 Hello everyone, I hope you had the chance to go through the last classes that I discussed many aspects of oxidation reactions. So today we will have briefly, first we will go through the few points of the last classes. For example, what we did was selenium dioxide based oxidations here and then we also did the sulfoxide sulfonate rearrangements. Also we did in the case of selenium dioxide based oxidations, conversion of ketones to diketones and also oxidation at the allylic position to go to ally hydroxy group. Then we also did the Sygoosa eto oxidation where enol silyl ether was converted to the corresponding enone. Now let us go to another topic today is that 1-2 keto transposition. What we discussed in say earlier conversions using selenium dioxide was, conversion of a ketone to a diketone. However, other possibility and other requirement that happens in cases of many synthetic transformations is how are we going to convert a ketone which is already present say here to the next position which is here. Now this is something very important and this is important mainly because what happens is in many of the transformations that are required is conversion of one keto position to the next keto position. For example, in this case if one takes an example of this kind and if one wants to carry out some transformations at say this position, one possibility of course is directly you can functionalize this carbonyl group. But if one wants to utilize both these car hydrogens here therefore, we need to convert this carbonyl group here to the next position here. Now once that is done, then you now have two possibilities. One of course is the functionalization at this center now and the other is also at this center. Now that you have a choice. So if somehow you can make this molecule, if somehow you have arrived at this molecule and if you want to convert the carbonyl group from this position to the next position, then what are the ways? So this is just a hypothetical example which I have shown. Similarly, there is another example like this where if you have blocked this particular position of the alpha position of the carbonyl group, then we can convert this ketone to the next carbon atom and once that is done, then one can also then functionalize here the position next to the carbonyl group. So this is called as one, two keto transposition ketone transposition. So what are the methods? There are of course several methods but we will take a one or two methods which are using the sulphur based or selenium based chemistry. One of the earlier methods was that one takes this carbonyl group here for example, which I have written 3 pentanone and if one converts into a tosyl hydrazone like this and treats with two equivalents of butyl lithium, then of course you will first deprotonate the hydrogen which is attached to the nitrogen and of course you will deprotonate the hydrogen with the second butyl lithium at the alpha position leading to this di anion that is you have an anion here at this carbon and you have an anion at the nitrogen. And when this is treated with any disulfide such as RS or SR which is can be any of the disulfides say diphenyl disulfide or dimethyl disulfide. So you can have a choice of the disulfide that you would like to use it. Now once that happens this anion of the this anion then reacts with the sulphur here and this breaks off and of course what you introduce is SR. I have shown here in methyl but we can have anything. Now you use one more equivalent of butyl lithium so that you generate now another anion at from the hydrogen that is abstracted to form this anion which is easily formed alpha to the sulphur and now that undergoes elimination the way it is shown here that the tussle group is a good living group and therefore it goes off with the movement of the negative charge to form this intermediate. Now this intermediate then loses nitrogen as it is shown here to form this vinyl lithium this is what is formed. Now during this process when the protonation occurs here then you get the corresponding hydrogen at this stage. Now this vinyl sulphide this is a vinyl sulphide that can be easily hydrolyzed with the help of mercury chloride Hg++ and aqueous acetonitrile. Essentially what is happening in this particular case is the SR group being soft the sulphur being soft interacts with the Hg++ and followed by the attack of water at this centre. So essentially what you have is a is a is a complexation of the mercury onto the sulphur making this bond relatively weak this particular bond relatively weak and then the attack of the water then allows the formation of corresponding enol and which is what is basically nothing but so you have ketone formation. So you can start with one ketone make the corresponding Tosselheiger zone here then treat with butyl lithium to form introduce essentially at the vinylic position a sulphur and once that has happened you can then hydrolyze it with the help of soft Hg++ in aqueous acetonitrile medium to hydrolyze the vinyl sulphide to the corresponding ketone. This is one of the earlier methods which is what is utilized and of course this is based on a reaction called Shapiro reaction which forms vinyl lithium. So Shapiro reaction is nothing but similar fashion as Tosselheiger zone is there. So once you have a Tosselheiger zone and you form a di anion from here. So you have a anion formation here and you will have an anion formation here if you simply remove the nitrogen from here. So what you have is a di anion like this and which is what then of course in these cases there will be lithium plus lithium plus and this loses the Tosselheiger zone Tosselate group as I showed earlier and to form this particular intermediate and this is the intermediate then that loses the nitrogen and then forms the corresponding vinyl lithium. Now this vinyl lithium is of course similar to the vinyl lithium that I showed here on the top except that in the top position we had SR group introduced here we do not have an SR group introduced. So if one starts with a ketone like this and prepares the Tosselheiger zone and then finally gets the vinyl lithium. Now you can react with any electrophile that is possible to react and that allows the introduction of say if you have an E plus as an electrophile and X minus of course as a counter ion then one can introduce the E here. So this is what is called Shapiro reaction. So basically the method that I introduced on the top is nothing but an application of the Shapiro reaction for the conversion of one ketone to transpose into the second position. Now Barry Trost also has reported an interesting way of doing this kind of 1-2 ketone transposition where he took a ketone and introduced simply an SR group adjacent to the alpha position next to the ketone at the alpha position without