 In this lecture, we will discuss arene sandwich complexes. The analogy to a sandwich is because an arene ring or a cyclic pi system is a flat is a flat object pretty much. And between two flat objects, you have a metal atom, which is sandwiched between the metal atom is sandwiched between the two flat objects. And so, we call this a sandwich complex. And one often encounters cyclopentadienyl sandwich complexes. And the next probably the most common sandwich complex is a arene complex, where you have a benzene or a benzenoid ring, which is sandwiching a metal atom. The first time, someone made a sandwich complex, although not intentionally was as early as 1919. And that is when heine mixed a Grinard reagent and chromium trichloride. And he isolated a compound that analyzed for bisbenzene chromium. Now, there are a lot of complexes, where the two arene rings or the pi layers are surrounding a metal atom. And the metal atom happens to be in the chromium metal series. And so, we will see little later why this is the case. And why it is stable when you have the chromium group metal atom in between the two sandwich slices. So, a more rational synthesis of course, emerged after the synthesis of ferrocene and analysis of its structure. Fisher realized that it is possible, it might it should be possible to make metal complexes, where a benzene ring would be used instead of the cyclopentadienyl anion, which is of course, aromatic and is having six pi electrons. So, the two systems cyclopentadienyl anion and the benzene ring should be iso lobel or similar enough to replace one another. So, the Fisher Haffner synthesis as it is called was carried out in 1955 and directed by Fisher. And it was simply a reaction in which aluminum powder and chromium trichloride were used in the presence of an aromatic ring. And that is the aromatic ring, there are some limitations will come to that. And what was formed in the reaction mixture was in fact the cationic form of the sandwich complex. So, this is the cationic form of the sandwich complex that was isolated as a l c l 4 minus salt. But then it had to be reduced, because it was in the chromium plus 1 state. It had to be reduced and this was done using any 2 s 2 o 8. And it generated a chromium 0 complex in fairly good yield. Aluminium trichloride in fact was used as a catalyst. And it also serves as a suitable way by which you can form a large anion. Very often the larger the anion, the better it is when you have a very large cation to stabilize the crystal structure. So, this is how the first Fisher Haffner synthesis was carried out. And there are some serious limitations for this Fisher Haffner synthesis. There are limitations in terms of the metal. It works for pretty much most of the metals in the transition series, but there are some surprising gaps. For example, manganese, manganese is a metal atom, which is not found in this series. And so also the vanadium series, only vanadium is a suitable candidate for a metal in this Fisher Haffner synthesis. So, you take the metal chloride M x n and then reduce it with aluminum, reduce it with aluminum in the presence of A L C L 3. So, that is the Fisher Haffner recipe for making a metal sandwich complex with arenes. Now, apart from the fact that you have some limitations in the metal, there are also some limitations on the type of aromatic rings that you can use for making the metallocene complexes that we are talking about. Whenever there is a substitution on the aromatic ring, if it has a lone pair, it tends to coordinate to the aluminum and it stops the A L C L 3 from functioning. And as a result you cannot use aromatic compounds, which have got lone pairs of electrons. So, that means the chloro group and the N R 2 groups cannot be used as substitutions. Any halogen or any amine cannot be used if the aromatic ring will become inactive. Now, the reason for this is given here. It is the coordination of such groups to A L C L 3. There is also another set of compounds, which will stop the reaction or cause complications. And that is the presence of alkyl groups on the benzene ring. Surprisingly, because this alkyl group and A L C L 3 are the right combination for doing a retro free geocrafts reaction. If you have a couple of alkyl groups, you could have isomerization reactions. So, you might end up with molecules, where you started out with meta or also and you might end up with the para substituted, the alkyl para substituted aromatic ring as the bread part of the sandwich. So, there are some limitations in terms of the Fischer-Haffner synthesis, but it is still one of the simplest and most convenient ways of generating aromatic sandwich complex, metal sandwich complex. One can imagine a reaction between an alkyne and here I have shown you an alkyne, a simple alkyne reacting with a chromium phenyl compound. That is C 6 H 5 thrice chromium reacting with this dimethyl acetylene to give you hexamethyl benzene complex to the chromium. In a sense this is a templated reaction, where in the coordination sphere of