 In this lecture, we will discuss a very important class of molecules called metal allyl complexes. These metal allyl complexes are indeed very complex, because they adopt a variety of interesting structures and they undergo a variety of interesting reactions as well. So, before we plunge into metal allyl complexes, let us just get the context in which we are talking about them. We started by looking at molecules where we had a formal metal carbon double bond. These are molecules like carbenes and also metal carbonyls, where you do have a metal carbon double bond, which can be written in the valence bond formalism. You can use a double bond to indicate the type of interaction between the metal and the ligand. The important thing is that there is only one carbon attached to the metal. Following that, we also discussed metal alkyl complexes. These are complexes where there is a metal carbon sigma bond and there is no formal pi bond between the carbon and the metal. So, these are typified by the example that I have given you here. Any alkyl compound, metal alkyl compound represents a case where you have only a sigma bond between the organic ligand and the metal. We also talked about metal allyl complexes, where two carbons are interacting equally with the metal atom. Although we do not indicate these interactions specifically, you can see that these are molecules in which two carbons are bonded to the metal. So, we write them as hapticity as 2 and indicated with eta 2. This eta 2 symbol indicates that two carbon atoms of the ligand are interacting with the metal atom. Today, we are going to talk about metal allyl complexes, which are interesting because they have two forms which are predominantly found in the literature. One in which the three carbon atoms of the allyl moiety are interacting with the metal. In almost equidistant, the metal is almost equidistant from all three carbon atoms. In other words, carbon atom 1, 2 and 3 are interacting equally with the metal. So, the hapticity is 3. So, we indicated with the eta 3 notation, eta with the superscript 3 to indicate that three carbon atoms are interacting with the metal simultaneously. The same molecule can also interact with the metal through only one carbon atom. When it does so, we have a eta 1 allyl species. This is also an allyl compound, but this is a eta 1 allyl, where there is only a sigma bond between the metal and the carbon moiety. The double bond, which is present on the carbon species is not interacting with the metal. We shall see specific examples of these molecules as we go through this lecture. So, in the context of the ligand space that we are examining in organometallic chemistry, you will notice that the sigma molecules are all listed above the central line, where carbon monoxide is falling. So, if you can see this central line and all the molecules, which have only a sigma bond or a sigma and a pi bond are indicated above this line. All those which are interacting with the metal with only a pi fashion are indicated below this line. So, these are the type of ligands, where you have a sigma bond as well as a pi bond in the molecule. So, today we will be discussing the allyl complexes in organometallic chemistry. As I mentioned earlier, allyl complexes are in fact very complicated because of the way in which they interact with the metal. There are a variety of ways to make them and today we will concentrate on the ways by which we can make these allyl complexes. Later on, we will talk about the reactivity of these metal complexes. So, today we will try to cover the ways by which we can make these allyl complexes and the type of structures they adopt in solution and in the solid state. First, let us talk about a simple substitution reaction. In these reactions, which two examples are given here in this slide. If you treat a metal halide with an allyl magnesium bromide, which is a simple Grinard reagent, allyl bromide with magnesium turnings will give you allyl magnesium bromide. This allyl magnesium bromide is predominantly C 3 H 5 minus. In other words, you can think of it as if it is C H 2 C H C H 2 minus. This allyl anion is what we are talking about is a predominant species that is formed when you have allyl magnesium bromide. Although it may not be in an isolated form, that is the net species which reacts with the nickel bromide. So, it displaces a B R minus and so what you end up getting in this reaction is M G B R 2 as at the end of this reaction magnesium bromide would precipitate out and a bis allyl complex of nickel is formed. Now, the complex that is formed can be written in this fashion. The anionic allyl groups are indicated here as being coordinated to the nickel atom in a sandwich fashion. Although we draw it in a flat two dimensional representation like this, the molecule is in fact like a sandwich and we will talk about this in a few minutes. The same reaction can be done with a variety of metal halides and this allows you to prepare homoleptic metal allyl complexes. Homoleptic again let me repeat, there are molecules in which there is only one type of a ligand. So, in this particular