 diene and polyene metal complexes are a natural extension after discussing olefin complexes. So, in this lecture, what we will do is to look at some of the diene complexes that are readily available and also look at the structure and bonding in these systems. If you look at the ligand space that is available for organometallic chemistry, you will notice that they can be roughly divided into two halves, especially the ones which are having a metal carbon bond. So, these are the ligands which are available when you have a metal carbon bond and this is indicated by this large cloud. The compounds that support organometallic chemistry and support metal carbon bonds, but do not generate a metal carbon bond inherently such as phosphines and nitrosyls are for another class and these are indicated by the smaller cloud. So, if you take the ligand space which has only metal carbon bonds, you can divide them roughly into two groups. One group is the one which has got an even number of carbon atoms bonded to the metal and these are alkenes, alkynes, dienes and trienes. So, it is this group that we are talking about and in the second group, we can talk about a series of ligands where we have an odd number of carbon atoms bonded to the metal and these are listed on the top. In between is the unique carbon monoxide ligand and some substitutes for carbon monoxide and also some compounds where you have only the sigma bond and those are the alkyls and this is indicated by metal alkyls, vinyls and aryls etcetera. So, today we will talk about the chemistry of metal polyolefins, metal dienes mostly. If you look at dienes, these are just examples of compounds where you have two double bonds strung together and it is quite common in organometallic chemistry. If you look at the molecular orbitals of the pi system that is there in a diene, you will notice that the most stable or the lowest energy pi bonding orbital is the one which is labeled as psi 1 here. We will label it as psi 1, it has the phase of all 4 p orbitals the same and so you have no node in the pi orbital manifold. Secondly, you have a system where you have a single node and that is denoted here as psi 2 and then the third molecular orbital has got two nodes and these two nodes are indicated here, these two nodes are indicated here and then the orbital which has got three nodes and that is the highest energy orbital that would be molecular orbital that would be generated, but that would be vacant. So, let us take a look at the metal orbitals that are available for interaction with a simple polyene system, the diene system which is shown here. The metal orbitals can be conveniently drawn in this particular fashion, you have if you take a 3 d element then you have the 4 s orbital which is available. So, this is the 4 s orbital which is available and then you have the 3 p, the 4 p orbitals which are available and then you have 4 p orbitals which are available and these are 4 p x, 4 p y and the 4 p z orbitals and they are higher energy than the corresponding d orbitals and these are the 3 d orbitals which are drawn here. You will notice that we have drawn the d z squared x, y, x z and y, y z, but we have not drawn the d x squared minus y squared orbital that is there of course, we can note that down as d x squared minus y squared orbital, but it is something that does not have a proper overlap with the butadiene orbital, we will see that in a minute. So, let us take a look at how these orbitals can be matched up, interestingly if you draw it in this particular fashion you will realize that it is exactly in the same order in which they match with the butadiene orbitals. So, here are the butadiene orbitals, the 4 s orbital and the 4 p z, 4 s 4 p z and the 3 d z squared all have the right combination to overlap with the totally symmetry combination of p orbitals which form the lowest pi bonding orbital. So, that is the psi 1 and that will be filled with 2 electrons. So, you have 2 electrons here and for butadiene you have 2 p total of 4 pi electrons and so you have 2 electrons here also. So, this combination of s p and d z squared has to be empty in order to interact favorably with the filled psi 1 orbital. So, this is the filled orbital, this has to be the metal orbital has to be an empty 1 in order to form a favorable combination. So, invariably it is going to be the 4 s and the 4 p which are most suitable if you have a late transition metal. If you have a early transition metal of course, where the 3 d manifold is also relatively empty then you can utilize the 3 d 3 d z squared for interacting with this orbital. But notice what happens when you move on to the next higher lying orbital. This is your homo orbital and the highest occupied pi molecular orbital on the butadiene has got one node and so you need to choose an appropriate atomic orbital on the metal atom. So, this has to be either the 4 p y if you orient the butadiene