 So, good morning and welcome back to NPTEL lecture series on Classics in Total Synthesis Part 1. So, we have been discussing many total synthesis and today we will discuss one of the classical total synthesis of a natural product called Langefolin. So, this Langefolin is a seschiterpene as you can see here it is a hydrocarbon and it was obtained from pine resin as an oily liquid and the name itself was derived from where from it was isolated, it was isolated from your pine species called Pinus Langefolia. So, that is why the name was given as Langefolin and if you look at this molecule from synthetic point of view the major challenges are first of all it is a tricyclic compound 1 and it has 2 quaternary centers, it has 2 quaternary centers and 3 chiral centers and what is more important was here there is no functional group other than the double bond, only you have a double bond as a functional group other than that you do not have any other functional group and as it is a hydrocarbon again not only isolation of this molecule but also synthesis of this molecule is quite difficult because of its volatility. There are 3 total synthesis which I am going to discuss today on Langefolin and the first synthesis I will talk about is reported by E.J. Coray and then second synthesis I will talk about was reported by Opulsar and the third one was reported by W.S. Johnson. First let us start with the total synthesis reported by E.J. Coray. So, how did he do? What are the real retrosynthetic disconnections he has cleverly used to make this Langefolin? So, if you look at this molecule obviously the first retrosynthesis first disconnection should be to introduce the double bond and it is easy is not it? So, you can easily think of making this double bond as the last step that is if you can have this ketone if you can synthesize this ketone then it should be possible to do Wittig reaction or you can add a Grignard followed by Dehydration should give the natural product. So, now the next problem is you have one functional group sometimes when you look at a natural product if the natural product does not have a functional group then what one should do is you should introduce a functional group because always you need a functional group in the product so that you can use that functional group as a handle for further disconnections as well as functional group transformation. Though this particular stature has a functional group that is carbonyl group but still Coray felt one functional group is not sufficient it is better to introduce one more functional group. So, that is how he introduced another functional group which is also a carbonyl group and the reason for introducing another functional group is if you look at this the relationship between these two carbonyl groups are 1,5. So, when you have 1,5 di ketones which I am sure you would have seen in synthesis of Wieland-Mischer ketone and other enones it can be prepared by Michael addition reaction. It can be easily prepared by a Michael addition reaction. So, here I have given a numbering so you can see 1, 2, 3, 4, 5. The way I have given this numbering was I am going to redraw this molecule as well as do a disconnection. So, the disconnection is suppose if I remove this if I disconnect this bond what you will get is this one. So, now you are going to generate an anion here that will be enolate that enolate will add to this double bond. Basically it is a Michael reaction. Now you have a 6, 7 few string. So, whenever you want larger ring or medium size ring one reaction which you can think about is ring expansion reaction. Say normal rings are 5 membered and 6 membered. So, if you want to go beyond so one can think about ring expansion. So, what Coray thought? So, let us start with 2 6, 6 few strings. So, you can see here one 6 membered and another 6 membered. Now he wanted to carry out a ring expansion reaction to get 7 numbering. So, what he thought if you know he wanted to use a well known penicol, penicol rearrangement and the problem with this diol if you have this 1, 2 diol and if you want to carry out a penicol, penicol rearrangement if you want to carry out a penicol, penicol rearrangement then this tertiary alcohol only will be protonated. This tertiary alcohol will be protonated. So, what you have to do for the ring expansion you have to make the secondary alcohol. You have to make the secondary alcohol as the leaving group then the 7 membered ring formation is possible. And this can be easily obtained from Peele and Mishir Kit. So, when I discuss the real synthesis I will talk about how this ring expansion was thought and how he successfully could do it. So, his synthesis actually started with Peele and Mishir Kit. It is a racemic one then the ketone you have 2 ketones one is the 6 membered ketone other one is a enone. Between enone and ketone one can selectively protect the ketone with ethylene glycol in the presence of acid. So, now you have protected the 6 membered ketone here. So, now you can carry out the Wittig reaction on the enone. So, simple Wittig reaction gave this exocyclic double bond with a methyl group. So, the stereochemistry as you know it is a mixture does not matter. So, next step is the dihydroxylation you have internal double bond and an external double bond. So, one could selectively dihydroxylate the external double bond with osmotic oxide. So, now he has got basically the key precursor for the ring expansion in 3 steps from known Vela Mishir Kitone. So, now the key problems which I mentioned that is you have to do the ring expansion. When you want to do the ring expansion as I said the protonation will take place only at the tertiary alcohol. So, if you use any acidic condition for the