 Organometallic chemistry primarily involves the chemistry of organic fragments which are coordinated to the metal. When we are talking about the organic part of the organometallic compound, we notice that there are often changes in the reactivity patterns of the organic ligand. Today, we would like to consider some of the changes and see if we can understand the changes that have come about on the ligand on the basis of the electronic structure of the molecules that we have considered so far. So, let us go through this. We will give a few examples as we go along and see if we can glean some principles which are behind these reactivity changes. So, in this first slide I have shown you what happens or what is the feature that we noticed when we considered a metal olefin complex. Primarily, we have depletion of electron density in the pi bond and that weakening of the pi bond is seen by a lengthening of the carbon-carbon double bond distance and so that implies that we have lost some electron density from the bonding orbital. At the same time, we have also populated the pi star orbital or the anti-bonding orbital of the olefin. The result of these two factors should be what I have listed on the right side of the screen. Namely, you should have nucleophilic addition of the double bond, nucleophilic additions on the double bond. If there were electrophilic reactions of the free ligand, then these should be suppressed in the complex. In addition, a combination of the two factors that we have mentioned namely the promotion of electron density from the pi orbital to the pi star orbital should resemble the excitation of an olefin from its ground state to its first excited state. So, some although we have not transferred one electron completely as in the case of photochemical excitation, it should be possible in the case of complex ligands to see some type of photochemical reactivity. So, today we will take a look at some of these reactions and see how we can rationalize them on the basis of our expectations. In the first example, I have shown you the generation of olefin complex, which is complex to the olefin is complex to a ion 2 center. In this olefin complex, you have the attack of a dimethyl malonate anion, the anion of dimethyl malonate, which attacks the double bond from the side opposite of the ion in such a way that we transfer the two electrons which are present in the pi bond on to the ion directly. So, much so that we now form an alkyl complex. Because we started with a cationic system and we attacked it with an anion, the final product is a neutral product. There is no charge on the complex and we have started with an alkene complex and ended with an alkyl complex. So, here is essentially the key step, which is the attack of a nucleophile R minus on the double bond. This is exactly what we expected on the basis of the first effect that we saw namely depletion of electron density from the pi orbital of the olefin. There are several reactions of dienes and olefins, which involve oxidation, net oxidation of the alkene, which is complex to the metal. Here I have shown for you a reaction of cyclohexadiene, which in the presence of palladium acetate gets oxidized to 1, 4 hexene or acetoxyhexene. In this 1, 4 acetoxy compound, what we notice is that the net result is addition of acetoxy groups in the 1 and 4 positions of the diene. So, this molecule has in fact added two OAC minus groups, essentially added OAC minus groups at these two positions and palladium has been reduced from palladium 2 plus to palladium 0. Now, this type of a reduction, although this type of a reduction of the palladium and oxidation of the olefin is a, although this is the net reaction, the individual steps involve a nucleophilic attack of the OAC minus on the palladium complex, which is coordinated to the diene. So, here I have pictured for you the individual steps and individual steps indicate that a diene coordinated to palladium is attacked by the OAC minus the acetate anion generating a pi allyl complex. This eta 3 pi allyl complex again undergoes reaction with an acetate anion and generates the diacetoxy compound. So, here we can see that the important step or the key step involves the attack of a nucleophile on the coordinated double bond. If you notice the net reaction that is happening in this whole process is the oxidation and we can make this a catalytic process by bubbling oxygen through this reaction mixture, because palladium 0 can be readily oxidized to the palladium acetate, can be oxidized to palladium acetate in the presence of oxygen very often with copper 2 plus as a catalyst and the presence of acetic acid to form palladium acetate again. So, although this is an oxidation reaction, the key step involves a nucleophilic attack on the coordinated diene. So, we see that the attack of an olefin, the attack of a nucleophile on a olefin is a process which is not favored when the metal is not present, but when the metal is present the depletion of electron density from the pi molecular orbitals enhances the possibility of the acetate attacking the double bond. Now, this type of an attack on a coordinated pi system happens in both neutral olefins, dienes and also in allylic pi systems. Now, here I have shown for you what happens when you have a alkene which has got an allylic hydrogen. So, here is a allylic hydrogen that we are talking about and in the course of interacting with