 Hello everyone, once again I welcome you all to MSP lecture series on trans-metal chemistry. This is 25th lecture in the series. In this lecture I shall give you the preparation of important and useful metal complexes keeping the catalysis and also further interesting organ metallic compounds in mind. So let me begin with some compounds of palladium. So for example if we just look into platinum metals such as ruthenium, palladium, platinum, gold, rhodium, iridium these compounds are quite extensively used in homogenous catalysis for a variety of organic transformations. Then how to make some of these compounds and what are the properties of these compounds that can be used as precursors for further substitution and making desired compounds with desired ligands to guide the reactions in the name of homogenous catalysis. So you cannot use any complex that comes to your mind for catalysis well defined ligand should be there and well defined metal should be there and also precise oxygen state has to be chosen and such compounds has to be chosen and what kind of coordination number should be there to begin with and once if one or two ligand leave or dissociate what kind of compound we are going to get whether that can be a catalytic precursor or not. So all those things we have to keep in mind and when we keep all those things in mind and we come up with an excellent ligand system and then we need an appropriate metal precursor for further substitution reaction to make the desired compound. Simply you cannot use any compound from that point of view I shall give you the preparation of very very important compounds. Let me begin with palladium chord chloride you must have heard about palladium chord chloride. So chord is cycloctadiene it is very easy to write it is a diene pyridentate ligand so this is chord cycloctadiene. Then how to make this compound the starting compound is palladium chloride you take palladium chloride and palladium chloride if you take as such is insoluble in organic solvents and also it is insoluble in water as a result what you should do is you add HCl once you add HCl it forms palladic acid something like this and this is soluble now. So this is the first step take unhydro-spalladium chloride add HCl and then generate this species here once of generating this species treat this one with C8H12. So this is C8H12 of course this reaction does not need any heating or anything this reaction can be carried out at room temperature in ethanol. So you should remember in future if I write RT means it is room temperature what we get is if you are curious to know the structure this is how the structure looks like this is palladium chord chloride this is very useful for example you take this compound let us say treat with two equivalents of triphenyl phosphine in dichloromethane this two triphenyl phosphine molecules would replace this bidentate ligand then we get a compound like this. So this cycloctadiene being a diene has a characteristic smell and this compound does not have any smell but the moment you treat with two equivalents of triphenyl phosphine and then this chord is liberated when the chord is liberated you can see the smell coming characteristic smell of olefin comes out so that indicates that reaction is progressing and also there can be color change this is a bright yellow color and this will be slightly yellow or even pale yellow color. So visually you can monitor and also if you are careful you can also the through smell you can make out that reaction is progressing of course once this compound is made you can crystallize and take NMR and NMR should show a single phosphorus resonance indicating this one and in case if there are two isomers are there then you should be able to see two chemical shifts 31p chemical shifts indicating cis and trans isomers. So with this reaction usually you get a cis isomer so this is the utility and not only this one if you want to use any other ligands bidentate ligands you should be able to use conveniently starting from palladium chord chloride. So another important compound is bisbenzone nitrile compound so here this is very similar to the previous one instead of cycloctadiene we are using phcn ligand here this is a nitrile so this lone pair is there this lone pair goes to the metal that means this forms a very weakly coordinated compound and but moderately stable bright yellow powder we can handle even in atmosphere without any problem. So how to prepare this compound take this palladium chloride here and add excess of and reflects you get this compound here of course one can also write like this fashion but it is always appropriate to write ligand in such a way that the donor atom is facing the metal center this is not correct one should write something like this. This is also very useful compound in this case also you take again a bidentate ligands consider for example a bidentate ligand such as this one this is called bisdiphenyl phosphino methane it is a methyl derivative this is a bidentate ligand you take this one and two benzonate this will come out and you get a compound like this a chelate compound. So this is how one can make some of this very labile metal precursors and conveniently without losing this compound or further decomposition you can get quantitative yield of compounds depending upon what kind of ligands you are using and if astro benzonate is not available it should not be a problem you take this palladium chloride and take dry astronitrile freshly distilled astronitrile or methyl cyanide and excess reflects it for about 2 to 3 hours you will see the formation of bright yellow color solution that is due to bis astronitrile complex formation this compound is much more unstable compared to benzonate as a result whenever we need this can be generated in C2 and without isolating one can use it if you want isolate this one as bright yellow crystals it has to be stored under nitrogen in air tight bottles. So this is another important precursor and sometime what happens you there is no need to isolate this one take palladium chloride and reflects in dry astronitrile for 2 hours and then add ligands whatever the ligands you want to add add and substitution would be complete and you can quantitative conversion of bis astronitrile compound into corresponding compound for example you take here add you can get this compound so like this. So these 2, 3 are very very important compounds and very easy to make commercially if you try to buy they are very expensive but on the other hand palladium chloride is little cheaper and one can conveniently make this compounds and also even yields are quantitative so you start with palladium chloride and since you are increasing its molecular weight by 2 fold you can see start with 1 gram you can get 3 to 4 grams. So that is the advantage of learning this preparatory methods to prepare in your own laboratory instead of buying all those things commercially. Next consider one more compound this is tetrachys triphenylphosphon platinum compound this is one of the very useful platinum 0 compound here a D10 system and D10 system means you should remember that it has tetrahedral geometry similarly one can also make palladium compound this is most useful palladium 0 compound this is an anti-electron species with D10 in homogeneous catalysis and its commercial available many reactions carbon-carbon cross coupling reactions are hydrogenation and many other reactions people use this