rearrangement to take place then ring expansion will not take place. So, what you have to do you have to go you have to modify. So, what one can do is if you can make the secondary alcohol as a good living group secondary alcohol as a good living group then this bond can migrate, this lone pair will come here and this bond can migrate that will lead to the formation of 7-membered ring. So, that is what he did. So, what he did was he took this diol and then made the secondary alcohol as a good living group. So, by converting that secondary alcohol into tosylates. So, now you have made the tosylate then it was easy. So, he treated with the perchlorate in the presence of base like calcium carbonate then as I said this bond migrated and gave the 7-membered enone. So, now if you look at this enone the double bond is still not in conjugation. The double bond is not in conjugation. So, for the key intramolecular Michael reaction to take place for the key intramolecular Michael reaction to take place first this double bond has to come here then you have to remove the keto. So, what he did first he removed the keto to get back the keto. Now the double bond was isomerized with triethylamine and he did you just isomerize the double bond. So, that gave the key precursor for the intramolecular Michael addition. Now, if you look at this molecule. So, we have to draw in such a way that you can see that this carbon is closer to this carbon. So, that is very important. So, I draw this in such a way now you rotate this you rotate this by 90 degree you rotate this structure by 90 degree. So, now you have this structure this structure I ask you to visualize like this bicyclic system. If you look at this bicyclic system you can see here this is this is the cyclohexanone the cyclohexanone here is written like this. Now, this hydrogen is alpha you can see that is then this methyl is alpha S. Now, you have the 7 ombre ring S that is beta you have the 7 ombre ring. Now, when you look at this molecule you can see carbon X is near to carbon Y that is the beta carbon of the Evo. So, this tells that from this confirmation you can make out that it is possible to carry out an intramolecular Michael reaction and that is what in the did and treated with strong base it generated anion and Michael reaction worked very well to get this bicyclic calm. So, now you have got the core structure. So, with two carbonyl groups so 1, 5 diketone and as you know what needs to be done is you have to introduce one more methyl group and remove this carbonyl then you have to do petic reaction on this carbonyl group. So, these are three things left for the completion of longifolding. One introduction of methyl group, second removal of the carbonyl group, third conversion of carbonyl into a double bond. So, first it was easy to introduce the methyl group with a very strong base sodium titl group then quench with methyl iodide. So, you could introduce the methyl so the dimethyl group was put in proper place. Next you have to remove this carbonyl group selectively in the presence of other carbonyl group. So, between these two carbonyl group and this one is sterically you know hindered. It is sterically hindered position. So, it is easy to protect the other carbonyl group. So, it was protected as diethane derivative and as you know when you have a diethane derivative you can reductively remove so that you can get the corresponding CH2. So, before that the carbonyl the other carbonyl was reduced because that should not interfere in the removal of diethane. So, once you reduce the ketone to alcohol then the diethane was removed with sodium and hydrogen and to get the CH2. So, now what is left you have to oxidize the secondary alcohol to ketone then convert the ketone into double bond that will complete the total synthesis of lunch fully. So, that is what he did. So, treatment with chromium trioxide oxidize the secondary alcohol to get the ketone. So, then the Wittig reaction actually was not successful. So, he has to do the other method that is normally you can use either Grignard or organolithium species. So, he tried methyl lithium addition to this ketone. So, that gave the tertiary alcohol and that was treated with thionyl chloride pyridine to get the exocyclic double bond and that is how he completed the total synthesis of lunch fully. So, if you look at the synthesis reported by Coray, this synthesis started with the commercially available and well known Veylon Mishir ketone it is a bicyclic compound and the key reactions which Coray used were intramolecular Michael Michael addition and the formation of 1, 2 diol first formation of 1, 2 diol selectively on one of the double bond followed by rearrangement ring rearrangement of the 1, 2 diol to get the 7-membered ring. Overall, he took about 15 linear steps with an overall yield of 2.8 percent. So, that is quite decent considering the natural product and this was the first total synthesis. So, now we will move to the second total synthesis which was reported by Wolfgang Opulser in 1978 and he used a very, very interesting photochemical reaction followed by group fragmentation as key reactions to construct the core structure of logifoline. And as usual the first disconnection was same as the case of E.J. Coray that he disconnected the double bond and the ketone can be easily converted into the double bond using methyl lithium and followed by dehydration and this ketone again you can see he introduced another ketone like what E.J. Coray has done and the next step is the one which he literally you know is different than what E.J. Coray has done. Now, his idea is to use a grub like fragmentation. So, if you see this tetracyclic compound his idea is so if we do the ring