case you have only allyl ligands and zirconium because the oxidation state is plus 4. You have 4 allyl molecules which are attached to the zirconium. Nickel is divalent and so we have 2 allyl groups on the nickel. The nickel group in the periodic table, nickel palladium platinum form very interesting allyl molecules, allyl complexes and I will show you a structure of allyl palladium complex to start with. So, that you can appreciate the type of interaction that is there between the metal and the allyl moiety. So, this is the complex in which we have a metal atom which is sandwiched, which is sandwiched between 2 allyl groups and you will notice that the allyls are staggered. In other words, one case the central atom has got a hydrogen pointing upward and in the other case the central atom has got a hydrogen pointed downwards which I am indicating with an arrow. So, here is one hydrogen pointing upward that is the central carbon atom and we can label all the atoms and you can see that the 3 carbon atoms are equally bonded to the nickel which is present here. So, this is a case where you have the 3 carbon atoms almost equally bonded to the central nickel and you can you will be able to see these distances in a moment. I will measure them this is 2.026 to the terminal carbon 2.026 is a distance between the terminal carbon and the central carbon is at a distance of 2 angstroms. So, you can see that they are almost the same distance the central carbon atom and the terminal carbon atom and you can see that the 2 allyl groups are in fact in a sandwich like fashion between the central metal atom here. So, you can appreciate this 3 dimensional structure and the way in which the allyl group is interacting with the metal. You will also notice that the allyl group is almost planar if I have aligned it in such a way that it is almost planar all the hydrogen atoms and the carbon atoms are in a approximate plane and between these 2 planes the metal atom is centrally located. The carbon atom that is in the center is slightly elevated or slightly away from the metal atom or so it appears, but you can see from the distance that it is in fact almost equally placed from the metal as the terminal carbon atoms. So, these are some very interesting molecules that are formed by the metal allyl species and you can have with zirconium you will have 4 of the allyl groups it will be arranged in a tetrahedral fashion. The only difference is that because it is a tetrahedron there are more steric crowding, but nevertheless zirconium being a large metal is able to accommodate all 4 allyl groups around it. So, let us just get back to our presentation here. Let us move on if you instead of doing a simple substitution reaction we can also do a trans metallation reaction. I distinguish the substitution reaction from the trans metallation reaction purely because in the trans metallation reaction I have a allyl tin group which is a main group allyl complex which is sigma bonded where the allyl group is sigma bonded to the tin. This allyl group is not heavily polarized as allyl minus and tin plus and it can react with the palladium chloride. So, in such a way that you can transfer one allyl group to the palladium. When you do that and you reacted with 4 molecules of a triphenyl phosphine you end up with a very interesting complex where you have palladium coordinated to allyl group and a triphenyl phosphine moiety and a chloride. This molecule has been studied extensively because it reacts with nucleophiles and it allows you to carry out a nucleophilic substitution on what is apparently an allyl group which is originally allyl almost like an allyl minus and in spite of that it can undergo reactions with nucleophiles. The metal allyl usually should be the, if you want to do a trans metallation reaction the metal allyl should be little more electropositive than the metal to which you want to transfer it to. Similarly, the metal to which you want to transfer it to palladium, if it is a palladium chloride then the chloride group is transferred to the group which is transferring the allyl group. So, that should like the chlorine better. So, in this case that is a tin. So, this turns out to be an exchange of two groups and that is why we call it a trans metallation. Now, we will just take a look at the palladium allyl triphenyl phosphine complex because that is also a system where you have 3 allyl groups pointed towards the metal atom and here is the molecule which has got a triphenyl phosphine. So, the phosphorous atom is shown in bright orange here. The phosphorous atom is bright orange the metal itself is the metal is present in the metal is present just in the as in the other case in the nickel case it is present in almost equidistant from the 3 carbon atoms which are interacting with it and the allyl group is almost planar. There is a small deviation of the hydrogen from the plane in which the 4 the 2 CH 2 units are present. So, the central hydrogen is moving away from this plane away from the metal atom. This turns out to be unique for allyl complexes one of the hydrogen atoms of the central hydrogen