 Hello everyone, once again I shall welcome you all to MSB lecture series on trans-traumatic chemistry. This is 50th lecture in the series, only 10 more are left. In my previous lecture at the end started discussion about the stereochemical consequences of substitution reaction and how even Werner contributed for understanding of the different isomers that can be formed if we consider optical isomers or geometrical isomers and once of the substitution is over whether we can get cis compound gives cis or trans compound gives trans or we get a mixture of isomers or what would happen if we consider optical isomers all these were studied based on cobalt 3 complexes because cobalt 3 complexes I mentioned they undergo substitution reactions very slowly as a result with simple devices one can measure the rate constant in that context cobalt 3 complexes and their chemistry looks to be very vital in understanding reaction mechanism. Still we use the same data started during Werner scale Werner's time to explain substitution reactions in square pattern complexes although lot of people have contributed much later. So now let us continue from where I had stopped I was discussing about how important to see the entering ligand and the leaving ligand positions and in what direction entering ligand is set in the second coordination sphere to come to the first coordination sphere doing substitution reaction. Let us consider these two cases here I have considered a trans compound here in the plane we have four ligands A and we have X, X is the leaving so in all cases in future X denoted ligand is the leaving ligand and then B will define the isomers conformation. So now this is a trans compound so now when we start substitution reaction in the second coordination sphere Y has an option of entering in a position opposite to the leaving group in that case what happens simply it comes here once when it enters either here or here what happens one of the A group would occupy the position vacated by X and then the cis compound would result can see here it can go any of this then you will be having a cis relationship between B and Y hence it is called cis compound the trans compound would give cis compound if the entering group is in the opposite side of leaving group on the other hand let us consider the same trans compound and if the entering ligand is in the same position or very nearby the leaving group in that case what happens there is no net change in the you know geometry conformation a trans compound would give a trans compound so that means substitution where the entering group Y in the second coordination sphere is opposite to the leaving group trans would give cis on the other hand substitution where the entering group Y in the second coordination sphere is adjacent to the leaving group then the trans compound would give a trans compound so this kind of analysis is seems okay for reactions that yielded primarily only one isomer if there are no preferred orientation of the entering group in the second coordination sphere would give a mixture of isomers if a if it has a preferential orientation that may be governed by sometimes the type of ligands we have in the surrounding and also the type of ligand coming and also the bulkiness steric and all those things matters so in that case what happens we would get a mixture of isomers Warner during that time concluded that as there is no way to predict the orientation of the entering group in the second sphere it is not possible to anticipate the stereochemical change that would take place during substitution this was his conclusion however with modern spectroscopic and spectrophotometric evidences and also analytical instruments many we have at our disposal so prediction of stereochemical changes may not be very difficult in homogeneous catalysis of course you must be knowing people use the directing groups and these directing groups play a major role in generating or synthesizing organic compounds or performing desired organic transformations so now keeping in mind that a noctahedral complex have a option of forming the product via dissociative pathway SN1 mechanism or displacement pathway SN2 mechanism also known as associative pathway so that means dissociative process for a noctahedral complex requires the formation of 5 coordinated intermediates we should remember the ligand goes what happens now we will be left with 5 ligands when we have 5 ligands it can assume one of the two well known geometries that is square pyramidal geometry or tetragonal pyramid and other one is more common one is trigonal bipyramidal geometry why when coordination number 5 is there all complexes in the intermediate have preference for these two geometries is a stable complexes are known with these two geometries and we have plenty of complexes among coordination compounds having both square pyramidal geometry in some cases and also trigonal bipyramidal geometries so that means the most preferred geometries are square pyramidal or tetragonal pyramid and trigonal pyramidal geometries so these two structures can be derived from the octahedral with little atomic motion that is another important aspect when we derive another geometry from the existing geometry we have to do movement of atoms or ligands with octahedral going to square pyramidal or trigonal bipyramidal with little atomic motion we can generate these two low coordination geometries and also such structures are in keeping with current theories of bonding in metal complexes and further we also have evidence through experiments that these two are the preferred geometries and also even the steric amyl consequences and after performing some reaction of isolating various optical or geometrical isomers that can also tell you about the geometry switch