 transition metals have been used for hydrogenation reactions for a long time. After the renaissance of organometallic chemistry in the 1950s, it became obvious that a single site catalyst or a well characterized organometallic catalyst for hydrogenation could be discovered and used and this became reality quite soon after the renaissance due to the discovery of ferrocein. So, today in this lecture we will discuss the catalytic hydrogenation reaction and a closely related reaction which we might call as catalytic isomerization of alkenes. Now, why do we really need a catalyst for hydrogenation? If you think about the reaction for example, two bond formation in the case of butadiene and ethylene reacting together, that is the classic reaction where you have a 4 plus 2 cycloaddition reaction. You do not seem to be requiring this catalyst and in fact there are several reactions which do not require a catalyst and would produce and would proceed under fairly normal conditions. But hydrogenation is one reaction where the reaction will not proceed unless we have a catalyst and usually this is a transition metal catalyst. Now, if you look at the molecular orbitals of hydrogen and ethylene this becomes familiar, this becomes easy to understand. Here I have shown for you the highest occupied molecular orbital of hydrogen and the highest occupied molecular orbital of ethylene. You can see that these two can be overlapped rather easily. So, that you can have a non zero overlap interaction and in this case both are filled orbitals. This both orbitals are filled and as a result the overlap of these orbitals would actually lead to a repulsive situation. What you need is to have an overlap between the filled orbitals of one reactant with the unfilled or empty orbitals of the second reactant and vice versa. This is in fact not possible because if you look at these two orbitals that is the lumo of ethylene and the homo of hydrogen you can see that there will be zero overlap and the same thing holds good for the filled orbital of ethylene and the empty orbital of hydrogen. So, you can see that this reaction cannot proceed in a simple fashion by a cycloaddition type reaction unless you have a catalyst. So, what does the catalyst do? Die hydrogen has to be activated and that is what we have shown just now and the catalyst basically has filled orbitals which have pie symmetry. They have electron density in d orbitals which are capable of pumping an electron density into the sigma star orbitals of hydrogen. So, the sigma star orbitals are the empty orbitals of hydrogen and they are populated and that in turn leads to weakening of the hydrogen-hydrogen bond and the metal hydrogen bond is formed leading to activation of hydrogen. There are very few catalysts which are non transition metal which are capable of activating hydrogen. For that matter even die oxygen is more readily activated by a transition metal than by a simple organic compound. That I should say is till date and this can obviously change if one discovers a more suitable catalyst which is completely organic. Let us back up a little bit and talk about the ways to activate hydrogen. There are three ways in which we can activate hydrogen using a transition metal complex. The first one of course is the oxidative addition of hydrogen to give a dihydride. In this case the metal which is in the 0 or in the n oxidation state goes to a plus 2 oxidation state and then it has got 2 bonds to hydrogen. So, you just have to change the oxidation state of the metal to plus by plus 2 and make it a dihydride. It is also possible to do a homulative cleavage of the H 2 bond by using a M M bond to carry out this reaction. Here the M M bond is a weak bond and the H H bond is a strong bond. But because the M H bonds will compensate for the formation of the for this reaction for the loss in the bond strength between the two hydrogens is possible to push this reaction from the right left to the right. It is possible to take compounds which are metal metal bonded and make the metal hydrogen bonded systems. The third class of activation belongs to those where there is a heterolithic cleavage of the H H bond the dihydrogen molecule. This is usually carried out by an electrophilic metal system which reacts with H 2 and generates M H plus. The hydrogen is the other hydrogen is then reacted with the base to produce B H plus. The classic example of course is copper 2 plus and if it reacts with H 2 you can see that it can generate C u H plus and H plus ion which can be mopped up with a base. So, if you have a base in this reaction medium this would then generate B H plus. So, you can see that there are variety of ways by which you can activate hydrogen. Now, the formal oxidative addition of hydrogen as we had mentioned in the class on oxidative addition is in fact change in the oxidation state, but not in the charge on the metal itself. You should keep this in mind throughout this lecture also because we will be frequently shifting the oxidation state of the metal, but that does not necessarily imply that the metal base the increase in the charge on the that does not mean that the metal increases the charge that is present on the complex. Now, several of these complexes are in fact in equilibrium with dihydrogen complexes. So, here is one example where there is an apparent activation of dihydrogen and you have a tungsten complex which is in equilibrium between a dihydride and an H 2 complex. So, the second class of molecules which undergo oxidative cleavage of the M M bond one of them is shown here where you have a manganese bond cleaved with hydrogen, but these reactions require slightly higher pressure of hydrogen and in this case about 80 atmospheres of hydrogen is required in order to carry out this reaction. So, H M and CO 5 is a mono met mono hydridic metal complex which can carry out reactions which the simple system cannot carry out. So, there is another molecule rhodium rhodium bonded system the way it is written it does not appear that there is a rhodium rhodium bond, but it is the two rhodium atoms that are bonded together and this rhodium rhodium bond is cleaved and rhodium hydrogen bond is formed. So, these molecules