 Hello everyone, welcome to today's lecture. I hope you had opportunity to go through the last class material and we can have a brief recap of what we did last time. So we discussed the dissolving metal reductions in the last class where we had seen that reactions with lithium in liquid ammonia or say sodium or potassium in alcoholic solvents like ethanol or magnesium in ethanol or some reactions with zinc in acetic acid. These are the reactions which allow the reduction of ketones to the corresponding alcohol and we discussed in great detail how these reductions allow the thermodynamically more stable trans product in cyclic cases. We also saw that the reaction proceeds via radical anion and this radical anion can allow the coupling to take place and we can form the C-C bond easily and we can get a di anion of this type which can lead to the formation of 1 to diol. We also saw that titanium 0 permits the coupling of the carbonyl groups to form the corresponding olefin at one stretch in one go but also we saw that titanium trichloride and DME together and then in the presence of zinc copper couple leads carbonyl groups to form the corresponding diol. So one can easily carry out the reaction at relatively low temperature with this system here and one can get the diol. If one heats the same reaction by heating condition then of course one can get the olefin. This is what we discussed in the last lecture that titanium 0 if it is done under normal McMurray coupling olefination conditions then directly we get the olefin. On the other hand if one wants to diol to form the pinnacles to form these are basically pinnacles and if that is what is required then the reaction can be carried out at relatively low temperature and with titanium 0 form from TICL 3 DME crystalline complex with the reduction with zinc copper couple. So the same zinc copper couple either we can also use potassium metal as such and with the TICL 3 it allows deoxynation of visceral diols to the corresponding olefin if the reaction is carried out at a high temperature. That means if the reaction is run in T, reflecting THF with excess of potential potassium metal the diols are essentially converted to the corresponding diolate and they react with titanium to form the same intermediate which is implicated in the McMurray coupling. We will discuss that in a minute how the 2 are converging into the same intermediate. But if we start with acyclic systems such as here we can see that there is not much of stereoseleactivity that is observed. This is the equation 1 and trans diols deoxynage as readily as the corresponding cis isomer. For example here is a mixture of cis and trans. So if this is beta oriented for example and this is a mixture so we can expect both cis and trans diols but then a mixture of cis and trans diol directly gives the corresponding olefin. This is what indicates that the mechanism must be involving in a non-conservative fashion. So you have a diol which is stereochemically properly oriented. However it leads to the 33% formation of cis olefin and 45% formation of the trans olefin. So when the reaction is carried out in such a way that the visceral diols are allowed to react under the same conditions in which we had originally formed the diol. But now we start hitting then the corresponding olefins form. So these two cases are very good indications of how the reactions may proceed and therefore it will reflect on the mechanism of the reaction. So a lot of work has been done by McMurray and others and how does this diol give the olefin. So essentially first thing is that there is an oxophilicity of the titanium that allows the formation of the olefin because there has to be a tendency of the oxygen of the diol to interact with the titanium. So there are three such possibilities which have been implicated. One is that if we start with the diol like this one to diol it could form a cyclic transitions or intermediate of this type where only one titanium is involved and therefore we have two oxygens of the diol interacting with one titanium and that is in titanium 2 form. That is if we take the diol and have only one titanium 0 giving two electrons and then combining by the loss of hydrogen. So we can get this type of cyclic 5 member intermediate or a species of this type. The other possibility is that each of the hydroxy group interacts with the titanium and forms an intermediate of this kind where the two oxygen titanium bond this particular part of the molecules are away from each other. So this also can lead to the formation of the olefin here and this also can lead to the formation of the olefin here. And the third possibility is that there is a basically simple coordination that means the OH or the O- that is formed during the reaction simply coordinates simply attaches itself to the titanium 0 part and then of course there is a transfer of the oxygen to the titanium. Of course when we talk about these and if there is a formation of this kind of species on the surface of the titanium 0 there will be titanium 1, there will be titanium 2 and of course titanium 0. So what these possibilities suggest that we can expect such intermediates which are likely to form. How do we rule out one or the other and see which exists or which forms the best way. Now many experiments have been done and this McMurray has written a review which you can check it here accounts of chemical research in 1983. So what happens in here is that McMurray took two different types of rigid and locked alcohols which can easily come from this Norborn system. And one of them is cis diol and the other one is trans diol. So obviously one can expect that the intermediate that we discussed where we had this oxygen and oxygen and then coming onto the same titanium 2 coming and forming this type of intermediate one can expect that to happen with the cis diol but it was found that both cis and the trans diol they are equally reactive and they form the corresponding olefin without any problem. So it is very clear that the formation of such intermediate is very unlikely because both of them they react equally facile. If there was a difference then of course the trans would have reacted slower than the cis because formation of such fine member intermediate would be better and easier with cis diol. The other reaction that they carried out was this cis diol and the trans diol and it was found that the cis diol reacts very easily and the trans one does