 Hello everyone, it is with pleasure I welcome you again to MSB lecture series on transformative chemistry. In my previous lecture I started discussion on classification of ligands by donor atoms and after giving little bit of introduction I started discussing about metal hydride compounds that means using hydrogen as a donor atom. So let me continue from where I had stopped. So when I mentioned that hydrogen molecule or hydrogen is a sigma donor as well as a pi acceptor. So how that happens I have shown here. You can see here in this diagram I have shown how H2 molecule acts as a sigma donor. For example if you recall writing Lewis dot structure we put a pair of electron between two hydrogen atoms that is responsible for making H2 and this is what I am referring to. This is going to suitable empty orbital on metal to establish metal to hydrogen bond. So this is called sigma and then if you see the molecular orbit diagram that I am going to show in a couple of minutes you have filled sigma orbital with pair of electron that is responsible for making H2 molecule. And also you have high energy empty sigma star. The energy of sigma star is more or less comparable to the energy of non-bonding electrons present in a given metal complex whether it is tetrahedral, square-pan or octahedral. So in that case what happens electrons from metal orbitals can be promoted to sigma star and that is called back bonding. So this is how it acts as a back bond you can see here this is the sigma star so this will be like back bonding. So it becomes more clear in my next slide. So more details I am going to provide once again when I consider oxidative addition of H2 to a typical square-pan or complex. Let us look into how these interactions sigma donation as well as pi acceptor nature of H2 would weaken the HH bond. If you consider an isolated H2 molecule a typical HH distance is about 84 to 90 picometer. It is about 74 picometer for free H2 when it forms a complex with H2 as a neutral donor in that case bond distance increases to 84 to 90 the HH distance I am talking about. So that means HH distance in free molecule is 74 ppm in case of when it is binding to metal. So this distance I am referring to this will be in the range of 84 to 90 depending upon to what extent these electron pairs are drifted more towards the metal atom. So metals capable of strong back donation and with an accessible oxygen state plus 2 units higher can break the HH bond to give a dihydrate. That means you should recall whenever we want to perform oxidative addition on a metal metal should have 2 stable oxygen states with a difference of 2 electrons. That means for example if you consider a D0 system it should be readily oxidized to D2 system for example if you consider a metal in 0 valent state it should be readily oxidized to 2 plus state or if you consider M in plus 1 state then it should be readily oxidized to plus 3 state. So for example if you consider D10 system palladium 0 palladium 0 can be readily oxidized to palladium 2 in the same way if you consider rhodium or idiom D8 system is plus 1. So plus 1 can be readily converted to plus 3. That means whenever you want to perform oxidative addition you should consider a metal having 2 stable oxygen states with a difference of 2 electrons because this oxidative addition as well as reductive elimination process are 2 electron process. In that context if the metals are capable of strong back donation with an accessible oxygen state 2 higher than the original oxygen state that means iridium 1 to iridium 3 or palladium 0 to palladium 2 something like that or palladium 2 to palladium 4 in that case if H2 comes one can break HH bond to form 2 metal to hydrogen bonds or metal hydride bonds that is what it means. You can see one such example is given here you consider this molecule here we have tungsten complex here. So this one is in equilibrium with neutral as well as anionic one so of course one can always tilt the equilibrium either towards the right side or towards the left side by choosing appropriate ligands having right kind of donor and acceptor properties. Once again to make it clear I have shown here you can see clearly here a pair of electrons are there and this is going to metal through sigma donation and then the electrons present in its non-bonding are by orbitals that can be promoted to sigma star. So in this M O diagram it is very clear these 2 electrons are coming from H2 molecule and to the metal through sigma donation this is the sigma bond and then these electrons with pie symmetry would be going to sigma star this is back bonding. Now what happens the bonded electrons you are drifting away one thing so that essentially losing a weakens HH bond once again besides this you are also adding electrons to the sigma star so that means there is 2 way weakening effect one can envisage here one is taking the bonded electrons and then adding electrons to the anti-bonding both would result in