atom moves away from this plane not very much may be by a few degrees, but nevertheless it moves away. So, that the bond distance between the carbons and the palladium and the orbitals are oriented in the right way to have maximum interaction. We will talk about this interaction little later, but you can see that this is the molecule in which the 3 carbon atoms the 3 carbon atoms are indicated here in grey. The 3 carbon atoms are interacting equally with the palladium and the chlorine atom is coordinated to the palladium as well. So, let us get back to our presentation now. If you want to prepare an allyl moiety apart from trans metallation and apart from the substitution reaction that we talked about we can also carry out what is called an oxidative addition reaction. These are reactions which we have talked about little earlier. If you have a metal halide or metal anion and an alkyl halide the metal anion can carry out a nucleophilic substitution reaction on the chloride. In principle this is MnCO5 minus. So, we can talk about it as MnCO5 minus and sodium plus and this MnCO5 minus now makes a nucleophilic substitution reaction and SN2 reaction basically and it forms an intermediate which is a sigma allyl complex. So, in the sigma allyl complex that we have indicated here we have a carbon metal sigma bond and this double bond which is present on the allyl moiety. These 2 carbons are not interacting with the metal. We will see a crystal structure of one of these molecules shortly, but once this sigma allyl complex is formed you will notice that these 2 carbon atoms this double bond can give 2 electrons to the metal and on the metal you have carbon monoxide units. So, this carbon monoxide is also a 2 electron donor. So, one can do a substitution reaction where the carbon monoxide is replaced with this double bond in a internal attack. So, that is exactly what happens you have loss of carbon monoxide and the formation of an allyl manganese complex and this allyl manganese complex now has 3 carbon atoms bonded to the metal just as we have shown in the earlier crystal structures. So, in a initial step you do form a sigma allyl complex, but the sigma allyl complex is fleeting in nature. It is not stable it will quickly rearrange to give you the eta 3 molecule or the tri hapto allyl species which we are able to isolate finally. Now, it is possible to have only the sigma allyl attached to the transition metal in some complexes and if you want to see a structure of this I will now show you a structure of this molecule where you have only a sigma allyl species which is attached to the here is a platinum complex not the manganese complex that we talked about earlier, but here is a platinum complex which has got only the carbon atom the terminal carbon atom attached to the metal. So, here we will let us take a look at the distances which are present this is a platinum which is attached to the terminal carbon and that distance that distance if you measure this distance let us measure this distance again between the platinum and the carbon is 2.086 angstroms. So, this is the carbon which is attached to the metal whereas, in the same way if you measure the distance between the terminal carbon atom. Let me show this to you in this orientation. So, that you can see the two distances very easily here is the distance between the carbon atom which is attached to the platinum and that is 2.086 angstroms. The carbon atom which is further away which I have marked in yellow now that is at the distance of 3.874 angstroms. In other words the double bond these two carbon atoms which I have indicated for you in yellow now this carbon atom and the second carbon atom are not really interacting with the platinum. So, this is a clear example of a sigma allyl complex where only the terminal carbon atom is interacting with the metal and the two carbons which have the double bond are not interacting with the metal at all. So, the carbon carbon double bond is clearly indicated by a short 1.3 angstrom distance between the two atoms and the two atoms which are bonded through a sigma bond are close to 1.5 angstroms or 1.485 angstroms. So, it is very clear that only isolated double bond is present and carbon metal distance is close in one case 2.086 in the other case it is 3.8 angstroms. So, very clear indication that it is a sigma allyl complex. So, it is interesting that you can have a variety of different coordination modes between the allyl group and the metal atom and we have seen three instances one in which it is like a sandwich one in which it is pi bonded and attached to other ligands as well and the third example where it is only sigma bonded to the metal atom. So, because all these three forms of the allyl complex are equally accessible we have very complicated chemistries which can result from the interaction of these three forms. Let us proceed further with synthesis of metal allyl complexes from alkenes this time. These are some very interesting reactions that happen when you treat an alkene compound alkene organic alkene which has got an allylic hydrogen which can be abstracted by the metal