the palladium chloride, you can eliminate a molecule of hydrogen chloride and we went through this the various possible pathways by which this can happen in order to form a dinuclear eta 3 allyl complex. Now, this eta 3 allyl complex can is pictured here. In fact, it is something that could be written in this fashion also fully. Now, you will notice that this allyl complex is capable of undergoing a nucleophilic attack provided we make it more attractive for the nucleophile. How does one do that? Simply by replacing the chloride ion, the chloride ion is a negatively charged species and so it reduces the electron density requirement on the palladium. So, when you replace that with a neutral ligand, very often this neutral ligand can be Pph3 or a phosphine and this results in the system becoming a neutral species and eventually if you add two equivalents of triphenyl phosphine, it can even become a cationic species. This cationic species turns out to be more attractive for the nucleophile and the nucleophilic attack now takes place. Nu minus can attack one of the ends of the allyl group resulting in the formation of the nucleophile attached to the resulting the formation of a substituted alkene now, where the allylic position has been substituted, hydrogen has been substituted by a nucleophile. Very often the nucleophile has got a more electronegative atom like nitrogen or oxygen and that results in a net oxidation of the alkene in the allylic position. So, we have seen two examples here where you can have a nucleophile attacking an alkene either on the double bond or in the allylic position resulting in the formation of a new compound. Now, we also notice during the course of us studying the allylic complexes that an allylic complex is in fact a chameleon. A chameleon changes color depending on the surroundings in which it is present and the allylic compound depending on the oxidation state of the metal which is coordinated to it. It can change its property from an allyl radical to either an allyl cation or it can change itself to an allyl anion. In this picture itself the slide that is before you I have shown you three different examples palladium and nickel 0 tend to enhance the property of allyl group to make it look like an allyl radical whereas the ion compound makes it look like an anion and the ruthenium 4 species which is pictured here understandably makes it look like a allyl cation. So, it tends to be attacked by a nucleophile and this type of a changing in the reactivity of the allylic group is brought about by the ligand that is present and the metals oxidation state. So, two things are important oxidation state of oxidation state of metal and also the ligand L that is present on the metal. So, the modified ligand reactivity that we have seen here is a result of the general features of the metal olefin bonding. Now, we can also have additional reactions which are independent of the two factors that we have just considered. Consider for example, the protection of the double bond from the usual reactions of a double bond because it is now masked by the presence of an olefin. The loss of conjugation with a polyene is also another possibility. Now, we will see a few examples where these two aspects are probed. A classic example is the interaction of dichobalt octacarbonyl with alkynes. This reaction is an extremely useful and valuable reaction because the formation of the dichobalt hexacarbonyl complex of an alkyne is extremely facile. It takes place at mild condition under mild conditions. Secondly, the complex that is formed is reasonably stable and cannot be dislodged easily under normal circumstances. The double bond the distance between the two carbon atoms which host the triple bond is almost close to that of a double bond. We will also notice that the four atoms that form this unit turn out to be almost at the corners of a tetrahedron. So, it is almost like this where let us just mark out the metal. The cobalt is in these two points in space and the two carbon atoms are in these two points in space in the tetrahedron. So, this particular compound turns out to be a very useful way of protecting the triple bond from the normal reactions of the triple bond. So, if you have for example, as shown in this projection that is there before you, you have a double bond and you have a triple bond. Now, if you want to carry out reactions on the double bond without affecting the triple bond, it is possible to just complex it with the dichobalt hexacarbonyl and then react it with the reagents. There are two extreme examples that are shown here. One is the oxidation, the borohydride reduction, V H 3 addition and oxygen oxidation which results in the formation of the alcohol. So, essentially we have added in this case OH group and a hydrogen to the double bond, but if we had left the triple bond unprotected, it would have been, it would have also reacted with the V H 3. Now, at the end of this reaction, what one does is to add an oxidant like Fe 3 plus and the ferric ion is capable of oxidizing the dichobalt hexacarbonyl and cobalt 3 plus is generated and carbon monoxide leaves the coordination sphere of the metal. Now, you will notice that this is not really green chemistry. You end up destroying the protecting agent completely at the end of the reaction, but in cases where you do not have another option, this might be the only way to go. Here is another example where