weakening of HH bond if the back donation is extensive and pie acceptor nature is also extensive that eventually leads to the breakage of HH bond to form 2MH bond that means whatever we have something like this and you can see some sort of intermediate coming like this and then that would become something like this. So this is what a metal complex is capable of doing for H2 molecule when we use this kind of complexes in hydrogenation of olefins and other things in organic transformations and for example if you take a olefin such as ethylene and add H2 to under high pressure and high temperature and even leave it for days probably at the end you can see only about 10% or maximum of 10 to 15% conversion of olefin to corresponding hydrocarbon but in contrast using an appropriate metal complex and interact that with hydrogen molecule in presence of olefin you should be able to do it even at room temperature hydrogenation of olefin how it does this is how it does it first what it does is through a sigma bond establishment this electron pair is taken away from H2 to weaken HH bond and then more electrons from metal are added to the sigma star and then your bond order is totally destabilized and eventually this bond breaks and once this bond is broken and olefin comes and sits here and then it is what opposite to beta hydride elimination happens and then hydrogenation process proceeds so we will discuss those things say once again later what are the methods we have at our disposal to characterize a metal complex when we have M H bond 1 H NMR comes very handy and you should remember 1 H NMR we can see anywhere from 0 to minus 60 ppm for non-detend complexes that means if you put if you make metal hydrogen bond for a metal having non-D0 electronic configuration that means if you have more than 1 to up to 8 or 9 electrons then then in that case for non-hydrogen complexes then 0 to plus 10 for D0 complex you should remember 1 H NMR signal for M H bond would appear between 0 to plus 10 for D0 system whereas for non-D0 complexes it can come anywhere from 0 to minus 60 because of paramagnetic effect and also shielding effect and once again if you have NMR active nucleus such as platinum or rhodium this coupling also comes very handy in characterizing these compounds and also sometimes if you have polygons which are also NMR active for example you have fluorine or even phosphorus that even P phosphorus I equals nucleus when I equals half and it is also 100% abundant so this also comes very handy in observing phosphorus to hydrogen coupling and when you look into IR spectroscopy M H stretching frequency for a typical terminal hydrogen appears around 20 to 100 to 1500 centimeter minus 1 but can be weak therefore under label of course this is always a weak absorption you will see as a result what happens it is under label you cannot really depend on IR data to know whether M H bond is formed or not or it is broken or not or it is disappeared or not and sometime deuterium labeling can also help because in that case what happens 1 H NMR does not show any signal and once again if you have single crystals of molecules containing M H bond one can go for single crystal X-ray diffraction but hydrates difficult to detect because of minimum electron density we have and M H bond may be underestimated and if you want to really get a precise bonding parameters one has to go for neutron diffraction it is most preferred one but not readily available and large crystals are required here unlike X-ray diffraction. So early trans-metal hydrates are acidic in nature for example you can see the pKa value will tell you late trans-metal hydrates can be quite acidic especially with low valent metals when the metals are in low valent state they can be quite acidic if you can see here H2 FeCO4 is there pKa is 4.4 and here iron is in well H2 is there iron is plus 2 state and when you look into this one pKa of a typical acid is 4.75 it is almost comparable and then if you go for this one where we have CO4 and CO and H this one pKa is 0.4 that means it is quite acidic and see here very similar to we have in trifluoro acidic acid like that. So that means pKa values will give some idea about to what extent metal bound hydrogen is acidic or not that also helps in using such compounds in catalytic transformations. Now let us look into various methods that we have at our disposal to make some of these compounds having MH bond whether we can use directly H2 or whether we can use some reagents so that we can perform reaction under mild conditions one such method is protonation. Let us take FeCO4 di anion and add H plus here we can make anionic and then if you add one more we can make neutral complex here. So this is from hydride sources or hydride donors this is from hydride sources hydride donors. So this through protonation now let us look into if we have some hydride sources or hydride donors how we can perform reaction to make metal to hydrogen