and in this case it is a palladium 2 complex palladium pdcl4 2 minus which is interacting with the allyl group. It first forms a olefin complex and this olefin complex is indicated here. Here is a olefin complex only the double bond is interacting with the metal, but because you have a hydrogen which is adjacent to the metal. You can have what is called abstraction of a hydrogen and that abstraction of a hydrogen can result in a formal oxidative addition resulting in this palladium complex which is formally in a plus 4 oxidation state. Now, the oxidation state of plus 4 is not common with palladium but nevertheless it has been observed. So, it is possible that this reaction goes through an oxidative addition and then it can rapidly eliminate a molecule of hydrogen chloride and that is exactly what is represented here. The isolated complex is a palladium chloride dimer because allyl group is negatively charged species. You can see in this molecule that palladium has retained the plus 2 oxidation state reverted back to the plus 2 oxidation state. We started with palladium in plus 2 and then we formed an olefin complex and in the olefin complex we had an oxidative addition, abstraction of the palladium and that results in a species which is palladium in the plus 4 oxidation state. You have 2 chlorides, 1 hydride and 1 allyl species. So, this species rapidly loses HCl that is does a reductive elimination and forms the palladium 2 complex which we have indicated here. So, this is one way in which we can understand the formation of an allyl complex from an alkene. Let us talk about an alternative way in which we can understand this reaction. Here I have formed the double bond. You have an olefin complex which is formed between the palladium and the alkene. After it forms the palladium olefin complex, the hydrogen which is present on the allylic position, this position it can be removed from the molecule as H plus. In other words, the pair of electrons which are held between the hydrogen and the carbon are donated to the palladium and that is indicated here and the chloride leaves from the coordination sphere of the metal. You should remember that palladium first had 2 extra chlorides it was PdCl 4 2 minus and in the initial coordination of the alkene we lost 1 chloride and that is what is indicated here. So, we have a mono negative species to start with and if you lose an H plus and simultaneously lose a Cl minus, this hydrogen goes off as H plus, this hydrogen goes off as H plus and this chloride goes off as Cl minus. We will be still left with a negatively charged species, but notice we have not increased the oxidation state of the palladium during the course of this reaction. As a result, what we have ended up with is a palladium 2 species in this case also, here also palladium is in the plus 2 oxidation state. So, you have an allylic group 2 chlorides and a net negative charge and as a result this molecule can now lose a Cl minus and form the dimeric species which I have formed in the previous case also. So, in order to form this beautiful palladium allyl dimer palladium chloro dimer, you need you can do it in 2 different you can understand it in 2 different ways. One case you do not need to do an oxidative addition and a reductive elimination and in the other case you do end up with an oxidative addition and the palladium 4 complex which rapidly eliminates HCl. Because topologically you cannot distinguish the 2 processes it is not possible to figure out exactly how the reaction is happening, but nevertheless the final product formed is the same. So, let us this the alternative yet another alternative way of understanding this is to think of it as a sigma bond metathesis. Now, after forming the palladium olefin complex one can think of this H as leaving as HCl. So, you have a sigma bond metathesis palladium forms a new bond between the carbon and the metal and a new bond between hydrogen and chlorine and the old bond that was present the old bond which was present between the carbon and hydrogen palladium and chlorine are broken. So, in the sigma bond metathesis 2 sigma bonds are broken 2 new sigma bonds are formed and hydrogen hydrochloric acid is eliminated and you form the complex. So, whether you think of it as a sigma bond metathesis or you think of it as a loss of a proton and a chloride either way you would end up with no oxidation state change on the palladium and the formation of a palladium allyl chloro complex. So, these are 2 different ways of understanding the reaction without an oxidative addition. Now, oxidative addition to palladium definitely happens in situation where you reduce the palladium 2 here again we start with Na 2 PdCl 4 palladium is in the plus 4 oxidation plus 2 oxidation state Na 2 PdCl 4 and you reduce it with carbon monoxide. Carbon monoxide or SNCl 2 tin chloride tin in plus 2 oxidation state can reduce palladium to palladium 0 and tin will get oxidized to palladium 4 sorry tin will get tin will get oxidized to tin 4 and will become SNCl 4 and palladium will get reduced to palladium 0. Now, palladium 0 can oxidatively add allyl