the metal which is chromium, the three dimethyl acetylenes are trimerized and you end up with the hexamethyl benzene as the aromatic ring system which is sandwiching the metal. So, you also need little bit of oxygen to carry out the reaction. It is not entirely clear how you would end up with unoxidized material, but you can readily reduce it in the presence of sodium thiosulfate to back to the chromium zero species. So, here is another rational probably a more rational way of making it, but it has got limitations because it will not work with benzene. If you have large aromatic systems which are annihilated rings, then it is well known that they stabilize electrons. You can in fact reflux sodium or lithium preferably a lower boiling lower melting alkali metal in the presence of an annihilated ring. It generates highly colored solutions, where the electron has been transferred to the aromatic ring system. So, if you have ARH as an aromatic system and lithium and you heat it, then you end up with Li complex to this ARH, where presumably the electron is in fact has been transferred from the lithium to the ARH. You can now use this as a reducing agent because it is a source of electrons and reduce metal halides. So, you take metal halides which are complex with conveniently complex with THF, you dissolve them in tetrahedra furan, then you can reduce them with this lithium salt. It is literally a solvated electron where the solvent is naphthalene right here. So, what happens is you end up with the reduced form of the metal in the presence of ARH. The reduced form of the metal is in a very reactive state and so it reacts with the aromatic ring fairly easily and generates a sandwich complex. Now, this method is most suitable for the early transition metals right next to chromium and molybdenum for which of course, the most number of sandwich complexes are known. So, you can see that reduction and substitution is also a fairly common synthetic methodology that is employed for making a variety of organometallic compounds. Anytime you need an organometallic compound, you realize that there has to be a low oxidation state metal and so you generate the low oxidation state metal using suitable reducing agent. If you have an organic ligand in the presence of this as the reaction is being carried out, you can capture the low valent metal. In this case, it is metal 0 in the presence of your organic ligand to generate the complex. So, it is not devoid of surprises. Here is a surprise which should have generated the bis naphthalene zirconium compound, but surprisingly it generates a very strange compound where you do not have an arene ring, but you have the naphthalene which is kind of lost its aromaticity in one of the rings. So, that two double bonds or a conjugated diene system is now coordinating to the zirconium. Now, one can imagine why this would have happened because if you have two naphthalenes surrounding the zirconium, you still do not have enough number of electrons and we will come to that in a moment. This can lead to an unstable situation and the formation of this trimeric tris naphthaleno zirconium. This here again zirconium is in the zero oxidation state and it has got 4 into 3, 4 pi electrons into 3. So, total of 12 electrons donated to the zirconium. Totally zirconium had 4 electrons and so this is a 16 electron complex. So, let us proceed further. The most convenient way or the most general method for making a sandwich complex between an arene ring and a metal turns out to be the co-condensation technique. The co-condensation technique very often abbreviated as CC, co-condensation technique works even when the electron count is not appropriate. So, even if you do not get an 18 electron complex by using the arene ring and the metal, you would still form a nice sandwich complex as a result of this reaction. This reaction turns out to be a reaction where you have the metal in the vapor state, the vaporized metal and the aromatic ring also as a vapor which is condensed on to the sides of the vessel. We will show a schematic diagram of it in a minute and these two are allowed to react at a very low temperature at the temperature of liquid nitrogen. This is typically liquid nitrogen temperature that is used and then one slowly warms up the reaction mixture and ends up with very nice sandwich complexes which are pictured here. So, this turns out to be an extremely general method. Here is a schematic diagram of this setup. You have a way to vaporize the metal under very high vacuum. This whole reaction vessel is kept under very high vacuum and at that temperature when you have a very high vacuum you can now pass the ligand. So, that the ligand is now coating the sides of this vessel. So, if you have a very thin film of the ligand coating the vessel and this you have done at a very low temperature. This is liquid nitrogen as I told you and a very high vacuum. You can evaporate a variety of ligands and then if you resistively heat metal like chromium then you tend to have vaporization of the