bond. Let us take WCOCl6 tungsten is in plus 6 state and treat this one with this lithium boride in presence of a tertiary phosphine such as it can give you mixed hydride phosphine complex. So this typical reaction involves hydride sources or hydride donors and if you look into synthetic methods one can use in laboratory for small scale preparation we can go for alcoholic KOH for example let us consider platinum complex such as this triethylphosphine dichloroplatinum a trans compound and treat this one with KOH in EtOH ethanol this is also called as alcoholic KOH this gives selectively monohydride another example is let us consider this complex when you treat this one with sodium borohydride in THF similarly one can also perform a reaction in aqueous medium using a reducing agent such as zinc consider rhodium 3 plus in aqueous solution it can give rhodium hydride of this type just by looking into it a type of anionic ligand we have and also the counter anion we have here you should be able to tell the oxidation state of rhodium I leave up to you to guess the oxidation state of rhodium in this case let me give some methods of preparation involving H2 gas take RuCl2 Pph3 thrice I gave you the preparation of this one in my previous lecture treat this one with H2 in prints of a base such as triethylamine so it leads to the formation of a very useful hydride chlorohydride complex plus triethyl ammonium chloride is formed one can also start from dimethyl zirconocene of course zirconium is in plus force state in this case when you treat this one with H2 at 60 degree centigrade and 60 atmosphere pressure it forms the corresponding dihydride should be careful here when something like this is written like this we have two zirconium to hydrogen bonds that means hydride so if simply put H2 then it indicates H2 is acting as a neutral ligand and now of course you can anticipate the product that comes out here two equivalents of methane would come out similarly if we take hexamethyl tungsten and treat this one with the hydrogen in prints of a phosphine such as a trimethyl phosphine here one atmospheric pressure of hydrogen is sufficient it can give you so here all methyl groups are coming out as methane. So how about with rhodium or iridium let us consider one such example let us consider Vasca's compound that you are all familiar with of course in this one iridium is in plus one state so it is here hydrogen it is a reversible reaction now oxidative addition takes place and we get dihydride complex so I will be elaborating this this particular reaction in more detail later when I take up oxidative addition reductive elimination reactions of coordination compounds these are the few examples or some representative examples in which one can see how some metal complexes with appropriate ligands can be treated with directly hydrogen gas under various pressures and temperatures suitable for that particular transformation to get metal hydrate bonds a typical reaction is something like this a general reaction I would write here a general reversible reaction will be something like this initially and then it changes to hydrate complex after oxidative addition process is completed so this is very very important when it comes to the hydrogenation process in homogeneous catalysis besides this we also use some of hydrogen species such as any BH4 also to make metal hydrate complexes example let me write couple of examples here you can see here copper is in plus one state and then BH4 minus BH4 minus is bridging and making two bridged hydrogen bonds here so that means borohydride can establish either single bond or triple bond or even double bond I mean you can have two bridging units or one bridging unit or even three bridging units that means if you take a typical one can have something like this something like this one can have or one can also have something like this so all these three types of examples are known if you ask me whether this is possible certainly this is not possible because you should remember in BH4 B is utilizing its SP3 hybrid orbitals which are disposed in this fashion that case what happens so visualizing this one is something very difficult on the other hand in this case it can have some sort of pyramidal still it can retain distorted tetrahedral geometry whereas in this case it is not possible as a result this kind of utilizing all four hydrogen atoms for establishing bridging something like this cannot be seen or it is not possible whereas these things are quite possible here so let me continue discussing more chemistry and also why we have to make these complexes of course we have seen in case of oxidative addition of H2 to iridium 1 to form iridium hydride complex and that can very efficiently do hydrogenation of unsaturated organic molecules and but can we go beyond this one and utilizing further organic transformation let us see in my next lecture until then have an excellent time reading chemistry.