 Hello everyone, once again I welcome you all to the MSP lecture series on transfer metallic chemistry. I started discussion on oxidative addition and reductive elimination reactions and in my previous lecture I showed a scheme in which I considered many substrate molecules which can undergo oxidative addition or which perform oxidative addition on a metal and let me continue from where I had stopped. So this is the scheme yesterday I wrote. You can see the difference between few substrates and their addition onto the metal centre. For example if you see hydrogen, hydrogen is added on sideways that is in a cis manner and then if you look into oxygen that is also added in a cis manner. So that means oxygen or hydrogen when they are added first they go and sit something like this and then they split into two anionic ligands. And then if you just look into ME3SiH trimethicillin it also follows the same sequence and then another one is these three but look into other four reactions here. So just to differentiate the chlorine or chloride ligand that is coming from the substrate molecule and also one that is already present I have given different colours. Let us look into this one here if you see this one is coming here and this one is coming here that means they are added trans positions on iridium moiety and then if you look into mercury chloride again same thing it is added here and here and then if you look into typical addition of hydrogen halide that also they are going to opposite ends and same thing is to in case of methyl iodide. That means why that is happening like that it is not just with these substrates. If you consider substrate such as SNCl4 this can also add like SNCl3 and Cl and then if you consider CH3SO2Cl this also adds this one separately or trans positions on iridium here or if you take CH3COCl these two fragments will be added after this bond is broken CCl bond is broken and if you take even this one so this will be added like that or if you consider this one this space and this is added and this is added separately and then they are all trans addition will be there. So with this now let us try to understand the different methods or mechanism through which one can perform oxidative addition reactions. I have shown here oxidative addition mechanism can be broadly split into two categories one is two electron mechanism so another one is single electron mechanism within two electron mechanism you can notice three different type of oxidative addition reactions one is SN2 one is ionic one is three centered cis addition. Of course the moment the term cis you see here you can correlate this one with we saw few substrates molecules in the previous scheme which are the substrate molecules which are added onto the metal in a three centered cis concerted manner is called three centered concerted addition. In case of single electron mechanism we can come across radical mechanism and also radical mechanism can be further split into chain and non chain mechanisms. So now let us try to understand these mechanisms one at a time to begin with let us focus our attention on three centered cis addition of course whatever the mechanism I have written here has to do something with the class of substrate molecules we are choosing you recall I made a statement that all substrate molecules can be divided into three classes one is non-electrophilic and the second one is non-electrophilic intact and the third one is electrophilic. So whatever the substrate molecules you are thinking that they can do or perform oxidative addition on a metal will fall into one of these categories. So now with this information let us look into the mechanism of addition through three centered concerted addition in which the two anion clients will go to cis positions on the metal center. So now that means the oxidative addition reactions can be classified into broadly three types one is three centered concerted additions nucleophilic oxidative addition of Rx that is polar molecules and radical pathways. So that means when we perform oxidative addition substrate molecules while adding follow one of these pathways. So let us consider three centered concerted addition here whether you consider HH bond whether you consider CH bond or CC bond all these additions are taking place in this type of mechanism they have very similar one also I wrote many other BH, SiH etc. Let me write a typical three centered concerted addition again considering Vaskos compound and simplest molecule H2 because it is very very vital in organic reactions. So start here iridium is in plus one state and this is a 16 electron species and start with it is a D8 species. Now this is 3 so iridium is in plus 3 and if you count electron it is 18 electron species and it is a D6 system. So that means this is what exactly happens here initially so this H2 before it is getting ready for oxidative addition it has to be pre coordinated to the metal center in this fashion and then it abstract two electrons from iridium to generate two hydride anions and they will be added so here oxidative addition is complete. So if you see here this intermediate this one we have D8 system we have still iridium is in plus one state and then this is an 18 electron species here this is an 18 electron species here ok this is fine. Now let us look into two aspects one is electronic aspect and another one is thermodynamic aspect so how this thermodynamics can explain formation of this two iridium hydride bonds by breaking one HH bond. So here you should remember one equation that you are all familiar with. So now we have to find out these entities from this oxidative addition reaction to calculate the value of free energy. So in this reaction first let us look into delta H0. So in order to calculate this one what we have to see is in the reaction if you see we have broken one HH bond and we have formed two new iridium to hydrogen bonds formation of two iridium to hydrogen bonds and then breaking of one HH bond we should consider here. For this one if you put the values that are available this is minus 16 kilo calories per mole or mole inverse ok. So now we have to calculate delta S of course if you look into that reaction delta S is large and negative because H2 gaseous one is disappearing and in that place we are getting a solid. This is very large and negative so this