 So, good morning and welcome back to this NPTEL lecture series on classics in total synthesis part 1. The last few lectures we have been discussing about total synthesis of natural products having 4-membered ring. So, now the next few lectures we will focus on natural products having 5-membered rings. So, when you talk about 5-membered rings, there are 2 classes of natural products which should come to your mind immediately. One is prostaglandins, the other one is tricunin based natural products. So, first let us start with prostaglandins, prostaglandins they were well known and during 1960s and then 70s got lot of attraction from synthetic chemists to develop new methodology for the synthesis of all the prostaglandins. The prostaglandins as you can see has a cyclopentane has a cyclopentane with 2 side chains, so this is the core structure of prostaglandin which we can call it as prostoniac acid, that is the core structure. So, most of these prostaglandins were discovered and reported in early 30s by one Euler and of course it was not easy to elucidate the structure of all these prostaglandins. It took about the more than 35 years to elucidate the correct structure of prostaglandin. And once the correct structure of prostaglandins were identified then lot of action take took place from synthetic chemists to synthesize this compound and I will at least talk about 2 total synthesis of prostaglandins, one by E.J. Coray, Nobel laureates and other by Gilbert Stark, okay. So, if you look at this closely, so they are all carbocyclic, oxygenated C20 molecules, okay and in addition they have 2 side chains, okay, so one having carboxylic acid at the terminal the other one is having methyl group, okay and how do you classify prostaglandins and how do you name it, okay. It depends on the number of different functional groups present in the molecule. The basic structure is same, cyclopentane with 2 side chains, okay. And you want to classify it was based on 2 things, one the functionalities present in the phymumbertary, okay. What are the functional groups present and how they are present, okay. What is their relationship and the second one depends on the number of unsaturation, number of unsaturation present in the 2 side chains, okay. Let us directly go into the nomenclature. So normally when you talk about prostaglandins, they write PG, PG as the first 2 letters. So the PG represents prostaglandin, PG means it is prostaglandin. Then you see PG A, PG B, PG C, PG E, PG F. What are these additional letter A, B, C, D, E, okay. PG A, PG B, PG C, all of them have cyclopentino, okay. You can see this phymumbertary is in the form of cyclopentino. Well, but the double bond position changes from A to B to C. When you go from A to B to C, you can see the position of the double bond also migrates. One is alpha, beta unsaturated ketone, then beta, gamma unsaturated ketone. The other one is also alpha, beta unsaturated ketone, but there are no chiral centers. These 2 are no chiral centers. It is a tetrasubstitute to come, okay. So first one is alpha, beta unsaturated ketone, but it is dye substituted, double bond is dye substituted, whereas the second one it is trisubstituted and third one it is tetrasubstituted. So they are called PG A, PG B, PG C, okay. All of them as you can see have one double bond, but the position of the double bond gives their name. Then what is D and E? Here the double bond, the double bond is replaced. You do not have double bond, instead you have a hydroxyl group, okay. And the hydroxyl group also it is beta hydroxy, beta hydroxy means it is like aldol, okay. Beta hydroxy ketone. So if the ketone is here and the hydroxyl is there, then it is PG D and if it is opposite, then this is called PG E, okay. PG A, PG B, PG C have cyclopentenone. PG D and PG E have one hydroxyl and one ketone. You can call it as hydroxyl, beta hydroxy ketone or aldol in the cyclopendane ring. Still the two side chains are intact, okay. Now we will come to the last category that is PG F. What is PG F? Here the ketone which was present in PG D and PG E, they are reduced or in other words you have two hydroxyl groups in the cyclopendane ring. They are called PG F. And you also see when you read literature, you will see PG F alpha, PG F beta. What does it mean? That means the ketone which you reduced here, if it is alpha, the stereochemistry is alpha, then you write PG F alpha. If it is beta, then you write PG F beta, okay. So that is how the names are given for prostaglandins. Then you also see PG F 1 alpha, PG F 2 alpha. What does it 1 and 2 mean? So that means the side chain, you have two side chains, it depends on the number of double bonds present in the side chain, okay. If they write 2 alpha, that means the side chains have two double bonds. If they write 1 alpha, that means it has only one side chain, okay. So this is