just a three dimensional model to illustrate how the flipping of the metal or flipping of the allyl position, allyl group, the vinyl group in the allyl, both of them are equivalent and lead to change of the sin and the anti hydrogens on the terminal position. So, this has serious consequences for understanding the NMR spectrum of the molecule. So, now let us take a look at the written text, what we have shown in two dimensions. So, allyl complexes are formally four electron donors. They are in the anionic method. If you use the anionic method, the carbon is the anion. So, you have a four electron donor and all three carbons are equally bonded to the metal atom. So, you have a hapticity of three and that is indicated by eta 3. In the neutral method, we consider that as a neutral group as an allyl radical. So, it becomes a three electron donor and in this case also the hapticity is three. So, it is eta 3 whether it is use the ionic method or the neutral method. But in the neutral method, you consider it as a three electron donor. The ionic method you consider it as a four electron donor. Now, what we just discussed is that in the fluxional state, allyl molecules are in equilibrium with the eta 1 structure when it is only bonded to one of the carbons of the three carbons which form the allyl group. So, when it bonds only through one carbon, then in the ionic method, it would only be a two electron donor. The ionic method requires that it is only a two electron donor and in the neutral method this would be a one electron donor. So, it is important to remember these change in electron counts as you go from the eta 3 position to the eta 1 position. So, here are the two flips that I talked about. One is the case where when the metal atom which is indicated here, the metal atom moves from the central position to one of the terminal positions. Here I have indicated it as if it is moving towards carbon 1 which bears the H S 1 and H A 1. The S 1 is indicative of the fact that it is sin with respect to the central hydrogen. So, it is this relationship which we are talking about and then it is called the S hydrogen. If it is anti, then we call it if it is on the opposite side, we call it the anti-hydrogen and then we label it as A 1. Now, when the metal flips, when the metal moves from the bottom plane to the top plane as we just illustrated in the three dimensional model, then the position of the S and the A get interchanged. So, now H S 1 is in the position where H A 1 was present and you can move the metal from this terminal carbon to the central carbon again and you will get a completely equivalent structure. So, these two structures are identical. The structures that are present on your right hand side or on your left hand side are two identical structures. The structures are only different because of the position of the metal either they are above the plane or below the plane. But when the metal flips from one side of the plane to the other side, what we do is to change the position of the hydrogens. Now, what happens in the following flip is that only one set of hydrogens are changed from the anti to the sin position. The carbon atom 3 which is indicated here, the carbon atom 3 has no change in the position of the two hydrogens. But nevertheless you will see that in most spectroscopic details, the two carbon atoms are almost equivalent. So, let us see what happens when you flip the vinyl group. If the allyl group flips and this is indicated by rotation around the C 1 C 2 axis. So, this is the C 1 C 2 axis and if you rotate it around the C 1 C 2 axis, the metal retains the position and in the with respect to the plane of the allyl group. But nevertheless you will notice that the two hydrogens which are on the carbon atom 1 are the ones which are changed now between the anti and the sin positions. So, if the metal is attached to C 1, then it is the S 1 and A 1 which get interchanged. So, it is possible either through an allyl shift or through a metal flip or an allyl flip or a metal rotation or an allyl rotation you will be able to interchange the position of the sin and the anti hydrogens. What is important is that we need to keep track of two, we need to keep track of the carbon to which the metal is attached. It is that carbon which will have the change in the sin and anti hydrogens. Let us proceed further now. Let us take a look at what would happen if you change from C 1 to C 3. Now, it is possible that the metal instead of moving from the center position to the right side instead of moving from the center position to the right side as we just discussed it could have also equally moved to the left side provided there are no substituent on the C 1 and C 3 which are different. So, if it moves through the left then you would end up with an allyl or metal flip through C 3 or through C 1 if it moves to the right. If it moves to the right you have a flip with a C 1 if it moves to the left you have a flip through the C 3. So, what will happen is that if there are no substituents on the allyl group all four hydrogens H A 1, H S 1, H S 3, H A 3 will all become equivalent as the metal flips back and forth from the top of the plane to the bottom of the plane or from left to right. So, this complicated flipping arrangement flipping that goes on with a position of the metal equivalences the four hydrogens. So, let us take a look at the spectroscopic features of the allyl group that are affected by this fluxional behavior. Both positions 1 