 Good morning and welcome back to this course on Classics in Total Synthesis. So far in the last 3 lectures we had a lot of introduction about organic synthesis, total synthesis and so on. So now we will actually go into the you know topics, first let us start with total synthesis of natural products having 3-membered ring. So when we talk about 3-membered ring, not only natural products have 3-membered ring as core structure but they also will have other rings. So what we will do, we will talk about how these 3-membered rings are made for these natural products and also other rings and in the end how they have done the total synthesis. So today's lecture we will talk about 2 natural products we are called Illudin-M and Illudin-C and both the natural products you can see here there is cyclopropane here, there is a cyclopropane here and in Illudin-C you have a cyclopropane this side okay. So normally when you talk about cyclopropanes what are the normal standard methods to make 3-membered ring? One is if you have a beta diketone, 1-3 diketone or beta keto ester then one can think of SN2 displacement reaction, double SN2 displacement reaction with 1 to dibromocompone, 1 to dihalocompone okay that is one of the easiest way to make a 3-membered ring. Then if you have a diesocompone, if you have a diesocompone then you can do a cyclopropanation on a double mole okay that is also quite simple and straight forward. Then one can think about Siemens-Mieter reaction so that was nicely exploited by Andrew Chariff in asymmetric cyclopropanation which we will discuss tomorrow in one of the total synthesis and if you have an alpha, beta and saturated ketone okay, if you have an alpha, beta and saturated ketone, sulfur elides particularly dimethylsulfoxonium elide can undergo a 1,4 addition and in the end it can form a cyclopropane. So these are the 4 standard methods in literature if you see widely used for making cyclopropanes. But as I said in this particular synthesis we not only talk about how to make cyclopropane but also how other rings can be made okay. So let us move to the real natural product so iludins you know there are many iludins here I show a few iludins like iludin A, iludin B, iludin M and S all of them you can see having 3 rings, 2 are fused together that is the 6 membered and 5 membered they are fused together and here the 6 and 3 they are spiro fused together okay, spiro fused and most of them you know they are very closely related only the functional groups are located at different places otherwise you know they have the common basic skeleton 6 membered, 5 membered and 3 membered and they are also oxygenated. In the first case you can see there are 3 oxygen and second case also you can see 3 oxygen whereas in the 4th one iludin B you see 4 oxygen. So they are all isolated from several fungi and as I said it has several degree of unsaturation and a unique tricyclic ring system. So these are the references if you want to see how they were isolated they are quite old and there are few more iludins you can see iludin C, iludin C2 and C3 okay they differ by presence of OH okay and again they have the same structure okay same core structure and iludinic acid where the CH2 OH is further oxidized you have the carboxylic acid and this is another natural product called 6 hydroxymethyl acyl fulvin. So the fulvin you can see here this is the fulvin subunit of this natural product. Then you also have oxidized version okay so you know there are 2 or 3 hydroxyl groups in iludins and some compounds these hydroxyl groups are further oxidized for example here the hydroxyl group is oxidized to ketone again you can see the same thing here the hydroxyl group is oxidized to ketone. And then they show a very good biological activity particularly it shows selective toxicity for tumor cells compared to normal cells and particularly the oxidized analog this is has been shown to be less toxic than iludin 2 mice actually okay. Now let us go through directly the synthesis of iludin M and C before that how these iludins are acting so there is mechanism action for iludins one should know so if you know that then it is also easy to make analogs of iludins. So what happens it undergoes you know 1, 4 addition basically when your compound has a Michael acceptor you can see that there is a Michael acceptor here okay alpha, beta and such a ketone then the DNA with a nucleophile can attack in a 1, 4 fashion you can see that it undergoes a 1, 4 fashion. Then if you look at your substrate you also have a cyclopropane and next to a carbon having hydroxyl group you have a cyclopropane next to carbon having a hydroxyl group. So what will happen once you do this 1, 4 addition the double bond shifts here okay you get already you see there are 2 double bonds okay there are 2 double bonds okay form. Now another DNA molecule when it attacks the cyclopropane as you know cyclopropanes are