the double bond is completely reduced to an alkane. We have added two hydrogen atoms, we have essentially added two hydrogen atoms to the double bond very conveniently in the presence of a more reactive triple bond. Once again, the presence of this dichobalt hexacarbonyl moiety is extremely effective in protecting the triple bond from any other reaction. Now, there are some other reactions which we will take up later, but now we will look at the reactivity changes that happen for an aromatic compound when it is coordinated to a metal center. We have considered the way in which a chromium atom, which is one of the favorite metal atoms for interacting with an aromatic ring, how a chromium atom interacts with the benzene ring. Now, although the reactivity of the benzene ring is well known in terms of its reactions with electrophiles for electrophilic aromatic substitution, we noticed during the lecture on the aromatic rings on the aromatic compounds of chromium, the bisbenzene sandwich complexes of chromium that the Friegelcraft reactions are not favored when it comes to these compounds. That is because the easy oxidation of chromium to chromium plus 1 and then chromium plus 3 and during the course of this oxidation, the benzene ring cannot coordinate any longer with the chromium and essentially the complexes destroyed, but there are other reactions which are enhanced in the presence of the metal atom. Here I have shown for you about four types of reactions that we can consider. First of the first thing that we would like to consider is a fact that you can do nucleophilic substitutions of a group that is present on the benzene ring. This is rather unusual because usually you do not have a nucleophile attacking the attacking the aromatic ring. This is definitely a no no in terms of reactivity because the benzene ring pi system is extremely stable and it does not allow for disruption of the benzene pi system, but when the chromium is attached to the benzene ring, the effect of the chromium is sufficient, the depletion of electron density from the aromatic ring is sufficient to allow for a nucleophilic attack. What is interesting is that there is a effect on not just the benzene ring which is present here, but also in the benzylic position in the alpha and the beta positions of substituents on the benzene ring. You do have some changes. One of them is the increased acidity by which we mean that the anion that is left behind. So, essentially we are talking about this molecule right here or this fragment right here. See, our reaction is minus. So, essentially we are talking about this fragment being formed and this anion if it is stabilized, if this is stabilized, if this is stabilized, then the activity in the acidity is increased for this molecule. How can this anion be stabilized? That is something that we will look at in a moment. You can also have increased hydrolysis which means that if you have the same species, let us just write the structure of that molecule. If you have C H plus and if this species is stabilized to a greater extent, then SN1 type hydrolysis of this molecule would be favored. Surprisingly, in both alpha and beta positions, it is possible to have increased hydrolysis, solvolysis of X groups and also the increased acidity of these hydrogens. This seems rather contradictory, but if you look at the electronic structure of the bisbenzene chromium, we will understand why. Lastly, we have a pi system. Usually the benzene ring has got two phases and both phases are equivalent, but when we coordinate a chromium atom to one side, obviously we have changed the reactivity and the protection of one side of the benzene ring from attack by any reagent. Let us first look at the source of anchomeric assistance. We said that the valence orbitals on the bisbenzene chromium were mostly the set of d orbitals, some of which had stabilized or given electron density to the pi star orbitals on the benzene. So, I have slightly stabilized that is the dx squared minus y squared and dx y. There was another group which was in fact interacting with the dy z and the dx z. Now, because these groups of orbitals are the valence type that is those are the most easily accessible orbitals, the dy z and dx z can accept electron density. So, any type of electron density change on the ligand which can transfer electron density to the dy z or dx z like the increased acidity that we just noticed where you have an anion generated in the alpha position. That anion can be stabilized by partial overlap with the dy z or the dx z orbital. Similarly, if you have a cation then that cation can accept some electron density from the dz squared orbital that we have here. So, this type of anchomeric assistance is easily understood by looking at the molecular orbital picture which suggests that the valence orbitals are readily capable of interacting with the anion or the cation that is formed in the alpha position. So, let us look at bisbenzene chromium itself. We noted for example, that the lithiation of the aromatic hydrogen is in fact not a favored reaction. You can you can and you do lithiate it, but then it is not as favorable as it is with pure benzene. This in fact is consistent with the electronic structure that we have looked at for the bisbenzene chromium. The chromium itself bears a positive charge and a slight negative charge results on the benzene ring. So, this clearly suggests that you