in such a way in this fashion then you will have to choose the 4 p y and the 4 d or the 3 d 3 d y z in order to interact with this filled psi 2. So, the filled psi 2 can interact with 3 d y z or the 4 p y. Now, that is assuming that the orientation of the metal is along the z axis and the butadiene itself is lying in the x y plane. Ligand is lying in the x y plane and the metal is along the z axis. So, this is your z axis. So, if the metal is lying along the z axis and if the butadiene is lying on the x y plane then you will have this particular combination. So, this is a filled orbital again the second pi molecular orbital is a filled orbital. So, you need to choose an empty metal orbital and so that will be the 4 p y or the 3 d y z in case the metal is a early transition metal and has got empty orbitals here also. You will notice that there is a slight difference between the type of interaction that is present in the butadiene and the S p and d z square orbital that we talked about first. If you rotate the butadiene around the z axis. So, if you carry out a rotation in this fashion if you do this particular operation nothing would happen to the bonding interaction between the metal and the butadiene. On the other hand if you take this combination if you rotate it along this axis that is the z axis what you will realize is that you will break the type of interaction the favorable interaction that is there between these two lobes. These two lobes have got perfect overlap now. So, you have a nice overlap here and a nice overlap here that is stabilizing and you will break this if you rotate the metal with respect to the butadiene. So, this is the pi type orbital molecular orbital that you generate. So, one is sigma where you do not break any interaction if you rotate the other is a pi because you will spoil the interaction and make it completely 0 if you turn it 190 by 90 degrees. So, let us move on to the third interaction here we have an empty orbital. This is an empty orbital and it is our first. So, we will call this the Lumo. So, the lowest unoccupied molecule orbital on the butadiene. So, the psi 3 so the psi 3 is an empty orbital. So, you need to take a filled orbital and invariably you if you want to have a stabilizing interaction between the metal and the butadiene. This is empty and it has to be interacting with a dxz orbital because the 4p is unlikely to be filled. So, the 3dxz orbital if it is filled then it can interact by pumping in electrons into the empty psi 3 orbital. So, again you will notice that this is a pi type of interaction because if you rotate it with respect to the along the z axis if you rotate one either the metal or the butadiene with respect to the other then you will end up raking the type of interaction that is there between psi 3 on the butadiene and the metal. So, this is again a pi type of orbital. The last type of interaction which is a type of a delta bond which can be formed between the butadiene and the metal is pictured here. This is the high lying empty orbital and this has got 3 nodes. This is a high lying empty orbital with 3 nodes and this can interact with a dxz orbital. This type of interaction is going to be very weak because your d orbital is not pointed towards the diene and it is on a plane parallel the 2 butadiene and the metal orbital are in 2 parallel planes. So, the type of interaction that would be there is like a delta bond if you rotate it by 45 degrees you would lose all overlap. So, this is a very weak bond at the best if it can if it is there and it is possible between the dxy and the psi 4. So, the dxy is filled the 3 dxy is filled then you can pump electron density into the psi 4, but this is likely to be a very weak interaction. So, these are the type of interactions that you can have between a butadiene and a metal fragment the metal orbital primarily. So, here I have shown you some pictures the type of orbital pictures that would normally would generate using a molecule orbital program like Gaussian. So, you have the 4 orbitals and you can see the nodes here very clearly. There is this single node that you have in between the 2 double bonds and you have 2 nodes in this case and the 3 nodes there are here. So, the butadiene orbitals that are available are kept in 1 plane and the metal is below that plane and it is interacting with the butadiene. So, if you look at let us take a look at the energies that one calculates for the butadiene and compare them with what you have with ethylene. So, if you draw cis butadiene the butadiene twists itself because it does not like the interaction that is there between the 2 hydrogens which are pictured here. So, what the what butadiene does is that it twists a little bit and these are the energies of the twisted orbital, but nevertheless they illustrate some key principles. If the butadiene is cis the highest occupied molecule orbital this is the highest occupied