metal. So, this vapor tends to go and lodge itself in the sides of the vessel and that is what will happen when you start heating this resistively. So, here you can see the red hot metal which is getting vaporized and it goes and deposits itself in the sides of the vessel which is at a very low temperature. So, now what happens is you have vapor of the metal condensing on to the sides of the vessel which already has the ligand the aromatic ligand and at that temperature when the metal has got nothing else as to satisfy its electron count to get an octet and you have only the aromatic species which is present in the medium. Then you tend to have a reaction very convenient reaction between the two which leads to the formation of sandwich complexes. Now, the interesting thing is you can even use compounds like chlorobenzene or dimethylaminobenzene as suitable substrates for this co-condensation technique because here you do not have the complication of requiring aluminum trichloride in the reaction mixture. So, you can mix the two species conveniently even if it is a very strong ligand for aluminum chloride here you can use them in the co-condensation technique and form nice sandwich complexes. Now, using the metal ligand vapor co-condensation technique you can also use ligands which are sterically constrained. Here are a couple of examples where you have the metal sandwich between two aromatic pi systems where the aromatic pi system is constrained either by an alkyl chain as we have it pictured here. You have an alkyl chain which is constraining the two pi systems or you have tertiary butyl groups. So, these are very bulky tertiary butyl groups which are present on the aromatic ring system and still you have managed to push in a chromium between the two ring systems. So, these are systems where you have difficulty in making these molecules with the traditional fissure hafness synthesis. But using the metal vapor co-condensation technique you generated the metal complex rather readily much of this work was done by one of the students of fissure who originally initiated the fissure hafness synthesis and that is Elchin Broeck who worked in Marburg and he was responsible for making a large number of metal sandwich complexes using the co-condensation technique. I mentioned that a variety of aromatic ring systems can be used. In fact, unrelated ring systems were very commonly used for these metal vapor co-condensation techniques and when there is a choice the metal tends to prefer aromatic ring rather than ring system which has got only partial aromaticity. So, here are series of ring systems where the aromaticity is more complete in the terminal rings ring and that is where the metal ends up. So, you can see that the metal prefers some of these rings which have which has got greater aromaticity than the middle ring which has got lesser aromaticity. So, this brings us to the end of the synthetic section for these molecules and now let us talk about the bonding and the structure in these molecules. Now, it turns out that the type of interaction between the metal and the benzene ring is best explained by using the molecular orbital theory or the molecular orbital framework. So, we will use that and what we are going to do is to use a technique called getting the ligand orbital ligand group orbitals and we will do that using a particular orientation of the metal and the ligand. Here I have shown you the M O treatment of the metal ligand in this particular orientation and you need to remember this orientation always if you have to understand the type of interactions that are present. The metal is present at the origin of the Cartesian coordinates. So, that will be at 0 0 0 and the two aromatic ring systems are parallel to the x y plane. So, one of them is displaced along the plus z axis and the other aromatic ring system is displaced along the minus z axis and it is parallel to the x y plane the ring system is parallel to the x y plane. So, let us take a look at the metal d orbitals refresh our memories. So, here we have the 4 s if it is the 3 d series the 3 d transition metal ions it will be the 4 s which has the 4 p x, 4 p y and 4 p z and the 5 d orbitals which are the 3 d series. So, here are the d orbitals also listed and now what we have to imagine is to imagine the presence of these orbitals at the origin of the Cartesian coordinates and then see how they will interact with the benzene ring. So, the frost energy level diagram is a convenient way of generating the various energy levels of an aromatic pi system. Now you will remember that the number of nodes has to keep increasing as you go up in energy. As you go up in energy in the frost diagram you go from 0 nodes to 1 node to 2 node to 3 node. So, even without solving the Huckel molecule orbitals you can in fact write down the various nodes that are available for benzene. Today we are going to discuss benzene and so here are the m m's of benzene a pictorial representation of the molecule orbitals of benzene. Now the pi m m's are formed if