not react at all. So what does it mean if we look at the confirmations of these diols we can see that the two hydroxy group in the trans diol would be here like this and the second possibility which we considered earlier was that each hydroxy groups coordinates to the titanium species here where the two of them are interacting separately. But then that obviously is not happening because this is not reacting. So there is no reaction here it means there is no product formation it means that this binding of the oxygens to individual titanium is not possible. On the other hand if we take the cis diol where we can anticipate that such a reaction will allow the OH here and the OH here both are cis to each other then of course they can go through the same intermediate as this. But obviously they cannot they need not go through this two different titanium binding. So these two possibilities are completely ruled out. Now the third type of possibility of the mechanism basically involves attachment of the oxygen to the titanium surface and that eventually leads to the formation of the olefins. Now for both the types of reactions that is coupling of carbonyl compounds to form olefins and conversion of diols to olefins can be considered and that is shown below here. The reaction takes place on the surface of the titanium particle and we can also explain the loss of stereochemistry by invoking a stepwise fragmentation of the bound diolate. Now what is happening is the titanium surface in which generally we show as titanium 0 but it also has titanium plus 1 and titanium plus 2 species. When a carbonyl compound or a di carbonyl compound of this kind where the two carbonyl groups are attached to the same molecule when such a molecule comes in contact with the titanium surface then there is a transfer of an electron from titanium to carbonyl carbon resulting into a radical anion of this particular type here as I have shown on the top and in a similar fashion there is a transfer of an electron from the titanium surface to the other carbonyl leading to another radical anion. Now these anions on the oxygens will of course will have an interaction or attachment to the titanium surface and the radicals which are on the carbons would couple to form a carbon carbon bond. Now since the titanium surface has lost an electron so we are showing it here as a titanium bond and of course then you have the oxygen titanium attachment. Now what happens is we then allow or we invoke or we propose that there is a carbon oxygen bond cleavage resulting into a radical formation and of course the oxygen radical would take an electron from the titanium surface and the titanium surface will become a titanium 2 for example from titanium 1 to titanium 2 and there is a titanium oxygen double bond. And then in a stepwise manner the second carbon oxygen bond then would break and leads to another radical here and of course oxygen radical. The oxygen radical will form the titanium oxygen double bond the way I have shown it up and the radical which is generated here would couple with this radical to form this particular olefin as the final product. Now this titanium oxygen double bond would lead to what is proposed to be a titanium dioxide eventually. Now obviously here the stepwise fragmentation has been suggested mainly because it allows you to lose the stereochemistry because if we start with a particular diol of a particular stereochemistry it is found that eventually it loses the stereochemistry and that is the reason why stepwise fragmentation has been proposed. Now if we take diol of this kind and treat it with potassium and titanium trichloride under the magma recoupling conditions then of course we expect that potassium diolate will form and it will have similar type of interactions on the titanium surface as we have discussed that the titanium 0 surface not only has titanium 0 particles but also other low valent titanium particles like titanium plus 1 or titanium plus 2 and therefore the interaction of the potassium diolate with the titanium surface leads to similar type of intermediates of this kind which then eventually lead to the formation of the olefin and of course titanium dioxide or similar type of titanium species. So this is how the mechanism is perceived and this is the mechanism that is believed to be operating in this particular magma recoupling case. What is the disadvantage of this magma recoupling is that the reaction is limited because many functional groups are not easily tolerable under these conditions. So they can get also get reduced because it is powerful reducing system. There is a ketones, thio ketones for example ketones, thio ketones, acyloins, the hydroxy ketones, bromohydrenes, oxidins that is epoxides, cyanohydrenes, allylic or benzylic alcohols and some other functional groups can also get affected by this particular reagent. So there is a disadvantage but at the same time the advantage is that we can easily carry out the coupling either in an intermolecular fashion or an intramolecular fashion. We can stop at the diol stage, we can also make an intramolecular coupling and get medium size ring or large rings and of course we can convert the diols to the corresponding olefins using the same condition but by refluxing or heating at high temperature under the same reducing system conditions. So there are advantages and there are also disadvantages but because of the simplicity of the reaction it has been used in making different types of olefins or diols using this reaction. We go for the next part of this particular dissolving metal reductions where we take say for example sodium or potassium or lithium liquid ammonia conditions and try to look at the reductions of alpha beta unsaturated ketones. We had seen the normal reductions of cyclic ketones or a simple ketone where reduction gives the corresponding alcohol. Here if we take an alpha beta unsaturated ketone like this and transfer an electron from lithium or potassium or sodium obviously first thing will happen is the transfer of the electron occur at the carbonyl carbon here. So you can anticipate to form something like this. Here you have an O- and you have an electron here which of course will go to this particular carbon atom and form this intermediate of this type. So we have a possibility of such a movement of here there is an electron. So you can anticipate such a resonance structure of the radical anion to form. And when second electron