molecule orbital is lying at minus 6.7 electron volts. Whereas, in the case of ethylene the energy is at minus 7.6 electron volts what this means is the 2 electrons which are lying in the homo can be donated to the metal more readily than the electrons on an ethylene. So, butadiene would in principle be a better donor not only does it have 2 pi bonds to donate electron density to the metal it is also able to donate it more effectively because of the higher lying energy of the homo the higher lying homo orbital. So, now let us take a look at the lumo you will notice that the lumo is at minus 0.1 electron volts in the case of ethylene. So, in the case of ethylene it is at minus 0.1 whereas, in a corresponding calculation 1 estimates minus 0.9 electron volts. So, this energy is lower the lumo the lumo is at a lower level compared to the lumo of the ethylene. So, what this results in is the fact that you can put in more electron density or rather you can put in electron density much more easily into the lumo of butadiene the cis butadiene then you can put it into the lumo of ethylene. So, cis butadiene is a better acceptor and at the same time it turns out to be a better donor. So, this is this has got some very important consequences in principle then as you go on from ethylene to butadiene to triene you will keep increasing the donor capacity of the in and you will increase the acceptor capacity also at the same time. So, it will lead to much better bonding not just a matter of interacting two parts of the ligand with a metal that of course leads to something like a chelate effect you have two double bonds interacting with the same metal you have something like a chelate effect. But in addition to that you also have the situation where you have better donor capacity and better acceptor capacity. So, this turns out to be a key factor in the study of butadiene or polyenes with metal complexes. So, let us take a look at what happens to the bond distances it is instructive to first take a look at what happens in ethyne in ethyne the bond distance is about 1.54 angstroms of 154 picometers as it is given here in this projection you reduce the bond distance to about 134 picometers in ethylene that is the double bond an archetypical double bond between two carbon atoms and then that further falls to about 120 picometers in the case of a triple bond. So, these are some somewhat like standards that we can use and look at partial double bond orders which will result when you start making conjugated olefins. So, there is a empirical relationship which has been found between the bond distance that is observed in the molecule and the bond order because the bond order is something that we compute and we infer from the bond distance. So, here is the relationship between the two it is in fact a linear relationship if you look at the carbon-carbon distance in the molecule of question this is a carbon-carbon distance in the molecule that we are discussing then if you subtract 0.18 p from 1.57 you can estimate this gives you an estimate of the partial double bond character. So, this is a double bond character that you have partial or fractional pi bond order. So, that tells you how good a pi bond you have between the two carbons which are separated by a particular distance. Let us take an example now let us take free butadiene free butadiene has got a bond distance of 134 picometers between the two carbons. So, that is 134 picometers between the two carbons and or rather 1.34 angstroms and you have as bond between C 2 and C 3 if this is 1, 2, 3 and 4 the C 2, C 3 distance is 148 picometers or 1.48 angstroms. So, this equation is actually given in angstroms. So, if you can calculate the pi bond order between the C 2 and C 3 is actually only 1.2 and the pi bond order between C 1 and C 2, C 1, C 2 is 1.98 which means it is almost the bond order is almost 2 between carbon 1 and carbon 2 and the carbon 2, carbon 3 bond order is 1.2 which means it is got a partial pi bond order of 0.2. So, we know that there is a delocalization between the four carbon atoms. So, we sometimes indicated like this. So, that is indicated by this partial pi bond order and that pi bond order is 1.2 pi bond order is 0.2 and the total bond order is 1.2. So, let us proceed further. Now, let us take a look at the complex that is formed by butadiene and iron tricarbonyl. This was probably the first tricarbonyl butadiene complex that was synthesized way back in 1938 long before ferrocene was discovered. This complex or this molecule was known. What is interesting is that the carbon-carbon bond links in this butadiene complex is almost the same. All carbon-carbon bonds are approximately 1.41 angstroms. So, this particular equation should be written as 1.41 angstroms. So, this is 1.41 angstroms equals 1.517 minus 0.18 p and this p is giving you the fractional bond order which is the pi bond order. So, we get a value of 0.59 