 Hello everyone, I would like to welcome you all for today's lecture. Before we go towards the new aspects of reductions, we would like to see once more what we did in the last class briefly. For example, we towards N we discussed LS and KS selectrides as are shown here and these were considered to be very statically hindered new reducing agents. And as I have mentioned many times that lithium ion based reducing agents are stronger than sodium or potassium based reducing agents because of the stronger binding of the oxygen with the lithium plus. Also we looked at how sterically hindered borohydrides these selectrides affect selectivity in a great way. For example, this is this example we considered last time and we saw that reduction leads to the formation of this hydroxy compound where the hydroxy group comes from the the reduction happens from the beta side and the orientation of the hydroxy group is towards the alpha side. And for this we did not consider under normal circumstances what we consider is trans conformation however we considered the cis conformation and how we saw that the steric hindrance leads to the formation of this particular molecule. So it is not only the steric hindrance that we need to see we also have to see the steric hindrance in conjunction with the conformation of the molecule that allows because in that case the C5H11 group as well as the methyl group were both equatorially oriented and thus the reduction is guided by the conformation of such molecules. And we also studied the sodium cyanoborohydride based reductions and we saw that since sodium cyanoborohydride is stable under acidic conditions up to pH 3 it was easy for carrying out reactions under acidic conditions. For example we also we consider reductions of enamines which are possible because the protonation allows the formation of this type of intermediate where the proton comes in here and of course positive charge will come here. So you have the lone pair of electron pushing this double bond and protonation and now under these conditions the reduction allows the reduction of the ammonium ion and sodium cyanoborohydride delivers the hydride at this particular position. And we saw many different types of reductions using sodium cyanoborohydride under acidic conditions. One of the last examples that we took was how to convert a carbonyl group to the corresponding hydrocarbon where both the hydrogens are basically coming from or during the sodium cyanoborohydride reduction. One of the ways as we have discussed last time was of course you convert the carbonyl group to a diethion protections and then you do the runny nickel cleavage where the carbon sulphur bonds get cleaved. On the other hand if we have to use sodium cyanoborohydride then we convert the carbonyl group to the corresponding tosyl hydrazone and this tosyl hydrazone under acidic conditions in DMF in the presence of sodium cyanoborohydride and the heat conditions gives the corresponding hydrocarbon. And we looked at the reactions of alpha beta unsaturated tosyl hydrazones such as this which undergoes rearrangement and forms this particular molecule where as I have shown here by mechanism that the reduction allows the shifting of the double bond due to this 6 member proton transfer or the hydrogen transfer. So such kind of reductions accompany with rearrangement. So this is something very interesting because this normally does not happen. So therefore we looked at various aspects of sodium cyanoborohydride which are very useful and one of the other important reaction that we also saw was the conversion of a primary amine to Nn dimethylamine which is highly synthetically useful using formaldehyde and sodium cyanoborohydride under acidic conditions. Now we move on to dissolving metal reductions. In dissolving metal reductions we have alkali metals, alkaline earth metals and some transition metals. For example, lithium, sodium and potassium in liquid ammonia can be used as a reducing system and we can also use magnesium in ethanol as a reducing agent system and zinc and acetic acid or zinc and proton with an acetic acid is also useful as a reducing agent in which the metals dissolve and the reduction occurs. Now these reducing systems allow reductions of many carbonyl compounds or sometimes they also lead to coupling of two molecules depending on conditions. It is a very useful reducing system and the reduction as well as coupling lead to products which are both important. Now if we look at the monovalent metals such as sodium or potassium in ethanol, they reduce carbonyl group to the corresponding alcohol. Whereas, bivalent metals like magnesium lead to coupling products. First we would like to see what exactly happens when a carbonyl compound is brought in contact with a metal either a monovalent or a bivalent. So for example, once this metal transfers an electron to this carbonyl carbon, there is a movement of electrons and that leads to the formation of this radical anion which is here. Now if this radical anion receives another electron from sodium and then undergoes protonation, so then we get a reduced product. On the other hand, if this particular radical anion undergoes coupling, so that is possible with magnesium. So it is basically undergoing coupling and this coupling leads to C-C bond formation. As you can see here that there is a C-C bond formation here and this leads to this 1, 2 di anion and that under acidic conditions or during the reaction receives proton and then we get the 1, 2 diol as the product. So this is the coupling product. One can of course see that this particular radical anion would be in resonance with another radical anion of this type. So there is a stability associated with it. So what we are trying to look at it is that this radical anion which is what is here, this radical anion when under proteic conditions as liquid ammonia gets proton to form this radical which then again receives an electron to form the anion here and then that gets protonated with ethanol to form this particular hydroxy compound where this hydrogen also comes in here. So now we can see that there are