 Yeah, good morning and welcome back to the NPTEL lecture series on classics and total synthesis part 1. So yesterday we were talking about total synthesis of zebrilic acid by E.J. Coray and that time we discussed up to the synthesis of this key intermediate by 3 different routes. So now we will discuss further reactions towards the completion of the total synthesis of zebrilic acid. So if you look at the molecule A, you can see already there is a diene. So what one has to do is you have to attach the dienophile. So how Coray has done? So he added this acid chloride and having a chloride at the beta position plus he deprotonated this OH with butylethium and then carried out this acylation. So now the key intramolecular 4 plus 2 cycloaddition reaction was done in the presence of propylene oxide. So which is normally you know proton sponge that gave you know very good yield of this pentacyclic compound okay. So how many rings are formed? Two more rings are formed normally Diels-Alde reaction gives one ring but when it is intramolecular reaction then you can get more than one ring. So you could construct two rings one and two using this key 4 plus 2 cycloaddition reaction okay. So if you look at this molecule so what is missing is in zebrilic acid you need one methyl group here okay then this CH2O should be oxidized to COH okay. So for that first it treated with LDA equivalent LDA is lithium diisopropyl amide. So now instead of one isopropyl group if you replace the isopropyl group with cycloexyl group so then you get this amide okay. This amide generates anion here okay generates anion here and then upon quenching with methyl iodide you introduce this methyl group okay. So that was the idea so the methyl group was introduced as well what happened you can see this HCl okay there was also an elimination of HCl because of the presence of base like lithium isopropyl cyclohexyl amide okay. So that generated a diene and also you could introduce the methyl group at that required position. So the next step is to remove the mom group okay so you have a mom group here that mom group was removed using Zim bromide then if you use POS gene it can form half ester okay. So one equivalent of POS gene and then this hydroxyl group you form the half ester that on treatment with phenyl methyl amine okay. This is a good chiral resolving agent so now you form this carbamate and one can resolve two diastereomers okay you resolve two diastereomers and these two diastereomers can be easily separated. This diastereomers if you hydrolyze the required one you get the corresponding hydroxyl group okay. So until the hydrogen hydroxyl group all the compounds which we discussed are racemic. After forming the half ester and then treating with alpha methyl benzyl amine and the resolving from now onwards you can see all are chiral real chiral centers okay. Then what you need to do? You need to oxidize this CH2O to COOH as well as hydrolyze this. So both were done in one step by treating with potassium hydroxide and sodium pythenate. So it cleaved and then you got a di carboxylic acid okay. There are four rings now you need only one more ring that is a lactone. This diene should form a lactone and one hydroxyl group then the total synthesis is accomplished. So before that if you have to do iodolactonization with this carboxylic acid then the other carboxylic acid should be protected okay. So that was selectively done by treating with triethylamine, tosyl chloride and methanol okay. Just think about mechanism. So what happens? It involves two reactions okay. Think about and then write the mechanism for this key reaction. Then MCPBA so you have diene this on treatment with MCPBA. If this double bond forms epoxide then it can open the epoxide if it opens from this side this will be 4 ombre ring but if it opens from this side that will be 5 ombre ring. So the 5 ombre is better than the 4 ombre ring so 5 exotate opening of this epoxide gives the corresponding 5 ombre lactone okay. But what we need is lactone here. You do not want lactone here but this is important so what happens? Now once you have this you treat with sodium hydroxyl methanol. Basically what you do? You hydrolyze this lactone. Now you got trihydroxycarboxylic acid 1, 2, 3 trihydroxycarboxylic acid having an ester here 4 functional groups okay along with 2 double bonds quite sensitive. Then you carried out the key iodolactylization okay. You have double bond and treat with iodine in the presence of sodium pi carbonate it forms the Iodolactone okay. Then what you need? You need to get a double bond prolygalic acid okay. Protect these 2 hydroxyls which are trans to each other as trifluoroacetate then treat with zinc. So when you treat with zinc as you know zinc will give 1 electron and it can eliminate to give corresponding double bond. So the lactone is formed and then you have the oacetate what you need is the ester should be hydrolyzed. So 10% sodium bicarbonate water the trifluoroacetate could be hydrolyzed to get the corresponding allylic alcohol. Then the methyl ester the hydrolysis of methyl ester is quite difficult because that is very unstable the carboxylic acid as the COO minus is formed it is quite unstable. So he has to use sodium thiopropylate to hydrolyze the methyl ester to get the natural product. So this is how E.J. Coray could successfully accomplish the total synthesis of gibalic acid in almost 40 years ago. And his starting material was from cyclopentadiene in one case and so betadiene in another case betadiene having a CH2OH