I have oriented them correctly along the x y plane the pi orbitals will be completely made up of p z orbitals. So, the p z on each carbon will be contributing to the pi molecular orbital framework of the benzene ring. So, here is a benzene lowest energy molecule orbital and that has got all six p orbitals aligned together. So, that they have the right all oriented in the same direction and in the same phase. So, that will form a very nice 0 node molecular orbital which is represented here. If you go to the next level if you go to the next higher energy level you now have a system where you have a plane a nodal plane which is cutting across the benzene ring. That is perpendicular to one another, but along the x or along the y axis. So, you would now have two molecular orbitals both of them having one node each and then the next molecular orbital will have two nodes. So, this will have two nodes and this will have three nodes and so on. So, let us proceed further. Let us take a look now at how we can match these orbitals one after the other. So, in order to do that I am going to switch the mode. I will switch the mode and show you how you can overlap these orbitals. So, here is the s orbital the 4 s orbital and if I place it in the center you can see how that it will have the right orientation to overlap with all the m all the 6 lobes of the p orbital. So, all the 6 lobes of the p orbital have the right orientation in order to overlap with the s orbital. So, the s is a suitable combination 4 s is a suitable combination and similarly the 4 p z if it is if the metal is placed at the center of the ring that also has got a suitable is a suitable combination and forms a non zero overlap with the lowest energy molecule orbital. And similarly you can in fact use the 3 d z squared orbital and that also has got the right symmetry to overlap with all the p z orbitals of the benzene ring. So, you can see that there are 3 combinations that we can make with the metal orbital and the lowest energy molecule orbital of benzene. Similarly, we can in fact overlap the p x orbital and here I have I am showing you how the p x orbital of the metal can overlap with the benzene ring orbitals. The highest occupied molecule orbital is degenerate and there are 2 in number and the benzene ring metal orbital combination that is suitable is the p x and this psi 2. We can also in the same way use the d x z. So, here I have shown you a combination of the d x z. So, if I place the metal in such a way that the d x z can now overlap with the psi 2 of the benzene ring. So, you can see that by suitable combinations of these molecular orbitals on benzene and the d orbitals and the p x orbitals you can form several molecular orbitals. This particular molecule orbital is very similar to the interactions that we showed here except that this now requires p y and the d y z. So, the d y z and the p y orbitals will be suitable for interacting with this molecular orbital. In order to show the interactions with the next set of molecular orbitals I want to change over to another view. Now, we are looking at the molecular orbitals from the top as if we are looking down the z axis. So, this is the orientation that we are using. Here is the y axis and here is the x axis. Now, my benzene rings are in the plane of the screen. So, these are the same molecular orbitals that we looked at earlier. Now, it is just that it is shown in a different color. Nevertheless, it helps you to imagine how psi 3 can interact with these molecular orbitals right here. So, here is a suitable combination. You can see this overlap is perfect when you have psi 3 and the d x y orbital. The d x squared minus y squared and the d x squared minus y squared orbital is suitable for overlap like this. So, you can see that psi 3 and this is psi 4, psi 4 and psi 5 have got suitable combinations with the d x squared minus y squared and the d x y orbitals. Now, you will also notice that these metal d orbitals are in the x y plane. So, they are in the x y plane and so the benzene ring is also parallel to the x y plane. So, these are weak interactions that are happening between the metal d orbitals and the pi orbitals of the benzene because although the pi orbitals of benzene are pointed towards the metal, the d orbitals are in the same plane that is in the x y plane. So, you can see very clearly that the type of interactions that you have between the metal and the pi orbitals of benzene can be visualized by using a simple overlapping scheme like this. This leads us to draw a molecule orbital scheme where you have interaction between the metal orbitals and the pi systems of the benzene. As I mentioned in the case of metallocenes, one should remember that there are in fact two benzene rings and they can be combined in two different fashions. In one fashion, you would have the lobes, the same faced lobes of the benzene ring pointing in the same direction. So, this is the orientation where they are pointed in the same direction and so the clear lobes of the P z are pointed