which means that the total bond order between C each one of these carbons is 1.59. You will remember that this was close to 1.98 in the case of free butadiene and it is decreased considerably to 1.59. We also estimated 1.2 for the bond order between C 2 and C 3 and now we are bringing it we have increased it to 1.59. So, what you realize is that the bond order between C 2 and C 3 increases and the bond order between C 1 and C 2 drastically decreases. So, this delocalization is brought about by the iron atom which is interacting with the butadiene. So, how does this come about? If you remember the Dover-Chat Duncanson model which we discussed in detail earlier, Dover-Chat and Duncanson model DCD model of bonding, you remove electron density from the butadiene in from the pi orbitals and push it into the pi star orbitals of the butadiene. So, let us take a look at the interactions that we discussed earlier. So, here we notice that we remove electron density from the bonding orbitals and the electron density flow is from the bonding orbitals into the empty orbitals and this happens for both psi 1 and psi 2, the two bonding orbitals of butadiene and we are pushing in electron density from the d x z into the empty lowest unoccupied molecule orbitals that is the psi 3. So, what we are doing is something like removing electron density from the pi bonds and pushing it into the pi star orbitals and so this results in a weakening of the C 1 C 2 bond and strengthening of the C 2 C 3 bond because the C 2 C 3 has got some bonding interaction between in the pi star or the lowest unoccupied molecule orbital. So, let us take a look at the structure itself, here is the structure, this is called as S cis butadiene because you have in fact cis geometry around the single bond. So, this is the S sigma cis butadiene complex that you have. So, you notice that the bond distance between C 1 and C 2 is 1.419, it is slightly shorter than the longer distance that is found between C 1 and C 2. So, let us take a look at the structure of this molecule in 3 D. So, this is the butadiene compound which is complex to iron, iron is shown in red and that is the atom in the center here at which the arrow is pointing and the butadiene itself is in a plane which is above the iron atom. So, this is the grey atoms here indicated here are the carbon atoms and you will notice that the grey atoms are the carbon atoms and they are in a plane, they are all lying in one plane and the two hydrogens at the terminal carbons are slightly twisted away from this plane. But nevertheless the four carbon atoms are interacting with the iron atom which is below the plane. The three carbon monoxides are in like a piano stool, they are like a stool for this small table which is formed by the butadiene unit. So, let us get back to the structure here. So, you will notice that we have a slight widening of the two carbons in this geometry and that we saw in the complex also. What we have done is remove electron density from the Psi 2 that is the bonding orbital and push it into the Psi 3 which is the lowest unoccupied molecule orbital. This is exactly what is going on when you have a photochemical reaction with butadiene. So, if butadiene absorbs light and there is an electronic excitation you are doing exactly the same thing that we are doing when you form a metal complex. So, here is an example where butadiene is taken and it is photo excited and the first photo excited state the lowest unoccupied molecule orbital has got one electron and the HOMO is also only having one electron because it is lost one electron to the LUMO and we will the excited state geometry is predicted to be exactly this moving in the same direction. We have lengthening of the C 1, C 2. So, earlier it was 1.36 angstroms and it has elongated to 1.45 angstroms. These distances remarked on the projection are in picometers. So, 136 picometers to 145 picometers. Similarly, the bond distance between the central carbon that is C 2, C 3 has increased from 145 picometers to 1 or is the bond distance between C 2 and C 3 has decreased from 145 picometers to 139 picometers. So, this is clearly indicative of the fact that you are pumping an electron density into the psi 3 orbital which has got bonding interaction between C 2 and C 3. So, let us compare the two extreme descriptions of how a metal can interact with the butadiene unit. In the first description, we just have a pi type interaction, the pi bond of the butadiene interacting with the metal in a simple sigma donor fashion. That means the two pi bonds donate electron density to the metal and this is the primary interaction. If this is the only interaction of the primary interaction, two pi electrons to the metal are given from each of the double bonds. So, total it is a 4 electron donor. This is a 4 electron donor. Now, it turns out that if the electron density that is donated to the metal is pumped completely