about the novel nature of prostaglandin. So now let us see what are the challenges these molecules provided for synthetic chemists to make this compound. And if you look at the PG F series, the PG F series have 5 chiral centers, okay. PG F have 5 chiral centers, 4 in the ring, 4 in the ring and 1 in the side chain. I will come to that later, but 4 in the ring, 1 in the chiral center and the 4 in the ring, they are contiguous, 4 contiguous chiral centers are present in prostaglandin. Then the hydroxyl group which is present in the side chain, it is little far away from the 4 chiral centers present in cyclopendane. So that means sometimes it will be difficult to use the 4 chiral centers present in the cyclopendane to direct the hydroxyl group or introduction of hydroxyl group at C15. Then when you look at the side chain, one side chain has cis double bond, the other side chain has trans double bond, okay. One side chain has cis double bond, other side chain has trans double bond. And the prostaglandins D and E have beta hydroxyl ketone that is aldol, okay. Once you have this aldol as you know, they are slightly unstable when you treat that with acid or base, okay. So one should be extremely careful when you reach that stage, you should not use acid or base, okay. And because you have beta hydroxyl ketone and diol and triols, so the synthetic strategies should have proper protecting groups and also you should also have orthogonal protecting groups. Your 2 different protecting groups should be introduced and cleaved at different times, okay. So these are the synthetic challenges one could expect while talking about total synthesis of prostaglandins. So as I said there are quite a few total synthesis but the first total synthesis was reported from Professor Coray's laboratory, even today that synthesis is considered one of the best synthesis of prostaglandins. Their retrosynthesis is based on few key reactions. First of all, his basic idea is he called it as bicycloheptane approach. That means he starts from this bicyclohe 2, 2, 1 system, bicyclohe 2, 2, 1 system having a CH2X here, okay. Now what he does, what he wants to do is you cleave this bond, okay. When you cleave this bond that should get converted into a nicely cyclopentadiene, cyclopentane and these 2 side chains, okay. You can see 1 and 2, 2 side chains can be easily introduced, okay. So this becomes 1 side chain and this becomes side chain 2. So what is important is you have to cleave this bond, okay. The cleaving that bond is very, very important and how you cleave accordingly you can fix the stereo centers of these 3 contiguous currents. That was his plan, okay. The whole thing involved 3 key reactions, 1, dill-salt reaction, 2, bare-villager oxidation and third one, aido lactation. These are the 3 key reactions which he used to synthesize prostaglandins, okay. Let us see his retrosynthesis. So when you look at this molecule here, for example, we start with PGF 2 alpha, okay. So 2 alpha that means 2 double bonds are there, isn't it? And alpha, this hydroxyl is alpha, okay, PGF. His first retrosynthesis is to disconnect the cis-double bond, okay. If you disconnect the cis-double bond, then what you should get is, so this is the vitic reagent. The other portion, this should be aldehyde, isn't it? This should be aldehyde. That aldehyde, the hydroxyl group immediately will cyclize to form lactol, okay or in other words, if you have this lactol, then one can do a vitic reaction to get this compound. But before that, you have to protect the hydroxyl group. Here the hydroxyl group is protected as THP ether, that is tetrahydropyranyl ether, okay. Now the lactol can be obtained, the lactol can be obtained by dibol reduction of this lactone. And if you have this aldehyde, this double bond can be obtained from this aldehyde by Wadsworth Iman's modification, okay. So this is very easy and that also will give you trans double bond. A simple vitic will give cis double bond and this Wadsworth Iman's modification will give you the trans double bond, okay. The next step is how to get this bicyclic aldehyde, bicyclic aldehyde, okay. So the aldehyde of course can be used as protected form of the primary alcohol. Then this lactone, 5 umbered lactone. So whenever you want to synthesize a 5 umbered lactone, again one reaction we should come to your mind is iodolactonization, iodolactonization that means if you have a double bond and if you have a carboxylic acid, if you treat with iodine or potassium iodide in the presence of sodium bicarbonate, first iodonium ion will form here followed by attack of this carboxylate, it will open up the iodonium ion to give iodolacto. But iodine can be easily removed with tributyltin hydride, okay. So what you need is this double bond and then carboxylic acid. Depending on the ring size you can have this either CH2 or