opening it is like push-pull system when you open this 4-membered ring that will lead to the formation of 7-membered ring. And when you have the 4-membered ring obviously 2 plus 2 photochemical cycloaddition reaction only should come to your mind and that is what he thought as the precursor for this tetracyclic compound. So, now you can see there is one double bond here and another double bond these 2 should undergo an intramolecular 2 plus 2 cycloaddition to give this tetracyclic compound. And this can be easily obtained from this you know this is same compound this can be easily obtained from this cyclopentene carboxylic acid. So, his starting material is quite simple and in few steps he thought he can easily achieve the synthesis of large poly. Now, let us see how he really executed his retro synthesis in the synthesis of large poly. So, he started with cyclopentene carboxylic acid and treatment with thionyl chloride converted the carboxylic acid to acid chloride and this acid chloride upon enamine reaction. Enamine reaction with this particular enamine derived from cyclopendron. Cyclopendron on treatment with morpholine will give this enamine. This enamine upon treatment with this acid chloride basically what you are doing is you are doing the acylation reaction. You are doing the acylation reaction next to the carbonate. So, once you have this 1, 3 diketone. So, this is 1, 3 diketone is not it 1, 3 diketone and this 1, 3 diketone you can selectively enolize. So, what he did he not only enolized but also protected the enol as CBZ. So, CBZ is a good protecting repress for MH2 and OH. So, now he protected that enol as enol OCBZ. The CBZ is nothing but benzyloxycarbonate and that can be easily cleared under hydrogenalysis condition later. Whenever it is required, you can use hydrogenalysis condition to clear that BZ group. So, this can be drawn like this as we have seen in the retrosynthesis. So, the 2 plus 2 cycloaddition work. So, when he tried this intramolecular photochemical 2 plus 2 cycloaddition reaction. So, which we can also call it as DMAO cycloaddition reaction. So, this reaction work well. So, exactly as planned he got this tetracyclic compound. So, now what is left is to cleave the protecting group. The CBZ should be cleaved and then followed by graph fragmentation should give the servant number tree. So, it was cleaved under hydrogenalysis condition. As I said it is nothing but this OCOO benzyl. So, as you know hydrogenation it will cleave the benzyl and then carbon dioxide will also will go and you get the corresponding OH. Now, this will open up to give the 1, 5 like this. Again as he planned the 2 plus 2 cycloaddition DMAO cycloaddition work well and also the ring expansion the graph fragmentation also worked very well to get the core structure. Now, if you look at this tricyclic compound this is the core structure of longevity. So, what is required is this should be converted into dimethyl group and this should be converted into CH2. So, it is easy. Now, you do a vitic on this particular less exposed carbonyl group as you know this is the other carbonyl is slightly hindered. So, it is easy to do a vitic reaction on the less exposed carbonyl group. So, you get the double bond then you do the cyclopropanation. You do the cyclopropanation on this double bond to get the cyclopropane. Once you have the cyclopropane as you know you can hydrogenate to introduce the dimethyl group. Cyclopropane is converted into the dimethyl group. Now, only 2 reactions are left one this hydrogen. This hydrogen should be converted into methyl group then you have to remove the carbonyl group. So, simple treatment with LDN methyl iodide. So, that is the only place it can go. So, methyl group was introduced then as reported by Coray do a methyl lithium addition to the ketone to get the tertiary alcohol and then treat with thionyl chloride pyridine you get the corresponding double bond and that is a natural product. So, that is Langefolin synthesis. And if you look at the synthesis reported by Opalser it started with commercially available cyclopentadiene 1 carboxylic acid. But what is more important was 2 key reactions which actually assembled the natural product core structure quickly one was D-Mayo photochemical 2 plus 2 cycloaddition reaction and the second was graph like fragmentation. Overall yield was 23% which is very good and the number of steps taken to complete this molecule was 11 steps. So, 11 steps with a overall yield of 23% is really very, very high. So, this is one of the best synthesis of Langefolin reported so far. Now, we will move to the third synthesis which was reported by W.S. Johnson. Actually here his total synthesis based on his 70-petre observation of his polyene cyclization. Johnson has reported several polyene cyclization which is like biomimetic cyclization he was inspired by nature's many cyclization. So, he has developed many biomimetic cyclization. So, during his report when he was working on this polyene cyclization. So, he was trying to cyclize this particular E9. So, his idea is this OH will be a good leaving group. So, then this double bond will migrate and this will attack and finally, a nucleophile will neutralize the positive charge generated here. So, when he treated this with Stani chloride what he got was as expected this cyclist product. So, the mechanism is simple as I said the OH was protonated. So, now the water is a good leaving group and followed by migration of the double bond and the triple bond. So, you can get the bicyclic compound. So, bicyclic compound with a vinylic carbocation. Now, the Cl- which is formed in C2. So, can attack and then get this bicyclic compound with a exocyclic double bond. Okay. Interestingly, in