towards the shaded lobes of the P z. So, this is suitable for combining with the P z because I can place the metal P z orbital in such a fashion that on the top face it will have the clear lobes interacting with the clear lobe of the P z and the shaded lobe interacting with the shaded lobes of the pi system in the lower plane. So, in the same fashion I can also interact this molecular orbital with the d z squared orbital because that will have the shaded lobes on both sides interacting with the shaded lobes of the metal atom. So, you can see that whether it is a s of the d z squared they will interact with this molecular orbital group of benzene ring and this molecular orbital group of the benzene ring where the phases are pointed in the opposite direction. That means the clear lobes are pointed towards the shaded lobes that will interact with the P z orbital. So, now in a similar fashion you can imagine the type of interaction that will be there between the P x and the d x z orbital here. So, you can mix and match these molecular orbitals in such a fashion that you can have all these five orbitals interacting with the s P and the d orbitals of the metal atom. So, this gives us now a completely complete molecular orbital picture an energy level diagram where we have interacted everything other than the highest occupied molecular highest unoccupied molecular orbital of the benzene ring. That remains non-interacting everything else has got some interaction on the other although some of them are weakly interacting in a delta fashion. So, they have got weak interactions, but all of them interact with one another and form this m o picture which have drawn here. You will remember that the benzene ring has got six electrons in each firing and so 6 into 12 electrons will be filling up up to this level. From the metal one now needs to fill up only the molecular orbitals that are not destabilized. So, that turns out to be six of these six electrons can be filled in these three orbitals which are primarily metal character, but nevertheless they have interacted with the benzene ring. So, they now molecular orbitals. So, they have the origin in the d x squared minus y squared and d x y which as I explained to you just now they are interacting in a weak fashion. So, their stabilization is not too much and then the d z squared. So, if you fill up these m o's now you will end up with six electrons here and as a result you will have the most stable system where chromium and two benzene rings are interacting with one another. If you want a d 6 system obviously you have to choose an element from the chromium group. So, this gives us a total electron count of 18 and so once again we encounter even in the bis-arein metal complexes an electron count of 18 is ideal for forming a stable metal complex. Sometimes the highest occupied molecular orbitals that is a d z squared can lose an electron and the system is still stable because the destabilization of the d z squared is not significant enough to make a great dent in the electronic stabilization energy. So, experimentally it has been found that this particular m o picture that I showed you is the most satisfactory in terms of explaining all the characteristics that people have observed. It is also interesting that the chromium bears a slight positive charge of 0.7 electrons and the benzene ring bears a small negative charge which means that the net transfer of electron density has happened from the metal to the benzene ring. Surprisingly usually one expects the ligand to give electron density to the metal. Surprisingly in this case we have transferred electron density from the metal to the benzene ring and this small transfer of electron density is because of the transfer of electron density from each one of the each one of the metal filled orbitals into the benzene ring especially even the delta bonding that we talked about earlier. So, one other factor before we look at some of the structures of these molecules is that the bond energy of this benzene chromium is 170 kilo joules per mole and this is significantly less than what we have observed for ferrocene which is stabilized by almost 260 kilo joules per mole. So, you can see that the stabilization of benzene is significantly less and this compared to ferrocene where you had significant electrostatic interaction between the metal and the metal and the ligand which is negatively charged. So, the ligand was negatively charged in the case of the cyclopentadienyl cases you have a large negative charge on the ring system and the metal very often was a 2 plus cation. So, this was a stable system whereas, in the case of best benzene chromium you have lesser stability. Let us take a look at some of this molecules now and I am going to show you two or three complexes where you have a benzene ring sandwiched between a metal atom. Here is this benzene chromium itself. This is chromium sandwiched between two benzene rings and you can see that they are perfectly eclipsed. You can see that the two benzene rings are