back into the pi star orbital, what you will end up doing is forming a bond between the metal and the C 1 and the C 4 that would almost look like a covalent bond. So, the other extreme description is to form a sigma bond between C 1 and the metal and C 4 and the metal. So, if this sigma bond is formed by complete pumping of electrons from the metal on to the terminal carbons, then you will have the description that you have a metallocyclopentene. So, here is a metallocyclopentene. So, these are two extreme descriptions. Either way, you have 4 electrons, 2 alkyl bonds to the metal from the terminal carbons. So, you have 4 electrons which are involved between the metal and the diene fragment and you also have a 4 electron interaction in the case of the pi bond description that I gave you initially. So, these two extreme descriptions are really not completely correct. What is true is something that is in between depending on the nature of the transition metal. You will have different amounts of electron density transferred from the metal on to the pi bond and so into the psi 3. So, you will have either a bond lengthening that corresponds to this metallocyclopentene geometry or geometry which is a very slight bond distortion which is indicative of just the pi bond giving electron density to the metals. So, here is an example which is a zirconium complex which has got 2 cyclopentadienyl units which we have not discussed earlier. But, let us just take them as a flat 5 carbon unit that is interacting with zirconium. Let us concentrate on the butadiene which is interacting with the zirconium. You will notice that the bond distance between the zirconium and the carbon. So, here is the butadiene. This is C 1, this is C 2, C 3 and C 4. So, these are the 4 bonds which are 4 bonds which are 4 carbons which are interacting with the zirconium and you will notice that earlier the 4 carbons were interacting equally with the metal atom. Now in this description or in this particular compound which is a metallocyclopentene you have a very long bond between C 3 and zirconium that is almost 2.59 angstroms. Whereas, the zirconium C 1 bond is very short that is 2.30 angstroms. So, a short distance between C 1 and C 4, short distance between C 1, C 4 and zirconium. And a longer C 2, C 3 and zirconium is indicative of the other extreme description which is pumping in of all the electrons from the zirconium on to the butadiene. So, much so that you now form a covalent bond between zirconium and carbon and you have a metallocyclopentene description. So, you can see that this is almost like a cyclopentene envelope which is common in organic chemistry only the flap is occupied by the metal atom. So, there are a variety of complexes where you can have both the description that I gave you just now which is a metalocyclopentene geometry and also the butadiene geometry. But in a few cases the butadiene is not present in the cis geometry, but it prefers to interact with the metal when it is in the trans geometry. This is reasonably rare. We will take a look at the 3 D structure if it is if time permits, but there are a few complexes where both the trans geometry and the cis geometry are co-existent on a metal atom. So, this is very common when you have a very early transition metal. So, in the early transition metal this type of interaction is common you have the trans geometry of butadiene interacting with the metal atom. So, let us now move on to another type of butadiene and now in this butadiene we have joined together the 2 hydrogens which were removed the 2 hydrogens which are present in the butadiene geometry and formed a covalent bond. So, that gives us this rather familiar butadiene compound which is cyclobutadiene and if it is a singlet if it is in a singlet state it will the most stable geometry is that of a rectangular geometry where the 2 double bonds are isolated. It is not a square cyclobutadiene, but a rectangular cyclobutadiene. You will notice that a similar description is good for the pi molecular orbitals. You have a completely symmetric molecular orbital which can be drawn in this particular fashion. There are no nodes in this pi system except for the node which is in the plane of the molecule and that is not seen in this particular orientation. Similarly, you have one node in the pi system and that is the highest occupied molecular orbital. So, now if you will notice that there is one node here and then the second one has also got a node but that node is perpendicular to the first node that we drew and that is the lowest unoccupied molecular orbital. Now, we can in fact take these molecular orbitals and track them with the metal orbitals just the same way that we did earlier. You will form a nice molecular orbital picture for the cyclobutadiene metal complex, but when the cyclobutadiene interacts with the metal it very often forms a square geometry. In fact, this is one of