 Hello everyone, I welcome you all once again to MSB lecture series on trans-metallic chemistry. I am sure you are enjoying the topics that I am discussing since previous lecture that is oxidative addition and reductive elimination reactions. And in my previous lecture I showed you even with an 18 electron species having coordination number 6, we can perform oxidative addition reaction and also showed you how in case of a trimethyl bisphosphine iodocomplex undergoes sequence of reactions to end up with an oxidatively added product having platinum in plus 2 state and 16 electrons. So now let us you know discuss another example here, very interesting example. Let us consider a rhenium complex for a change, this is anionic, if it is anionic you should remember that we have a pair of electrons are here so that can be readily abstracted by a ligand that is entering, the ligand chosen is this one. So now oxidative addition reaction happens plus Cl minus comes out. So now this species loose a carbon monoxide that means CO dissociation happens one thing and second point is now you can see what would happen to this eta 1 ligand. So this is native charge is there and this also comes means it can be shown something like this, this is allied. Now eta 1 is changed to eta 3. So now if we do electron count for this one, rhenium is in plus 1 state and now it is an 80 electron species, we can count we have if you go with neutral method where rhenium has 7 and plus we have here 8 electrons and plus this is a 3 electron donor now because this double bond 2 electron and one this in neutral method 1 electron donor 3, this is 80 electron species in neutral method in ionic method this is 6, 1 electron is gone 6 and of course 4 ligands are giving 8 and now this is 4 electron donor, allyl group anionic allyl. So it is 80 electron species. So this is another example. So anion here what we are considering is the best donor ligand is the best donor ligand. So in the case of an 80 electron complex only one of the 2 anionic ligands you can see here this also an anionic ligand this also anionic ligand usually the strongest binding generated from the oxidation will end up coordinated to the metal unless a separate substitution reaction occurs. Why this one because rhenium as a pair of electrons are there and now if the electron is taken it becomes carbocation you should remember it becomes carbocation C plus and Cl minus. So that is the reason that immediately abstract the pair of electrons at the end it appears like both of them are anionic and both of them are taking one electron each from the metal center. So these are typical examples where you can see how 80 electron complex undergoes oxidative addition reaction to give the desired product. And another interesting reaction is there that is called oxidative coupling. What is this oxidative coupling? Let me write a specific chemical equation so that you can understand. Let me consider a chromium complex. I am considering 2 ethylene ligands and 2 chlorides my favorite a bisphosphine. Now let us do electron count for this one this is cationic if it is cationic obviously you can tell chromium is in plus 3 state yes. If chromium is in plus 3 state we have D4 S2 so we left with 3 electrons this is a D3 system. If you count electrons now we have total of 15 electrons here. So you consider all of them 10, 2, 4, 6, 8, 10, 12, 12 plus 3, 15, 15 electrons species this is. What would happen to this one? Here this is not the ligand actually this is a methyl amino bisfine phosphine ligand that is the reason I said favorite ligand I am very fond of amino bisphosphines of course this bidentate coordination mode remains intact and these 2 chlorides also remain intact then what is happening here you can see here what is this one metallocyclopentane this metallo earlier I showed a metallocyclopropane now we ended up with a metallocyclopentane and now let us do electron count before we try to understand this reaction. Now chromium is in plus 5 oxygen state because positive charge is there 4 anionic ligands are there. Chromium is in plus 5 oxygen state as a result it is a D1 system and if it is D1 system you can count now 6 into 12 plus 1 13 electron system 13 electron system. So what is the prerequisite for an oxidative oxygen reaction oxygen state should increase by plus 2 and coordination number yes can change and come back does not matter not that important here your oxygen state plus 3 become plus 1 so this is a again an oxidative reaction but here we had 2 independent ethylene molecules are there and we lost a pi bond on each of them and in the place of 2 pi bonds what we have seen is a C C bond formation C C bond formation is there and that resulted in again di anionic ligand. So for this one if you consider so this is anionic this is anionic one thing is compared to other molecules we are not ending up with 2 independent anionic ligands whereas here we have a di anionic ligand and now this binds to the metal to complete the metallocyclopentane formation and in this process we can see metal oxidate increases by plus 2 and coordination number does not change does not matter so this is also a typical oxidative addition reaction but here since 2 independent