 Hello everyone. It is my pleasure once again to welcome you all to MSP lecture series on Transstitial Chemistry. This lecture is 41st lecture in the series of 60 lectures. Today I shall discuss about two important reactions which are very very vital when you want to use coordination compounds or organometallic compounds in homogeneous catalysis for a variety of organic transformations. If you are an inorganic chemist or if you are an organic chemist, if you are working in an industry or if you are planning for doing doctoral research understanding these two reactions are very very important. From that point of view I shall start discussion on two important reactions that is oxidative addition and reductive elimination reactions. Let me tell you some general information. Before that let me write a general reaction that represents oxidative addition. Let us take a typical metal complex. I shall tell you what kind of complex and conditions and all those things later. Let us consider a substrate molecule AB. My interest is to show you oxidative addition reactions which are reversible in nature at least in some cases. So if I reverse this reaction that is called reductive elimination. So that means there should be an equilibrium between oxidative addition reaction and reductive elimination reaction. And if the rate at which oxidative addition reaction happens to the same extent reductive elimination happens then that is called microscopic reversibility. And we will discuss more about those things as we progress with that one. Now let us consider what kind of metal complexes we should look for. Before we go for that one let us try to recall our understanding of main group chemistry and let us try to recall whether we have come across any similar reactions when we studied main group chemistry. So let me try one reaction that you are all familiar with. For example you take methyl bromide and treat with magnesium. What you get is methyl magnesium bromide. So if you see this one this is a typical oxidative addition reaction. Magnesium to begin with the 0 valence state and it goes to plus 2. And another reaction I will show you. Let us take chlorine and treat with PCL3 at appropriate reaction condition what we get is PCL5. So again here phosphorus from plus 3 state to go to plus 5 state and its coordination number if you say it was 3 now it has become 5. So remember these things. Now that means an element having 2 accessible oxygen states with a difference of 2 electrons is a must to perform oxidative addition reaction. Before I write point by point the conditions the nature of the ligands and nature of the metal. Let us look into a few more reactions. So let us consider a general equation here ancillary ligands. Ln means we have N number of ancillary ligands plus let us say E plus the electrophile comes and then we have a reaction something like this. So in this also if you see oxidation state increases by plus 2 and coordination number increases by plus 1 in this case. So this is a typical let us consider one type electrophilic attack of metals another specific example I will give you. Let us consider iron pentacarbonyl and treat this one with H plus what we get is H Fe C O 5 plus and then another one let us consider let us take Mn C O 5 minus add methyl iodide. So if you see this one of course you can also side wise try to write whether it follows 80 electron rule or not try to do counting of electrons. So here let us go with neutral method 1 plus 7 plus 10 so 18 electrons if you go by ionic method 2 plus magnesium plus 1 state 6 plus 10 so 18 electron. So if you think if it is oxidative addition why iodide was not added down to the metal because then it becomes 20 electron species or something. So you can understand why and it comes out. So these are about mononuclear reactions is it possible to perform a binuclear oxidative addition yes there are examples. So let us consider a bimetallic complex having a metal-metal bond and again add a substrate molecule AB. So in this case what we get is ln M, A and B they are added to two different metal ions. Of course if you consider having metal to metal double bond reaction follows the sequence one bond will be broken and now they will be added separately to two independent metal ions retaining one of the metal-metal bond. So for this one I will give you very nice example. For example if we shine UV light to this molecule what happens the bond this C bond is activated. So now if you see here oxidation state increases by plus 1 and coordination number increases by plus 1 for each metal center. Of course if you heat it back it can revert back to reductive this starting compound. If you want to know from where I have taken this one you can refer to this paper. So in this case if you see now this is broken and this is also broken of course that is the thing that happens a neutral molecule that splits into two anionic ligands and both of them are added oxidatively since it is a bimolecular process they are added to two different metal atoms whereas here also same thing happens but initially we had two bonds one of the bond is broken and then one of the bond is retained so that still it remains as a dinuclear species. So now with this information let us try to understand more and the definition of this reaction let us try to understand and what are the prerequisite in order to perform an oxidative adhesion reaction or reductive elimination reaction. So during this process of oxidative adhesion the oxidant state of a metal increases by two units. If you take a metal in 0 valence state it becomes plus 2 state if I take metal in plus 1 axis state it becomes plus 3 if you take plus 2 it becomes plus 4 or if you take plus 3 it should become plus 5. So that means a metal should have two stable oxidates with a difference of two electrons at least we should have two accessible oxidates with a difference of two electrons and in this process coordination number increases by two units. For example if you take four coordinated compound it becomes six coordinated compound if you take five coordinated sometime it can become seven coordinated or if you take three coordinated it become five coordinated and then the common for electron count most common and important aspect is this should not be 18 electron saturated system of course one can also perform oxidative addition on a saturated system having six coordination number and 18 electron that I can tell you but in general in order to perform an oxidative addition reaction a metal should have 16 electrons or less and it should be co-ordinatively unsaturated. I am going to show you some special cases also. So this is the typical reaction here I showed you already you take a complex n plus is there here and then oxidative addition n plus 2 increases by two units and we have taken x y and now this is a neutral molecule this breaks into two anionic ligands and they will be added as a result coordination number if increases by plus two in future if I write OA it is understood that it is oxidative addition reaction and also reductive elimination I will be writing abbreviating RE and oxygen state I will be using this one and coordination number I will be using these terms please remember I may not be expanding. Now let us look into what kind of ligands are substrate molecules we come across which are capable of performing oxidative addition reaction on a given metal. So there are three main classes of molecules that we can think