 So, good morning and welcome back to NPTEL course on Classics in Total Synthesis Part 1. And we have been talking about total synthesis of various alkaloids and we will continue our discussion on total synthesis of one more alkaloid, well-known alkaloid called reserpin. So reserpin and yohembine another alkaloid if you look at these two alkaloids you can find some commonalities between these two. They are pentacyclic compound and only reserpin has more substituents basically it has one more substituent here you can see that is a substituted benzoic acid attached to the hydroxyl group. And this was isolated from the dried root of Indian snake root actually earlier this reserpin was used for the treatment of snake bites, insanity etc. And you can look at this molecule how complex this molecule is. So the structural illustration took quite some time after it was isolated took almost 2 decades to arrive at the correct structure of reserpin and finally it was the correct structure was proposed in 1953 and it took 2 more years to propose the correct absolute configuration. A year later the first total synthesis of this complex alkaloid was reported by none other than the father of modern organic synthesis RB Woodward. Of course there are many synthesis of reserpin afterwards but in this lecture we will talk about how Woodward thought about and successfully completed the total synthesis of reserpin. So when you look at the molecule obviously it is quite complex and what are the challenges one can see when you look at this reserpin. And that time I am talking about 70 years ago when this molecule was isolated and then structure was proposed it was the most complex natural product isolated and then obviously when you have more challenging structure available many synthetic chemists were interested in developing new strategies for the synthesis of reserpin. Of course Woodward was the first one to complete the synthesis. If you have a closer look at this natural product you can see there are 5 contiguous chiral centers 1, 2, 3, 4, 5 there is one more here there are 6 chiral centers and in that 5 or in one ring this E ring has 5 chiral centers that makes this molecule quite complex. And more closer look at this natural product will suggest that there are 21 atoms 21 atoms put up in 5 rings okay. So these are the real challenges when you talk about total synthesis of reserpin. From synthetic point of view what Woodward thought was see this is indole ring isn't it? This is indole ring substituted indole and if you take this nitrogen also it is like triptamine okay. So he thought if we have to succeed in total synthesis of reserpin first thing is he has to focus on E ring okay E ring has 5 chiral centers okay. So that was the idea of Woodward as well as many people who followed after Woodward's total synthesis. So they wanted to make the E ring first then add D and then you add AB and C. So this is how most of the synthesis of reserpin were reported. And when you look at E and D ring okay D and E ring when you see the ring junction is cis the ring junction is cis and the E ring is 6 numbered E ring is 6 numbered D ring is 6 numbered. So when you have a 6 numbered ring which is attached to another 6 numbered ring and the ring junction is cis. So one reaction one can think of is dill-sol reaction as you know dill-sol reaction will normally give cyclohexenes and one can also properly plan and then think about having a cis ring junction okay. So that is what many people did and if you look at C D ring C D ring junction C D ring junction can be epimerized can be epimerized under acidic condition. If you look at C D ring junction the hydrogen is beta okay. So how these are all fixed in the total synthesis reserpin let us have a look. So let us start with Woodward's total synthesis. So though that time the retro synthesis was not known. So in this thought process what Woodward thought was he wanted to use a Bischler-Naprielski synthesis okay. So if you have this D ring then you can easily connect with the Indole using Bischler-Naprielski synthesis okay. And once you have this so this is a commercially available compound semi-toxicotophan and if you have E ring with 2 substituents then it should be possible to make this okay. That was the idea and this can be obtained from this aldehyde. So when you have this aldehyde and this amine it can form an imine okay. That imine if you reduce its sodium borohydride it will form secondary amine. That secondary amine will attack this ester directly to give this compound. So he simplified the target molecule that is reserpin to this highly substituted E ring of reserpin. Now if you look at this it is a PENTA substituted E ring with cis ring junction okay PENTA substituted E ring with cis ring junction okay. So now as I said he wanted to use Dealsol reaction as the key reaction. See when you talk about Dealsol reaction okay. Now you have to keep either diene or dienophile static okay. So then what you do if you are keeping diene static then the dienophile can approach the diene either from the top face or from the bottom face. So he started with the Dealsol reaction between this diene having a carboxylic acid