chloride. So, this is a simple reaction which will also lead to the chloro palladium allyl complex which we can represent like this. So, this is the dimeric species that we are talking about the same complex which we have been showing in the last 3 slides. So, you will notice that here palladium 0 has changed to palladium plus 2. So, this is an oxidative addition to palladium 0 and it is promoted either by an initial reduction of the palladium 2 to palladium 0 with carbon monoxide or with tin. Both of them are known to cause this reaction to proceed forward. Now, it is always possible to use palladium metal or any metal in a 0 oxidation state when you do a co condensation reaction with a ligand. Then you can make the allyl complexes and in the case of co condensation reactions the advantages the metal is present in a very highly reactive state. In this case we are using palladium as an example, but when you have a benzyl chloride. So, here you have benzyl chloride and you have a very reactive very reactive metal atom. You can undergo an oxidative addition and this is in fact a molecule which should have formed only a sigma complex because you have an aromatic molecule. You have an aromatic molecule and you have the possibility for oxidatively adding it to the halogen. So, this is what you would expect, but surprisingly the double bond which is present on the benzene ring just this double bond also interacts with the metal. So, you have an allyl complex with the benzene ring participating in the reaction and partly it will lose the aromaticity. Surprisingly there are many bonds which can do this. So, benzyl halides can oxidatively add to the metal atom. On oxidative addition one of the double bonds interacts with the metal atom to form a allyl compound in which it is a eta 3 molecule. The hapticity is 3 and it is looking just like other allyl complexes. Let us take a look at this molecule because this is in fact a very strange compound where you can have the benzyl interacting with the metal atom. So, we now take a look at the benzyl case. So, here is an example which has been taken from molybdenum compound. We will not worry about the remaining ligands on the molybdenum, but you can see that there is a very nice aromatic ring which is present. This aromatic ring is interacting in such a fashion with the molybdenum such that 3 of the carbon atoms this C 1, C 4 and C 14 all of them are interacting with the molybdenum in an equal fashion. So, if you want to measure the distance you can see that this carbon is at a distance of 2.2 angstroms. The first carbon the benzyl carbon is at a distance of 2.2 angstroms and I have indicated it here. The central carbon which is the C 4 is at a distance only at a distance of 2.3 angstroms. So, 2 carbon atoms are interacting almost in a equidistant fashion from the molybdenum atom. The third atom is at a distance of 2.48 angstroms. So, although it is slightly away you can see that all three carbon atoms are interacting with the molybdenum and it looks like a allyl complex. In fact, from the electron count you can see that this double bond is in fact interacting with the molybdenum atom. To show that there is bond alternation you can look at the carbon carbon bond distance of the bond which is trans or which is opposite this and that is clearly 1.42 angstroms and it is much longer compared to the other two double bonds. So, there is some bond fixation or the delocalization is completely destroyed. Two bonds are short 1.3 to 8 angstroms one bond is very long. So, that this really looks like a complex in which the allyl moiety is interacting one of the double bonds has decided to interact with the metal in such a way that the aromaticity of this benzene ring is destroyed. So, the gain that is there for this molecule in going from this aromatic system to this non aromatic system is because it can satisfy the molybdenum requirement for two extra electrons and those two extra electrons are coming from the pi bond which is present here. So, it is interesting that you can have this alternative way of interacting with a metal even if you have a aromatic system part of the aromatic system can interact with the metal and form a pi allyl complex. So, these are called pi allyl complexes or tri hapto species or eta 3 complexes all three nomenclatures are used in the book. So, let us move on let us take a look at another way of making allyl complexes and in this case what I am shown you is a system in which there is a formal oxidative addition oxidation state change and these are molecules in which there is a manganese in a plus 1 manganese is present in a plus 1 oxidation state these two are carbon monoxide units. So, what happens is there is a olefin complex and in this olefin complex there is an allylic OH group. This OH group can be protonated with a strong acid like HBF4 and if it protonates it and loses a molecule of water after protonation if it loses a molecule of water then formally this allylic moiety will become allylic cation. Now, all three carbon atoms of this allyl group all three carbon atoms which I am going to indicate with green atoms