and 3 are equivalent because the metal can move either to the left or to the right, but sin and anti can be rigid. Now, that is possible and so the NMR spectra is the only tool that can distinguish between the different fractional behaviors of the metal and the rotations that are involved. Very often one has to do a temperature dependence of the spectrum in order to understand what is going on in the allyl complex. So, here is a hypothetical NMR spectrum of a simple allyl complex. Now, what we have done is that we have dissolved it in DMSO and warmed the compound to about 40 degrees. Now, when you warm the compound to about 40 degrees what happens is that the two positions the position of the metal keeps changing very rapidly. So, instead of recording the spectrum at room temperature we recorded at a slightly elevated temperature and the metal keeps moving between C 1 and C 3 the two positions and then what happens is that A 1, A S 1 and A 3, S 3 all become equivalent. When they become equivalent there are four hydrogens and these three four hydrogens these four hydrogens are split into a doublet by the single hydrogen which is there in the center or which is the unique hydrogen which is labeled as H C. So, the central hydrogen H C appears as a quintet because there are four equivalent hydrogens which will mark in a slightly different color for you to follow. So, these are the four equivalent hydrogens and these four equivalent hydrogens split the central hydrogen which is marked in red and the central hydrogen appears as a quintet because of the four equivalent hydrogen splitting it. So, you can see that the spectrum will be in a slightly different fashion if you have a low temperature spectrum and in a solvent which does not coordinate to the palladium. So, here is the same compound dissolved in chloroform. So, that this flipping is reduced. We will just explain in a few moments why the flipping is reduced and because of the temperature the movement of the palladium is restricted. So, there are two effects that are going on. So, when the palladium movement is restricted you will have the anti and the syn hydrogens in unique positions. So, the anti and syn hydrogens are in unique positions and appear as two doublets. Two doublets the anti hydrogen in a higher field and the syn hydrogen at a lower field and both of them are split into doublets because of the unique central hydrogen. So, the central hydrogen which is here is responsible for making this anti hydrogens as doublets and the syn hydrogens as doublets. Now, what happens is because these two hydrogens have got different coupling constant with the central hydrogen. The central hydrogen appears as a triplet of triplets very often there is significant amount of overlap in the triplets. So, you would see much less than the nine line spectrum that you theoretically expect. So, this is the hypothetical spectrum that you get for an ally group which is bonded to the metal when the system is not equilibrating in a very fast fashion. Now, what happened when you dissolved it in DMSO was that the sigma allyl complex was slightly stabilized by the coordinating DMSO solvent. So, this facilitated the equilibration of the eta 3 to the eta 1 forms and the eta 1 form is a one which allows the rapid rotation of the palladium and equivalencing of the anti and syn hydrogens. So, there are two effects that are going on. One was the temperature the fact that we heated this allyl complex solution to a slightly elevated temperature and the second is the presence of a coordinating solvent which facilitated the formation of a eta 1 structure. So, allyl complexes are well known for the fluxional behavior for the fact that they can go from eta 3 to eta 1 forms which we have just demonstrated here. Also, for the fact that they can equivalence the anti and the syn hydrogen through rotation around the carbon carbon bond or the flipping of the metal from one position to the other position. So, now we proceed to some reactivity of the allyl complexes and today we will see time permitting three different types of reactions of the allyl group in and the metal. Now, here I have illustrated to you one reaction where palladium which is bonded to dibenzalidine acetone. Now, dibenzalidine acetone is a ligand which stabilizes palladium 0 and this ligand is bonded to two atoms of palladium. It is a fairly complicated structure and so suffices to say that two atoms of palladium are interacting with the two double bonds of the dibenzalidine acetone and there are three ligands which are present. So, we will put a bracket and put three of these ligands which are bonded to the two palladium atoms. So, this is a interesting structure which is formed by palladium and palladium is in the zero oxidation state. Now, this compound is capable of converting butadiene in a stoichiometric reaction to a complex which is indicated here where there are two allyl groups. Now, what has happened is that there has been a change in the oxidation state of the palladium. Palladium was initially in the zero oxidation state and in this compound formally it has converted to a plus 2 oxidation state. So, oxidative addition of the groups the organic groups has happened and also carbon-carbon bond formation has happened. So, this is a typical case where you have oxidative coupling of two carbon based ligands. Remember, we have already seen two different examples where you can have oxidative coupling. One was a case where you have a cobalt 1 