and benzo quinone where this double bond will act as dienophile. So from the stereochemical outcome what you do? You keep the diene static okay diene you keep the diene static. Now the dienophile approaches the diene from the bottom face there are 2 faces is not it? It can approach either from the top face or bottom face okay of the diene. This is the top face this is the bottom face. So first let us see when the dienophile that is quinone approaches the diene from the bottom face. So this is the transient state that will give you this product okay I will leave it for few seconds just to see whether this is the correct structure okay. This is what you get if the dienophile approaches the diene from the bottom face of the diene okay. And this can be return okay this can be return like this. Now another interesting thing is if you want to carry out any reaction on this double bond or this carbonyl or this carbonyl or this double bond the reagents will attack only from the convex face okay this is the convex face okay this is the convex face from this place only where reagents will attack because that is the least hindered okay. So now what I have done if you look at I have drawn this in the 2 dimensional form okay is it clear? Now what I have done is what I have done is I have rotated this compound I have rotated this compound and return like this okay are they same? Are they same? Yes okay they are same. So this is one way to look at the Diels-Aub reaction where the dienophile approaches the diene from the bottom face. Now what we will do? Same thing we will do where the dienophile approaches from the top face. So now you see your diene is static but the dienophile approaches from the top face okay this is the top face okay that will give you this product that will give you this product okay. Again can it be drawn like this? Can it be drawn like this? Look at this. So you have this cyclohexene diene okay that is here then the other ring with carboxylic acid is here okay. Of course as you know the Diels-Aub reaction the major product is end of product so that is why you see the whole dienophile is just above the diene. Now can you rotate this? Can you rotate this? How do you rotate? You go through this plane go through this plane and rotate it by 180 degree okay. If you go through this plane and rotate it by 180 degree you will get this. So this is the first step of the total synthesis of reserpene reported by Woodward. Now he treated with sodium borbutate ethanol so there are two carbonyl groups okay and this carbonyl group will have hydrogen bonding with carboxylic acid. So obviously one can selectively reduce this carbonyl group and I already told you here only the convex face is more open so the reagent will come from the convex face that means the hydride will be delivered from alpha. When the hydride is delivered from alpha the alcohol will be beta okay. So what you get is the beta alcohol okay what you get is the beta alcohol. Now if you take this compound and then treat with MCPBA you take this compound and treat with MCPBA what will you get? There are two double bonds double bond 1 double bond 2 which double bond it will epoxidize question number 1 question number 2 whether the epoxide will be alpha or beta okay. To answer our questions one is alpha beta and saturated ketone so that needs alkaline hydrogen peroxide the other one that is second alkene is electron rich alkene so you need normal paracids. So between these two only two will be epoxidized when you treat with MCPBA that is answer for question number 1 answer for question number 2 whether the epoxide will be alpha or beta as I told you the convex face is the free face so that means the epoxide will come from the alpha side. So when you do that this is what you get okay. Now you have a hydroxyl group and you have a carboxylic acid group. If you treat with acetic anhydride sodium acetate what will happen? This will form an ester cyclic ester that is called lactone okay this carboxylic acid and this alcohol will couple to form a lactone okay. Next you have few more functional groups and the enone if you reduce under MPB reduction condition that is mirbeen pond of valley reduction condition this carbonyl group will be reduced again alpha phase is more free the hydride will come from the alpha phase so you get beta alcohol. But what you get is this compound okay so that is the beta alcohol and does not stop there the beta alcohol now it attacks this carbonyl of the lactone and opens this the reason is this will give you a 5 umber lactone this will give you a 5 umber lactone whereas if you look at this this is a 6 umber lactone. So 5 umber lactone is preferred over 6 umber lactone and does not stop there okay after opening the lactone the intermediate does not stop there what happens this O minus attacks the epoxide this O minus attacks the epoxide and you get the corresponding ether cyclic ether and alcohol corresponding cyclic ether. So there is a 5 umber cyclic ether and alcohol okay this is what you get when you do when you take the lactone this lactone and then treat it under MPB reduction condition and of course if you see this hydroxyl group it is beta