geometry you know it has opted in the substitution reaction or which intermediate it has operated and hence we can also get information about mechanistic pathways whether it is dissociative or associative let us consider the reaction of cis or trans again ma4 bx through a square pyramid intermediate now it can take place without rearrangement because why enters the position vacated by x so that means basically what happens what we have is a so these four are a and this is a b and this is x and now what would happen is if I take out this one okay we generate a square pyramid we generate a square pyramid like this now this is blank and now here why enters the position vacated by x this assumption makes perfect sense as the central metal atom is easily accessible at this position and the formation of new octahedral complex requires no additional atomic motion this just goes and another one comes so this is what exactly happens square pyramid I take out this one and then it comes back x goes okay and y comes as per wellness bond theory if we consider wellness bond theory the hibernation that utilized by octahedral complexes d2 sp3 if it is inner orbital complex inner sphere complex or inner orbital complex in that case what happens this d2 sp3 one of the d2 sp3 is empty now without electrons because the ligand has leaving ligand has departed now it is projected outward okay it is something like this it is projected outward in this direction this is suitable for maximum overlap with the orbital with a pair of electrons coming from the entering ligand if the entering ligand is coming with a pair of electrons and this already empty and this is projected out this empty d2 sp3 so immediately electrons can be donated and if metal to ligand bond can be established so as per crystal field theory nucleophilic attack at this position is favored because of low electron density even if you go with crystal field theory this is a nucleophilic we are telling so in that case what happens since here electron density is less this is the ideal position for nucleophilic attack so now let us just look into this stereochemical changes that one can see in a dissociative mechanism in the dissociative mechanism as I mentioned earlier we can think of two possible geometries one is square pyramidal geometry and other one is trigonal bipyramidal geometry so here we can see in this case initially let us say the chances are 50-50 let us assume the chances are 50-50 for forming square pyramid as well as dbp when it forms square pyramidal geometry so this x departs and this position is vacant now and next y is coming here and this is established here this compound comes back here this compound comes okay so y comes here and this is b and this is a so what we get is a trans compound so that is what I have shown here and then on the other hand it can also assume trigonal bipyramidal geometry in that case you should know how to number 1, 2, 3, 4, 5, 6 and be consistent when you are labeling them you start with the axial top one and then go to the plane 2, 3, 4, 5 and then 6 and keep this numbering constant when you generate intermediates that is what is important so now with this one what you can do is you go for trigonal bipyramidal geometry so now so b is sitting here when the b is sitting here so y should attack y can attack from different positions for example it can come from here or it can also come from 1, 3 here when it comes and when it reverts back to octahedral geometry so this y will be cis with respect to b so in this case what happens in 2 cases when it approaches from 1, 2 and 1, 3 so we end up getting cis isomer on the other hand if the entering ligand comes from y here what would happen is so the relationship with b with respect to y would be always trans so we end up with trans that means if it takes trigonal bipyramidal geometry what we are getting is we get one-third trans and two-third one-third trans isomer and two-third what we are getting is cis isomers this is about trans compound way considered here now let us look into the cis isomer let us consider cis isomer in case of cis isomer again we can think of square pyramid intermediate again it is straight forward now what happens x is deported from here when the x is deported y comes here and then we get a trans compound trans compound would be there but when we go for trigonal bipyramidal now we have two options are there what are those two options b can be apical or b can be equatorial so b can be apical and this this one can be apical or this can be equatorial that means we have two options the chances of formation of these two trigonal bipyramidal intermediates is 50 50 50 percent and this is 50 percent together and this is of course overall if I consider this 25 25 percent and this is 50 percent because if I consider both are 50 50 of course at the end when we get the different isomers we should be able to tell whether it really followed this method or not so now when it goes from this method this isomer now you can see here again 1 2 1 3 position you can get whereas in case of 2 3 you get so this holds good again for cis with this tbp tbp this you can call tbp 1 and this is 1 and this is tbp 2 in this case why we have put it now and now we have 3 positions it can come here it can come here and it can come here so if it comes no matter from which direction it comes we get cis only so now you should be able to calculate the total ratio of what kind of isomers we are getting and what is the ratio of cis to trans with cis and also trans it is up to you now do once again repeat this exercise and try to say we are taking 100 percent of this molecule in terms of percentage you should be able to write how much percentage of cis we got how much percentage of trans we got when we started