 Welcome to today's lecture. What we discussed last time was basically the end part of the oxaziridine based reactions where we checked the reactivity of chiral as well as a chiral oxaziridines with say a chiral enolates and chiral oxaziridines with chiral enolates that means enolates coming from chiral substrates and the reaction with a chiral oxaziridines. And we saw how the stereochemistry of the hydroxy group comes based on the steric factors and of course polar factors depending on what substrate we take it. We also saw how camphor based reactions can be carried out to go to the chiral substrates with high enantiomeric purity based on high diastereoselectivity when we use the auxiliaries. Then towards the end we discussed the Barton reaction which is the oxidation at the unfunctionalized carbon. So we will now proceed with the Barton reaction and related reactions where the oxidation at unfunctionalized carbons can be carried out. As I discussed last time that it is more easy to do the reaction at the functionalized molecules which say for example if you have a carbonyl group here then alpha position is very easy to deprotonate and introduce an electrophile. But when we have a substrate of the kind that is shown here then of course if one wants to introduce some oxidize or introduce some electrophile at this position it is somewhat difficult. So what Barton did was basically starting from a substrate of this type here basically what he did was to take the corresponding hydroxy group here and react it with a source of NO plus such as NOCl and that allows the formation of this molecule in which the oxygen nitrogen bond can be easily cleaved in a homolytic fashion if we photolyse it. So homolytic fission so this is exactly what was done here. So we start with this particular substrate which is derived from the corresponding hydroxy group and idea was to introduce a functional group at this position. So we can prepare this and then photolyse it during the photolysis the oxygen radical is formed and of course you have a nitroxy radical. This alkoxy radical then takes up the hydrogen from here via this 6 member transition state cyclic transition state and generate a radical at a very remote position which is the delta position. So if one starts in numbering then you have an alpha then you have a beta then you have a gamma and then this is the delta position. So abstraction of hydrogen from delta carbon occurs because of the 6 member transition state here. Once this radical is formed then you are released with a nitroxyl radical and this nitroxyl radical then combine here with the carbon radical to form this particular substrate where the NO group has come to an unfunctionalized delta carbon of this substrate. So generally a 6 member transition state of chair conformation is favored though both or semi chair forms are also invoked. It is generally expected that in the transition state the carbon, hydrogen and oxygen atoms are approximately linear. So if that happens then obviously what one expects is that the transfer of the radical occurs from oxygen to a remote delta carbon. Now so the coplanarity is an important part. For example if one starts to form the corresponding oxygen radical from the corresponding OH via the ONO type of molecule then you generate a radical here like this. Now this radical obviously cannot form the 6 member transition state with this because you have 1, 2, 3, 4, 5. This is a 5 member transition state and not a 6 member transition state. The other possibility is that you have 1, 2, 3, 4, 5 and 6. So you have a possibility of a 6 member transition state so that you can generate the radical disposition. But the distance and the approach of the oxygen to the hydrogen if one looks at the model of this molecule it is not easy to form this radical because the distance between the oxygen and the delta carbon is too large for hydrogen abstraction. And therefore what happens is this particular cleavage occurs in this fashion that you break it and generate a radical on this particular carbon atom. So if one does not have a possibility of hydrogen abstraction then there is a possibility of this kind of C-C bond cleavage. So you have a C-C bond cleavage. Now because you have provided energy to the molecule and you have generated a radical so there is a cleavage if there is no abstraction of the hydrogen. If we use iodine in the reaction medium as we discussed last in the last part that we nitroso radical attaches to the carbon and then O-C NO bond is formed. If we use iodine for example this particular radical then interacts with the iodine and what is formed is basically this iodocompound and this iodocompound can allow the ether formation to take place or we can also oxidize these two corresponding aldehyde. So if we take this particular iodocompound and react it with a base then you generate a negative charge here that can attack to this carbon and form the corresponding tetrahydrofuran based product. On the other hand if we oxidize this alcohol to the corresponding aldehyde then we use a base then you can generate an anion alpha to the aldehyde and that undergoes an intermolecular reaction to form the corresponding cyclopropane. So if we use iodine then we have a possibility of a cyclopropane ring formation or an ether tetrahydrofuran type of ether formation is also possible. Now what can also happen is we can use some other reagent systems such as iodine and phenyl iododiacetate and also have somewhat similar type of reactions as we discussed. As you can see from here you have the proximity. There is a proximity of the two particular centers here you have O1, 2, 3, 4, 5, 6. So you have a 6-member transition state that can form. The iodine PHI OAC twice that is phenyl iododiacetate gives I plus and this I plus allows the OH to form OI and that undergoes cleavage. We will discuss this particular path little bit later. Now what can also do is that if one gets the corresponding molecule of this type when the NO dot has reacted at this particular center here so we get this. This is basically nothing but the first corresponding oxime because this molecule here is nothing but a nitroso and then that is in equilibrium with the corresponding oxime. So this oxime and this is basically