like double bonds okay it can undergo ring opening so when it attacks the cyclopropane then this hydroxyl group can come out as water thereby you make this compound as an aromatic compound okay. So that is the driving force for the overall mechanism of action of iludins okay. So even when you want to design something similar to this you should keep this in mind it is not only for iludins but also other natural products having similar skeleton. Now let us start with iludin M first the total synthesis of iludin M was reported by Frederick kinder almost 28 years ago okay. So it was a 6 step total synthesis and they used 1, 3 dipolar cycloaddition between a carbonyl elide 1, 3 dipolar cycloaddition between a carbonyl elide and cyclopentenol okay carbonyl elide and cyclopentenol as the key reaction to construct the tricyclic compound okay the key tricyclic compound was constructed by reaction between carbonyl elide and cyclopentenol okay. And of course the carbonyl elides are normally made from diazo ketone I will come back to that how carbonyl elides are made and what is 1, 3 dipolar cycloaddition before we actually move to the total synthesis. The other natural product which we will discuss under three number ring is iludin C. So this was reported 20 years ago by Reb and Fang and he took about 10 steps and with a high overall yield of 8.2 percent to make this combo okay. Here he used another 1, 3 dipolar cycloaddition in the earlier case carbonyl elide acted as a dipole dipole whereas in this case nitrile oxide then nitrile oxide was used as a dipole and nitrile oxide olefin cycloaddition as the key reaction to make the 5-ohm boundary okay. So this is the starting material and you can see the addition of litheo species to this ketone followed by nitrile oxide formation from this oxane and intramolecular dipolar cycloaddition gives this tricyclic compound and that can be converted into iludin C. We will discuss this in detail in the subsequent slides. Before that what I want to discuss is if you look at these two natural products there are key reactions and there are key starting materials. So how they are made okay that we will have some discussion so that it will be useful when you talk about the total synthesis. So what is the key reaction if you look at in the both synthesis the key reaction is 1, 3 dipolar cycloaddition okay 1 and what is the key intermediate which is used in iludin M and you will see in one case we have used nitrile oxide the other case we have used carbonyl elide. So what we do in the next few minutes is what is we will discuss what is 1, 3 dipolar cycloaddition and what is carbonyl elide so what one can do with carbonyl elide and we also see how nitrile oxide can be generated and what nitrile oxide can do. So first let us start with 1, 3 dipolar cycloaddition reaction okay and when you talk about 1, 3 dipolar cycloaddition reaction it is like Diels-Alder reaction but 1, 3 dipolar cycloaddition is known before 4 plus 2 cycloaddition that is Diels-Alder reaction is known. So 1, 3 dipole okay it is a 3 carbon unit okay having a dipole and it adds to an alkene okay these alkenes are called dipolar of fire okay in Diels-Alder reaction it is called diene and diene of fire okay here it is called dipole and dipolar of fire. So when they react together the product the reaction is called 1, 3 dipolar cycloaddition reaction obviously when you add 3 plus 2 you get 5 ombardering okay so that is a basic thing about 1, 3 dipolar cycloaddition. The first 1, 3 dipolar cycloaddition was reported in 1988 by Buchner okay the reaction was between this alpha, beta and saturated ester and diaso acetic ester okay. So that underwent a 1, 3 dipolar cycloaddition to give this 5 ombardering that was the first 1, 3 dipolar cycloaddition reported in literature. Then 10 years later so nitrons were discovered okay nitrons are also you know this is called nitrone so nitrons were discovered in 1898 and the early 20th century Werner and Burst discovered nitrile oxides okay. And when you talk about dipole 1, 3 dipole see this 1, 3 dipole can exist in two forms one it can be like allyl anion okay see if you have allyl group so this is like allyl anion but since it is an ally it is a dipole you should also have a positive charge like this so that is why either it can be like this or you can you know you can write another canonical structure okay. So this is one form of dipole. Another form of dipole is linear. So that is if you have a triple bond if you have a triple bond you can see that you have a triple bond then the triple bond the atom which is attached to one of the triple bond can bear the positive charge then adjacent oxygen or whatever it can be nitrogen sulphur the adjacent hetero atom should bear the negative charge. So these are the two types of dipoles one can see in the literature one is allyl anion type another one is propogyl anion type. So we are