with a cis lateral complex and also when we started with a trans actinatal compound involving both square pyramidal geometry as well as trigonal bipyretic geometry in case of cis 2 trigonal bipyretic geometries are involved along with square pyramid whereas in case of trans what would happen only one trigonal bipyretic isomer is involved and one square pyramid that you should remember. Let us look into optically active compound to see the consequences of substitution reaction on optical activity for this one best would be to choose a simple ethylene diamine complex of this type cis ethylene diamine complex and of course 1 b and x can be anything one is water one is chlorine or one is bromine one is chlorine any anything there should be different they are different then it exhibit optical activity. So then you can think of optical isomerism in this compounds when you do substitution reaction. Now we will consider like this so again it is very straight forward for that one I have chosen something like this okay this is one ethylene diamine and I have another ethylene diamine something like this. So now you can see this looks like ethylene diamine compound and these two are the ligands and this is b and this is x. So now what would happen is x departs here if the x departs okay and this position is vacant not much one can do it y can easily enter and this is the site where it can establish a bond with the metal and substitution reaction would be completed in this case what happens the cis compound we start with cis is there. So cis would give cis means retention you can see on the other hand when we think of trigonal bipyramid geometry so we get something like this let us assume this is the b and in now what would happen is positions I have already numbered you can see here positions already I numbered here and here let us look into the position from which y is entering. So y can enter from 3 5 it can enter from 3 5 here or it can enter from 3 4 so here. So we enter from here or in both the cases we get here it can enter or here it can enter in both the cases what would happen here it can enter and then here it can enter in both the cases what we get is cis product. On the other hand if it enters from 4 and 5 position 4 and 4 position is somewhere here. So now if it enters from here what would happen you get a trans isomer so that means with this one. So now we will see with other one again as I said this these two possibilities are 25-25% and this is also so total this is 50% and if this is 50% let us assume in this case this is 25 Tbp2 and Tbp1 is 25% just I am giving a hint for you to calculate the percentage of cis and trans isomers or retention or resimulation in this reaction this is how we should see and if I say exclusively it undergoes substitution reaction involving only Tbp then you can ignore this one just ignore this one and focus on these two in case of cis and when you go to the trans there will be only one Tbp and one square pyramid that is in case of non-optical other geometrical isomers. Now with other one with this one what would happen is okay this is something like this we are considering in this case what would happen is we have position 1 4 and 1 6 when it enters from 1 4 here and 1 6 here in both the cases what happens resimulation happens we get a mixture it appears very statistical but it makes sense it makes sense and if you happen to be a research scholar or if you are doing some MSc project certainly you can go for some of these complexes and you can examine what kind of isomers you are getting this is a wonderful experiment to look into it and analyze whether these assumptions whatever we are making with possibilities having two different geometries or something good or not. So in this case again so what we get is in case of a optically active compound says you can have square pyramid okay retention would be there no changes and the other hand when it undergoes substitution through two intermediates of Tbp Tbp1 Tbp so Tbp1 we get two isomers 2 by 3 and 1 by 3 trans we are getting here and cis we are getting here and here we are getting resimulation. So now if you think it is 100% you can segregate and write in terms of percentage what is it how percentage retention is there and how much percentage cis is there how much trans is there and how much resimulation. So that you can make yourself familiar. So now this is about dissociative pathway. So dissociative pathway I told you we can consider both cis complexes trans complexes and also optical isomers and involving both square pyramidal geometry and trigonal bipyramidal geometry. If the substitution reaction follows associative pathway or displacement method in that case we have to consider an intermediate having coordination number 7 and the best and ideal and symmetric geometry one can think of for the intermediate with coordination number 7 is pentagonal bipyramidal. And here now let us look into the stereochemical changes accompanying the reaction of cis and trans ma4bx with y as entering ligand involving SN2 mechanism through a pentagonal bipyramidal intermediate. So for this one again you have to label this way as I mentioned first you label the starting top axial one as 1 then come to the plane 2, 3, 4, 5 and then go to the lower axial 16. Remember this one and now here what happens we call cis attack and trans attack very similar to what we saw in the beginning. Cis attack means it will come to a position close to the leaving group, trans attack means it goes to a position opposite to the leaving group. You should remember or it can also be called as front attack and back attack. In case of cis attack what would happen why comes here and of course here the rate determining step is fission of entering ligand to metal bond. So in