all the three carbon atoms can now interact with the metal atom with the manganese in a eta 3 fashion. So, because this is an eta 3 interaction and formally we talk about allyl as allyl minus the carbon is always considered as more electronegative than a metal you end up with a oxidative addition reaction. Because manganese is in the plus 1 oxidation state in this molecule after protonation and loss of water molecule you will end up with a cationic species and this cationic species now has got two negatively charged molecules or ligands one is C p minus and the other is the allyl minus. So, as a result what you end up with is a manganese in plus 3 oxidation state. So, here manganese is in the plus 3 oxidation state and you started with manganese in plus 1. So, you formally carried out a oxidative addition reaction a similar situation happens when you protonate a diene complex here is an iridium complex which is butadiene unit is coordinated to the iridium here also iridium is in the plus 1 oxidation state and after the reaction after protonation iridium goes to plus 3 oxidation state. Now, in order to show this reaction exactly how it is reacting I have indicated this double bond this is a C H 2 unit here this C H 2 unit if it is protonated it will become a methyl group and that is that methyl group is now indicated initially it was indicated in green C H 2 and now it is become a C H 3 group and once it becomes a C H 3 group you will notice that this carbon can equally interact with the iridium and all three carbon atoms the two carbon atoms that were part of the diene and one carbon atom that was part of the in both all three of them can interact with the iridium and so now you have the possibility of an allyl complex. So, here also it is a eta 3 allyl complex and iridium because you have protonated it you need to add a plus charge to this molecule and if you count the number of anionic groups present on the iridium there are already two anionic groups there is a plus 1 charge. So, iridium is in the plus 3 oxidation state. So, these are cases where you can do an oxidative addition with a proton oxidative addition with a proton resulting in a change a formal change in the oxidation state of the metal and the formation of an allyl molecule. So, in both cases you get a iridium 3 sorry iridium 3 and manganese 3 species and both cases you have a eta 3 coordination. Now, there is another anomalous reaction which we should talk about and that is the formal addition of a HCOCO4 that is a formal addition of a metal hydride a classic example is HCOCO4. We have discussed this earlier, but hydrides are anomalous because many times they will react as if they are protons the hydrogen reacts as if it is a proton. So, HCOCO4 is a strong acid in fact and if this metal hydride formal metal hydride is added to a diene it ends up informing a eta 3 allyl complex it ends up informing a eta 3 allyl complex and here I have a cobalt example you will notice that when you add the hydrogen the hydrogen can be added to the diene to the diene. So, that it forms this allyl complex, but the metal group can be either oriented away from the central hydrogen. So, if you talk about the central hydrogen as a unique hydrogen now the metal group can be added in such a way that it is pointed in the same direction. So, this metal group is now sin with respect to the central hydrogen or it can be anti to the central hydrogen. So, this results in the formation of two different molecules either a sin or an anti molecule. One last example where you have a hydrogen transfer which results in the formation of allyl molecule. Here you have a metal group and one of these hydrogens can be transferred to the nickel such that you form an allyl molecule. This molecule is in fact a dynamic system at temperatures above 40 degrees you have the formation of a eta 3 allyl complex. If you cool the molecule if you cool the reaction system and then you end up with a olefin complex. So, this tells you how the energy required for transferring the hydrogen from the alkene moiety from the allylic position of the alkene to the metal is very very small. So, you are able to transfer the hydrogen very easily and form a eta 3 allyl molecule. So, here is yet another example which is formally an oxidative addition. In this particular case there is no negative charge, but still C P C O C O 2 is a electron rich species. This is a electron rich species and this electron rich species does a nucleophilic attack on this allyl and as a result the iodide goes out and it forms a species forms a intermediate in which you have eta 3 allyl coordinated to the cobalt and iodide is a counter ion. But during the course of this reaction iodide attacks the cobalt again this iodide moiety attacks the cobalt again and displaces the carbon monoxide. So, that you end up with end up with a cobalt neutral compound. This is a positively charged species and now you have a completely neutral molecule in which cobalt is in the plus 3 oxidation state. Here also it is an oxidative addition, but nevertheless earlier we looked at a manganese compound which is