beta to this carbonyl. So it is a good leaving group under this condition otherwise hydroxyl is not a good leaving group it undergoes elimination to give the correct corresponding alpha beta unsaturated lactone okay understood. So in one reaction it is MPB reduction condition how many reactions are taking place and what is more important is those days NMR was not there crystal stretcher was not there okay with just UV IR melting point they could assign correct stretcher for many such interesting transformations okay products arising out of many such interesting transformation. So now when you treat this with sodium methoxide and methanol so sodium methoxide will add in a 1, 4 fashion again the alpha phase is the free one so because of convex phase so the metoxy will come from the alpha side and while quenching the enolate also this hydrogen also will come from the alpha side okay. So in few steps you could get this complex tetracyclic stretcher starting from benzo quenol using Diels-All reaction as the key reaction okay later he also used a very simple method okay he improved the method which he already used and got the same intermediate in few steps how instead of carboxylic acid what he did was he used ester his idea is when you do the Diels-All reaction when you do the Diels-All reaction now instead of selectively reducing only this carbonyl why do not you reduce both carbonyls why do not you reduce both carbonyls so when you do that directly this ketone when it is reduced it will form the 5-ampered lactone okay. Now you have the allylic alcohol okay this allylic alcohol can be functionalized but the problem is how do you know you got only the 5-ampered lactone and not the 6-ampered lactone how do you know that you got only the 5-ampered lactone and not the 6-ampered lactone okay those days you know you can use IR and 5-ampered lactone where it will come and 6-ampered lactone where it will come. So 5-ampered lactone you get around a very strong peak around 1770 whereas for 6-ampered it is between 1730 and 1740 you got only this because the IR peak was around 1770 so that conform that you got only the 5-ampered lactone. So once you have this lactone now if you treat with bromine and methanol if you treat with bromine and methanol again the bromine will attack only from the alpha side once the bromine attacks from the alpha side that is bromonium ion the free hydroxyl will attack and it forms the cyclic ether okay. Now if you treat with sodium methanol you get the compound which we discussed in the previous slide if you look at the stereochemistry if you look at the stereochemistry you see the bromine is alpha and metoxy also is alpha bromine is alpha metoxy is alpha. So what does it mean so do you think we are talking about not SN2 reaction SN2 means it should be beta the metoxy should be beta but what is happening is here the sodium metoxide acts as a base as well as nucleophile as a base it eliminates HBr it eliminates HBr and then it gives this intermediate. Once you get this intermediate sodium metoxide now acts as a nucleophile okay then it undergoes one for addition and you get this product. So this is the same intermediate you saw in the earlier slide but that took more steps here it took only you can see 1, 2, 3, 4, 4 steps you could make this compound starting from benzoquium okay. Once you have this next you have one more double bond and treat with NBS water if you have a double bond and then treat with NBS water it will give bromohydrate it will give bromohydrate where will the bromine go where will the hydroxyl come obviously the bromine will be alpha that is the first step is in the bromonium ion formation. The bromonium ion will come from the alpha side then the hydroxyl group which opens the bromine ion will come from the beta side. So this is what you get okay. Now if you oxidize the secondary alcohol if you oxidize the secondary alcohol using chromium to oxide aqueous acetic acid you get the Keto okay. Here it did another very interesting reaction to get the precursor for the totals in this sorosirpene the key reaction is see you have a bromine and you have a carbonyl group what you did it treated with zinc acetic acid the zinc acetic acid is known to give one electron is not it. So first it opened the cyclic ether at the same time it also opened the lactone starting with donating electron to carbonyl group when this happens carefully you see the arrow which I have written okay then what you get is I will leave it for few seconds so that you can understand how does it happen okay. So you get essentially you get a cyclohexenone and if you look at this E ring if you look at the E ring all the 5 chiral centers though it is a relative steric chemistry it is not asymmetric synthesis it is a relative steric chemistry all the 5 chiral centers are fixed all the 5 chiral centers are fixed okay. Now what you did the carboxylic acid was esterified with the diazomethane and the free hydroxyl was acetylated in 2 steps you find this okay and for the intermediate which you want what you need is you need to cleave the double bond and you should get a aldehyde here and you should get an ester. So that is very simple you