conditions under which we can allow such reductions to take place. Non-protec solvents lead to more of coupling products. It is very clear that the protonation of the anion should not take place until it undergoes coupling. So if we want coupling products to form we have to preferably use a non-protec solvent. We can always protonate towards the end of the reaction during the workup. Now for this coupling to take place we need to use magnesium or magnesium iodide this mixture or magnesium amalgam or aluminum amalgam. These are generally the metals or the systems that are usually used. When we use bivalent magnesium the ketile this is what is called ketile or the radical anion is basically stabilized of course and reduction is slow and coupling is dominant that is because you have two molecules of this radical anion or the ketile they come in contact with the magnesium II plus and they bring into the closeness and thus there is a very easy possibility of coupling of these two radicals leading to the formation of this particular C-C bond and then the protonation allows the 1, 2 diol to form. So this coupling is easily possible in a non-protec solvent. This is also something that is used when we are trying to dry THF when you take a THF and try to dry by using sodium as a metal then we use a small amount of benzophenone into this THF suppose you have 2 liters of THF and use a 1 gram or less than a gram of or even less than that of benzophenone which and of course you use sodium wire into this to make sure that the moisture is not present. Now how do you make sure that the moisture is not present that all the sodium all the moisture which is present has reacted with the sodium and that is the reason why we use a small amount of benzophenone which allows to form the radical anion or the ketile and this is what is form similar to the ketile that is here. Now since there are 2 phenyl groups attached the radical here is highly conjugated and because of this conjugation this gives a dark blue color and so if we see that when we take THF and which is already pre-dried with some other like calcium chloride or something and then you add sodium wire into it you press sodium wire into it and to flux it and add a pinch of benzophenone if the blue color comes that means all the water has reacted. Sodium has a preference of reacting with water first and then it can react with the benzophenone. So this is a basically an indicator which is used which I am sure many of you would then later on use it in your practicals in during your research or PhD. Now sodium and alcohol reduction with cyclic ketones is also very interesting because if you have a cyclic ketone like two methyl cyclohexanone and react it with sodium and ethanol we can see that there are two possibilities one is a trans product and the other is a cis product. So the trans product is formed in 99 ratio is to 99% and of course the other one is formed in 1%. So as you can imagine the confirmation that if we have the hydroxy group like this the methyl will go up like this and this is what is the trans product. On the other hand if we take the cis product then we can see that we will have the hydroxy group and the methyl group oriented in this fashion. So obviously thermodynamically this is more stable. So why is it that the more stable trans product is formed is because if we consider this cyclohexanone's confirmation the when the radical anion is formed by the transfer of an electron from sodium the preference of the oxygen is to adopt equatorial orientation like this in radical anion or other intermediates. That is because the orbital containing this particular radical is relatively small and the carbon oxygen bond is a bit and the oxygen is bigger and therefore it prefers to remain in an equatorial position which then gets protonated and then again sodium gets transferred and eventually the hydrogen comes in. So this is the reason why the reduction of cyclic ketones the OH group generally prefers to come to the equatorial side. Reduction of ketones with dissolving metals and low valent transition metals in the absence of protic solvents or a protic donor proton donor leads to coupling product this is what we saw when we looked at the magnesium based coupling. Now there is also something very useful and interesting reaction which is called as McMurray coupling basically it is a pinnacolic coupling or reductive coupling that leads to van der Dijol's in a very effective manner. Now apart from magnesium amalgam and aluminum amalgam that what we saw even low valent titanium chromium or samarium iodide can also be utilized towards coupling. So there are a variety of methods or approaches to allow the coupling to take place. Now why is it that the coupling is important because if we allow the carbonyl groups to couple of course we can get 1 2 diols but under certain conditions they also lead to the formation of corresponding olefin. That means if we have a carbonyl group something like this here and we allow the coupling followed by elimination of the hydroxy groups we can have the 2 molecules from a corresponding olefin. This is what is called McMurray olefination or McMurray coupling. So we will discuss that in detail now McMurray olefination somewhere in under 70s this particular reaction was established and developed by McMurray where he used titanium 0 which was derived from titanium trichloride and lithium aluminum hydride or reaction with potassium or lithium as electron transfer. So basically TiCl3 was reduced to the corresponding Ti0 that is titanium 0 using either a reducing agent or a metal electron transfer from potassium or lithium. And that allowed say for example cyclopentanone would give such a molecule and this type of molecules are very difficult to make under normal conditions and therefore it has a lot of importance. For example we can also take an aldehyde even if it is an alphabeta unsaturated aldehyde it undergoes coupling and forms such a large molecule very readily by using this titanium 0 derived from titanium trichloride and lithium aluminum hydride. It is also possible that we can conduct a