the first molecules that was predicted before the molecule was synthesized. This molecule was predicted before the compound was synthesized and the paper was published in 1956 by Orgel and Higgins. They predicted that it is possible that you have a transition metal complex between cyclobutadiene and this particular complex would have the geometry of having a flat cyclobutadiene interacting with a metal atom. So, right after the prediction it was possible for people to search for this complex, try and make it and sure enough they were able to make the cyclobutadiene complex. The metal carbon interaction was quite strong and this was a stable system. It could be isolated and characterized very readily. The bond distance between the carbon atoms was quite delocalized compared to what you would expect for cyclobutadiene. Cyclobutadiene is not stable in the free state and it readily dimerizes. So, you do not have experimental structures for free cyclobutadiene bond distances, but you have calculated distances and these are shown here. As I told you, the singlet is the one which would have not conjugated but isolated double bonds. So, when it forms the complex, you have a very nice interaction. Let us take a look at how one can make these molecules. It is possible to make them by using a compound which is in the low valence state. In this particular example, we have nickel tetracharbonyl. Nickel tetracharbonyl can be interacted with dichlorophenyl. 1, 2 dichloropsychlobutene and that gives you nickel 2 complex which has got cyclobutadiene coordinated to the nickel. So, here you have nickel interacting with cyclobutadiene and nickel is in the plus 2 state. This is again a fairly stable complex. If you want to synthesize the ion complex that we just described, then also you can start with this cyclobutene dichloride that we have pictured here. So, this is a very good starting material for making the cyclobutadiene complex. You treat it with Fe 2 C 1 9. In this example, you have an ion 0 complex of the cyclobutadiene interacting with the pi system. So, in this example, it is a plus 2 system. Either way, you have a very stable complex. Instead of nickel, we can also make the corresponding palladium analog and these compounds have been characterized fully. So, here is the example where you can make the palladium complex. If you treat diphenyl acetylene with the palladium 2 chloride salt, then you form this very nice palladium complex which has got 2 cyclobutadiene units which are coordinated to the palladium and a palladium chlorine palladium bridge connects the 2 systems together. So, you will notice that one can in fact form a very nice cyclobutadiene complex and these cyclobutadiene complexes appear to be pseudo aromatic. So, if you treat them with acetyl chloride, that is typical Friglcraft acylation conditions. You end up with an acylated product of the cyclobutadiene. So, this would be interesting. This is interesting because cyclobutadiene is supposed to be anti aromatic or non aromatic depending on which geometry you choose. If you treat that with acetyl chloride, you have a typical aromatic reaction. What we have done is, you have taken an anti aromatic molecule and converted it into a pseudo aromatic molecule by interacting it with a material. So, this is typical of several cyclobutadiene complexes. One can see some aromatic type reactions for the compounds. Let us take a look at the NMR spectra because NMR spectra, especially the carbon 13 NMR spectrum is very characteristic of the electron density on the carbon atom. So, in the case of ethylene, the carbon 13 resonance occurs at 123 p p m and the proton NMR of the ethylene hydrogen is at 6.5 p p m. Now, when you form a butadiene compound or when you form an ethylene compound, what happens is that the carbon 13 resonance shifts down to 64 p p m is reduced to 64 p p m, which is clearly indicating that you have a low field resonance for the free ethylene and a high field resonance for the complex form. What you have in ethylene is a pi system that results in the carbon 13 NMR peak position at a very low field. This is a low field. It is a high chemical shift p p m value, but it is a low field. What happens is when you put a metal atom next to the ethylene, then it goes to a high field. It is a high field. It is a low value of p p m. So, this is indicative of the fact that you have a large amount of electron density around the palladium. This electron density is shielding the carbon atom from seeing the effective magnetic field. So, you need to apply a high field in order for the resonance to occur. Now, a similar situation happens when you have cyclobutadiene complex. It appears that you have a large amount. You have in the case of iron tri carbonyl complex chemical shift of 40 p p m for the terminal carbons and 85 p p m for the internal carbons. What happens is when you form the cyclobutadiene complex goes to a