can carry out reactions which dihydrogen itself will not carry out. The third class of activation is a heterolytic cleavage of the M L bond and the heterolytic cleavage is actually carried out with the help of a ligand or a base. Sometimes it is the ligand, sometimes it is the base and which is present in the medium and it is shown here. In this example for a palladium complex which is a saline derivative and this can readily add on hydrogen. One hydrogen is attached to the metal and the other hydrogen which is a proton is attached to the phenolate of this system. So, much so that you have a four coordinate molecule in the end, but one of these original metal oxygen bonds has been broken and OH bond has been formed instead and there could be only a weak interaction between the palladium and the oxygen. So, let us talk about reactions that depend on the formation of a dihydride. These are systems where the oxidation state of the metal changes by plus 2 units. So, you have M 0 going to M plus 2 and many cases these dihydrides carry out hydrogenation reactions very readily. The classic example of course belongs to the Wilkinson's catalyst which most of you might be familiar with because this is the most popular example and probably the first example of a hydrogenation catalyst that was discovered, but more importantly this catalyst can be used to carry out asymmetric hydrogenations and that is the reason why it is extremely popular. Now, the basic Wilkinson's catalyst is also readily prepared in the laboratory starting with rhodium chloride rhodium 3 chloride rhodium 3 chloride and ethanol. Just reflecting these two compounds together generates the rhodium 1 complex which is pictured here. So, in the presence of p p h 3 you just have to reflux the 2 and the rhodium 3 becomes rhodium 1 because of a reduction reaction which we will show in the subsequent slides. The p p h 3 stabilizes the rhodium 1 species which is formed and you have this d 8 system stabilized in a square planar geometry. Now, this molecule readily adds on hydrogen. So, even if you mix if you take a solution of this Wilkinson's catalyst and expose it to di hydrogen at room temperature and atmospheric pressure 1 atmosphere pressure of hydrogen you end up with a color change in the solution and you have the formation of a di hydride. Now, this di hydride can lose one of its ligands the p p h 3 and have a vacant coordination site on the road to rhodium atom. So, multiple steps are compressed in order to make this fit in a single slide what you should realize that there are 2 steps which are involved loss of p p h 3 and addition of hydrogen leading to the oxidative addition of di hydrogen. As we saw in the lecture on oxidative addition molecules like di hydrogen add in a cis fraction. So, you have a cis addition of di hydrogen which is pictured as intermediate 1. This intermediate 1 has a vacant coordination site which can take up a molecule of ethylene. Ethylene when it is coordinated to rhodium this rhodium atom which is now in a plus 3 oxidation state formally it is in a plus 3 oxidation state and the insertion of a hydrogen or a hydride species from the rhodium on to the ethylene if you have the reverse of a migratory insertion reaction. So, you remember what happens when you have a insertion reaction you have the nucleophilic component going from the metal on to the organic component and a new metal carbon bond is formed. So, you have the formation of a metal rhodium carbon bond between the 2 species that I have just underlined and that is what you have here and the hydrogen has added on to the terminal carbon. So, you have an ethyl group attached to the rhodium now and this species which is also a rhodium plus 3 center. So, you have not changed the oxidation state you have only done a migratory insertion reaction which has converted the neutral ethylene molecule into an alkyl moiety. This molecule is now well set up to do a reductive elimination. So, elimination of H s indicate that with a different color now if you eliminate these 2 groups if you eliminate these 2 groups from the rhodium plus 3 center you will end up with a rhodium 1 center which has got 2 p p h 3 units and 1 chloride the 3 ligands that I have underlined for you. So, this would lead to a rhodium 1 center which can readily add on hydrogen di hydrogen to give you again the catalytically important intermediate which I have labeled as 1. So, the first step is a simple substitution step where p p h 3 has been replaced by ethylene. So, that is the substitution reaction. The second step is in fact oxidative sorry the first backup the first step is the oxidative addition of hydrogen to this catalytically active intermediate which gives you intermediate 1 and then there is a substitution reaction where ethylene goes on to a vacant coordination site and this system undergoes a migratory insertion reaction. This is a migratory insertion reaction and following that there is a reductive elimination reaction. So, you do have an oxidative addition and a reductive elimination usually the oxidative addition is the first step and the reductive elimination is the last step in between you would have a substitution and a migratory insertion or just a substitution and a subsequent reductive elimination. So, this is the general pathway for most catalytic reactions and we will see that in the subsequent slides also. However, the addition of the ethylene could be preceded could precede the addition of hydrogen so here is another catalytic system which is also based on the rhodium 1 catalyst. But in this particular instance the olefin coordinates to the metal first before the oxidative addition of hydrogen. This complex is formed is the one that is available in the laboratory or in chemical companies which you can purchase and the advantage of this complex is that if you add hydrogen to it it is capable of being hydrogen the ligand is being hydrogenated in the first step in order to generate a very active form of the catalyst and so S O L really refers to the solvent. So, this is your solvent here and the solvent is something which is easily substituted by the ethylene in order to give you the olefin complex in the first