perfectly eclipsed. The difference between the two benzene rings is in fact 3.21 angstroms and since they are perfectly planar it is convenient to measure just the distance between the two carbon atoms as 3.21 angstroms. All six carbon atoms are perfectly bond or symmetrically bonded to the chromium atom. So, you have this very nice symmetrical structure between the benzene ring and the chromium atom. So, let us look at one more structure where you have a cation this time and here is a dimethyl amino group you have a dimethyl amino group. This is sandwiched between dimethyl amino substitution in the aromatic ring. Here is a C 6 ring system and this C 6 ring system has got two dimethyl amino groups which are shown in blue here and these two ring systems are sandwiching the chromium atom. This is again a chromium atom here. I will show the label. This is again a chromium atom which is sandwiched and it has got two of these ring systems and the structure of this molecule has been collected for the oxidized species and it has got a counter ion which is not shown in this structure. So, let us take a look at one more molecule which is this in this ring. So, this is molecule it is there are two atoms of biphenyl. Now, biphenyl is just C 6 H 5 C 6 H 5. So, you can see the two phenyl rings connected together by a single bond and that is the single bond which connects these two carbon atoms which I have highlighted for you. Now, there is one ruthenium atom which is sitting in between the two phenyl groups and you will notice that ruthenium has got eight electrons. So, if in order to fall into the chromium group you have to remove two electrons from the ruthenium. So, this is a di cationic molecule where two counter ions are present in the crystal structure, but nevertheless this is identical to what we would have generated if chromium was present. Once again you see that the C 6 rings are eclipsed whereas, the two other benzene rings are not oriented parallel to each other. They are at a slight angle with respect to each other. So, these are three structures just to illustrate the fact that you can have very nice complexes where the metal is sandwiched between the two aromatic rings. So, let us proceed further now. We have looked at the structure of these aromatic molecules. Let us take a brief look at the properties. You have once again based on the molecular orbital energy level diagram that we saw prediction of the magnetic properties of these molecules. You would have for the titanium species a diamagnetic molecule and indeed you observe a diamagnetic species and you have zero magnetic moment, experimental magnetic moment and predicted magnetic moment is also zero. Whereas, for the vanadium you would have one electron and that indeed is what is expected and what is observed. The bisbenzene chromium itself you would expect zero magnetic moment. It should be diamagnetic and in fact it is diamagnetic as it is predicted. Many of the other molecules you will notice they have methyl groups on the aromatic ring and that gives you kinetic stability for the aromatic ring system that you have. So, the most stable system that you have is for the chromium. Other metals do form sandwich complexes, but they are not as stable as the bisbenzene chromium that was first isolated and characterized. So, let us take a look at a brief look at the reactions of these arene complexes, arene sandwich complexes. If you treat bisbenzene chromium with n-butyl lithium which is known to metalite aromatic rings, n-butyl lithium will metalite aromatic rings and generate lithiated aromatic rings. You find the same reaction happening with bisbenzene chromium also. So, here is a product that you would isolate for bisbenzene chromium. You end up with both the monolithiated species. You do get the monolithiated species and the dilithiated species. So, this is typical of these reactions because the two rings seem to behave independently and there is not much communication between the two. Unfortunately, the ALCl3 fritill craft reaction that we were able to carry out with ferrocene is not possible in this instance. So, although you have a six membered C6H6 which is aromatic coordinate, aromatic that ring system coordinate to the chromium, you do not get the, you do not isolate the acetylated ring. Instead, you have you end up with chromium being oxidized and you end up with acetylated ring. This is a system which is a mixture of several compounds and you do not get the expected acetylated compound. So, this suggests that the aromaticity is not completely retained in these ring systems. You do have considerable loss in aromaticity and we saw that some negative charge has been transferred from the metal to the aromatic ring system. So, it is indeed less, it should be less aromatic than what you expect for a Huckel 4 N plus 2 pi system. So, there is another reaction also which suggests that you do have some loss in aromaticity and that is a fact that the benzene ring, especially if it is annihilated can exchange. So, there is a quick