addition to this product, he also got another product. In addition to this product he also got another product. So, that product is like this. That product is like this. So, he proposed a simple mechanism for this. So, first it forms the vinyl carbocation. So, before the chloride attacks, before the chloride attacks, what happens? If you redraw this molecule, if you redraw this molecule like this, you can see here. So, this is the five-umbal ring. Okay. You can see this is the five-umbal ring and this is a seven-umbal ring and you have exocyclic double bond with a carbocation. What he proposed was this double bond if it migrates and neutralizes the vinylic carbocation. Okay. Then what you get is this carbocation. Okay. Now, once you have this carbocation, simple attack of water on this carbocation should give this compound. Okay. Now, if you look at this structure, if you look at this structure. So, this is a core structure of longevity. This is the core structure of longevity. So, then he thought by choosing proper starting material, it should be possible to synthesize large-poly using this serendipitous observation. Okay. So, with that, he wrote a proper retrosynthesis as usual. The first disconnection was to have the carbonyl group and this can be obtained from this internal double bond. Okay. So, through functional group transformation which I will discuss while talking about the total synthesis and this in principle can be obtained from this alcohol. So, if you look at the previous slide, so this is what is the intermediary. Okay. And now, if you look at this, this can be obtained from this cyclopentenone with this appended side chain. Okay. So, this can be easily obtained from this substituted cyclopentenone. Okay. This cyclopentenone can be easily obtained from cyclopentenone and aldol reaction with acetone. Okay. So, that is the starting material. So, he started with this compound, then carried out a 1, 4 addition. Okay. 1, 4 addition with this derivative. Okay. After the 1, 4 addition, the resultant enolate was quenched with acetyl chloride. The resultant enolate was quenched with acetyl chloride to generate the enol acetate. Okay. Or in other words, the enolate was trapped as enol acetate. So, after trapping the enol acetate, if you want to generate the enolate, what you can do, one can treat with methyl lithium. So, methyl lithium will attack the acetate and generate the lithium enolate species. Okay. Then when you add bromine, that bromine will be quenched here. So, that is what happened. So, first you generate the lithium enolate by treating with methyl lithium. Then you add bromine. You introduce the bromine next to the carbonyl group. So, once you have the bromine introduced next to carbonyl group, one can easily introduce the double bond through elimination. So, it was easily done under mild condition. So, now the key starting material is thread. So, what you should do? You should reduce the carbonyl group selectively to get only the allylic alcohol, not fully reduced compound. So, LAH reduced that enone to the allylic alcohol and that set the tone for the key Lewis acid mediated cyclization to get the longifold in core structure. Then it treated this compound with typhloracetic acid and as expected, we got this carbocation and that was quenched with water to get the corresponding tertiary alcohol. So, this compound is very simple. You can see already has a dimethyl group. You have to remove this hydroxyl and somehow you have to migrate this double bond to here as well as you have to introduce a methyl group. So, it was very easy for him to synthesize the core structure. Now, his challenges are remove the hydroxyl group, push the double bond to exocyclic and introduce a methyl group. So, these are three more things to do in the synthesis or longifolding. So, first he removed the hydroxyl group with zinc bromide and sodium cyanoborohydrate. So, that was done easily. Next, he has to migrate the double bond. So, that was done by treating with para toluene sulfonic acid. He reflects to get the exocyclic double bond. Now, you have exocyclic double bond, but how will you introduce a methyl group here? It is not straight away possible from the double bond, you can introduce a methyl group. However, it is possible if you osanalyze or if you cleave the double bond to ketone. So, with tritinium oxide and sodium peroiodic acid, he could get the ketone. Once you have this ketone, the next step is to introduce a methyl group, LDA and methyl iodide. You introduce a methyl group and followed by like Watt, Coray and Opalser has done. Add methyl lithium to get the tertiary alcohol followed by treatment with thionyl chloride and pyridine. So, he could accomplish the total synthesis of longifolding. So, if you look at the total synthesis of longifolding, though it was reported 1975, he started with simple commercially available cyclopentanone which upon aldol reaction with acetone, it gave two isopropylalanine cyclopentanone that was used as the key starting material and what is important was the serendipitous observation during the polyene type cyclization to give a minor product was used as a key step to construct the tricyclic compound in one step. One step formation of the tricyclic compound using this polyene type cyclization was the key step and overall his synthesis was achieved in 11 steps and with an overall yield of 26.6 percent. So, the overall yield of 26.6 percent is very impressive and all the three synthesis are really unique and classical in nature and these three synthesis reported in 60s and 70s considered as real classical total synthesis. So, with this I will stop here and then we will discuss more total synthesis in the next class. Thank you.