going to talk about carbonyl elite and we are also going to talk about nitrile oxide so carbonyl elite belongs to this and nitrile oxide belongs to this okay. So nitrile oxide is used for synthesis of elurine C and the other one is used for elurine M okay and when we talk about carbonyl elite how carbonyl elites are generated first of all what is an elite? Elite is a substance where you will have carbon having negative charge and the adjacent atom that is the hetero atom having a positive charge is not it carbon atom should have negative charge the adjacent hetero atom should have positive charge. So now if you take a diaso compound and then if you have a carbonyl group either within this curve within the substrate or you add extra okay then this is basically it will form a carbene is not it once the nitrogen goes it will form a carbene and immediately the lone pair on the nitrogen will attack so that it will become positive charge and this will become negative charge okay. So that is how carbonyl elites are generated and for generation of such carbonyl elites you need transition metals okay there are two transition metals which are used routinely to make or generate such carbonyl elites first it goes through carbonites then it forms the carbonyl elites what are those two metals one is copper other one is rhodium okay these are the two metals you can see in the literature routinely being used to form carbonyl elites. So now let us see one example how carbonyl elites are made say for example if you take this carboxylic acid having an ester now when you treat with oxalyl chloride it forms the corresponding acid chloride subsequently when you treat this with disomethane it forms the corresponding disociate. Now if you look at this it has a carbonyl group which is in the same substrate okay and if you add a rhodium metal dirodium tetracetate okay dirodium tetracetate it forms this rhodium carbinoid okay it forms this rhodium carbinoid okay now what will happen this is like carbene but it is not really carbene rhodium carbinoid then immediately this lone pair can attack and then form the corresponding carbonyl elite okay so this is what it will form once the carbonyl elite is formed then you can do dipolar cycloaddition with various dipolar files. For example if you do this with n-phenyl mellimide then you get this 1, 3 dipolar cycloaddition attack you can see. So this is the dipolar file and here originally you had the dipole so that undergoes dipolar cycloaddition to give this bicyclic system okay likewise nitrile oxide, nitrile oxide can be generated from 2 different starting material one is oxime other one is from nitro compound. So once you have an oxime treat with any chlorinating agent when you treat with chlorinating agent first a monochlorination takes place then followed by elimination of HCl in the presence of bases like triethylamine you get the corresponding nitrile oxide. Then the second method which is also widely used is nitro alkyl here what you need is a dehydrating agent. So one of the most widely used dehydrating agent for making nitrile oxide is phenyl isocyanate. So the phenyl isocyanate removes water molecule from this and gives nitrile oxide. So once you have nitrile oxide then as I said you can do a dipolar cycloaddition with any alkene. So if you use trans alkene and you can see the trans products and if you use cis alkene you will get cis products. Now once you made this if you use ranninical so ranninical is known to cleave this NO bond. Ranninical is known to cleave this NO bond and afterwards aqueous workup will also hydrolyze the imine to give it. So now if you look at this product what is this product? This product is nothing but aldol. So indirectly what you are doing is you are doing an aldol reaction with starting with nitrile oxide and alkene that is it undergoes a 1, 3 dipolar cycloaddition followed by reductive cleavage with ranninical. So nitrile oxide and carbonyl halides are routinely used in 1, 3 dipolar cycloaddition and one can also use LIH. LIH what it does it not only it cleaves a NO bond but also reduces the C double bond N. So you get the corresponding amino alcohol. So with this introduction on 1, 3 dipolar cycloaddition, carbonyl elide and nitrile oxide now we will go to the real total synthesis of eludin M and eludin C. First let us start with eludin M. So when you look at this eludin M as I said the it goes through carbonyl elide and 1, 3 dipolar cycloaddition. So this final target molecule eludin can be obtained from this tetracyclic compound in 2 to 3 steps. How? One if you reduce this if you reduce this you get a hydroxyl group but before that you have to open this. How you do? If you treat with a base then it can open up the oxo bridge. If you treat with a base you can open up the oxo bridge. So you can see this double bond can be formed and this will become hydroxyl group that hydroxyl group can