negatively charged and it underwent or it attacked allyl halide displays the chloride. Here we have a neutral molecule which is carrying out to nucleophilic attack on an allyl halide which is a allyl iodide species. So, a variety of reactions have been shown. These are examples which we will talk about later where allyl molecule is formed from a larger ring system. So, we will skip this here is a species where you again have a diene and instead of adding a hydride instead of adding h plus or h minus we are adding an R minus. This R minus can be R M G B R and this R M G B R essentially does a nucleophilic attack on the double bond here and it results in the formation of an allyl molecule. So, there are a variety of ways probably more than 10 ways we have considered in which we can form an allyl molecule. Now, what we have to look at is the structure and bonding in these molecules. I have drawn out the allyl system which is a 3 carbon system in which there are 3 pi bonds or 3 p orbitals which are not utilized in bonding. So, if you look at this particular system you will notice that the 3 p orbitals can combine together to form 3 molecular pi molecular orbitals out of which this is the lowest energy orbital in which all 3 of these lobes have the same sign. Now, the best way to illustrate this is to is to show is to overlap these molecules overlap these orbitals here on my on the right side of the screen you have the metal orbitals on the right side of the screen you have the metal orbitals on the left side you have the allyl group and the allyl orbitals and the best way to understand which orbitals would overlap with the metal orbitals will overlap with the pi orbitals of the allyl group. One has to just move these groups of these groups to the right position and see if they will form an overlap a positive overlap and you will notice that the 4 s orbital can form a nice overlap with new if the allyl group is present let us say in the x y plane. So, if the allyl group is present in the x y plane and if the metal is present in the z axis at this position if the metal is present in the minus z axis along the z axis, but in the minus part of the z axis then you can see very clearly that the s 4 s 4 p z and the 3 d z 3 d z squared orbital all of them are capable of interacting with the lowest energy pi molecular orbital. So, the lowest energy pi molecular orbital will have 2 electrons which will have 2 electrons and these 2 electrons can interact or can be donated to the 4 s which will be m t the 4 p z and the 3 d z squared if it is m t as well. So, you can realize that if you are talking about an allyl anion as we formally talk about there are 2 electrons in this orbital also. So, these 2 electrons can be in fact interacted with or these 2 orbitals on the metal the 4 p x and the 3 d x z both of them can interact with the with the second pi molecular orbital which will also have 2 electrons if it is an allyl anion. So, the 2 donor orbitals on the organic system the 2 pi molecular orbitals on the organic system can donate electron density to the metal orbitals which are indicated here. Now, comes the highest energy molecular orbital which is the third pi m o and this is invariably m t. So, whether you talk about an allyl anion or an allyl cation the last highest occupied molecular orbital is always m t and that orbital can interact with the 4 p y orbital and the 4 p y orbital can interact and the 3 d x y z orbital can also interact. The 3 d y z orbital is the one in which the 2 the 2 lobes are pointed are going in and out of the plane of the screen. So, you can see some the second lobe is almost hidden by the front lobe. So, these 2 orbitals can interact nicely with the m t orbital now because you can have electron density on the 3 d y z. The 3 d y z can in fact donate electron density back into the m t orbitals of the pi and this will lead to stabilization of the pi system through back donation of electron density. Now, it is also possible that we talk about the allyl group as a allyl cation in which case the 3 d x z can also donate electron density back into the allyl group. So, there can be a lot of give and take the electron count on the metal electron count on the metal is important electron count on the allyl group is also important. You will remember that we generated the allyl group starting with an allyl alcohol by protonation and elimination of water. In that case it is very obvious that we are talking about an allyl cation which is coordinated to the metal. In that case we can say confidently say that the 2 orbitals both orbitals will be m t. But formally we always talk about it as if it is an allyl anion and so the back donation from the metal to the allyl pi star orbitals must be happening only from the 3 d y z. Now, the 2 d x squared minus y squared and the d x y orbitals do not have suitable combinations with the allyl orbitals and they are not interacting with the metal they are not interacting with the allyl group. But all other groups all other the s p and the 3 d orbitals are capable of interacting with the allyl moiety. You will also realize that these 3 