do have increased electron density on the benzene ring and so it is not likely that it will undergo a nucleophilic attack. So, that is true for attack with n-butyl lithium. Now, we will look at how we can change the scenario and make this benzene ring more reactive. So, let us consider two different pathways. If we change the metal for example, in this projection that we have here in front of you, we have changed the ion to ion 2 plus. We have changed the metal from chromium to ion. So, in order to make an 18 valence electron complex, we have oxidized the ion to ion 2 plus. So, here is a complex which is isoelectronic with the bisbenzene chromium, but because the metal now has got a positive charge, a plus 2 positive charge, it now becomes easy to attack the benzene ring and form a complex where the r minus has in fact attacked the benzene ring and destroy the aromaticity and you have a pentadienyl complex which is coordinated to the ion. The net charge on the complex has now changed from 2 plus to plus 1. What is interesting to note is that once one ring is attacked, it becomes less reactive and the second r minus in fact attacks the second ring which was which is still aromatic and that leads to a bis pentadienyl complex which is shown here. So, it is clear that by modifying the metal system, by enhancing the electrophilic nature of the metal, we can in fact change the nature of the benzene ring which is coordinated to the metal. Now, we will move on to another way of enhancing the nucleophilic attack and that is by attaching very electron poor ligands as supposed to the benzene ring. In the previous case, we had bis benzene chromium, we have 2 benzene rings which are attached to the chromium and now what we have done is we have converted the chromium benzene ring system to chromium tricarbonyl rings, chromium tricarbonyl molecule and this as we know is a piano stool complex or a half sandwich complex and because the 3 carbon monoxides are highly electron deficient, they are capable of behaving as pi acids. So, electron density flow is in the direction of the carbon monoxide, the electron density on the benzene ring is moved over to the metal to a greater extent. In the following slides, we will approximate the M CO 3 unit as M with 3 lines, each one of these lines has got a carbon monoxide attached to it and so this will make it simpler for us to draw the systems as we go along. So, here is the first example, a fluorobenzene which is coordinated to chromium tricarbonyl, remember that each one of these lines represents a carbon monoxide. So, chromium tricarbonyl coordinated to fluorobenzene undergoes a nucleophilic attack with a secondary amine and surprisingly a carbon fluorine bond which is quite strong has now been broken and amine has been attached to the benzene ring. This would be impossible in a simple aromatic system, we have been able to carry out a nucleophilic substitution reaction on a benzene ring using a simple secondary amine purely because of the presence of this chromium tricarbonyl unit, which withdraws electron density extremely well and as a result it is able to break the carbon fluorine bond through the intermediacy of course of a sigma bonded intermediate complex, we will come to that in a few minutes. So, this in fact turns out to be a general reaction, you can have a variety of leaving groups and we have put a generic X here, you can have a variety of leaving groups and these groups can be replaced by a series of carbon nucleophiles and all these carbon nucleophiles many of them are pictured here have got at least one electron withdrawing substituent. So, that the negative charge on the carbon can be stabilized to start with. So, they stabilized carbon nucleophiles and these nucleophiles now attack the aromatic ring very readily generate something like a Miesenheimer complex, where you have the negatively charged species coordinated to the benzene ring forming a cyclopentadienyl intermediate and then subsequently the X group or the leaving group is detached and you have the formation of the substituted product. At the end of this reaction again we have to oxidize the if you want the free aromatic compound one has to oxidize the chromium with either iodine or seric ammonium nitrate. Seric ammonium nitrate is often abbreviated as can, but this is serium 4 plus which is now capable of oxidizing the chromium 0 to chromium plus 3 through one electron oxidations and this results in a convenient substitution reaction. It is a two step elimination addition elimination reaction that is happening the R minus first attacks forms an intermediate which is then undergoing elimination process. The person who studied this extensively was Martin Semmelhak who looked at how we can functionalize organic aromatic species by simply using a chromium tricarbonyl unit. The intermediate as I told you has been even crystallographically characterized it is basically a pentadienyl cyclopentadienyl complex which now is in a eta 5 mode it is in a eta 5 mode and the carbon that is undergoing substitution is lifted out of the plane of the 5 carbons which are coordinated to the chromium and the negative charge is present completely on these 5 carbons and that is coordinated to the transferring electron density to the chromium. The subsequent oxidation can be carried out on this step which leads to the elimination of either even a hydrogen or a x group. If an x group is