be oxidized. Second the elimination. The elimination will give you the required double bond. So this is the precursor for eludin. It takes maybe 2 or 3 steps. Then if you look at this carefully this is nothing but a carbonyl elide and a dipole. So this will attack here and this will attack so that will form the 5 ombre ring. Now your job is how you generate this carbonyl elide. This carbonyl elide can be easily generated from the diaso ketone. Now let us see how this diaso ketone is prepared. That is very easy. As I said whenever you have beta keto ester or 1, 3 di ketones then one can use a SN2 displacement reaction. So for example if you take methyl acetacetate or ethyl acetacetate then treat with base like potassium hydroxide in the presence of phase transfer catalyst 1, 2 di bromethane will give the cyclopropane. So it is easy to introduce the cyclopropane. Since you are using potassium hydroxide the ester also gets hydrolyzed and you get the corresponding carboxylic acid. It is a well known reaction and it works very well. Now the acid the carboxylic acid can be easily converted into diacyl ketone in 2 steps. One treat with oxalyl chloride it forms oxalyl corresponding acid chloride. Second step you treat with diosamethane you get the corresponding diaso ketone. So basically as you look at in this scheme this methyl acetacetate is very cheap it is available you know plenty in ton scale one can buy. In 3 steps one can make the starting material which is required for making the carbonyl elate. That is the first fragment. For the second fragment what you should do you should start from ethyl adipate. So ethyl adipate again it is a commercially available compound and in one step one can make this 2 methyl cyclopentanone with ester at alpha position. Now you reduce the ester as well as the carbonyl group that is ketone to get a diol. You have primary alcohol and secondary alcohol. What we have seen is you need a dimethyl group. So this CH2OH can be selectively tosylated in the presence of secondary alcohol. It is easy to tosylate. Now if you treat with LAH the tosyl group will go and you introduce the methyl group. So now you have 2 methyl groups what you need is you have to oxidize the hydroxyl group, introduce the double bond and also introduce the bromine. So simple oxidation a Jones oxidation of course one can use other oxidizing agents. Then treat with bromine in the presence of lithium chloride and DMF and you heat the bromination followed by dihydrobromination takes place that is how you introduce the double bond. Now you introduce the double bond now you need one leaving group at this position because that is how you introduce the double bond. So that can be done under you know peroxide condition NBS and the peroxide you can easily introduce this. So now you have made both the fragments one diacycetone and enone what you need to do take the diacycetone treat with dirodium tetra acetate and cyclopentenone. So then it will undergo first the formation of carbonylilide then immediately it will undergo one free dipolar cycloaddition reaction. So take the diacycetone and then treat with rodium acetate you get the corresponding carbonyl and then it forms the carbonylilide and you treat with cyclopentenone and it undergoes the one free dipolar cycloaddition to form this compound very simple straight form. Then once this is formed you need to of course this can be drawn like this also. Once this is formed what you do you have to add a Grignard reagent you have to add a Grignard reagent. So now there are two ketones one as well as here between these two this is sterically less hindered this is sterically less hindered so Grignard addition takes place at this ketone and you get the corresponding tertiary alcohol. Next step is the removal of not only removal of this oxygen oxo bridge but also when you treat with base potassium hydroxide and methanol it undergoes an SN2 displacement of this compound SN2 displacement of this bromine so you get a methoxy group and this one opens and you get a hydroxyl group understand. This opens and you get oxygen bridge and the bromine undergoes SN2 displacement with methanol. Now if you oxidize there is only one alcohol which can be oxidized other one is tertiary alcohol and during that condition elimination also takes place and you introduce the alpha beta unsaturated ketone. Now there are two ketones I should say both are enones and selectively one can reduce one of them but it is very difficult one can think of reducing one of them selectively but it is very difficult so when you treat with LIH it reduces both however when you oxidize when you try to oxidize this alcohol you get iludin M as the major products and the other one can be recycled you know you can again reduce it again you can reduce it you get the dihydroxy compound the dihydroxy compound can be oxidized with the mixture of iludin M and this compound. So