and the cobalt 1 was bonded to two acetylene units and this got converted to a cobalt 3 system and this cobalt 3 system had a metallocyclo pentadiene structure. So, a new carbon-carbon bond was formed. The new carbon-carbon bond is indicated by this new position here. There is a carbon-carbon bond forming here and at the same time you have a change in the oxidation state of cobalt from plus 1 to plus 3. So, this is one type of an oxidative addition and oxidative carbon-carbon coupling reaction that we saw. Another system was the metathesis reaction itself. In the metathesis reaction we had a carbene and an olefin and these two combined together to form a metallocyclobutane. So, this was a metallocyclobutane that was formed because of the formation of two carbon of a carbon-carbon bond and at the same time oxidation state change of the metal from say let us say 0 to plus 2 in this case. So, there are several examples of oxidative carbon-carbon coupling and today we will look at the allyl based carbon-carbon coupling reaction which will lead to very interesting molecules. So, here is the example where we have let us just number the carbons because it is difficult to keep track of them. Here is 1, 2, 3 and 4 these are 4 carbons. Here is another set 2, 3 and 4 and let us mark the third one in a different color all together. So, here is 1, 2, 3, 4 carbons. So, we have a two carbon-carbon coupling and total of 3 butadiene units which have been combined together and some carbon-carbon bonds have been formed. So, let us take a look at this reaction in detail. It is interesting that it is only the nickel group which is very commonly seen to promote this particular reaction where butadiene units are stitched together. So, this is the product that is formed in a stoichiometric reaction between palladium 2 D B A 3 and 3 equivalents of butadiene. So, the same reaction if the reaction is not done in a stoichiometric fashion it could be done in a catalytic fraction and one can isolate the cyclo dodeca trine. So, this cyclo dodeca trine is formed by the carbon-carbon coupling reaction that we just discussed and it is formed as a result of stitching together 3 molecules of butadiene. So, having said this let us just proceed to the famous nickel based reaction. The same reaction can be carried out by nickel and this was extensively studied by Wilke and co-workers and Wilke was able to generate a very efficient catalyst for this process which he called as naked nickel. This naked nickel was called naked nickel because it seemed to have no ligands. There are no ligands around the metal atom or apparently no ligands around the metal atom and what he did was to generate a nickel compound N I C L 2. Which was then reduced with it was then reduced with A L R 3 which was the reducing agent and that reduced to nickel 0 and this naked nickel was capable of converting butadiene to the cyclo dodeca trine complex. In the palladium complex we isolated either the cyclo dodeca trine or an allyl complex but in the case of nickel the complex that was isolated by Wilke had 3 olefins coordinated to nickel. So, this is a 16 electron complex. This is a 16 electron complex which was synthesized by Wilke. It was not thermally very stable but it could be isolated and characterized and what was interesting was that 3 molecules of butadiene have been stitched together in a sequential fashion in order to form a very symmetrical complex which is bound to the nickel and nickel is again in the nickel 0 state. So, he could either take this allyl nickel which will as we will show currently it can become composed to give you naked nickel or he used nickel in the plus 2 state and reduced it with tri alkyl aluminum. So, in the same reaction could also be done in a catalytic fashion in which case you do not isolate the nickel trine complex which we just showed earlier but you isolate only cyclo dodeca trine. This is a one step synthesis of cyclo dodeca trine which is C 12 with 3 double bonds in a cyclic symmetrical fashion starting with a simple butadiene molecule. So, let us see how this catalytic reaction happens. Let us imagine in the first step we can easily picture the elimination of hexadiene 1 5 hexadiene starting with a bis allyl moiety. Now, this just results from a reductive elimination of the 2 allyl groups the 2 allyl groups combining together. So, that you form the cyclo 2 allyl groups combining together to form the 1 5 hexadiene and that is this molecule right here. Now, this will as I mentioned earlier it will produce the very reactive form of nickel which we can write in a tentative fashion as 2 butadiene molecules attached to each other in such a way that it forms the nickel 0 18 electron complex which is pictured as A. Now, if you treat this molecule if you treat this molecule with another molecule of butadiene it induces a coupling reaction where if you watch this reaction closely you have a total of 8 carbons. You have a total of 8 carbons and what you have done is that you have oxidatively added the nickel 0 the allyl group form the allyl group in the coordination sphere of the nickel and at the same time form the carbon carbon bond between the terminal carbons. So, let us imagine that you form a bond between these 2 carbons and that is this bond right here. So, you have 4 carbon atoms and we will mark we will mark them as 1 2 3 4 and 4 and this is exactly what we have here and atom number 4 dash and 4 have combined together to form a carbon carbon bond. So, 