CH2M, okay. Normally 5 umbered and 6 umbered work very well. This how will you get it, okay. If you look at here the advantage is very creative. This carboxylic acid and this hydroxyl group, this carboxylic acid and hydroxyl group, if you connect it, if you connect it, then that will give you this 7 umbered lactone, okay. What you have done, this CH2COH you are connecting with this one, okay. Now how will you get this lactone? What are the reactions we know to get lactone? One of the simplest and straightforward reaction to get lactone is to get from carboxylic acid and alcohol. But that is what you want here, okay. But this lactone one can get it from Bayer-Williger oxidation. If you do a Bayer-Williger oxidation of this ketone that will give you this 6 umbered lactone, okay. And this as soon as you look at this molecule you can see a cyclohexene, okay. You can see a cyclohexene. So cyclohexene, again, next very important reaction which would come to our mind is Diels-Alder reaction. That in principle should give you the diene and the ketene equivalent. As you know ketene cannot be used in Diels-Alder reaction as dienophilic, ketene undergoes dimerization, they are unstable. So normally people use ketene equivalents, either nitro, ethylene or alpha-chloroaclone nitride. So this is called ketene equivalent, okay. One can use ketene equivalent to get the corresponding ketone, okay. Now let us see the synthesis. So for the synthesis, first we started from cyclopentadiene, okay. Cyclopentadiene, and when you treat this sodium hydride, you know it can generate anion. Then cyclopentadiene and anion is aromatic, isn't it? So you can easily generate anion. Now quench with methoxymethylchloride, that is CH3, methoxymethylchloride is CH3OCH2Cl. So this is the leaving group, okay. Then this will attack here and your chloride goes. Once the chloride goes, what you get is CH2OMe, okay. This is what you need. Here there is one small problem. The problem is, so this is cyclopentadiene, okay. Once you have cyclopentadiene, it can undergo various 1-5 hydrogen shift, okay. So if you are doing this reaction above 0 degrees, above 0 degrees, it can undergo 1-5 hydrogen shift to form these 2 dienes also, okay. So that means not only while making this compound, one should do the reaction below 0 degree and remove the solvent also below 0 degree. But also when you do the Diels-Aubreaction, you should do below 0 degrees. So those days as you know, the Diels-Aubreactions were done at high temperature, you know, either in seal tube or reflecting in benzene or reflecting in toluene and so on. So now if you have to use this diene, this particular diene without isomerizing to other 2 dienes via 1-5 hydrogen shift, then next step that is the Diels-Aubreaction should be done at 0 degrees or less. So what he did, he took this compound and then treated with alpha-chloroachloronitrile in the presence of copper fluoroborate. The copper fluoroborate helps to do this reaction at very low temperature. You can do it sub-zero and when you do that, you get the corresponding pi-cyclot-221 adduct, pi-cyclot-221 adduct. So now you got this chloroachloronitrile adduct as you know this is a synthetic equivalent of ketene. So next step is the hydrolysis of the acrylonitrile adduct to get corresponding ketone, okay. So as we have seen in the retrosynthesis, he could successfully make this bicycler-221 ketone, okay. The next step is to carry out Bayer-Wiehleger oxidation. So when you use MCPBA, okay, when you use MCPBA, there are two possibilities. One, it can epoxidize the double bond, two, it can undergo Bayer-Wiehleger oxidation. So between these two, Bayer-Wiehleger oxidation takes place because if you look at this double bond, the CH2OME is just above the double bond and that protects the double bond from attack by MCPBA, okay. So that is how when you do the Bayer-Wiehleger oxidation of this ketone with one equivalent of MCPBA, you get this bicycler-321 lactone, okay. Then obviously next step is open this up to get the corresponding hydroxycarboxylic acid. So that is very easily done by alkaline sodium hydroxide. So now you got the hydroxycarboxylic acid. The important feature of this reaction is these two CC bond and CO bond, they are cis to each other. They are cis to each other. So same thing you can maintain. Whereas if you look at this CH2OME, this is opposite to this CH2. So that way you can see this is beta whereas these two are alpha. So the stereochemistry also is correctly fixed though it is resuming but relatively if you see they are trans to each other, okay. So now you have hydroxycarboxylic acid. Then next step is the iodolactonization. So the iodolactonization can be done by treating with potassium iodide and sodium bicarbonate. Sodium bicarbonate remove the hydrogen of carboxylic acid, potassium iodide forms the iodonium ion. And the