ligands are coupled we call it as oxidative coupling. This oxidative coupling is happening with non-electrophilic intact ligand we are taking 2 alkenes so oxidative coupling can be anticipated when you are considering non-electrophilic intact ligand such as alkenes and alkenes and of course it goes with your imagination and if you think of such reactions in some organic transformation and organic synthesis certainly you can succeed and of course you have to choose right kind of ligands as anchoring ligands and also how they nicely control the steric and electronic properties of metal on which you are performing oxidative addition reaction. Now let us look into some kinetic data that is obtained from studies that I have mentioned here and just look into the type of ligands we have and type of reagents and also what kind of rate we are seeing and what is happening to delta H and delta S these are very very important and especially when we are considering electrophilic reagents or non-electrophilic reagents in one such example I am going to show you in one of my lectures. Now let us consider here reactant is H2 and in this case O2 and in this case methyl iodide and in all these cases methyl iodide and also you can see phosphines have been changed here we have in first three cases phosphines are there we have a different kind of phosphines are there and if you just look into the rate what would happen is the rate is quite high for electron rich species electron rich species are here you can see the rate is very high if you just look into this case here when you go from chlorine to bromine to iodide what happens the rate is decreasing it is because of steric effect. Then this CH3 comes out as a carbocation is added on to the iridium through axial position targeting dz square orbital in that case what happens because of iodine relatively bulkier and it prevents the free axis of dz square to methyl cation as a result what happens in this case readily decreases with increase in the size of the anion otherwise it is very interesting data we have always you can just look into these things and also look into analyze this later I will show you how one can assess a reaction through thermodynamic aspects whether that reaction that can undergo oxidation or not in case if it undergoes oxidation whether we can use it in a catalytic reaction. So those things I shall tell you as we progress now I have given another chart here and here I have shown if you perform oxidative addition reaction with metals having different electronic configuration what would happen after oxidative addition and also what would happen to geometry and specific examples and to which group these metals belongs all these details I have given here for example if you take a D10 species essentially I have to go for group 11 copper gold or silver I have considered gold gold 1 can gold 3 to gold 3 when you perform oxidation reaction in that case what happens to start with linear geometry is there that should change to square planar and in some cases even they can have tetrahedral geometry that can also change to square planar and 2 ligands would come out in that case because we started with 4 and we end up with 4 so that 2 ligands that were present on metal should come out a typical example is tetrakis triphenyl phosphine palladium or platinum in that one before oxidative addition happens 2 ligands would dissociate to generate a photon electron species to which this substrate molecule is added that results in the formation again a square planar complex and 2 ligands would come out. Example metal 0 to metal 2 palladium and platinum group 10 similarly when you consider D8 species that was changes to D6 to electrons and if you consider Vasca's compound or group 10 species with plus 2 state nickel palladium and platinum or group 9 metals in the plus 1 state rhodium or iridium in this case this can happen D8 to D6 16 electron species becomes 8 electron species and then they will be having octahedral geometry one such is palladium or platinum going from plus 2 to plus 4 or rhodium and iridium going from plus 1 to plus 3 or palladium and platinum going from 0 to 2 also quite possible many square planar complexes of platinum and palladium in plus 2 state show this kind of oxidative addition and if you have trigonal bipyramidal geometry 5 in that case what happens one of the ligand has to leave so that it would attain octahedral geometry of oxidative addition and in those cases one can think of when you have palladium 1 and going to palladium 3 or platinum 1 going to platinum 3 again group 10 metals and D7 to D6 in this case what happens if you can see this one electron process since metal has two oxidates with a difference of one electron in that case a typical oxidative addition reaction cannot be performed on the other hand we can consider dimeric species as I showed you and in that case you can perform for example if you consider cobalt case cobalt 2 to cobalt 3 it becomes and square pyramidal geometry changes to octahedral geometry and now square pyramidal geometry when it changes to octahedral of course we have a bridging ligand will be there and then two octahedral would change to two octahedral in that case one ligand