4 and 4 dash combined together to form a carbon carbon bond and it is 1 2 and 3 which make it an allyl group. Now, you can imagine the transfer of an electron to from the nickel to the butadiene unit such that it forms 2 allyl groups on either side 1 electron on either side and that results in the formation of an allyl group and at the same time results in the formation of a radical species. So, if you want to write this step wise this is what we would write this is a radical which is formed and this is the second radical that is formed and this is the 4 dash carbon that we are talking about this is the 4 carbon atom number 4 which we are talking about and we now form a bond between the 4 and 4 dash and that is the bond which we have indicated with an arrow here. So, it is possible to now form the oxidatively added species and this is induced by the addition of a butadiene molecule which we have pictured right here. So, this is what is going on what you end up with is an oxidative coupling reaction when you converted 2 neutral molecules to allyl groups and at the same time formed a carbon carbon bond. It is interesting that you can now add on this butadiene unit to form a 12 carbon chain the 12 carbon chain will have the 2 allyl groups interacting with the nickel atom again. So, here you have the 12 carbon atom chain which is interacting with nickel and you can also do the same thing in a slightly different fashion and you can generate this intermediate which is shown here. Now, the cyclo dodeca trion complex can be formed by a reductive elimination reaction which happens when you heat this reaction mixture to slightly more than 40 degrees and then it forms the cyclo dodeca trion. The same reaction could be catalytic in nature and it can generate your naked nickel and that is the reaction which is shown here. Now, in order to show the reductive the various steps that are involved let us just go through this reaction again once more but we will highlight the oxidative additions and the reductive eliminations. In fact, the first step is the reductive elimination and that is this step that we have shown here this is the reductive elimination step. So, where you have combined together 2 allyl groups and formed a diene unit and at the same time you have formed you have done a substitution reaction which has resulted in the formation of a diene molecule which is coordinated to the nickel 0. So, the diene unit now combines together in an oxidative coupling reaction. This is the oxidative coupling reaction and this coupling reaction is the one which gives you the 8 carbon unit which is shown here this is the 8 carbon unit. Now, to this 8 carbon unit we can do an insertion reaction. The insertion reaction merely converts this allyl group to another allyl group by adding this carbon atom here which we had labeled as 1 dashed and we are combining it to the terminal carbon of the butadiene which is let us convert that into yet another color so that we can follow that. So, this is atom number 1, 2, 3 and 4. So, what you would end up with is 4, 3, 2, 3 and 4. So, what 2 and 1 the atoms that are labeled in blue now are now the newly formed newly formed newly introduced butadiene. So, the new bond has been formed between 1 and 1 dashed this is the green 1 dashed that was present and that we will indicate it as 1 dashed, 2 dashed, 3 dashed and 4 dashed. So, we earlier had 4, 3, 2 and 1. So, this is how the reaction proceeds we first have a reductive elimination combined by a substitution and an oxidative coupling followed by an insertion reaction and then that gives us a molecule which can do a reductive elimination and the reductive elimination can happen through in a variety of ways and one of them is to retain the double bond between atom numbers 2 and 3. The blue 2 and 3 can now become if I write it in a slightly different fashion I can. So, we can have this written in this fashion. So, now we have the reductive elimination between these two positions this will have a double bond and this will have a double bond also and so you will end up with the formation of a cyclo dodecatriene. So, this helps us to follow how we can this helps us to understand how we can carry out a variety of oxidative additions and reductive eliminations including insertion reactions to stitch together these three molecules of butadiene. Now, what followed after these reactions was discovered what followed was the discovery that you can add some other ligands apart from butadiene in such a way as to turn the reaction or to steer the reaction in to a different channel all together. Now, what we have seen so far is the oxidative coupling reaction and the first step in these reactions are also the oxidative coupling reactions and what we end up with is coupling of two butadiene units and that is what we have done here two butadiene units have been coupled together to have the allyl group and the new carbon carbon bond that is the new carbon carbon bond that has been formed. If you have a ligand that will reduce the reactivity of this system then you do not have the reaction with another butadiene unit earlier this L was a butadiene now it is not butadiene. So, this is not a butadiene and what it does is to slow down this reaction to the coupling with a third butadiene molecule and if you do the coupling with the or if you do the reductive elimination in such a fashion that you eliminated to form a cyclobutane moiety then you get one four divinyl cyclobutane this is