dissociation of the benzene ring and recombination and if you have a different aromatic ring system in the in the medium, there are two aromatic ring systems that are pictured here. One is phenanthrin and so you get a diphenanthrin chromium complex instead of the naphthalene that you started out with. You can also have this annihilated system reacting to give you a different bis aromatic complex because naphthalene tends to go away and you have these ring systems where you have greater degree of aromaticity. So, whenever a ring system is partly aromatic it can be replaced by a more aromatic system, but nevertheless we can see clearly that the net complex that is formed has got less aromatic character than the ligand itself. So, before we proceed further we should mention that the 18 electron system that we worked out for chromium is the ideal complex. So, here is a case a complex which we saw in a crystal structure. It is a ruthenium 2 plus system which is sandwiching which is being sandwiched by two aromatic 6 pi electron systems and here are the two 6 pi electron systems which are stabilizing the ruthenium 2 plus. If you if you remove or add electrons then the system tends to show characteristics of destabilization. So, here is a situation where if you add two electrons to this plus two electrons and this can be done reversibly. When you add two electrons the system behaves in such a way that the two electrons in one of the double bonds of the benzene ring move away. So, that there are only four pi electrons donated to the ruthenium. If you remember ruthenium has got a ruthenium in a zero oxidation state has got eight electrons. So, if one aromatic ring gives six electrons then you need only four pi electrons to reach the 18 electron magic number. So, this pi electron can be moves away. So, that it is no longer coordinated to the ruthenium. So, you have only these four atoms which are coordinated to the ruthenium and it forms an electron system. So, you can remove two electrons from this complex and go here where both are eta 6 is eta 6 and eta 6. You can also add two electrons and change the electron count. So, it is very clear that this ring system is unstable. In fact, you can add sodium borohydride which is a H minus source and because it is two plus it will readily add two hydrogens. Here is a complex which is formed as a result of adding two hydrogens. The two hydrogens remove the double bond and convert them into single bond and form two C H bonds. As a result you have four pi electrons on the top ring and these four pi electrons plus six pi electrons from here result in a stable eta 4, eta 6 geometry for the ruthenium. So, it is very clear that it is possible to shift between the aromatic ring system to the non aromatic ring system when you have an electron count that is not the expected 18 electron count. In fact, the X-ray crystal structure of this molecule reveals that this double bond is completely moved away from the plane of the of the six membered ring. So, you do not have conjugation with this four pi electrons. I should also mention that the X-ray crystal structure however is dynamic that means this C 1 and C 2 rapidly interconvert with C 3, C 4 and C 5, C 6 positions. So, there is a rapid rotation around this bond resulting in NMR structure that suggests all carbons are equivalent. So, we will come to that later in a later lecture. Now, in the remaining few minutes we will talk about hetero sandwiches. By hetero sandwiches I mean the bread part of the sandwich which we had pictured which I had pictured as a slice of bread sometimes can be large and sometimes it can be small. So, for this lecture we were talking about hexagonal slices. We can also have four membered rings or we can have five membered rings in combination with six membered ring. So, this type of sandwiches where you have five and six and four and four are sandwich complexes which can be formed by metal electrons which have the suitable number of metals which have the suitable number of electrons. So, all we have to do is to do the electron counting properly. So, if you want to figure out what would be the ideal complex to form a C P M C P cyclopentadienyl C 5 H 5 M C 6 H 6 then we just need to remember that six electrons are given by the arene and in the neutral method five electrons are given by C P. So, that leaves seven electrons which has to be given by the metal if one has to attain the magic 18 electron rule. So, that metal would have to be manganese. So, metal which has got seven electrons in a zero oxidation state is manganese. We are using the neutral method to do the electron counting here. So, cyclobutadiene metal C 6 H 6 we can ask the question what should be M. Cyclobutadiene will give you four electrons and similarly six will come from the arene. So, eight electrons will have to come from the metal and so the metal will have to be in the iron group or it has to be a metal which gives eight electrons. So, here are some few examples of metal sandwich complexes which are formed by various