one and this compound can be made again easily. So let us look into first platinum compound preparation for this one we are using potassium tetrachloroplatinate that is K2 Pt Cl4 take this one add 4 equivalents of triphenylphosphine in presence of alcoholic koH that is for reduction purpose and this has to be heated to 60 degree centigrade remember temperature is very critical it should not go beyond 60 to 62 degree centigrade if it goes beyond what happens the yield will reduce. So take a mixture of potassium tetrachloroplatinate 4 equivalents of triphenylphosphine in presence of alcoholic koH what you get is what are the other bright products I am writing them also. So this is the balanced equation for the preparation of tetrachys triphenylphosphine platinum starting from potassium tetrachloroplatinate. Another important compound of platinum metals is ruthenium complexes. Rhythm is in plus 2 state here because 2 anionic ligands are there one is Cp1 is chloride then how to make this compound for most of these compounds we use trichloride trihydrate you should remember in case of iridium ruthenium and rhodium we are using plus 3 compounds MCL3 3 H2O they are ideal ones and in case if they have NH2O using appropriate method you have to generate first anhydrous salts and then you make the trihydrate compound. These trihydrate compounds are quite reactive and convenient to prepare a series of interesting organometallic compounds and also coordination compounds which can be further used in various applications. One such important application is again homogenous catalysis. For this one for the preparation of this ruthenium 2 complex having 2 triphenylphosphine and 1 C cyclopentadienyl group and 1 chloride one has to start with ruthenium trichloride trihydrate 2 equivalents of cyclopentadiene. So this should be freshly distilled. So for example if you take cyclopentadiene and if you keep it at room temperature for a prolonged time so it undergoes dimerization to form cyclopentadienyl dimer. So this dimer has to be cracked into monomer simply by distilling it you can convert into monomer immediately you have to use it that means you have to generate quickly you have to distill quickly and then you have to use it otherwise what happens you will end up with a dimer and that does not react with this metal precursor. So take this one and treat this with 5 equivalents of triphenylphosphine. So take ethanol and reflex it so you get this one along with this you also get for this purpose what we are doing is we are using excess of 1 equivalent of triphenylphosphine that leads to the formation of phosphine oxide and this acts as a reducing agent here and one thing you should remember one can write something like this there is no harm but to be precise when we have this aromatic groups with different hapticity it is always ideal to write that ligand in the front. So best way to write is in this fashion this is the correct way of writing. So this also useful compound for example in this one one can also replace 2 equivalents of triphenylphosphine with a bidentate ligand because entropically that is a favored one with chelate ligand. So you can replace this one or if you are using excess what happens you can use a polar solvent so that you can ionize RUCL and it becomes cationic compound. So if you use nonpolar solvent you can simply substitute 2 triphenylphosphine with the desired ligands bidentate ligands or monentate ligands which are better sigma donors compared to triphenylphosphine or if you want to use slightly excess to even get it of chloride you have to go for polar solvent for example take this compound and add 2 equivalents of bidentate ligands let us say I am adding 2 equivalents of DPPM. DPPM is diphenylphosphonamethane I used in previous palladium complex. So use this one in methanol and if you reflex it basically what happens it becomes ionic complex and then you have eta 2 DPPM and eta 1 DPPM and then you have chloride here. So something like this happens how this molecule looks like you can see here in this one 2 ligands are there and chloride is here it becomes an ionic complex so one can make it and how to know that this has 2 ligands one is a monentate ligand one is a bidentate ligand 31 PNMR would tell you again if you look into 31 PNMR spectrum of this complex 3 signals it should show for example these 2 are collated so they will be coupling with this one to show a triplet and then this would be coupling with this one as well as this one depending upon which one is larger either it can be a triplet of doublets or doublets of triplets and then this is very far from this one it may not couple with this one simply it can show a doublet and assignment and interpretation will be very easy and one can easily know simply by recording 31 PNMR spectrum whether you made your compound or something else has formed. Of course once this lectures are over and most of the aspects that I want to discuss are completed I shall give you some hints about 31 PNMR and how to understand interpretation of data using various multi nuclear NMR. So now let me give the preparation of another interesting compound of ruthenium this is also very useful compound how to prepare this one in the same way as I described for the previous one starting with Rucl3H2O here simply we are taking excess of triphenylphosphine and reflecting this mixture in ethanol that leads to the formation of a ruthenium 2 compound. So here the role of excess of triphenylphosphine we are using is to act as a reducing agent. So now let me tell you about a couple of rhodium complexes rhodium 1 complexes to make Wilkinson compound similar to Wilkinson compound or other bisphosphine bound complexes of rhodium plus 1. So these two precursors I am going to write are very very important. So rhodium that is for the first one is rhodium called dimer if you are curious to know the structure of this one this is how it looks like this is rhodium chlorocarbon dimer and of course if you can see here when we do reactions it symmetrically cleaves in this fashion to generate immediately a coordination site and then that coordinate site can be occupied by appropriate ligand and if you further force the reaction using drastic conditions you can also eliminate chord here and in its place another two ligands that means four electron donors can come into the picture so that means it is quite versatile either it can retain chlorobidges and substitute for chord or it can also breaks symmetrically to generate a vacant coordination site on each rhodium and also it can be substituted. So two options are there while using this one. Then how to make this compound here? Here again we are taking rhodium trichlorate trihydrate and then we are using cycloctadiene and then we are using ethanol as a solvent and also as a reagent and we are using sodium carbonate a base dimer already I have written the structure here. So this is how you can make this rhodium chlorocycloctadiene and dimer very useful compound this one and also from this one one can also generate chlorocarbonate dimer also or one can also use another method of directly reacting rhodium trichlorate trihydrate with carbon monoxide very interesting reaction you can perform solid state reaction I shall tell you details about the preparation of that one and also how to further react is to get some important derivatives I shall tell you in my next lecture until that have an excellent time ready chemistry.