 Hello everyone. I would like to welcome you all for today's lecture. We will briefly look at what we discussed last time. We started looking at the oxidation of alcohols using chromic acid and we considered the possibility of abstraction of hydrogen in from the intermediate that is involving chromium based intermediate either in an intermolecular fashion such as this. So the base picks up the proton from here in an intermolecular fashion and then the oxidation occurs in this fashion. This was one possibility that we considered after the alcohol reacts with the chromic acid and it forms this intermediate. Alternatively the same intermediate can also undergo an intermolecular proton abstraction and that can happen in this fashion so that the same intermediate that is formed. So from chromium 6 we started and we get the chromium 4 as one of the intermediates and of course the aldehyde comes from the corresponding alcohol. So this is what we discussed last time. Now the abstraction of proton in an intermolecular or intramolecular fashion is something that we need to establish. It is very difficult to do that with chromium based intermediates because the reactions are very fast and it is very difficult to carry out such reactions but whether such oxidations involve intermolecular or intermolecular proton abstraction has been studied using sulfur mediated oxidations. Now we look at sulfur mediated oxidations and one of the most important and very popular oxidations involving sulfur based intermediates is sohan oxidation. So in the sohan oxidation what has been reported is that the DMSO that is dimethyl sulfoxide reacts with oxalyl chloride. This is the acid halide from oxalic acid. So it is called as oxalyl chloride and this oxalyl chloride reacts with the DMSO at low temperature to form certain intermediate which then reacts with the alcohol in the presence of triethylamine at minus 65 degrees in dichloromethane at solvent and eventually leads to the formation of the ketone. Now this was reported in 1978 by Swern and of course his collaborator Komura. So this is what is a popularly known as Swern oxidation. Now this oxidation is actually based on a well-known oxidation called Moffitt-Fitzner oxidation which was discovered in 1963 which involved reaction of DMSO. So all of them are based on DMSO based oxidations and they react with dicyclohexyl carbodimide. This is the structure of dicyclohexyl carbodimide as you can see that the carbon atom here is flanked basically by two nitrogens by double bonds and therefore this particular carbon atom is highly electrophilic particularly under the presence of catalytic amount of an acid such as phosphoric acid. Now this was discovered in 1963. Based on this the oxidations of alcohols were reported and subsequently in 1964 Barton reported the use of Phosgene which is a gas at room temperature with boiling point being 8.3 degree centigrade in place of DCC. However the use of Phosgene is not very convenient and therefore it did not become very popular this reaction. But the Moffitt-Fitzner oxidation was actually carried out by lot of people in their oxidation endeavours. Now the intermediates which are involved in these kinds of oxidations are very closely related to another oxidation known as Kornblum oxidation. So what are those intermediates? We will look at those intermediates a bit later but the first let us see how the mechanism of the Moffitt-Fitzner oxidation and very closely how the Barton oxidation has been also looked at it. So as I mentioned the dicyclohexyl carbodyl mite which is DCC is having two nitrogens and one of the nitrogen gets protonated with the phosphoric acid that we use it here and it forms an intermediate or the protonation. So if we write cyclohexane as Cy then what it would look like it would look something like this and this gets protonated and then you have a positive charge on the nitrogen. This makes this carbon much more electrophilic than the original DCC because the positive charge here on the nitrogen would like to get neutralized when the alcohol attacks on it. So if or the DMSO attacks on it. So first DMSO attacks so you have a DMSO here your methyl here and the methyl here the DMSO O- attacks on to this carbon. This is how it is shown here. So when DMSO attacks on to this carbon which is now having a nitrogen protonated so this is the intermediate that is going to come out. This is a very crucial intermediate and again now the second nitrogen gets protonated and again forms the same ammonium ion like what we have written here but now on the other nitrogen and again this carbon becomes very electrophilic and at the same time now the alcohol reacts on to the sulfur and this carbon oxygen sulfur bond breaks and the new carbon oxygen bond is formed because the carbon now is electrophilic because of the protonation of the nitrogen here and therefore what is proposed is that the dihydrogen phosphate ion which comes out after the loss of proton from here in both the cases the