molecules and this synthesis involves reacting cyclopropenyl ring system cyclopropenyl bromide with nickel tetra carbonyl and that gives you a molecule in which it is a dimeric complex where you have the two cyclopropenyl units sandwiched in a ni b r ni 2 b r 2 unit. A more easy to understand system is formed by reacting this dimer with pyridine and then we isolate a monomeric complex which can be treated with thallium cyclopentadienyl. As I had mentioned earlier cyclopentadienyl thallium is a good source for C 5 H 5 minus and it now forms a very nice sandwich complex where you have a five membered ring on the top and a three membered ring on the on the bottom. So, these will give you five electrons and three electrons respectively and the nickel gives you ten electrons. So, this forms a nice 18 electron complex. A similar situation happens with cyclopentadien and this of course, turns out to be a rather unusual synthesis where an alumina cyclopentadien and alumina cyclopentadien is reacted with lithium metal to reduce it in the presence of nickel bromide to give you a elimination of this group to give you two four membered rings and this also gives you four plus four eight electrons. So, nickel is an ideal element because it has got ten electrons it is 3 d 8 4 s 2 and so that will give you ten electrons and a nice cyclopentadienyl complex. Cyclopentadienyl complex is formed where you have two bis cyclopentadienyl groups. So, it is bis cyclopentadienyl this cyclopentadienyl nickel complex that is formed. Now, can we go beyond six and the answer is yes and here is an example where you have vanadium group metals interacting with the vanadium group metal interacting with the pentavalent vanadium halides and you already treated the metal with C 5 H 5 minus and you treated with C 7 H 8 which is cyclopentadien and then you remove H minus using you remove H minus using the tritile cation. Then you end up with a metal complex where you have a five membered ring on the top and a seven membered ring in the bottom and this gives you a heterosandwich structure where vanadium is conveniently located between these two ring systems. If you do the electron counting you will notice that this in fact is not an ideal system it has only sixteen valence electrons and here you have another case where you end up with another complex which is in the titanium group MCL 4 that also reacts with C 7 H 8 and it forms a sixteen valence electron complex with cyclohebbitrion units coordinated to the metal atom. So, here there is again bond which is present between the metal and the center of the ring. In fact, each of these carbon atoms are bonded to the metal and we do not draw it because then it becomes very congested. So, the metal just is shown with a bond to the center of the ring to indicate that all of the atoms are bonded. So, here again P H C 3 plus removes a H minus. Now, beyond 7 it is it becomes difficult in order to form sandwich complexes or cyclo cyclic pi systems conjugated to the metal, but when you increase the size of the metal itself like we go to the lanthanide or the actinide systems then it is possible to generate cyclo octatetraene complexes. Here the cyclo octatetraene has been reduced prior to this reaction with two electrons. So, you have C 8 H 8 2 minus you will remember that C 8 H 8 2 minus is now aromatic it is a 10 pi electron system that reacts with uranium tetrachloride to give you uranicene that uranicene has got eight uranium carbon bonds on the top and eight uranium carbon bonds on the from the bottom ring. So, this is a very interesting complex where uranium because of its large size is able to react with 8 pi electron cyclic system. So, in summary one can say that the 18 electron rule holds good even in these complex molecules where you have pi systems interacting with the metal atom. The 18 electron rule stems from the fact that if you fill up more electrons than 18 the molecular orbital scheme you tend to fill up the electronic energy levels which are higher in energy which are the ones which have been destabilized. So, it is comfortable for the molecule to stop at 18 anything more than 18 is bad, but as we have seen in this lecture there are several systems where the 18 electrons need not be filled and you can have 16 electron systems and they will still be stable. It is also possible to have some reactions of the benzene ring, but not as much as we saw in the cyclopentadienyl case. Cyclopentadienyl is more robust it interacts with the metal and at the same time carries out is able to withstand Friedl-Krabs reactions. Here we do not have Friedl-Krabs reactions, but we can have litheation reactions and lastly we can make hetero sandwich complexes where the pi systems on both sides need not be the same. One can have a cyclopropenyl pi complex pi system on the top of the sandwich and one can have a 6 membered ring in the bottom of the sandwich. So, you can have a variety of very interesting molecules and we will look at some of the properties of these molecules in a future lecture.