cross coupling that means we take say for example cyclopentanone and we take acetone and we use say lithium and titanium trichloride. What they have observed is of course a formation of this type of cross coupled product that means one molecule of acetone and one molecule of cyclopentanone they react to form one molecule of this which is formed in around 50% and of course cyclopentanone can also couple with itself and that forms 26% of the product and of course remaining will be acetone coupling with itself also can form. So this kind of cross coupling is also very important because even if it is 50% yield it is not easy to get and therefore it is a useful alternative for getting such olefins where the two parts are from different alkytones. It is a powerful and versatile method for synthesizing a wide variety of alkenes this reaction can also be drawn in an intramolecular fashion which is very interesting because you can derive you can make a large number of cyclic molecules and not only just intramolecular fashion reactions but also they can lead to medium and large rings and that is something very much useful. For that you can of course use many reducing agents such as what we discussed lithium aluminum hydride, potassium or lithium but they also have this zinc copper couple with TICL3 or even zinc and titanium tetrachloride is used essentially the idea is to get titanium 0 which allows an electron transfer to take place. This intramolecular variant of the reaction to form large or medium size ring is very useful as you can see we can take such an example where we have already two double bonds which are present here and we have towards N a ketone and an aldehyde which then is allowed to react as you can see that this is the part which has come and this is the part which is the aldehyde carbon and this is the one that is the ketonic carbon and rest of the part of the double bonds and the other parts of the molecule remain as they are. So this molecule is called humiline and its synthesis has been done using McMurray coupling in a very easy way and therefore once we have the double bonds in a proper orientation as we can see here we have got the during the coupling we have got the trans product in this particular case where there is a medium size ring. What is the mechanism of such a reaction? It occurs in a heterogeneous fashion on the surface of the titanium 0 particles. So essentially what is suggested is that we have here a titanium 0 which is present here as a surface and there is an adsorption of the carbonyl group we will discuss the mechanism in a minute but everything happens on the surface of the titanium 0. Now titanium dioxide is formed as a byproduct after the reduction is over. It is also possible that the reaction may proceed via carbinoid and or other nucleophilic intermediate but commonly accepted mechanism which is generally believed indicates a binding of the di anion we will see how the di anion is formed to the titanium surface and followed by generation of alkene and loss of TiO2. So it is generally on the surface of the titanium 0 with where the reaction occurs. Now what is happening is this is the simplest way of indicating that we have a carbonyl group here carbonyl compound and the titanium 0 transfers one electron similar to the ones that we discussed with sodium or potassium or lithium or even magnesium. So the radical anion is formed by the transfer of an electron from the titanium 0 here and we obviously go through this ketyl radical or which is called as radical anion and then they couple to form this 1, 2 di anion. This di anion then binds to the titanium 0 particles and actually it is more of a coordination. So if this is the surface of the titanium 0 then we get this type of binding or coordination on to the surface through the oxygen of the di anions and as we can see that we can cleave this particular carbon oxygen bond where a di radical is formed radical on the carbon and radical on the oxygen and in a similar fashion it will also happen on the other carbon oxygen bond and they will couple to form the olefin that there is a C-C bond formation and the released oxygen which is now sticking onto the titanium surface basically leads to the titanium dioxide formations. So this particular type of mechanism we will discuss a little bit more in detail when we talk about another reaction where we can take this di anion which is also possible that we can trap it as a diol 1, 2 diol that is if we logically see that if this di anion is forming then it should also be possible to trap it as a 1, 2 diol and then we can see that 1, 2 diol also leads to the corresponding olefin under the same conditions. It was found by McMurray that a combination of TiCl3 and this is dimethoxyethane 1, 2 dimethoxyethane you have this is what is DME and this forms a blue colored crystalline complex. So that particular complex and zinc copper couple was used to form visceral diols by from the two ketones. DME was used as a solvent at room temperatures and the diols could be isolated. At higher temperatures then the olefin starts forming that means that during the coupling obviously once the radical anion is formed the diols can be formed and do form as intermediates which decompose at high temperature. So from this particular medium sized di aldehyde can be converted to a medium sized alcohol where cis trans ratio is 25 to 75. If we have a smaller ring formation needed then of course we get cis diol as the product and as we discussed it such a reaction proceeds via radical anion of this type. So we will stop it at this stage today and take it up next time further aspects of this titanium base reactions McMurray coupling but as you can see that we can very easily allow the coupling of aldehydes or ketones and stop at the stage of diol and also if required by heating we can convert that into corresponding olefins. So you can study these some references I have already mentioned for example here this particular tetradone letters in 1989 where this diol formation was first reported and then we get ready for the next class. So thank you see you next time.