protons are basically lost. So you have dihydrogen phosphate and ion that picks up the hydrogen from here that picks up the proton from here and the negative charge of the oxygen from the alcohol then reacts with the sulfur and then this bond breaks. Thereby now the phosphoric acid is again regenerated by the abstraction of this particular proton and of course the nucleophilicity of the alcohol becomes larger and that leads to this particular intermediate and this intermediate as it was proposed earlier is that again base which is present in the reaction medium you have this dihydrogen phosphate and that can pick up a proton from here and then oxidation can occur in this fashion to give the ketone and then as you can see the dimethyl sulfide goes off and of course you regenerate the phosphoric acid. Now the only thing what happens is that when this dicyclohexyl carbodimide is protonated and then of course you lose the oxygen from the DMSO then you form a carbon oxygen double bond and this is what leads to the formation of this dicyclohexyl urea. This is a urea having two dicyclohexyl parts. So this is something which is an important by product but then this particular by product does create some problem which we will talk in a minute. But here what I would like to mention is that this particular intermediate which is shown in blue and another intermediate this shown in this blue also are very crucial intermediates. Now as these crucial intermediates and this dicyclohexyl urea are the ones which are worth considering. Now what happens during the reaction that this dicyclohexyl urea becomes difficult to remove while purification of the ketone is being done. So many people did not want to prefer to use dicyclohexyl carbodimide because of the difficulty in separating dicyclohexyl urea. So originally there was no base that was used therefore the reaction took a bit longer time. But later on they started using a trifluoroacetic acid and pyridine as a combination in which trifluoroacetic acid protonates the dicyclohexyl carbodimide and of course pyridine acts as a base and thus the reaction gets facilitated. In general the DMSO is used in excess and of course as you can see that the main role of the DMSO is of course to get activated by dicyc. So the dicyc activates the DMSO forming an intermediate of this type where you can say that E is a part of the dicyclohexyl carbodimide and therefore you have a positive charge here. This is the actually role of DMSO that the oxygen of the DMSO gets activated by the DCC and then your alcohol attacks onto this carbon and this goes as a living group. We will look at it these intermediates a little more carefully. And the Barton's oxidation with phosphine is also somewhat similar. So there the DMSO attacks on phosphine and the phosphine is nothing but this particular compound where there are two chlorines attached to a carbonyl group and this is what is at the phosphine and it is acting like an electrophile and DMSO attacks onto this carbon and leading to the formation of this intermediate. Now this intermediate of course these reactions have to be done at low temperature and once this intermediate forms it is expected that such an intermediate can lose carbon dioxide and of course chlorine and forming after the loss of carbon dioxide. This intermediate here and this intermediate now has the sulfur as positively charged and there is a living group which is a chlorine. Therefore the alcohol attacks onto this particular intermediate and leading to another intermediate of this kind. Now if you look at very carefully whether you consider this as an intermediate or this as another intermediate and the second intermediate is this kind. So this intermediate is very similar to this particular intermediate and this intermediate is also similar to this intermediate or this intermediate or this intermediate because you have a living group on the sulfur which is connected to the oxygen in this particular case or in this case you have a chlorine as a living group. So both these crucial intermediates which I have been mentioning right from the beginning whether it is a Moffett-Fitzner oxidation or the Barton oxidation are similar in nature. Now if we go back and look at the literature so basically these intermediates which I have mentioned as crucial intermediates of two different kinds are essentially related to yet another very important oxidation which was the basis for all the oxidations involving sulfur is known as cornblum oxidation. So it was a Nathan cornblum who actually discovered this is in 1959 and this particular oxidation is also based on sulfur and of course all the sulfur based oxidations involving DMSO should definitely give credit to cornblum. So what was the cornblum oxidation? The cornblum oxidation involved the reaction of say primary Tosylate or a bromide of this kind. To start with of course the reactions were carried out using very good living groups on a primary carbon atom. The reason was that in these