take this compound and then treat with osmium tetroxide when you treat with osmium tetroxide you get the diol. The diol now if you treat with pariodic acid followed by treatment with the diazomethane so this side it will become aldehyde and this whole thing will become carboxylic acid followed by treatment with so diazomethane it will become CO2Me. If you look at the retro synthesis or whatever synthesis Woodward has planned he planned using this is not it. So he wanted to treat that aldehyde with metoxy tryptophan and then reduce it with sodium borohydride to get the lactum. So you took this and then treated with the corresponding amine it formed the emine then reduction with sodium borohydride methanol reduce the emine and that also spontaneously cyclized with the ester to form the cyclic lactum okay. So C ring is ready D ring is ready already AB ring you started with the corresponding with oxy tryptophan okay. Now we have to make the C ring C ring is the last ring to be made and then C ring as one chiral center and based on the earlier reports the chiral center is when you want to generate normally you get the other one okay. So POCl3 sodium borohydride that these are the standard conditions used for Bischler and Aprialski reaction so it forms this eminium intermediate then addition elimination takes place to give the corresponding eminium. So C ring is formed in C2 you reduce with sodium borohydride you get the complete pentacyclic structure of wood reserving there are two things missing in this one the acetate should be replaced by OCO and then aryl group and the second thing is in reserpene this particular chiral center has beta color this particular chiral center has a beta color but this is the most stable conformation this is the most stable conformation. So how you can change the stereochemistry okay if you draw a 6-membered chair like conformation you can see this is how you can draw the 6-membered chair conformation for what you got in this reaction but as I said for reserpene this is the stable conformation where you can see this hydrogen is beta whereas here this hydrogen is alpha. So you have to change that how do you change one you can think about nitrogen inversion. So if you do nitrogen inversion okay if you do nitrogen inversion everything will change every conformation so this you know if it is equatorial metax is equatorial yesterday is equatorial acetate is equatorial when you flip it it will become axial and same thing here so the whole indole unit it will come axial. So this is nitrogen inversion which can be done by treating with acid okay. Now afterwards if you reduce it then that can come here okay that can go to this one. So this conformational equilibrium can be utilized okay so first you treat with acid so it flips then during this process you are locking the conformation you should lock the conformation then only it will not go back so these two so one is ester another one is acetate. So if we can lock this conformation through a lactone formation through a lactone formation then this is what you will get then followed by just open it you get the most stable conformation of the surface okay so what you did you whatever you got this is what you got when you did the total synthesis this is what you got where you have alpha hydrogen here but what you want is beta you took this compound and then treated with potassium hydroxide methanol potassium hydroxide methanol acetate will get hydrolyzed and ester also will get hydrolyzed you get a carboxylic acid and hydroxyl group. Now to freeze the conformation what you have to do you have to make a 5 umbered lactone between these two you have to make a 5 umbered lactone between these two so if you treat this with DCC as I said it undergoes ring flipping and you get the lactone and you can see this one also this side also whole thing underwent ring flipping. Now if you treat with acid now if you treat with acid the whole thing whole thing will undergo ephemerization and you will get this conformation. Now if you look at this hydrogen it is beta okay from alpha you brought it to beta next is just cleave this lactone cleave this lactone with sodium ethoxide methanol so when you do that you get this compound okay. Now if you look at these three substitutes methoxy hydroxyl and ester they are all in axial position so that is not stable that is not stable so what it will do it will undergo ring flipping when it undergoes ring flipping this is what you get because everything will be in equatorial position now okay and same way this also will undergo exactly opposite ring flipping now still you see this hydrogen is beta earlier it was beta axial here now beta equatorial okay so that is how he cleverly used acidic condition and the locking the conformation followed by acidification to get that then what you do you have the free hydroxyl group attach the trimethoxy benzyl chloride that is a side chain present in resulpin so that is how the total synthesis of resulpin was completed successfully by Abadut world okay this is really very very important total synthesis reported in 1960s one of the classical synthesis and it is taught in almost all textbooks okay thank you