present then one does not have to do the oxidation at this stage. You will just have elimination of the x group whereas, if you have do not have an x group you can carry out the nucleophilic attack and subsequently carry out an oxidation on the intermediate. So, there are two different situations if you have an x group which can leave then you have an intermediate which is present in this fashion and this molecule will lose an x minus. This is one possibility and the other possibility is that when you do not have an x group or a leaving group still the aromatic ring is capable of undergoing a nucleophilic attack and the hydrogen can be released by a subsequent oxidation step and that oxidation gives you the substituted aromatic ring. This is a very convenient way by which variety of aromatic rings can undergo substitution reactions, but very often they have to be stabilized carbonyls in order for this reaction to occur. So, we have looked at a variety of reactions where you can have nucleophilic substitution and we will look at now the increased acidity of the carbon which is adjacent to the aromatic ring. Here I have shown for you the CH 2 group a CH 2 group present adjacent to the benzene ring and if you treat it with sodium hydride it conveniently forms the anion and that anion undergoes a nucleophilic substitution the CH minus conveniently undergoes substitution at the carbon leading to formation of an alkyl group which has another bromine at the third position. So, that comes again one more hydrogen can be removed from this adjacent carbon because of the increased acidity and that leads to the formation the convenient formation of a 4 membered ring adjacent to the aromatic ring. So, this illustrates how you can have increased acidity of a carbon which is adjacent to an aromatic ring coordinated to CRCO 3. Now, let us look at enhanced hydraulic reactions here I have a chromium tricarbonyl which is coordinated to a indenial system and saturated indane type system where you have a hydroxy group in the alpha position. Notice that this molecule this molecule can form the CRCO 3 compound this itself is not a single isomer you can have a mixture of two isomers the carbon which is at this position is capable of being chiral, but we take the racemic mixture and interacted with CRCO 3 we can form a two different compounds. The two different compounds are the one where the hydroxy group is present in this orientation and the other is a second isomer where the hydroxy group is present with the CR in the same side of the CRCO 3, but pointing away from the viewer. You will notice that these two are in fact capable of being separated because they can be separated either on a chiral column by chromatography or with another resolving agent, but this is come about this type of a separation is come about because you have this large group the CRCO 3 that is coordinated to the benzene ring. Now, you can also have another set of isomers where the OH group is coordinated away from the chromium. Now, the more favorable isomer is the one that we have shown here. So, it is possible to easily separate out one chiral in indenol and that can be oxidized very easily to form a chiral indenol Now, you will notice that an indenone is a flat molecule there is no possibility for chiralty, but when you have a CRCO 3 attached to it. It now turns out that it is chiral and you can understand this if you consider the fact that although the indenone carbon is flat it is a flat molecule, but on one side of this flat molecule you have a chromium tricarbonyl unit on the other side there is no other group. So, although the indenone the ketone is an sp2 hybridized center it becomes a chiral center purely because it is oriented with respect to another large group which is coordinated to the adjacent benzene ring in such a way that you have two different phases for the ketone. So, this is a convenient way of making a molecule chiral although there is no chiral center at the point of reference. Now, it is possible to do the same thing with this isomer and then you would get a different the opposite enantiomer of this chiral indenone which is coordinated of course to a CRCO 3 Now, what is more interesting is that if the chiral indenone was oxidized and the chromium tricarbonyl released you would end up with a flat molecule which is a chiral, but instead in this chiral version of this complex indenone. Now, you can carry out several reactions in a selective fashion here I have shown for you the abstraction of a hydrogen from the beta position as I mentioned earlier even the beta position is capable of being stabilized by the chromium atom. So, there is enhanced acidity at the beta position and once the anion is generated this anion carries out a nucleophilic substitution on the methyl iodide resulting in the formation or exclusive formation of one isomer and in a flat indenone you would not have been able to do this selective transformation. Here I have a second chiral center which is generated, but now the chiral center is generated selectively because the indenone carbon itself the keto carbon in the indenone itself is chiral and it has handedness associated with it resulting in the formation of a single optically active form. Now, you can carry out more reactions you just oxidize it with air in the presence of light you end up with this methyl substituted indenone. You can also reduce the chiral indenone