in equilibrium with each other and one can cleave it under different conditions of oxidizing or reducing way by which you can cleave the oxime to the corresponding ketone. So what we have done is we have started with the this kind of substrate coming from say here OH here and after the going through the nitrate ester what we have formed is the functionalization at the remote delta position. Now one can also do that if a oxime of this type if R is equal to hydrogen then we can also do the dehydration that means if this R is hydrogen here instead of any alkyl group or phenyl group we can do the dehydration here under acidic conditions and heat it and one can get the corresponding nitrile which can be hydrolyzed to the corresponding acid and then if we use acid here and heat it and remove water then one can get the corresponding lactone. So even lactone formation can take place. So you have several such possibilities depending on what is the structure of the starting molecule. So you can get delta functionalization as I mentioned alpha beta gamma and delta a functionalization at delta position. So we can the proximity is an important part and how are we going to generate this oxygen radical which can form a 6 member transition state that is also an important part of it. Now let us take another example here for example you can have a hydrogen here and there is a hydrogen here and if this substrate is photolyzed it is very clear that we can get the corresponding radical here that radical attracts the hydrogen radical form here and then eventually the oxime is formed at this stage. So what is happening is from here is very clear that you have the possibility of such a radical to form and this radical then leads to the corresponding next radical which is here and then this is attached by the NO radical to form the corresponding N double bond O which then is in equilibrium with the corresponding oxime and this is what the substrate that we have got it. This is the product that we have got it. Now if we carry out under the acidic conditions the Beckmann rearrangement then we can have rearrangement then we can get this particular lactum. So this is a very interesting method for functionalizing the carbon atom which is remotely placed and then due to the proximity of the oxygen radical coming from this particular part of the molecule and the corresponding hydrogen which is approachable one can functionalize at the unfunctionalized carbon atom. In a similar fashion like one of the previous examples I took is that we take this OH group here use nitrocyl chloride pyridine functionalize it to make this particular OH group into the ONO part of it and then from photolyze it and eventually this particular hydrogen can be converted to the corresponding oxime and this oxime can be hydrolyzed from here to the corresponding aldehyde that aldehyde is will exist as this lactol. Now this particular lactol has been converted to the corresponding such 4 member ring as you can see that there is a 1, 1, 2 then 1, 2, 3 and 4 member ring this is the 4 member ring that is formed here which is what is this 4 member ring here then there is a methyl group here this is the methyl group here and of course then the other part of the molecule is eventually transformed into this particular side chains and this is a molecule named as grandisol which is an insect pheromone that has been synthesized by using this Barton's photolytic reaction of functionalizing at the unfunctionalized carbon atom. So basically it was dependent on the functionalization of this particular molecule at this methyl group at an unfunctionalized carbon. This Barton reaction has also been utilized in many steroidal molecules as you can see that corticosterone acetate here is basically having a hydroxy group at this position and you can convert into the corresponding ONO group here, fertilize it and then either this particular methyl group can take the hydrogen or even this between the two of them this particular methyl group appears to be more easily approachable if one looks at the model then we can come very clear and then we can make the corresponding oxime and then again hydrolyze and make the corresponding lactol which is what is an aldosterone derivative. So basically what has been done is this particular methyl group which is remotely placed but accessible for the 6 member transition state to form and that allows the aldosterone derivative to form. So it is a very easy way of converting a corticosterone to an aldosterone derivative. On the other hand, if this particular oxygen radical which is formed from this cleavage of this particular oxygen nitrogen bond that gives the radical by abstraction of this particular methyl group here leads to the radical formation on this particular methyl carbon and that can undergo interestingly that can undergo attachment to this carbon and this can move to this carbon forming three membered cyclopropane ring and a radical on this carbon. So basically what is happening is you have a radical that is going to form on this particular path and then you have a three membered ring which is like this. And now this reacts with the NO radical to form this oxime via the nitrogen molecule to form the corresponding. I can show in this fashion that this is how it is going to form. So this appears to be a side product but it is an interesting side product because such a reaction is possible if the other radical is formed. Now there is another a very celebrated example of converting one of the steroid molecules called lanosterol to the corresponding molecule which is called a cycloartinol. If you look at the two molecules you can see the difference, the only difference that you can see is that there is a cyclopropane ring here whereas that is not the case here. So there is a double bond which is present here and there is a methyl group that is present there in the lanosterol whereas that is not the case otherwise rest of the path is exactly same. So this one was very easily converted the lanosterol was easily converted to cycloartinol. In a very specific fashion by Barton. Now how it is done is that you take the lanosterol as is here and this is the final molecule. Now what is done is this molecule is somehow converted to this particular