lower value of p p m and that is to a higher field. That is indicative of the aromatic nature of the cyclobutadiene ring system. Now, we can ask the question, can we in fact make a triene complex? Now, there are two problems which we have to encounter when we make a triene complex. First of all, there are two hydrogens which are at the terminal positions. So, this causes a steric interaction which is unfavorable when you try to make a triene complex where the metal atom is interacting with three double bonds. Now, one can circumvent this by forming a carbon-carbon bond. You remove the hydrogens and form a carbon-carbon bond and that gives you a benzene complex, a benzene complex with a metal atom. This is in fact a known system. So, this is a known system. The other possibility is to introduce another carbon atom such that the steric repulsion is removed. So, this is also in fact known. This is also possible and you can form a molybdenum tricarbonyl complex where the three double bonds are interacting with the metal atom. Each double bond will contribute two electrons. If you remember your electron counting, molybdenum has got six electrons. Six electrons are given by the triene and six electrons are given by the three carbonyl atoms. So, you end up with a 18 electron system. So, we can make metal triene complexes. In these complexes, you find that the bond distances are very clearly indicative of isolated double bonds. It is not like an aromatic system. You have removed the aromaticity in the ring and by introducing this extra carbon. So, you do not have extensive delocalization. You have this bond distances which are indicative of long bonds between the third carbon and the second and the third carbon. So, here you have if this is the first carbon that we are talking about, this is the second, this is the third. The carbon atom two and carbon atom three are separated by a long distance and carbon atom three and four are having a short distance. So, it is as if three isolated double bonds are interacting with a metal atom. So, three isolated double bonds are interacting with a metal atom and there is a long bond or only a single bond between the carbon atoms two and three. So, these are not aromatic systems. So, it is interesting to note that when you put three double bonds together, you can form a complex, but that is not like an aromatic system. Whereas, in the case of cyclobutadiene, you moved towards an aromatic system. Although you had an anti aromatic complex or anti aromatic compound organic compound, you have made it aromatic. So, here is a way by which you can make another zirconium compound. In this particular example, I have taken two cyclohypnotrienes and interacted them with zirconium because zirconium has got only four electrons. This will be a 16 electron complex. This will be a 16 electron complex. What you have done is reduced zirconium from the plus four oxidation state, reduced zirconium from the plus four oxidation state. In the presence of a neutral ligand C 7 H 8, you end up generating what you have here as a metal trine complex. Now, let us proceed further. Since aromatic benzene ring metal complexes have to be discussed in great detail in the remaining time, let us take a look at what happens when you have cycloctatetraene. Cycloctatetraene is a system which will have four double bonds. Can they interact with the metal atom? The answer to that question is yes, but the type of complex that you form can vary a lot because you have a large ring system. The distance between the metal and the carbon turns out to be very large unless you have a metal atom like uranium or a very large metal atom. A 5 D transition element or a 4 D transition element, you cannot form a bond which looks like a cyclobutadiene metal complex where the carbon atoms are in a single plane. Invariably the carbon atoms turn out to be in two different planes. Here is an example where the metal atom is interacting as if it is interacting from the top and bottom of the cyclobutadiene, but the cyclobutadiene, cyclo octatetraene, this is a cyclo octatetraene where the metal is interacting from the top and bottom. It looks as if it is sitting on the same position, but in fact one metal is interacting with the carbon atoms which are pointed downward and the metal on the top is interacting with the carbon atoms which are present on the top. So, this is an example that is being isolated and characterized. Another example is one where the cyclo octatetraene is like a step and the metal can interact from the top face and in the bottom face. That example is also known. It is also possible to interact only three of the double bonds. So, you can form a step structure, but now the step is only half step and one double bond is kept non-interacting with the metal atom. So, examples for all of these types of interactions are known and they are crystallographically characterized. Let us take a