react the chiral indenone itself with alkyl magnesium bromide and then oxidize it and then you get a chiral indenol which is again now possible only because you have carried out the reaction in this circuitous manner where you have isolated a chiral indenone and here there is yet another reaction where after doing the alkylation you have carried out reduction of the indenone to generate two chiral centers now. So, here is one chiral center and here is another chiral center and because we have started out with the chromium which was inducing chirality and then later removed it we have accomplished several reactions in a chiral fashion. So, this set of reactions that we just talked about is because of the steric protection of the chromium tricarbonyl. Now, we look at reduced aromaticity and the way in which protection of the double bond results in enhanced reactivity of other double the other unsaturated bonds in the system. Nikolausene is a 20 valence electron system and so if you want to reduce it with sodium and ethanol you can do so add two hydrogens easily with sodium and ethanol and those two hydrogens end up creating a eta 3 allyl nickel complex and eta 3 eta 5 eta 3 eta 5 system which is now 18 valence electrons. So, this turns out to be a case where the double bond which is not welcome to the metal is allowed to react with hydrogen easily and so forms the saturated carbon carbon bond and also generate a 18 valence electron species. Now, this can also this also reacts with C 2 F 4 which is which is understandable because now you have the formation of you have the formation of 18 valence electron system again. So, this is this is a 18 valence electron system which is formed as a result of reacting it and here again you can see that two carbons are isolated from the rest of the aromatic system during the course of this reaction because the eta 3 eta 5 form of Nikolausene is more stable than the 20 valence electron eta 5 eta 5 system. So, C 2 F 4 readily reacts in a 2 plus 2 reaction 2 plus 2 cycloaddition reaction resulting in the formation of a bi-cyclo compound which is again 18 valence electron stable nickel metal complex. Now, it is possible for us to rate the amount of electrophilicity that is present in these coordinated compounds Davis Mingos and Green formulated what is popularly called as a DMG rules. So, the DMG rules come from the three people who worked on understanding theoretically as well as experimentally the reactivity of various coordinated groups. So, the question is if you have multiple coordinated organic compounds which one would react first the Davis Mingos Green rules are easy to understand they suggest that a nucleophilic attack will happen preferentially at even electron compounds. So, that means if it is a C 2 or a C 4 then as a eta 4 or a eta 2 compound will react faster than a eta 3 compound. Secondly, nucleophilic addition to open coordinated polyenes is preferred or closed polyene compounds this is these two rules can be readily understood and we will explain them in the following slides. And lastly we should say we can add that the even polyenes the terminal carbon atom is the one which is most reactive these can be readily understood from the molecular orbital picture which is what they studied in order to arrive at these rules. So, here are three examples that are very popularly given in this example I have a molybdenum complex with a net positive charge and a nucleophile in this case an H minus a hydride a hydride is attacking this molecule. And the question is which one would it prefer the aromatic ring as we have noticed does not lose its aromaticity significantly. And the diene which is coordinated to the molybdenum is the second even electron system that is coordinated to the molybdenum and that is the one that undergoes the attack. And as we have said in the third rule the terminal carbon atom is the one which undergoes attack. And this can be seen from the fact that the loom of the butadiene is the one the empty orbital on the butadiene is the one that will have a larger coefficient on the terminal carbon atom. And as a result you have attack on the C H 2 which then becomes a C H 3 group. Now you have a bis allyl complex and because you have attacked it with an anionic group you end up with a bis allyl complex which is neutral. So, the net charge on this complex is 0. So, one can see that the use of rules 1, 2 and 3 allow you to predict exactly the product that would be formed in this complex reaction. For odd polyenyls attack at the terminal carbon atom is not the favored position. So, this is this might be surprising but again based on the fact that you have in the loom, if you have an allyl anion the loom is the highest occupied the highest energy molecule orbital. And that highest energy molecule orbital has got a large contribution from the central p orbital. And as a result it is more likely to have attack in the middle of the allyl group than in the terminal position. Whereas, if you deplete electron density from the allyl group and make it look like an allyl cation then a terminal attack is more favored because the middle carbon atom now no longer contributes to the molecule orbital. So, here is example where rule 3 and rule 4 are employed. You have the possibility of two odd systems and the closed polyenyl this is eta 5 cyclopentadienyl unit or an eta 3 which is open center. And it is the eta 3 open center which