is minus 30 entropy units or this is equivalent to minus 125 joules per mole per Kelvin. Of course temperature is 25 degree centigrade. Now if we incorporate this to calculate delta G what we get is minus 16 minus 298 into 30 of course to convert that divided by 1000 what we get is minus 7 kilo calories per mole. So what does it indicate? If you look into delta G it is very small it is very small that means if the delta G value is anywhere between plus or minus 10 to 15 kilo calories with appropriate reaction condition one can think of reversing that reaction that means the way we add H2 through oxidative addition we can get rid of H2 through reductive elimination ok. So that means reaction may be made reversible under appropriate condition so that means by using an appropriate reaction condition we should be able to reverse this reaction. So that means if once if we get an idea that we can reverse the reaction then the catalytic application automatically comes into the picture that means here this reaction can be used so that means whether we are bringing back the same H2 or we are taking the H2 source to bring back the complex that was underwent oxidative addition that is a different thing ok always one can do that. What one can get from this thermodynamic calculations is this reaction can be reversible. Now let us look into the kinetics involved in this one of course here the rate depends on the concentration of complex as well as H2 substrate molecule here this rate constant is consistent with the concerted process you can see here this is the reaction coordinates I have shown in this one to start with where we have iridium one complex here and then of H2 is added it forms a transient species where it is still it is a neutral ligand binding through these two electrons between two H atoms and then and enthalpy here is for the this one is about 10 to 12 kilo calories per mole and then once oxidative addition is completed we get this iridium three complex here this is quite stable and the difference is about minus 16 kilo calories per mole. So that means basically any reaction in this one overall delta G is about minus 7 kilo calories per mole so this thermodynamic aspects tell us about the reaction and whether it is exothermic or endothermic and if the delta G of that particular reaction or conversion is well within reaching well within 10 to 16 kilo calories per mole it can be made reversible. Now let us look into how for example I had mentioned earlier you take a olefin and add hydrogen under high pressure in a closed reactor and then heat it for several hours but still hydrogenation do not proceed completely and probably you can get 5 to 10 percent of olefin converting into hydrocarbon whereas the same reaction one can do it in presence of a metal compass such as Vasca's compound in some cases you can achieve 100 percent hydrogenation at lesser temperature under less hydrogen pressure with ease so that means something conventionally it is not possible how that can be done using metal complex that means we have to look into this thermodynamic aspect as well as electronic aspects and also the bonding and how metal initiates breaking of HH bond it is not easy so that means here I will show you molecular orbital picture here this already I showed you when I was talking about the classification of ligands by donor atoms under hydrogen ligands. So here this H2 molecule having two electrons between two HH and then that interacts with the appropriate metal here so say DI system in this case what happens these two electrons represents the sigma bond between metal to hydrogen H2 molecule here and then it depends whether we retain this as a neutral ligand or how quickly or how slowly this is split into two anions to form two metal hydrate bond can be seen here in this case once a pair of electrons are donated from through sigma bond what happens HH bond would be weakened okay so HH bond something like this is there it starts elongated okay after coordination because now the electrons are not really deciding between these two to hold this H together H atoms together instead what happens electrons are moving towards the metal through sigma donation it is elongated now because of back donation sigma star is pulling electrons from one of the orbitals here it is say XZ we have taken one you can take XZ YZ or XY one of those things and to begin with this is the HOMO for metal highest occupied molecular orbital and now this is shared between sigma star by generating a five bonding and anti-bonding orbitals so now basically what happens we are taking out a pair of electrons between two hydrogen atoms through sigma bonding not only that one that weakens the bond on in the same time we are also adding electrons to the sigma star anti-bonding orbital so that means these two kinds of bonding affects HH and eventually HH bond is broken very nicely so this you cannot imagine when we are simply taking in a conventional reaction and high olefin and H2 when the metal is placed the moment H2 comes near the metal what happens electrons between two H are pulled and the electrons from metal is added to sigma star two kind of loss suffered by H2 results in the breakage of HH bond completely and thus oxidation would be completed so that means sigma donation from field H2 to sigma orbitals to metal orbital and then by back donation from metal orbitals to H2 sigma star orbital this eventually results in HH bond cleavage so these two and of course the next question is is it possible to still use H2 as a neutral ligand and by preventing its oxidative addition yes it is possible then we have to consider following aspects metal with positive charge and metal should be in higher access state or containing good pi acceptor ligands but cannot back donate well so that means when the metal is positively charged or metal has higher positive charge it is a reluctant pi donor and again metal is in higher access state it has electron deficiency as a result it is a reluctant pi donor and also if you have very strong pi acceptor ligands already present on the metal like carbon monoxide they take lion's share of metal pi orbitals pi electrons towards it so that less number of electrons are left for back donation to sigma star of hydrogen so that means in that case what happens metal cannot do back donation well to sigma