carbonyl group. How can we do it? Obviously if we take this double bond here and this is the allylic position and this is the allylic position. So if we use a strong oxidizing agent we can introduce two carbonyl groups on both the sides one here and one here. This particular position is sterically hindered position because of the methyl group and is also known to be sterically hindered of a steroid molecule. But then we have to take care of this double bond we also have to take care of the hydroxy groups. So it was done in a very specific manner by protecting this hydroxy group also by protection of the double bond and finally reducing the corresponding double bond which is here as well as the carbonyl group that is going to come here. So basically it is a very specific way of introducing the carbonyl group at this particular position by a few steps but that was possible. Once that was done then without worrying about the protection and de-protection of the other groups. Once the carbonyl group was introduced here it was reduced and they got the corresponding hydroxy group like this which was through the ONO that is nitrosyl chloride and pyridine base and then fatalysis in the presence of iodine led to the formation of the corresponding CH2I. So this particular methyl group which was having similar orientation as the hydroxy group which is beta this is also beta oriented and therefore the abstraction 1, 2, 3, 4, 5 and 6 the hydrogen here and then when can prepare the corresponding radical in which in the presence of iodine leads to the corresponding CH2I. And now once this is happened then they used a strong base to de-protonate this particular hydrogen here and then this negative charge attacked on to this particular carbon and then iodine goes as a leaving group and eventually the cyclopropane ring is formed here. So this is a very interesting celebrated example of conversion of lanosterol to another molecule which is cyclo-artinol. Of course it is involved many steps and protection and de-protection but it was a very brilliant example of the power of the Barton reaction. Now the alkoxy radicals can also be generated from hypo halides. So if we can take the hypo halide here and we do the photolysis here then we can generate the corresponding radical. So if that radical is formed here a radical is formed here this radical can pick up the hydrogen from here to form the corresponding radical at this tender and of course you can get the attachment of the chlorine from the substrate or via chlorine radical and of course you can get this. And now we have a base which de-protonates and of course then that attacks on to this carbon and then one can get the tetrahedra furan. One can also start to use hypo iodides that can be generated from the alcohols upon reaction with iodine in presence of lead tetra acetate or mercury oxide or iodobenzene diacetate. This is an example of that under these conditions at 40 degrees. For example, you can have the generation of oxygen radical from this OH and this is beta oriented. This is also beta oriented and then one can through the iodine mediated reaction can also have a connection between this particular oxygen and this carbon to form this 5-membered ring. So this is a 1, 2, 3, 4 and this 5. So this very interesting such molecule having a tetrahedra furan unit embedded in the steroid molecule by this reaction. So this example is an extension or related reaction to the Barton reaction by using hypohellite hypoiodide. Now how do we make use of the hypohellite based chemistry in specific deprotection of a benzene ether? Now if we take a compound of this kind which has four different types of benzene ethers, a benzene ether is nothing but an OCH2 phenyl group. Now if we have these four types of benzene ether present in this particular molecule and if we want specifically this particular O benzene group to be deprotected to form this diol, then this particular chemistry of hypohellite based chemistry and the photolytic condition allows the formation of this ketol which can be hydrolyzed under acidic conditions to give this particular diol which is what is the result of a specific deprotection of this particular benzene group. Now what happens is that when this particular molecule is allowed to react with an iodo thalymide, then OH group here interacts with the iodonium and here to form the hypoiodite of this type and now when we do the photolysis then this particular iodine oxygen bond undergoes cleavage to form the oxygen radical here and of course iodine radical will go away and this oxygen radical intramolecularly picks up the hydrogen from here as a hydrogen radical and becomes an OH group here at this position here like this and generate a radical here at the benzylic position because this radical is then stabilized by the phenyl ring also it is stabilized by the oxygen. Now this particular radical then reacts with the same hypoiodite which is shown here and picks up the iodine radical from here and generate oxygen radical which is the same as this particular species and once this hypoiodite reacts with this radical leading to this iodo compound then there is a intramolecular closure of this particular part of the molecule leading to the ketal of this type. Now with this ketal of having three benzyl toxic groups can then be hydrolyzed under acidic conditions because these benzyl ether groups are stable under acidic conditions but this ketal is not stable and undergoes hydrolysis to form the corresponding diol. So it means that we have started with a monohydroxy compound having four types of benzyl oxyethers and eventually we get a diol and with three benzyl oxy groups intact. So this is how a specific deprotection of a benzyl ether is done under these conditions using hypoiodite based chemistry under photolytic conditions. This particular work has been reported in this journal Chemistry European Journal in 2000. So we will stop it at this stage today and take some aspects of the Barton reaction specifically the beta cleavage the cc bond cleavage in some synthetic transformations and then proceed further. We will stop and at this stage and we I expect that you will go through these things which I have taught today and I will see you the next time. Thank you.