will undergo reaction. And we also note that the eta 3 center will undergo attack in the middle position. And sure enough a methyl anion which attacks the allyl group attacks it in the middle position and forms an allyl complex which is pictured here. Once again you have a cationic system which becomes a neutral molecule at the end of the reaction. Now, odd polyenyls if they undergo reaction they undergo reaction at the terminal position if you have very electron withdrawing groups on the metal atom. So, NO plus NO plus and CO or pi accepting ligands and they modify the reactivity they modify the reactivity of the molybdenum. So, much so that now the methoxide attacks the terminal position because the allyl has been depleted of electron density. This now behaves as if it is a allyl cation. In the previous example we also had a metal which is in the same group. But now we have a cationic system which does not have electron withdrawing substituents. Instead you had electron donating cyclopentadienyl anions. And so the attack was in the middle carbon. Here I have two electron withdrawing substituents a nitric NO and CO. And this turns out to be electron depleting and the decreased electron density results in an attack at the terminal position. So, the altered reactivity of ligands is also seen in some other instances where you have a Diels-Alder reaction. For example, the Diels-Alder reaction that is happening here is happening between a hetero alkene and a cyclopentadiene. But when you have a group like a pH group then there is a possibility of having an endo and an exo product. And the endo product where the pH group is close to the double bond would be favored. In fact, if you carry out this reaction in the absence of any other reagent, if you carry out this reaction in the absence of any other reagent you get a 7 is to 1 ratio of the endo versus exo. And this is in spite of the fact that the phenyl group would be sterically encumbering the no-bonoil system. But the electronic effects, secondary electronic effects which favor the endo product can be modulated if the X group is coordinated to a metal pentacarbonyl unit. You will note that these thiocarbonyls and the selenocarbonyls readily coordinate to a metal pentacarbonyl. And when the coordinated thiocarbonyl reacts with the cyclopentadiene then it ends up with the same molecule except that the fact that the metal pentacarbonyl is attached to the X group. Now, it turns out that the reactivity changes and how does it change? Now, you have a greater amount of the exo product. You have greater amount of the exo product compared to the endo product. This change in reactivity has been brought about primarily because of the MCO 5. Now, the rate of the reaction also decreases, but how do we know that the metal is a cause for this? So, I have already told you that it is possible to have this is chromium or tungsten which is coordinated to the X group. And if we change the tungsten gives you a ratio of 1 is to 7 where 7 is a amount of exo 7 times exo product is formed. And if you have chromium instead of tungsten then you have a change in this ratio a clear indicator of the fact that the metal plays a role in modulating the reactivity of the hetero alkene which is undergoing a Diels-Alder reaction. So, there is another large section where the catalyst which is carrying out the or promoting the reaction has a chiral center and in which case the metal is able to transfer this chirality transfer chirality on to the product. Even when the product when the reactant and the product are not having any chirality it is possible to transfer chirality from the ligand. Now, this is this turns out to be an extremely large chapter. So, we will consider them together in a separate talk. However, let me just summarize what we have discussed today the attach the presence of a metal on a unsaturated organic molecule alters its reactivity. And how does it do? So, primarily you have enhanced and directed nucleophilic attacks. We saw that not only there is enhancement there is also directionality in the nucleophilic attack because of the presence of the metal atom which protects one side of the pi system and allows attack from the other side. We also noted that the oxidation state of the metal can enhance the nucleophilic attack. If it is more positively charged then the nucleophilic attack is enhanced. And ligands like CO and NO can also enhance the reactivity because they withdraw electron density from the metal. So, these two factors are basically once which are modulating the enhanced nucleophilic attack that we talked about in the first point. Now, if you have a double bond or a triple bond it is possible to protect it from further reaction by using a metal. This turns out to be useful especially for systems where you have a triple bond. And further the reduced aromatic in a cyclopentadienyl system was observed when you have a 20 valence electron system. This of course is a special case, but illustrates the principle nicely that you can have reduced amount of reactivity when you have a metal not just enhancement of reactivity. Lastly we saw several examples where there can be pi facial selectivity because of the presence of a metal atom. And we also noted in passing that the metal the use of metal in carrying out these reactions although it turns out to be a useful way of doing it it is not necessarily green chemistry that is going on.