this case what would happen is we get in intermediate here cis attack what happens it comes very close to the leaving group you can see either come here or it can come here it does not matter both are same and then when x departs it will revert back to octahedral geometry and we get the trans compound because this is with respect to this one we are taking. So now Bx were trans now By are going to be trans so we get a trans product and then this is I mean both are equally possible when you are performing substitution reaction and if it follows SN2 mechanism involving pentagonal bipedal geometry 50% chances for cis attack and 50% chance for trans attack because here there is no preferred orientation of the entering group in the second coordination sphere that we should remember. So in that context we should consider both of them equal probable geometry and now in the trans attack what would happen is it comes to a position trans to the leaving group here when it comes to the trans in leaving group it has to be cis with respect to the B here. In that case what happens the once x is eliminated and then when it reverts back to octahedral geometry the relationship between B and y would be cis so we end up getting cis product here. Similarly this is with respect to trans let us consider the cis compound here in case of cis complex we are referring to these 2 cis and again equal possibility therefore cis attack and trans attack in the cis attack what would happen is it would come here absolutely very clear cis would give a cis but in case of trans attack we can see a different products coming here you can see here I have marked here in case of cis what are the positions it can attack which is mutual cis disposition with respect to the leaving group here. So for example if it comes at a position 1 4 1 4 you can see here 1 4 is this one if it comes here 1 4 and 3 4 3 4 and then 4 6 this is 3 4 is this one and 4 6 is this one here in all these 3 positions when y is entering its disposition would be cis with respect to B as a result what happens we get 3 cis we are getting. On the other hand in the trans attack if it attacks from 4 5 position 4 5 position is this one 4 5 position if you see between these 2 A it is entering as a result what happens it will be trans disposal with respect to B we get trans compound. So that means here also we are getting 3 by 4 cis and then 1 by 4 trans we are getting here and then here what we got is here everything is cis here. So that means if it is 50 and this also 50 you can I will tell you how to calculate the ratio this is 50 percent this is 50 percent and then 50 is cis now here and in this one 50 divided by 4 50 divided by 4 is this is 12.5 12.5 that means this will be 12.5 and then that will be 37.5 I would say will be cis. So that means in this one a cis can give 87.5 cis and then 12.5 trans. So considering 50 percent possibility of cis attack and 50 percent possibility of trans attack again I am telling you this is with normal mono-dentate ligands if you have some bi-dentate ligand some of the positions are blocked for the entering ligand as a result what happens it will be having now minimum orientations or preferred orientations to do cis attack or trans attack. It is very interesting very interesting you take tri-dentate ligand you take bi-dentate ligand and also you take sometimes tetra-dentate ligand in that case how many isomers we can get very interesting you can you can go with different ligands and also sometime bulky ligands and try to establish the stereochemical changes that accompanies this kind of substitution reactions keep working on those things and the problems also one can look for in standard textbooks that deals with reaction mechanism with this let me stop this lecture before I conclude let me go with one optically active isomer that using displacement method here. So now again you can see in this case a typical same complex whatever I consider in dissociative method I am also considering here it is ethylene diamine is ethylene diamine and then say this is something two different mono-dentate ligands. So in that case again cis attack is there cis attack retention will be there because cis comes and you end up with a cis product and the trans what happens again retention will be there trans in case of trans attack in the other one here you come across four three different possibilities you should remember in case of optical isomer as I mentioned we have three options here this is one one and this is two and this is three of course there I wrote combined together I wrote and I showed you whereas here I have just divided them for better understanding. So in each position I have there also I could have written that many so because there showing the position would be very easy and that is the reason I did not do it in case of trigonal bipedal geometries whereas here it is little bit not straight forward that is the reason I have shown here. So trans attack can happen in three different thing with three different positions of relative positions of B and X and ethylene diamine one is retention and when four and five when it attacks from four and five here a trans compound is compound so optical activity is lost and then in this case what happens inversion happens when it attacks from one four position okay one one four position here in that case what would happen is inversion happens so that means we know now what kind of stereochemical changes that accompanies when we start with an optical active compound like this through displacement or associative mechanism okay so with this I conclude the discussion as stereochemical consequences of substitution reactions in octahedral complexes and in my next lecture let me talk about electron transfer reactions or redox reactions.