making the Tosselheiger zone. So if one takes this kind of any keto group here say like this and introduce by means of say a base and a RS, SR then you can introduce here SR and now you reduce it to the corresponding alcohol by means of say sodium borohydride and then you can eliminate it in terms of dehydration by say Tocic acid, Peratollin-Sulfonic acid and you heat it. So this elimination will give you this SR and once this SR is formed one can also do hydrolysis either by Hg++ or and water or as it is shown above is TiCl4 in acidic acid. So one of some of the maze by which you can hydrolyze it and one can go to the corresponding ketone. So this is the method that is also followed it is relatively easy by simply reduction followed by dehydration and then rehydration or forming the hydrolysis of the corresponding vinyl sulfide to the corresponding ketone. So this is another method by which one can transpose a ketone to the second position. Now there is an interesting reaction which is also required many a times is how are you going to convert an enone of this type to another enone. This is a 1, 3 enone transposition. So you have 1 position, 2 position and 3 position. Now you have inverted it is 1, 2 and 3. Except that here we have introduced another R group at this position. So it is not a simple 1, 3 enone transposition but it is a 1, 3 enone transposition with an addition of an R group. So if one starts with this enone here and introduce an R group such as this at the position onto the carbonyl carbon by reacting it with R lithium say you have alkyl lithium or aryl lithium or whatever kind of R group that you want to introduce. If you react it with a carbonyl group then it will form the corresponding allyl alcohol with an R group at the carbon holding the OH group. So now this allylic alcohol which is a tertiary allylic alcohol when it is reacted with say PCC which is pyridinium chlorochromate. So you have here pyridinium chlorochromate and then what happens is that it forms an intermediate where a tertiary alcohol will be reacting with the electrophilic chromium here. But then since R group here introduced R group is not a hydrogen anymore because we have used R lithium which is either alkyl or aryl. So that means no oxidation will occur at the of no oxidation of the hydroxy group will occur at the carbon because it does not have the any hydrogen left. So as a result there is some type of migration or some type of rearrangement that occurs where the intermediate eventually comes gets transformed in such a way that it reaches to the third carbon here or the vinylic carbon or the other end of the double bond which has one hydrogen. So this is what is called as a Dauben-Michino rearrangement. Now how does it happen is basically these are the kind of intermediates that are formed. So what happens is if you have an allylic alcohol of this type then when the chlorochromate or any chromium based reagent reacts with this hydroxy group. So in supposing if you have the pyridinium chlorochromate so if you have an R group here an OH group here. So this OH group will react with pyridinium chlorochromate and this is the intermediate that will form. Now you have an hydrogen here. So this part has come by reacting the tertiary alcohol to the pyridinium chlorochromate or any chromium reagent. Now this particular oxygen then interacts in this fashion and this breaks off to form an intermediate of this kind where the oxygen of the chromium species now has come to this particular position. Now since there is a hydrogen here then of course oxidation of this particular species occurs in a fashion something like this. Suppose you leave it as it is then you have O, Cr, Cr and O, O, O and then you have a hydrogen here. So this will undergo oxidation and of course you will then be left out with the corresponding ketone where there is a double bond of course and R group. So this species which I have shown here is similar to this species and that undergoes oxidation to give to the corresponding enone. So what we have done is we have transformed an allyl alcohol to this intermediate which undergoes a rearrangement of course it can also go back from here to here or here to here that is why there is an equilibrium. Same thing can happen here and this intermediate or this intermediate can be shown as somewhat like this where the chromate is coming out and then this species undergoes oxidation in the way I have shown here to the corresponding enone. So this is one of the very popular methods of converting an alpha beta unsaturated ketone with an extra substitution to the corresponding 1, 3 transposed enone. Now if one wants to do simple 1, 3 enone to the corresponding inverted 1, 3 enone or transposed 1, 3 enone say for example you have this and you want to convert into this where the ketone position is now changed from this position here ketone was here now we have made it here. So one another method which was published in 1980 is that we reduce the ketone to the corresponding to the corresponding alcohol here and by lithium aluminum hydride and then you acetylate it by means of acetyl chloride and pyridine or we can use acetic anhydride to form the corresponding acetate. When this allyl acetate is reacted with phenyl acCl what one forms is an intermediate of this time type chlorine here and Se phenyl. So essentially the double bond has interacted with phenyl Se plus and Cl minus to form this intermediate reacts this particular reagent reacts to form this where selenium comes in on this position and Cl comes on this position. Now when ozone is reacted ozone is allowed to react to this intermediate what is formed is is selenoxide and you have one hydrogen here and this undergoes elimination to form basically this. So you have an elimination here to give this double bond and then this double bond is reacted with 90% formic acid. So you get hydrolysis in this way with the loss of acetate and this undergoes then loss of hydrochloric acid to form the corresponding ketone like this. So when formic acid and water and the acidic condition reacts with this particular compound there is an addition of water followed by that means you have water you have water reacting with like this and this going like this under the acidic conditions and of course you lose acidic acid. So you form the corresponding this intermediate which is unstable intermediate and loses hydrochloric acid to form the corresponding enone. So we will stop it at this stage here and then catch it up for the next topic in our next lecture. So you go through these important transpositions of 1, 2 ketone transposition and enone transposition by both the ways one is substituted one as well as the other non substituted one and we will then look at some other oxidative transformations later on. So thank you.