is transferred by the metal here we can expect that a di anion is formed and this di anion on protonation gives the corresponding enol and this enol will obviously be in equilibrium with the corresponding ketone. So under these conditions what is found that the reduction of an alpha beta unsaturated ketone with metal liquid ammonia system leads to the formation of the corresponding saturated ketone that means the only double bond that has been reduced. This is a very important reaction and especially it is very useful when there are bicyclic enones or in a system like steroids and terpenoids we look at the reductions and it helps in designing the stereochemistry of the molecule. So you have a system like this here say you have a ketone of this type and then how does the reduction occur. Now obviously we expect the reduction should give the corresponding ketone but more importantly what is there is what is the hydrogen that is coming here what is the stereochemistry of the hydrogen at this stage that is what is most important. And in that respect lot of work has been done and Gilbert Stork at Columbia University has devised a rule that the reduction product is the most stable of the two isomers with the newly induced beta hydrogen being axial to the ketone ring. For example if this particular hydrogen which is what is expected to come after the reduction it prefers to remain axial. So this particular hydrogen is actually oriented in the ring that contains the ketone. You will see how does this happen. So there are essentially three possibilities. So we can write that bicyclic ketone into this particular conformation and as we can see that one of the left side ring is obviously half chair and the other one is expected to be chair and therefore we have written like this particular way half chair and the normal chair and once the electron is transferred to this particular carbon atom the anion the di anion which we had discussed earlier would be something like this here and we can expect that the anion at the junction would be axially oriented. And such an intermediate when it gets protonated or when the we can also use either a proton as an electrophile or we can use any other electrophile which we can term it as E plus then the electrophile comes in this fashion and what we are expecting is and that is what Gilbert Stark suggested that the proton or the electrophile would come in the axial orientation. Now we can write in a slightly different way that is we can write that the conformation of the di anion maybe somewhat like this where it is cis orientation. For example here we had this R group axial and this anion is also axial and therefore 1, 2 di axial would be trans relationship. But now we are talking about this being axial and this being equatorial therefore we are talking about a cis decalin type of system. We can also write slightly differently another decalin which is also cis and in this particular case for example this was axial here this was also axial and this is equatorially oriented. The anion is equatorially oriented. So we can write flip the rings in such a way that is R1 is equatorial, R is equatorial which were earlier axial and since the anion was equatorial now it has become axial. So we have basically flipped we have got another conformation but cis conformation and of course cis decalin system but in a different conformation. So we expect these three types of orientations one being this number one here which is trans decalin type then we have second which is cis decalin type and the third also cis decalin type. So these are the three possibilities we exist and here of course the the electrophile comes in an axial fashion and here as you can see that this will give an axial electrophile and this will give an equatorial electrophile. So let us see which one would be the best. So we have this number 2 here you have number 3 here and we have number 1 here. If we look at number 1 and number 2 if we take these two cases they are stereoelectronically favourable that is because the anion p orbital either here or here for example and you consider the pi orbitals of the double bond this is the double bond this is the double bond. If we look at the pi orbitals so they are all planar and this is also essentially planar. So there will be a proper overlap and stereoelectronically these are favourable confirmations. So the only other problem is that when the electrophile either a proton or any other electrophile when that approaches as you can see that in the second case where the R group here is axially oriented and when the electrophile approaches from the equatorial side we are talking about 1, 2 axial equatorial which is cis and there will be a steric hindrance. In addition to the steric hindrance that of course one has it here but that is anyway existing in all the cases. So when the incoming electrophile comes in and approaches this particular anion you have a steric hindrance offered by the R group which is at the junction. Whereas that is not the case in this case here because the incoming electrophile is coming from the axial side from the below and has nothing to do with the R group or R1 on the top. On the other hand if we take the third confirmation here like this we can see that the anion there is no sterile electronic overlap possible. If we look at the anion which is axially oriented and that is actually orthogonal to the double bonds p orbitals or the pi orbitals which are here. So basically there is no overlap between the p orbital containing anion and the pi bond there is no overlap at all. So stereo electronically it is not possible and also if you look at the approach of the electrophile to this particular anion then also you have there is a cis interaction or the 1-2 orientation will be like a axial equatorial. This is going to be axial and therefore there will be a steric hindrance here. So the third one is the worst among the three that is both stereo electronically as well as sterically it is not feasible and therefore the only way reaction occurs is via the confirmation of type I which leads to the formation of the electrophile coming in from the axial side and that is what is in conformity with the Stokes rule. So we will stop it at this stage and we will take up other aspects of reduction in the next class. You can study these reductions carefully and we will discuss it during our question answer if you have any questions too. Thank you and bye.