has to come out a common ligand bridging ligand and there are examples of oxidative addition with metals having D6 in that case they become D4 for example octahedral and take go to 7 coordinated complexes and then with the losing one ligand or it may be anionic so example rhenium 1 to rhenium 3 7 group 7 and D4 to D3 again chromium 2 to chromium 3 again octahedral to octahedral chromium 2 to chromium 3 and also there are examples with metals having D4 they can change to D2 for example molybdenum and tungsten okay that is the reason to make you familiar I chose examples having different metal ions and having different ligands on them okay just go through all these reactions I wrote for better understanding of oxidative addition reaction. Now I have given another interesting data here this is about Vascus compound which is undergoing oxidative addition with various substrate molecules here and of course what stretching frequency I have given in the first one is for a neutral starting compound that is typical rhodium chloro carbonyl bis triphenyl phosphine compound that is Vascus compound for that one stretching frequency for CO appears at 1967 and when after performing oxidative addition in case of oxygen it changes to 2015 this is the difference in the stretching frequency with respect to this one so that means increasing you should remember here okay stretching frequency is increasing for carbon monoxide after you perform oxidative addition and also it is increasing in this order when you are changing O2 to D2, D2 to HCl, HCl to methyl iodide, methyl iodide to tetrafluorohethylene to iodine to chlorine so that means you should try to understand how these ligands differ in terms of sigma donor abilities or electron pooling abilities it is always anticipated that stretching frequency should increase when a particular complex having carbon monoxide and glucose oxidative addition here in case of Vascus compound what happens here in this case the it appears like back bonding is very extensive when the back bonding is very extensive what happens CO bond becomes weaker as a result stretching frequency decreases you recall the free carbon monoxide frequency that comes around 21 something now it is 1967 you can see a considerable back bonding is there from iridium to carbon monoxide pi star so as a result setting frequency is 1967 now metal become electro positive so metal has lost two electrons because of oxidative addition in that case what happens metal is not a good pi donor or it reluctantly gives electrons to the pi star of CO through back bonding as a result what happens it is increasing and if you see here they are more electronegative and they are anionic ligands when they are anionic ligands what happens they pull more electron density as a result less and less electrons are available for back bonding as a result consistently you can see here in the same sequence stretching frequency is increasing here so least back bonding is there in case of chlorine because it is a more electronegative group and it pulls electron density and you may ask me can they give electrons through sigma donor and pi donors no not in this case especially with early metals which are not really axophilic late metals which are not really axophilic that is not going to happen here so in these compounds you should remember these are mere sigma donors but not pi acceptors but when they pull the electrons they pull electron density because of their electronegativity generate cl minus you should remember cl2 we are taking it becomes 2 cl minus when it becomes 2 cl minus two electrons are pulled out very strongly from the metal as a result less electron density is left on the metal so that metal is a reluctant pi donor and hence stretching frequency increases and of course in the next couple of slides I am going to show you different type of oxidative addition that one can perform on this Vasca's compound and I have taken all these from Lauri Vasca's paper published in 1961 in JAGS this is the famous Vasca's compound and this is Lauri Vasca let me write a few reactions to begin with to make you familiar with oxidative addition and thereby I can also allow you to distinguish between the type of ligands and how they are added and why they are added and how that can be explained simply I write l here you should remember l equals or let me write here l equals triphenyl phosphate now let me start with oxygen next let me consider important reaction addition of H2 here next let me consider another important reaction in terms of its utility in organic synthesis trimethyl silane we consider now methyl iodide and last one more reaction I would write here Hx so now I have written about 7 reactions in all 7 cases 7 different type of substrate molecules are performing or undergoing oxidative addition on iridium 1 and in all these cases you can see iridium ended up with plus 3 state and having 18 electrons and having 6 coordination number. Let me discuss all these reactions in my next lecture until then have an excellent time reading chemistry and understanding oxidative reaction and thank you for your kind attention and see you in my next lecture to discuss these things in a more elaborated manner thank you.