cases the reaction requires the use of DMSO in the presence of a base like sodium bicarbonate at high degrees 150 degrees. So in these cases what happens is the DMSO acts like a nucleophile and reacts at say for example here and in SN2 fashion and leading to an intermediate of this type here as I have shown it here. And in a similar fashion if you take it here then of course you will have a similar intermediate like you have O, R, CH2, CH2, CH2 and then of course you have oxygen and then you have the DMSO reacting like this and of course OTS will go as a living group here. So in all these cases as you can see the living group once it leaves the carbon the DMSO oxygen O minus attacks and this is the carbon and of course this is the carbon where DMSO is attacked. Now this intermediate which I have shown it here is very similar to the crucial intermediate that we have talked in Moffitt-Fitzner oxidation and the another one is of course this one. So this is related to this particular intermediate. So this intermediate here is very closely related to this intermediate as you can see that there is a carbon that holds a hydrogen, the carbon that holds a hydrogen here and of course you have oxygen then you have a sulfur positively charged and of course you have the two mythiles. So the base can pick up the proton as I have shown it here and the oxidation completes by the loss of diameter sulphide as we have discussed in other cases. So this particular intermediate as shown here is similar to the intermediate that shown here even in Barton's oxidation. So both in Barton's oxidation as well as in Moffitt-Fitzner oxidation the last intermediate is very similar to the intermediate that is formed in cornworm oxidation. Except that in the case of cornworm oxidation the first intermediate directly comes with having an oxygen here because the DMSO is the only as a nucleophile that is acting whereas in the other cases like Fitzner Moffitt-Fitzner oxidation or Barton's oxidation we activate the DMSO to form electrophilic sulfur to which alcohol attacks and gives the similar intermediate as we see in the cornworm oxidation. Apart from cornworm and Moffitt-Fitzner oxidation and Barton oxidation, Parik and Doring also reported in 1967 a reagent using pyridine sulfur trioxide and in a similar fashion the oxidation takes place. So DMSO interacts with this pyridine sulfur trioxide complex where now you have a electrophilic sulfur another electrophilic sulfur and the leaving group is this pyridinium ion. So the O minus attacks on to this sulfur pyridinium ion goes and this is the intermediate that is formed to which the alcohol attacks where the pyridine takes up the proton from here. So pyridine can take up a proton from here and of course you generate the anion of the alcohol which goes and attacks on to this and then you have this intermediate that is formed of course pyridine is still present there and then pyridine can pick up a proton from here and then oxidation can take place with the loss of dimethyl sulfide. So as you can see it is very similar to the oxidation that we have discussed earlier and this is another crucial intermediate here and this is another crucial intermediate. So as we have been naming all the crucial intermediates they are similar to this is the Moffet-Fitzner oxidation first crucial intermediate this is the second crucial intermediate in baron oxidation this is how the one intermediate forms and then rearranges to this crucial intermediate and this is the second crucial intermediate after the alcohol attacks. So in place of DCC first gene pyridine sulfur trioxide many different electrophiles are used such as acetic anhydride, trifluoroacetic anhydride and of course trifluoromethane sulfonic anhydride. So you have CO2O or you have CF3SO2 twice O. So basically these are the various kinds of intermediates which are used and of course this is the reference that you can see which gives the details of these oxidations. Now if you look at it very carefully what we have seen is that DMSO based oxidations in general can be written up like this that DMSO reacts with an electrophile whether it is DCC for gene or pyridine sulfur trioxide or anything of that sort you get basically activation of the DMSO by the electrophile and leading to this intermediate here to which the alcohol attacks and forms this intermediate from where the aldehyde or the ketone gets released along with diamethyl sulfide. So essentially what is happening is whether the EX is phosgene, DCC, pyridine sulfur trioxide or oxalicluoride, acetic anhydride or trifluoracetic anhydride any other electrophile and of course the intermediate is comparable with corn bloom type of oxidation here because this is the crucial intermediate which then undergoes loss of proton and along with diamethyl sulfide here to form the carbonyl group. Now whether the reaction is involving an abstraction of a proton in an intermolecular fashion or an intermolecular fashion was basically studied in detail by