orbitals interact with the allyl group in a sigma fashion. These 3 groups interact in the sigma fashion and so the allyl moiety can freely rotate about the z axis. So, if you think about the allyl moiety as sitting in the x y plane then the metal can rotate along the z axis can spin about its axis freely. Because the sigma there is a strong sigma bond which is formed between the allyl group and 3 orbitals on the metal. However there are 2 orbitals which are interacting in a pi fashion and that is the orbitals which are marked as 3 d x z and the 4 p x. These 2 orbitals if you rotate the allyl group with respect to the metal atom along the around the z axis. So, around this axis if we rotate it if you rotate the metal atom then what will happen is that we will lose the second p the pi orbital interacting with the 4 p x and the 3 d x z. So, these 2 orbitals will stop interacting or rather the overlap will go to 0 and the interaction energy will become 0. So, the extent to which these 2 interaction these pi interactions are important determines how much you can rotate the allyl group with respect to the metal atom. Just like we discussed the olefin case. Now having discussed the structure of the allyl molecules we must talk about the fluxionality and the reactivity. Let us take a look at the fluxionality of allyl complexes. We will discuss this in 2 dimensions when you show this figure and then we will try to illustrate it with a small video clip when we look at how this allyl group can react. So, here is a metal which is shown interacting in pi fashion and the plane in which the allyl group is present. This is the x y plane that we are talking about and this is the z axis. So, this is the z axis that we are talking about. So, when you rotate the metal atom with respect to this plane and you rotate the metal with respect to this along this axis along the z axis we can keep the sigma interactions intact the pi interactions will fail. Now it is possible for the metal pi a slight movement towards one direction. So, it can move towards a high carbon atom 1 and the carbon atom 1 is a 1 which is bearing h a 1 and h s 1. The a indicates the fact that it is anti to the central hydrogen. Central hydrogen is marked as h c and h a indicates that it is anti to the central hydrogen. S indicates that it is sin to the central hydrogen. So, this is anti and this is sin to the central hydrogen. So, these two carb, these two hydrogens will come closer to the metal if the carbon one is interacting with the metal in a sigma fashion. So, this is a pi allyl eta 3 system and this is a eta 1 system. So, we can go from a eta 3 to a eta 1 system fairly rapidly by a simple movement of the metal along this axis. So, that let us just mark this out. So, if you if the metal moves in this direction then it can go from the eta 3 to the eta 1 fashion. This can go back and forth very rapidly this eta 3 to eta 1 migration can happen very rapidly and it can go either to C 1 or it can go to C 3. So, we let us just mark this as C 3. So, either you can go to C 3 or it can go to C 1 both of them are equivalent and as a result you can see how the molecule can go from sigma complex to a pi complex and can go to this alternatively. In addition what happens is if the metal rotates around this bond if the metal rotates around this bond then from the minus z axis it can move to the plus z plane and if this happens then the allyl group is intact in the same position allyl group is present in the x y plane. But then the metal is moved from the minus z position to the plus z position. So, the metal can come back to the central position and so it will become a eta 3 system, but now the metal is in the plus z position. So, this type of a metal flip can result in a new complex and you will notice that because we have maintained the sigma bond what was originally H A 1 has now become has now come to a different position it has come it has formally become sin to the central hydrogen. So, the position of the sin and the anti-hyrogens have got interchanged as a result of this flipping of this of the metal from one side of the allyl group to the other side. So, this is called a metal flip and in the following lecture we will talk about how the allyl group can flip. So, there are two things that we are talking about one is a fact that the metal can go from an eta 3 geometry to a eta 1 geometry that also leads to a fluxional behavior and it will rapidly inter convert the two carbon C 1 and C 3. There is another fluxional behavior and that is rotation of the metal or rotation about this carbon bond carbon C 1 C 2 bond about the C 1 C 2 bond. If you have a flip then the metal goes from the plus z to the minus z or the minus z to the plus z and as a result you have interchanging of two hydrogens. The two hydrogens in which to which the metal was attached will now get interchanged and what was anti will become sin and what was sin will become anti with respect to the central hydrogen. So, you have a fairly interesting fluxional behavior and because these are rotations they can happen in the time scale of 10 power minus 13 to 15 seconds.