in another another method. The key reactions used by Coray in the synthesis of gibalic acid or 4 plus 2 cycloaddition, hydrolactonization, epoxylactonization and magma recoupling. So these are 4 key reactions which he used in the total synthesis of gibalic acid which involved 31 longest linear steps and with an overall yield of 0.5%. Such a complex molecule he could successfully accomplish synthesis and that was one of the classical and most difficult synthesis reported in the literature. So now we will move to the second total synthesis reported by Yamada on the same molecule and what are the key reactions he has used in the total synthesis of gibalic acid? Of course once you have a lactone one reaction which will be common is hydrolactone. So he used again the same hydrolactonization as the first key disconnection to get the gibalic acid then he wanted to cleave this bond. If you have an allylic ester, allylic carbonate one can cleave that under either palladium catalyst condition or metal ammonia condition to get the corresponding double bond. So he thought he can use that to get the corresponding double bond. This of course can be obtained from this alcohol. Then his idea was to do like this if you have a bicyclic compound. If you have bicyclic compound, his idea was if you do was analysis you get 1, 1, 2, 3, 1, 3 diketone then you can open this 4-womber ring. So you can open this 4-womber ring and then cyclize that will give this 5-womber ring. Opening followed by McMulley type coupling you will get that ion. So that was the key reaction and this he thought can be obtained by metal ammonia and some functional group transformation. So that actually took him all the way to this dyeing and this dyeing of pipe. So there are few reactions which are common between E.J. Coray and Yamada's total synthesis of gibalic acid. One, Diels-Ald reaction. Two, you could have seen already that iodolactonization and three he wanted to use a coupling reaction to form this 5-womber ring. In the case of Coray he used McMulley coupling and let us see what Yamada has used to make this 5-womber ring. So his total synthesis started with methyl pyruvate and nomenical condensation with melanon nitrile gave the dyeing of pipe. The dyeing of pipe is ready and the dyeing is very easy to make from anisoldehyde. If you have anisoldehyde then you can make this in one step and once the dyeing is ready do the Diels-Ald reaction. So when you do the Diels-Ald reaction you can see the quaternary methyl group. Quaternary methyl group is fixed and what you do not need is one extra cyanide. You have to hydrolyze this cyanide to ester at the same time you have to remove the extra unwanted cyanide. So that if you treat with potassium hydroxide, so what will happen? The cyanide will be hydrolyzed to carboxylic acid. Since it is dicharboxylic acid one will be decarboxylated at the same time this ester also you know during hydrolysis ester will be hydrolyzed. So you get directly the dicharboxylic acid. And once you have this dicharboxylic acid you can see here 1, 2, 3, 4, 5. Intramolecular Frederkauf's acylation may be possible on the aromatic ring. Did you do that? Let us see. So before that he made anhydride. He made anhydride of this dicharboxylic acid then he tried this Frederkauf's reaction. So if you have anisone and treat with acetic anhydride so that can undergo Frederkauf's reaction at para position. So same thing here he carried out the Frederkauf's reaction and then cyclized here to get the corresponding 5 ombre ring, 5 ombre ring and carboxylic acid. Then you treat with para-nitroperbenzyc acid, para-nitroperbenzyc acid what it can do? There is only one possibility that is if you have double bond the double bond will be epoxylized. So you get the epoxide and the epoxide will not be like that since you have carboxylic acid so that will open the epoxide and you get hydroxy lacto okay. Then protect the hydroxyl as TMS ether it is a transient protecting group and you cannot use it for long time because TMS is a very labile protecting group okay. So you protect that hydroxyl as TMS ether then carry out the Kore-Czechovsky reaction that means you convert this CH2 into an epoxide with one carbon extra okay CH2 minus and so Kore-Czechovsky reaction gives the epoxide. Basically as you know in this place what you need is COOH okay COOH. So this epoxide when you treat with BF3A3 it will undergo pre-arrangement. So epoxide will open up and it will form enol and that enol is nothing but corresponding aldehyde okay. Then reduce the aldehyde to corresponding primary alcohol with sodium boride methanol and protect that as mom ether, protect that as mom ether. Now the transient protecting group that is TMS is cleaved with t-buff to get the secondary alcohol and protect that as same ether with treating with hewnicks base and then same chloride. Now you have protected the sim and you have protected the primary hydroxyl as mom what you need to do is hydrolyze the ester okay. When you do that you can get the corresponding carboxylic acid but at the same time what you do not need is this oxygen. If you hydrolyze you get only carboxylic acid and hydroxyl group. When you do not need the oxygen then you have to do hydrogynolysis. The hydrogynolysis suppose if you have a benzyl ether what will happen you will remove the oxygen is not it? RO CH2 RO CH2 phenyl group what will happen the CH2 phenyl will go likewise so this is the benzylic carbon okay. Hydrogenolysis will cleave this so you get hydrogen here and this whole