the scientist named Torsell here in 1966. What he did of course as we have seen in general that DMSO gets activated with an electrophile forming this intermediate to which the first crucial intermediate to which alcohol attacks and forming this second crucial intermediate from where either this proton can get abstracted essentially to lose diamethyl sulfide and give the ketone but then is there a possibility of removing this proton first because this carbon is next to the sulfur which is positively charged. So this is what the scientist or the chemist Torsell did it. He took the intermediate of this type from any of these activated DMSO activated based intermediates and reacted with the alcohol that contain CD2 OH that means it is a deuterated alcohol. So once deuterated alcohol reacts with you have two deuterium here onto the carbon atom and when the base was added to it base. So the base allowed the oxidation to take place where of course one would expect that this oxidation would take place and you get the corresponding ketone with a deuterium but the release diamethyl sulfide led to the formation of the compound of this type in which there was a deuterium here and not this that means if only such oxidation is taking place as I have shown here then we would expect the hydrogen to be retained at the two ends of the sulfur as CH3 and CH3. But what was observed was at one end it was CH3 and the other end it was CH2D. So the mechanism that was proposed that the base picks up the proton from here generating negative charge of this type and this undergoes an intermolecular abstraction of the deuterium and then it forms the corresponding ketone and of course the diamethyl sulfide containing one deuterium. Yet in another experiment what he did was he took the DMSO which was having two CD3 groups attached to it and of course you activate with any electrophile that we have discussed so far and this is the crucial first intermediate that is formed to which now normal alcohol is added not a deuterated alcohol because the deuterium is now incorporated in the DMSO. So this is the intermediate that one would expect to reform. Now if this reaction occurs in a similar fashion as we have discussed that the deuterium the proton or the deuterium that is present on to the carbon atom attached to sulfur takes gets deprotonated then of course the base will pick up the proton from here and of course you will get an intermediate of this kind. Had it been directly then of course you would get back the DMSO as it as the diamethyl sulfide but if in an intramolecular fashion if the proton abstraction is taking place then of course one can generate an intermediate of this kind and which then loses the proton from here in an intramolecular fashion and then leads to the formation of the aldehyde as normal expected but then the diamethyl sulfide which has come out contains the hydrogen rather than the deuterium because the other possibility which we can think about it is of course without involving intramolecular activation or removal of a proton we can think that if it is an intermolecular hydrogen then of course the base can pick up the hydrogen from here and one can get the oxidation of course but then one would get RCOH plus the DMS that is formed should be this carbon here. So this is how it should form but what was observed was that the hydrogen gets incorporated into on to the carbon atom that means the mechanism is an intramolecular removal of the hydrogen from the intermediate that is the crucial intermediate that we talk about it. So having established this particular type of intramolecular abstraction of a proton then of course many more oxidations have been reported and it was finally Cori and Kim who also utilized conceptually similar but somewhat different starting materials for example one can also get this intermediate which we call it as from the crucial intermediate which is what is comes from the diamethyl sulfide and here you have a leaving group here. So this is the intermediate that is expected to form and derived from the DMSO to which alcohol reacts and forms the next intermediate that allows the formation of the ketone or aldehyde depending on what alcohol one uses but then you have this intermediate which is what is formed. So in order to get this intermediate of course you have an X-. In order to get this intermediate one can also start not from DMSO but one can also start from say for example dimethyl sulfide and react with some electrophiles such as halogen and generate an intermediate of this type here which can also act as a good intermediate crucial intermediate to which alcohol can react and of course can form the intermediate of this kind which eventually gives the aldehyde or keto. So this was developed by Cori and Kim which we will discuss in our next class more in detail. So you please go ahead and complete Lee look at whatever I have discussed today and get ready for the next class till then bye and see you next time. Bye. Thank you.