of to perform oxidative addition reaction with metal centers they are non-electrophilic, non-electrophilic intact and electrophilic. What is non-electrophilic means these molecules neither contain electronegative atoms or themselves function as good oxidizing agents that means they are neither oxidizing agents or contain electronegative atoms except for H2 they are often considered to be non-reactive substrates these molecules generally require the presence of an empty orbital on the metal center for coordination prior to being activated for oxidative addition by breaking the bond to generate two anionic ligands. For example, H2 you take CH bond you take SIH bond or SH bond BH bond NH bond SS bond and also CC bond and if you consider among all these substrate molecules H2 is by far the most important for catalytic applications and followed by SIH bond and then BHNH and SH bonds of course hydrocellulation hydroboration hydroamination when we talk about these kind of reactions we have to look for SIH bond BH bond NH bond etc and of course CH bond activation and functionalization is very very important and every group which are working in organic chemistry methodology or whatever they always try to look for a better method for CH activation however this method using homogeneous catalysis is not really practical because of various reasons attached to utilizing these kind of catalysts for CH activation most of these findings are confined to synthesis in small scale in under laboratory conditions. So now what is non-electrophilic intact type of molecules these molecules may or may not contain again electronegative atoms but certainly they contain a double bond or a triple bond so that means one needs a metal center with an empty orbital with having 60 in electron count or even lower is much better in order to pre coordinate the ligand if it is a double bond it should coordinate prior to undergoing oxidative addition or a triple bond it should coordinate something like this unlike most of the other substrate molecules that break a single bond between those two entities to generate two separate anionic ligands upon oxidative addition these ligands I am talking about non-electrophilic intact ligands always have a double or triple bond and only one of the pi bonds is broken leaving the sigma bond intact or a another pi bond along with the sigma bond intact in case of alkyne molecules. So the ligand does pick up two electrons from the metal and becomes a di anionic ligand typical intact ligands that can perform an oxidative addition without fragmenting or oxygen alkenes and alkynes if we have electron withdrawing functional groups on the alkenes or alkynes that would enhance their electron withdrawing ability so that the transfer of electrons from the metal to the ligand is very facile that means oxidative addition reaction will be much faster. So next the third class is electrophilic one is non-electrophilic next one is non-electrophilic intact the third class of substrate molecules are electrophilic these molecules do contain electronegative atoms and are good oxidizing agents they are often considered to be reactive substrates and these molecules do not require the presence of an empty orbital even 18 electron count is also okay and as I mentioned they do not require an empty orbital on the metal center in order to perform the oxidative addition reaction for example all halogens dimeric X2 where Cl2 Br2 I2 or we can think of polar molecules like Rx, alkyl halides, aryl halides or Hx and O2 also the most common substrates used here are alkyl or aryl halides and also Hx whenever it comes whenever we perform reaction when it comes to the coupling reactions. So that means the common substrates that are used in general are alkyl or aryl halides when it comes to the application of these metal complexes for coupling reactions cross coupling reactions. Let us consider a simple 16 electron count platinum complex and see how an oxidative addition reaction happens and what kind of ligands we are considering you should be able to tell you can see here platinum is in zero oxygen state because all are neutral ligands and of course now it is a D10 system and coordination number is 3 and next electron count electron count is 16 10 plus 6. It is better to write down all these points so when you perform a reaction so that understanding would be very better. Now let us consider an alkyne so during oxidative addition reaction one of the phosphine ligand comes out. So now you can see one of the triple bond is broken here and then it becomes dianionic unlike electrophilic ligands here the bond between these two dianionic sides that are going to metal through oxidative addition are retained by a single bond and a double bond. So we have still a sigma bond and pi bond is there one of the pi bond is broken. This is called metallocyclo propane and now let us do electron count. So platinum is in plus 2 state so it is a D8 system and coordination number is still it is 4 and now it is a 16 electron species. So in this case platinum is oxidizer from platinum 0 to platinum 2 so hence the D electronic configuration changes from D10 to D8 and generated a new dianionic unsaturated alkyne ligand in this case only one of the alkyne pi bond is broken. If you take an alkene then since we have only one pi bond that will be broken and carbon carbon sigma bond is retained. Now the question is is it possible to perform oxidative addition on any 18 electron complex. I mentioned that we can also perform oxidative addition reaction on 18 electron complex. How? Let us look into it by taking a right and appropriate example. I am choosing again a platinum complex of course one can also show like this. Here I am taking a bidentate ligand such as a DPP diphenyl phosphinoethane. So now let us consider look into this one. Here 4 anionic ligands are there as a result platinum is in plus 4 state. If it is in plus 4 state it has to be a D6 system and then if you do electron counting we have 12 plus 618 electrons. By ionic method you can do quickly platinum plus 4 6 are there and all are 2 electron donors now 12 plus 618. Now what would happen is it appears like oxidation is happening. So now what happens is if you heat it one of the iodide comes out and then platinum will be going to plus 4 state. So here already written no need to write here. Now if you see here this is platinum is in now in plus 4 state and then it is a 16 electron species and still D6 system. So 5 into 210 plus 616. Now if you see this one this is a 14 electron species and platinum is in plus 2 state. So 8 plus 6. Now this iodide which was here of course this will be added to here. So now now oxidation is completed it is a D8 system under 16 electron species. You can see we started with 18 electron species and one of the anionic ligand was eliminated to generate a cationic complex here and that is the 16 electron species. And then in this process reductive elimination happens and ethane molecule is coming out and then now here electrophilic attack happens here and hence oxidation is completed. So yes it is possible to perform 18 electron it is possible to perform oxidative addition reaction on an 18 electron complex also and this is how one should be able to understand the sequence of reactions. So let me give you a couple of more examples in my next lecture. I am sure you are very attentive listening to my lecture. Thank you for your kind attention. See you in my next lecture.