one four divinyl cyclobutane which is pictured right here and this can be generated depending on the type of ligand that you put in this particular position. But in the same reaction depending on the ligand you can also generate a slightly different reactivity and that is instead of forming the bond if instead of doing the reductive elimination between atom numbers three and three dashed you can do it between the two atoms one and one dashed. The two butadiene that are combined are combined in such a way that you do a reductive elimination between atom numbers one two three and four. So, these if you now come do the reductive elimination in or the coupling between one and one dashed you will end up with cyclo octadiene. So, cyclo octadiene is formed by coupling of two butadiene units two butadiene units combined together in such a fashion that you can form cyclo octadiene. You can also do it in such a way that you can combine it to form if you form a bond between three and three dashed then you would form divinyl cyclobutane. And yet another modification is to do it between three and one dashed and in which case you end up with a cyclohexene. The cyclohexene can be generated if you do the coupling between atom number one and atom number three. So, this is the type of reaction change that happens when you have a ligand which slows down the reactivity. Now, you might be wondering what would be a ligand which will slow down the reactivity and the reaction was slow down in the presence of phosphines. When PPS3 was used it was possible to isolate different amounts of cyclo octane, cyclo octadiene this should be an octadiene and cyclo octadiene one four divinyl cyclohexene, cyclobutane and also vinyl cyclohexene. So, all three products were formed in the presence of PPS3. So, here R was PPS group. So, PPS3 was able to catalyze these three reactions Heimbach's contribution was a fact that he generated what are called ligand reactivity control maps. So, he showed that depending on the ratio of the ligand in a metal depending on the ratio of the ligand in a metal it is possible to form different amounts of the three products. In the absence of the ligand that means in the absence of triphenyl phosphine in this graph what we are discussing is the triphenyl phosphine. If you have triphenyl phosphine which is very small that means almost 10,000 molecules of the metal caps catalyst with respect to the ligand that is why log of these two is minus 4. So, if you have 10,000 molecules of the metal catalyst the nickel catalyst with respect to PPS3 then you would end up exclusively with the formation of cyclo dodeca trinium. If you increase the amounts of the ligand and let us say you have 100 moles of PPS3 per mole of nickel then you end up with mostly the cyclo octadiene. So, cyclo octadiene is formed when you have 100 moles of the PPS3. If you now increase the amount of PPS3 to 10,000 with respect to the nickel then you end up with vinyl cyclohexene. So, vinyl cyclohexene is formed when you have very large amounts of PPS3. PPS3 by nickel ratio is 10 power 4 then you get only cyclohexene if you have if you want cyclo octadiene you have to carry out the same reaction in the presence of close to 100 equivalents of PPS3 little less than 100 equivalents of PPS3 with respect to the nickel. So, nickel is present only in catalytic quantities and the butadiene is present in a very large amount. In fact, it is less than 1 percent of the nickel catalyst with respect to the butadiene. 1 is to 170 is the ratio between the nickel and the substrate which is the butadiene unit. So, this is a very interesting ligand reactivity control map which allows you to synthesize or isolate one product with respect to rather than a second product. So, one of the reaction channels is facilitated when you have very large amounts of the ligand or you can eliminate one product all together by removing the presence of the ligand as when you want to synthesize cycloadedicotriene. So, what he also showed was the ligand was critical when you want to synthesize a particular molecule. So, if you change from PPS3 to the phenoxyphosphine. So, here you have two this phenoxyphosphine units. So, this is PPSOPH2, two phenyl groups have been replaced by phenoxy groups. So, this phenoxyphosphine is capable of changing the way in which the same reaction happens and here the ratio you can see that the ratio of the metal to ligand is different for the generation of the cyclohexene and the cyclo octadiene. Needless to say one has to painstakingly do the reactions at various concentrations of ligand to metal in order to generate such ligand reactivity control maps. But this gives you a very clear idea that the nature of the metal, the nature of the ligand and the ratio of the ligand to metal is critical in order to generate the product that you want. So, here we have also indicated the three products in perspective. So, I have shown you all three if you only want to join the two butadiene units together and for form a cyclic compound then you need to stop the reaction in the first stage itself and that is what you do. You have to combine it in such a way that you form a bond between 3 and 3 dashed and that is what gives you divinyl cyclobutene and you can also do it between one and one dashed and then you get cyclo octadiene and you can also combine it between 1 and 3 and then you get vinyl cyclohexene. So, depending on the carbon atom which does the reductive elimination you would be able