thing will come out as carboxylic acid. The cleavage of the lactone was done under hydrogynolysis condition to get carboxylic acid. Now reduce that with LAH to get the primary alcohol and protect that as mom ether so you get dimom. Then you have aromatic ring okay that aromatic ring should be reduced to get the ketone. So metal ammonia reduction gives diene and this diene upon hydrolysis you get the corresponding ketone and this double bond is still intact okay. Then potassium carbonate methanol isomerizes the double bond to give alpha, beta and such a ketone then carry out intermolecular 2 plus 2 cycloaddition with allene okay. So that gives this bicyclic compound. Osanalysis gives ketone this while doing osanalysis itself in the presence of methanol and sodium bicarbonate it gives alpha, beta, gamma okay or you can write the other way okay alpha, beta, gamma the ketoester it forms how it forms once you have this diketo. Osanalysis will give the diketo isn't it? That opens up because this 4 ombre ring is trained so this attacks this carbonyl and it can open up. So that will give the corresponding ketoester. Once you have the ketoester again you reduce with potassium and liquid ammonia. Potassium in liquid ammonia it will give 1 electron. So that 1 electron will go to ketone then that ketone can cyclize here to give the corresponding hydroxy compound okay it is a dihydroxy compound. First it will form hydroxyl here and then here it will be ketone. That ketone also further reduced under the potassium liquid ammonia condition to get the diol. Swan oxidation will give ketone then protect the rigid hydroxyl as mom ether then do the vitic to get the corresponding exocyclic W bond okay. Now you can see these 2 rings are done the third ring is also done what you need these 2 should become carboxylic acid okay. Then the hydro lactation has to be done and also you have to introduce a hydroxyl group on this particular ring. So t-buff will remove the sem group okay you get the hydroxyl group that sem is nothing but trimethyl saline ethyl methyl chloride okay. So mesolate then do the elimination with dBu to get the W okay. Then remove all the mom groups with HCl methanol you get the triol. One is tertiary alcohol other 2 are primary alcohols. So you can oxidize the primary alcohols under swan condition to get the aldehyde then further oxidation under pinnick condition you get the dicarboxylic acid then you carry out the iodolactanization. But before you do iodolactanization as you know you have 2 carboxylic acids only this hydroxyl group of carboxylic acid should undergo iodolactanization that means you have to protect this. So selective protection is difficult so all 3 were protected the hydroxyl and then 2 carboxylic acids were protected as mom ether then hydrolyze. So what happens you get back the dicarboxylic acid then you treat with iodine and sodium bicarbonate and that undergoes iodolactanization to give this iodolactanization. Then treat with dBu you introduce the double mom. Now you protect the carboxylic acid as mom ester because having free carboxylic acid is not good because free carboxylic acid can create more trouble. So you protect this as mom ester then treat with LDA. So what does LDA do? LDA it can generate anion and opens the carboxylic acid to get a diene as well as carboxylic acid. Now if you treat with MCPBA case this will undergo epoxidation and followed by opening of the epoxide you get hydroxylactone. Then the hydroxylactone if you can cleave this bond you will get the carboxylic acid at the same time the double bond should be intact. Cleave the CO bond as well as the double bond should be in same place. How these two can be done? If you have a close look at this molecule that is nothing but allylic acetate okay. So such allylic acetate can be cleaved under hydrogenalysis condition okay. Before that you hydrolyse the mom group okay simple water peridine mom group because it is ester okay. So ester hydrolysis give carboxylic acid then as I said you can use hydrogenalysis condition or equivalent to that is metal ammonia okay. So before you do the metal ammonia treat with the LDA because you have two hydroxyl this and this. So that will become corresponding Olythio derivative. Then you do metal ammonia that will cleave the CO bond followed by acidification you get this compound okay. Now you do the iodolactonation once again you get this and remove the mom group with HCl and followed by treatment with DBU okay. So DBU will eliminate the iodide to get the double bond and that is how MRO completed the total synthesis of allylic acid. To summarize so total synthesis of Gibrillic acid was accomplished by Yamada and his co-workers in 1989 and his total synthesis was started from melanonitrile and methylpyruvate to prepare the dienophile okay and the diene was prepared from anisaldehyde and a intermolecular dienosol reaction was the first key reaction and he also used osinolysis, he also used iodolactonization, epoxylactonization as key reactions to complete the total synthesis of Gibrillic acid. So overall he took about 42 steps to complete the total synthesis of this complex molecule and the overall yield of this synthesis is 0.3% considering the complexity as well as sensitivity of the natural product 42 steps is okay and overall yield 0.3% is also acceptable okay. So now we will move to total synthesis of other natural products in the next lecture okay. Today we completed two total synthesis of very complex natural product called Gibrillic acid okay. Thank you.