that is how the iludin M was synthesized where a carbonyl elide was used as the key intermediate and 1, 3 dipolar cycloaddition was used as the key reaction. So now we will move to the next natural product called iludin C and here the idea is to use a nitrile oxide cycloaddition. So if you look at this compound this enone it can be prepared from this 5 ombre ring. So if you cleave this enone bond you get a CH2OH and this also will get hydrolyzed to ketone followed by elimination one can get this alpha beta unsaturated ketone. And this one can be easily obtained by this intramolecular nitrile oxide polypene cycloaddition reaction. And this in turn can be obtained from the corresponding hydroxyl amine. Subramine source and you get this and here these 2 fragments addition of this lithiospecies to this ketone will give you the precursor for the nitrile oxide formation. Now let us see how these 2 fragments are made. First you can start with the corresponding ketone or you can make it as enol Taseether. Basically what you are doing is a Wilsmeyer AUK reaction with POBR3 and DMF and you get this alpha beta unsaturated aldehyde with a bromine atom at the beta position. So standard Wilsmeyer AUK reaction. Only thing is the ketone you start with enol ethyl. Then once you have aldehyde convert that into oxyne simply by treating with hydroxyl amine you get the corresponding oxyne. Now you have bromine and that bromine should be exchanged with lithium so best way to do is tertiary butyl lithium I will come to that later and before that the other fragment it is very easy to make from the corresponding 1, 3 di ketone acetyl acetone okay. Then you do cyclopropanation you get this it is also commercially available. Then reduce one of them since this is a symmetrical compound one of them can be selectively reduced to get the hydroxyl group. Now this ketone can be you need a double bond isn't it. So either this ketone can be converted into double bond or you can do dehydration to get the double bond. There are two possibilities. So what they did this hydroxyl group was converted into iodide okay using this reaction. Then DBU eliminated that to get the corresponding double bond okay. So this is how I know in three steps one could make this starting material. Once you have these two fragments the next step is to generate the lithium species and add to this ketone okay. How do you generate normally lithium exchange can be done with n-butyl lithium or with tertiary butyl lithium. So they used three equivalents of tertiary butyl lithium you should know why three equivalents of tertiary butyl lithium is required. Two equivalents are required to exchange the bromine and one more equivalent is required to remove the OH. So it adds to this ketone and you get this compound which is a precursor for the dipole of cycloaddition intermediate that is nitrile oxide. Now you treat this with chloramine T okay chloramine T is a chlorine source. So chlorination followed by elimination you generate this nitrile oxide. So once you generate this nitrile oxide then it is ideally the double bond is ideally situated for the nitrile oxide only feels cycloaddition reaction. So that takes place to give your tetracyclic compound okay. So very high yielding reaction you can see 99% yield okay. Then as I said you have to cleave the NO bond and hydrolyze the imine. So that can be done with ranonical okay ranonical cleaves NO bond and hydrolyzes and you get the beta hydroxy ketone okay it is an alder like that beta hydroxy ketone. Once you have this beta hydroxy ketone the next step is to introduce a double bond okay. So that is very easy convert the hydroxyl group into a good living group. So here the good living group is mesyl group straight with mesyl chloride triethylamine you convert the hydroxyl into mesylate then upon treatment with DVU the elimination takes place that is how you synthesize eludin C. Again if you look at this whole process the number of steps is only 6 in 6 steps total synthesis of eludin C is accomplished. Though in both cases the cyclopropane is formed by double displacement reaction the focus was more on the other rings making the 5 umber and 6 umber and both cases a dipolar cycloaddition was successfully used. So to summarize if you see in the first case carbonylilide was used as a dipole 1, 3 dipole and 1, 3 dipolar cycloaddition reaction with cyclopentenone gave this tetracyclic compound which in few steps was converted into eludin M. In the second case where you can see where nitroloxate precursor was made in 2 steps from this ketone and the bromine and intramolecular nitroloxate cycloaddition gave this intermediate which was converted into eludin C in a couple of steps okay. So these are the 2 natural products having cyclopropane we discussed and tomorrow we will discuss another natural product having more cyclopropane and then we will move to 4 umber okay. Thank you.