to control the size of the ring and the product that you want. If you use carbon monoxide he showed that you can in fact do a reductive elimination where you have one CO which is inserted in the during the insertion process and that seems to be fairly understandable. If you use high pressure of carbon monoxide you can introduce a carbon monoxide instead of adding another molecule of butadiene. So, that is what is happening here. So, you get a ketone and as one of the products you can stop the reaction literally poison the reaction and stop it from proceeding further by very simply adding triethylphosphine. Triethylphosphine is a very strong ligand and it stops the reaction at the stage where you get the cyclo dodeca triene and the nickel which is coordinated to the three double bonds in the cyclo dodeca triene in a symmetrical fashion. Pph3 when it is present in small amounts can do just the cyclization reaction. In the same reaction he also showed that if you added hydrogen a high pressure of hydrogen you can hydrogenate the three double bonds which are present in the molecule. You can form in fact the linear molecule where all three double bonds are all four double bonds are completely hydrogenated. So, it is possible to make a variety of interesting molecules starting with simple butadiene and the intermediates involved are formed through a carbon-carbon coupling process where the allyl group has literally behaved like a radical which is what we showed when we two butadiene units are combining together you form a bis allyl complex and a radical on each of the terminal carbon atoms and these two radicals can combine together and form the carbon-carbon bond. So, instead of having a radical type reactivity as what we have shown just now is it possible to make the allyl radical or the allyl species behave as a cation CH35 plus or as a CH35 minus. This really depends on the D electron count and the total charge on the complex. Both of these seem to be playing a role and we will illustrate a few examples where you can have such react three changes of the allyl group. Briefly, we will just go back to the molecular structure where we have shown the various MOs of the allyl group. If the allyl group is filled up to the second MO then what you have is an allyl anion. Now, the electron count on the metal can by pumping in electron density into the second MO you can have the behavior as if it is an allyl anion. Now, we if you remember the preparation of allyl molecules we use two different types of preparations to major classes of preparations one of which was a case where we took the allyl alcohol coordinated to a manganese for example and treated it with HBF4. HBF4 is a strong acid it converted the OH group to a positively charged system and then it generated a water molecule and then at least a formally allyl cation which is coordinated to the manganese. But because of our electron counting nomenclature we would think of the species as if it is a allyl group which is coordinated to the manganese where the allyl is got where the allyl is a four electron donor. A similar situation happened when you took the butadiene molecule coordinated it to iridium one and this iridium one species can be protonated again and it also generates an allyl species. It is clear that in these two reactions the allyl group would like to behave or is a electron deficient and would like to behave as an allyl cation. There are other examples example which I have shown right here where the metal atom is in fact a palladium 2 plus unit where the palladium electron count is d 8 and it pumps in electron density into the allyl group. So, you have in fact the generation of a species here for example, you have a p d c l 2 allyl moiety with a net negative charge. So, here the allyl group will probably behave as a allyl anion. Now, we can change the reactivity of this allyl anion by substituting one of this chlorides with the p p h 3 and what we have to do is to take this allyl moiety and eliminate a chloride and add a triphenyl phosphine. When you add triphenyl phosphine it forms a complex which is having a net positive charge because you lose the two chloride ions which are coordinated to it. When you do that then you can treat it with a variety of nucleophiles. A typical nucleophile which you can add is a diethyl malonate which is a diethyl malonate has been deprotonated to form the sodium salt and this typically adds on as a nucleophile to these allyl groups. So, it will behave as if it is a allyl cation which is needs an electron rich center to react with. So, here you have a different type of reactivity and the same situation happens when you have c p f e c o 2 c h 2 allyl here again the allyl group is eta 1 in nature. Here it is pictured here the picture is shown this is a eta 1 allyl and it behaves as if it likes positively charged centers. So, just as the palladium chloride compound was a system where you have a lot of electrons on the metal in the ion case also this is an 18 electron system and it has excess electron density which is pumped into the allyl group and so this allyl will behave as if it likes positively charged centers. In fact, this can react with aldehydes and with other electron deficient carbon centers. So, you can alkylate species using this allyl group. Now, here is another example where the allyl group will behave as a positively charged center. Now, it is clear that the molybdenum complex that I have written here is positively charged