intramolecularly the carboxylate attacks and you get the corresponding iodolacton. So as I said that iodine is not required because you need only the lactone and not the iodine. So it can be easily removed by treating with tributyltrinohydride and AIBN. So before that one should protect the hydroxyl group as acetate. So you protect the hydroxyl group as acetate, then treat with tributyltrinohydride, you get the lactone. That lactone is called corease lactone, okay. This is one of the very important lactone in the total synthesis of prostaglandins, okay. Other than many people made this lactone by different method and some people used this lactone and then made prostaglandins but this corease lactone is well recognized in the total synthesis of prostaglandins. So once you have this now the next step is to elongate one of the side chains, okay. Next step is to elongate one of the side chains. So you have CH2OME, okay. If you use BBR3, BBR3 is known to cleave OC bond, okay. So it will cleave the metoxy, okay. You get the corresponding alcohol CH2OME because it is SN2 displacement, it is easy to cleave the methyl. So once you have the CH2OH, next step is to oxidize. So you can oxidize the CH2OH using chromium trioxide, pyridine complex. So you get the corresponding aldehyde. Now you can do the Wurzwerthiemann's Wittig reaction to get the trans double bond, okay. Now the trans double bond is fixed, okay. What is the next step? You have to reduce the ketone, okay. You have to reduce the ketone in the presence of acetate and lacto. So zinc borohydrate successfully can reduce the ketone, particularly alpha-beta-unsaturated ketone. So initially, initially you could reduce it to the corresponding allylic alcohol but what he got was mixture. After he developed, you know, other reagents, for example, CbS reagent, Coray, Bakshi, Sibata reagent to get exclusively only one niacinamide, okay. A lot of chemistry was developed using the prostaglandin total synthesis project. Then the acetate to be removed. Why acetate should be removed? The next step should be to attach the second side chain, isn't it? So second side chain means you have to reduce lactone to lactol. Then do the Wittig. But the problem is if you reduce with a di-ball, if you reduce with a di-ball, not only the lactone will be reduced but acetate also will be hydrolyzed, okay. Not only the lactone will be reduced, acetate also will be hydrolyzed. So you have to hydrolyze the acetate to get the diol and re-protect that. Now you protect the hydroxyl group, both hydroxyl groups as THP ether. Though THP ether is not a good protecting group, for the simple reason that the THP will give additional chiral center. See, THP is nothing but, okay, you can see it will create additional chiral center, okay. But if you are using for one or two steps, it is okay. But for a longer sequence, do not use THP, okay. So now after protecting this as THP, next is to reduce the lactone to lactol, okay. So that was easy to get the lactol. Once you have the lactol, the lactol is nothing but your aldehyde and OH, isn't it? Lactol is nothing but aldehyde and the hydroxyl group. Then you can do the Wittig, okay. The simple standard Wittig when you do on this lactol, you get the corresponding cis alkyl. So now you see, we have already introduced the trans double bond, now we have introduced successfully the cis double bond. So what is left in the synthesis of PG E2 alpha and PG F2 alpha, 2 means 2 double bonds are already there, okay. E1 has ketone, F both are hydroxyl group. If you remove the THP, if you remove the THP, then you get directly PG F2 alpha. So that was done with an acetic acid water, you take this compound, take with acetic acid water at about little bit higher ambient temperature, you get PG F2 alpha, okay. Now you want one of the hydroxyl group to be oxidized, one of the hydroxyl group to be oxidized to ketone, but it will be very difficult to oxidize one of them, okay. So how do you do? Before you remove the THP, not this one, this one to be oxidized. Before you remove the THP, oxidize the hydroxyl to ketone, okay. Before you remove the THP, oxidize the hydroxyl group to ketone. Then in the second step, you remove both THP. So that will give you PG E2 alpha, okay. So from the same intermediate, that is corease lactone, one can easily make PG E2 alpha PG F2 alpha. This is not a chiral one, this is a tracemic one and for the chiral one, what he has done is here, a chiral dielsol reaction, asymmetric dielsol reaction he has done followed by even the reduction of ketone to alcohol, he has used CBS reduction to get only one isomer. I will discuss that how he has done asymmetric version of prostaglandine in the next class and I also will talk about Gilbert Stark's total synthesis of prostaglandins using a very interesting radical cyclization as the key step, okay. Thank you.