look at some of these structures because it would be difficult to explain them the nature of these structures otherwise. What I have here is an example where you have 8 carbons. Let us take a close look at this at the central ring. The central ring has got 8 carbons, but 4 of the carbons interacting with this metal atom right here. So, I will label the metal atom. So, that is a titanium that is interacting with 4 carbon atoms here and the other 4 carbon atoms of the cyclo octatetraene, the other 4 carbon atoms interacting with another metal atom which is on the low half. So, here is one titanium which is interacting with 4 carbons and here is another titanium which is interacting with another set of 4 carbons. The cyclo octatetraene has been divided by the titanium equally and one of them is having 4 carbon atoms and 2 double bonds and the other titanium is having the other 4 carbon atoms and 2 double bonds. So, in this example you also have interaction of the second cyclo octatetraene in such a fashion that it is almost looking like a flat ring and oriented such that the left side ring is in a plane. In a similar fashion you can also orient the right side ring to be in a plane and you can see that all 8 carbon atoms are interacting with the titanium metal here in the center. So, this is one example. Let us take a look at some of the other examples where it functions like a tub. So, here is an example where you have a platinum. The platinum has got 2 metal groups, the platinum has got 2 metal groups. Here is one metal group with a carbon atom and here is the second metal group. These 2 metal groups are in a transposition with respect to the double bond which is present and this double bond is like this hatch structure that I described to you. So, a tub geometry is formed by the cyclo octatetraene. The bottom part of the tub is interacting with one platinum atom which I am showing here and the top part is interacting with another platinum atom. So, you have a very interesting tub structure for the cyclo octatetraene and that is what we showed here that you can have a tub structure. This is the tub structure that we are talking about. So, the tub structure is the example that we see here. This is the tub structure where you have the 2 platinum atoms interacting with carbon atoms which are on one side of the cyclo octatetraene. So, let us go back now with the cyclo octatetraene structure. So, this is the tub structure. So, it is possible to form cyclo octatetraene complexes, but these complexes tend to be varied. You have either a planar structure or you can have a complex in which the double bonds have been divided up with between 2 metal atoms in such a way that whatever forms a bond is with either isolated double bonds or with conjugated double bonds. So, in one instance the titanium was interacting with 2 double bonds that were conjugated and in other instance the platinum was interacting with 2 isolated double bonds. This is in fact common in organometallic chemistry. You can either take here I have shown you cyclo octatetraene. You can take cyclo octatetraene in the 1, 3 cyclo octatetraene form that is indicated here. This is a 1, 3 structure and you can also have a 1, 5 structure. Both of them can interact with metal atoms and depending on the metal you will either form a 1, 3 diene metal complex or a 1, 5 diene metal complex. As I told you earlier this is a better donor. So, depending on the electron density requirements of the metal atom you can either have the 1, 3 structure or the 1, 5 structure. It is also dependent on the size of the metal atom. A large metal atom would like to interact with the cyclo octane in a 1, 5 fashion whereas the smaller metal atom will like to interact with the metal with a 1, 3 cyclo octatetraene system. So, it is possible to make these complexes. I will now show you some of the key features of what we have discussed. So, let us take a look at the key features that we have discussed today. The synthesis of metal diene complexes or metal polyene complexes is fairly easy and follows the same route that one has to use for making olefin complexes. Usually one takes substitution of the ligands especially carbon monoxide can be substituted and a metal diene complex can be generated. Because the diene can be conjugated then it can be like a chelate you can form these compounds fairly readily. You can also have reduction of the metals and it is convenient to have reduction of the metals from higher oxidation state in such a fashion that you can have back bonding between the metal and the ligand. Very often the dienes are useful as starting materials and the structure of the complex can often be explained using the DCD model or the bond the Dewar Duncan's and Chat model. So, the complex often looks like the excited state olefin and that is what we have shown in the bond distances. So, this concludes our talk on the diene metal complexes.