and it also has a nitric oxide which is coordinated to the molybdenum and this compound is electrophilic. In fact, it will love to react with a species which has extra electrons. So, this will this loves electrons and so it would like to react with the species which has extra electrons. In the previous example I have shown you a complex which loves positively charged centers. Here is an allyl group which is positively charged. In fact, it is interesting that you can now react these two species together and if you react these two species the actual reaction was carried out using a methyl substituted allyl group on the molybdenum. You could have either the exo or the endo compound of the allyl group which is present on the molybdenum. Now, you can react it with the eta 1 species that we talked about. What is interesting is that you now have the formation of a seven carbon systems system. Here you have four carbons and I had three carbons here and these two centers combined together to form a carbon carbon bond. What you end up with is a seven carbon diene system which is formed as a result of interacting these two allylic complexes. One was the eta 1 electron rich. This was electron rich and here was an electron poor and this is electron poor and positively charged center and this is an electron rich three carbon center. These two combined together to form a diene molecule where a new carbon carbon bond was formed and this is the carbon carbon bond that was formed as a result of interacting these two units. This molecule now has got two isomers because the two metals can be present on the same side of the diene or on opposite side of the diene. You should remember that the molecule the alkene system is flat. Although the other carbons which are attached to the alkene here are sp3 centers but the two double bonds can be in a plane and the molybdenum can be in one side and the ion can be on the opposite side. So, you have both syn and anti forms of this cis and trans forms of this molecule and you can both of them were isolated together in the reaction mixture. So, what we have shown up to now is the fact that you can have a variety of different reactions between the allyl group and electro files or nucleophiles. Now, we will take one metal complex which seems to or one metal center ruthenium here. So, this compound is capable of carrying out both electrophilic attacks and nucleophilic attacks. Now, here is the example you have a ruthenium two allyl complex which is formed by treating ruthenium carbonyl molecules ruthenium carbonyl molecules with an allyl acetate. You can generate or an allyl bromide you can generate this allyl complex which can be isolated and characterized. A stoichiometric reaction is what I have shown on the top of this transparency. Here is a stoichiometric reaction where you end up with allyl group adding an electron adding to an electron poor carbon center which is the aldehyde. So, this behaves as if it is electrophilic. This C H O is electrophilic and it reacts with this allyl group. Now, the same reaction can be carried out in a catalytic fashion by using a very small amount of ruthenium carbonyl and adding an aldehyde and allyl acetate. Allyl acetate reacts with ruthenium to form the allyl complex and that allyl complex now adds on to an electron poor center because it is electron rich. Now, this is the ruthenium two center which makes the allyl group electron rich. You can see that this reaction is something that produces two different isomers. People have been able to control the stereochemistry of this carbon carbon bond formation by using modifying ligands on the ruthenium. So, here is the structure of the ruthenium allyl compound which in which there is a water molecule in the crystal structure instead of the acetate group. So, this is just to show you how the structure might be present. We can also have on the same type of in the same in this reaction. We can also have addition of a nucleophile on to the allyl group and this can be done with a ruthenium cyclopentadienyl molecule in which case the ruthenium goes from the ruthenium is in a plus 4 oxidation state. So, this allyl group behaves at as if it is electron poor because it is attached to a ruthenium four center. So, the ruthenium four center now reacts with a nucleophile which is a heterocyclic ring system which attacks the allylic moiety. So, this attacks this carbon and eliminates the acetate, but this whole thing is facilitated because the ruthenium four allyl moiety is generated in situ during the course of this reaction. You have a new carbon nitrogen bond being formed and this carbon nitrogen bond is formed in the allylic position because of the formation of an allyl molecule. So, you can have a variety of reactions which are which are possible because of the coordination of the ruthenium and basically the allyl group behaves both as an electrophilic center and as a nucleophilic center. The allylic complex is almost like a chameleon where the just like a chameleon changes color depending on its surroundings. The allyl group changes color changes its electrophilic or nucleophilic nature depending on the metal which is attached to it. So, to conclude today's talk we can just say that the metal controls the reactivity of the allyl group and the ligand can modulate its reactivity. So, the based on the type of ligands that one can add on to the metal and also the oxidation state of the metal one can change the reactivity of the allyl group.