step. This olefin complex is then reacted with hydrogen and it is in the second step that the oxidative addition happens. In the Wilkinsons catalyst the oxidative addition happened in the first step and in the subsequent step you had the substitution reaction. So, these two have been interchanged in the Schrock Osborn hydrogenation catalyst catalysis. So, if you now add on the hydrogen you end up with a rhodium 3 intermediate this is also rhodium 3 intermediate. There is a net charge of plus 1 which has been indicated on the rhodium. So, this plus 3 oxidation state is because of the two hydrogens and the plus 1 charge on the complex itself. Now, this complex can also undergo a migratory insertion so that you can form a new carbon rhodium bond and that is what you have here. The same ethyl group can now be eliminated as ethane by elimination of this hydrogen and this ethyl group. So, you will regenerate the catalytically active species which is present on the which is you will regenerate the catalytically active species which has got to solvent in its coordination sphere. So, that you can do subsequent substitution reactions fairly readily. Now, hydrogenations which are carried out like this especially with rhodium 1 are typified by these characteristics. First of all these reactions work very well. If you have if you have unsubstituted or sub or a olefin which has got a small substituent when you have large substitutions on the olefin. Then the reaction becomes very sluggish and the rate at which hydrogenation happens in various complexes is shown here and the rate increases in this direction. If you have only mono substituted olefins then the reaction is extremely fast. If you have 1, 1 disubstituted olefins then the reaction is faster than when you have 1, 2 disubstituted olefin and if you have trans or 1, 2, 3 substituted olefins the reaction becomes extremely sluggish. Now, the rate differences are significant enough that you can in fact hydrogenate very simply a mono substituted olefin in the presence of a trisubstituted olefin which is at the bottom of the series. This is very advantageous when you want to do hydrogenation in a selective fashion. So, that is why rhodium catalyzed hydrogenations are extremely useful in the laboratory. Another important factor that one has to be born in mind, one has to bear in mind is a fact that cis hydrogenation takes place. So, here is a case where you have hydrogenated using deuterium and cis hydrogenation gives you the erythro product exclusively and this is extremely useful once again if you want to carry out reactions in a stereo specific fashion. So, let us take a real life example. Here is a steroid which has to be hydrogenated and there are there is more than one reducible group and if you use the rhodium catalyst that we have just discussed in the previous slide you end up hydrogenating only these two places. In other words the double bond which is here is not affected. So, the double bond which is here is not affected in the during the reaction. So, this is a very useful catalyst because of the selectivity and it can be used for carrying out many complicated synthesis. Here is another example where you have unhindered double bond and trisubstituted double bond and the trisubstituted double bond is left unaffected during this whole reaction. If you use palladium and di-hydrogen for carrying out the same reaction you end up selectively hydrogenating the double bond which was in this position and that is because the conjugated double bond is selectively hydrogenated by the palladium whereas the rhodium because of steric reasons it is hydrogenated only in the terminal position where there is less hindrance. Now, let us move on to hydrogenations which are catalyzed by mono hydride catalysts. Now, the mono hydride catalysts are typified by a few examples where the H metal or the hydrogen metal bond is existing in the reaction before the olefin is added. Now, I have shown for you two examples one is a rhodium example and another is a ruthenium example. The rhodium example has in fact a very slight difference compared to the example that we studied earlier. The only difference being that the rhodium hydrogen bond is already there and instead of the chlorine that was present on the Wilkinson's catalyst. Now, these complexes are extremely active and at room temperature and pressure you can carry out hydrogenations of terminal olefins and the advantage is also highlighted here many of the reducible groups which are present on the molecule will be left unaffected at the end of this reaction. So, here is a catalytic cycle if you take the ruthenium complex which has the mono hydride then you can carry out substitution reaction the loss of a phosphorous ligand and addition of an olefin leads to this phi coordinate intermediate. This is the phi coordinate intermediate here is capable of adding doing a migratory insertion of the hydrogen on to the olefin migratory insertion of the hydrogen on to the olefin. So, that is this group which is added on to this olefin and a metal carbon bond is formed. So, the metal carbon bond is formed between these two centers leading to this alkyl moiety and subsequently you now have an alkyl complex which can undergo oxidatively undergo a reaction with hydrogen which leads to this molecule which has got a di hydride attached to ruthenium. You can reductively eliminate these two species the hydrogen and the alkyl group in order to regenerate the catalytically active intermediate. So, the difference in the catalytic cycle is minimal. The main advantage of the main difference is the fact that you need to coordinate the olefin in the first step even before you do the oxidative addition. This leads to a strong steric influence in the whole reaction unsubstituted olefins are reacted or react much better than substituted olefins. Now, the catalytic cycle that we have gone through just now is common for many systems. We will just illustrate or give two more examples of how mono hydrides might act as catalyst. Here, I have shown you a copper hydride which specifically hydrogenates a conjugated enone and converts it into a simple ketone. So, the way in which this hydrogenation is