star of hydrogen so stable eta to H2 complex can be generated so that means yes it is at wheel you can either stabilize H2 as a neutral molecule binding to metal center or you can split that one to generate two anions for completing oxidative addition reaction. I will show you a few examples here you can see in these cases here you can see HH bond is about to break here and whereas here HH bond is retained still H2 is coordinated in eta to fashion whereas in this case of course we have both terminal as well as hydrates and as well as molecular hydrogen binding so that means as I mentioned here it is 7 coordinate and 8 18 electron di hydrate this one so here these phosphines are not very good pi acceptors compared to carbon monoxide as a result what happens and pi acceptor ability of hydrogen is more or less comparable to those of tertiary phosphines as a result what happens it also fights for a lion's share in the back bonding as a result what happens more electrons come here and then HH bond cleaves so that whereas in this case we have better pi acceptor ligates here carbon monoxide which will take lion's share and leaves very little or no electron density for the donation from metal to H2 sigma star as a result what happens this H2 is retained so that means we have to choose what we are looking for whether we want to break HH to add oxidatively to generate two hydrides or we have we want to retain as a neutral that is in our hand we have to choose right kind of fancy ligands with sigma donor and pi acceptor properties and also suitable suitable oxygen state on metal so that we can always tilt equilibrium either way. So now let us look into CH activation this is one of the most important reactions in organometallic chemistry from the economic point of view and of course here when we try to do CH activation we can come across two type of activation one is intermolecular CH activation it is very rare and other one is intramolecular CH activation very common cyclometallation happens here and also this intramolecular CH activation is responsible in many reactions. Let me show you one such CH activation so if I want to perform CH activation here what I should do is first I have to vacate some coordination sites on the metal for that one best way would be to shine UV light that means you do photochemical reaction under this condition H2 is eliminated to generate this transient species now you can see reaction this species is ready for oxidative addition now we bring an appropriate molecule having CH bond an appropriate solvent this is a typical CH activation reaction here. This is intramolecular CH activation and then one can also think of intramolecular CH activation that can only happen in some cases where we are using special type of molecules or there is some way we can generate that kind of intermediates. So let me show that one that is considered a platinum complex we have these two groups of course before we perform any further reaction one of the tertiary phosphine should go and of course this will be an equilibrium situation you can recall this is called some sort of very similar to beta hydrogen elimination so tertiary phosphine is in solution now what happens this ME4 so this CH2 is there this will abstract this one so this CME3 will come out and we end up with this is one example for intramolecular CH activation. So now we should look into some aspects here whenever we perform CH addition retention of configuration takes place at C if CH bond approaches metal side on always in a three bond concerted addition always this substrate at approaches in a side on position something like this that is shown here something like this it is coming here so in this case it forms a three membered intermediate that is the reason we call it as three centered concerted addition and then what happens it breaks here and oxide addition is completed. So in this case what happens easier for all especially for aryl groups because usually with 3D, 4D, 5D if you go metal to aryl bond strength increases and also compared to alkyl aryl bonds are more stable because of reinforcement of metal to carbon bond because of delocalized electrons present on carbon more favorable for 4D and 5D metals because they have higher metal to carbon bond strength and intramolecular is common intermolecular is rare requires free side at the metal to occur. Now let us look into CC additions industrially important reactions there and of course the cracking of unstrained carbon-carbon bonds is still limited to hydrogenous reactions and no matter how much we talk about homogeneous catalysis in the present scenario the contribution towards total catalysis from homogenous catalates is only about 16 percent. The 85 percent of the catalytic reactions are heterogeneous in nature whenever we talk about petrochemical industries are cracking to generate a whole bunch of organic molecules still extensively heterogeneous catalysis is used and strained only sometimes CC addition is possible especially when we have very strained molecules where CC bond is very strained because of very congested 3 or 4 membered rings are there in it. Strained CC bonds undergo readily accelerated addition to remove the ring strain. One such example let us consider here so what I am telling is cracking means CH activation means you take something like this and then this is called really cracking so it is not easy to do this kind of reaction using homogeneous catalysis and even if you do it probably the yields are so low and if you master this one there is very good scope in industries to perform this kind of reactions at lower temperature especially with cheaper metals. If we can do something with iron cobalt or nickel yes we are making fortune. Then I was telling you about strained ring let me give an example how strained rings can break CC bond and that can be added on to the metal I am considering here well known complex here I am sure you are all familiar with this compound. So take this compound here metal complex in presence of at least two equivalents of pyridine and then this molecule here this is an example for strained ring adding on to the metal through oxidative addition. So let me discuss more related to this topic in my next lecture until then have an excellent time reading thank you for your kind attention see you in my next lecture.