 Yeah, good morning and welcome back to the NPTEL lecture series on Classics in Total Synthesis Part 1 and this is Pramayati Bombay. So we will continue our discussion on total synthesis of natural products today. In the last class, we talked about total synthesis of an alkaloid called perhydroestrionicotoxine by E.J. Kauray which involved you know two important key reactions, one bottom reaction and the second one is Beckman-Reyland, okay. So today we will talk about one of the most complex alkaloid synthesized in 20th century. So that molecule is called strychnine. I am sure many of you might have heard the name strychnine is it was you know one of the most complex molecules isolated in 1918 and this is a structure of strychnine and in fact is one of the first alkaloids to be first complex alkaloids to be isolated. It was isolated from sickness next swamika in 1918 and it was considered as one of the dangerous natural product because if you consume anything more than 50 milligram of this natural product it is fatal, okay. It is so dangerous, so bad and this poisonous material should be handled very, very carefully. In fact many novels and poetry those who are familiar in reading English novels know that there are many stories where strychnine was given to poison and kill people, okay. So it has been routinely used in many novels. From the complexity point of view if you look at this molecule what are the challenges one can expect from synthesis point of view. First of all how many rings are there, see 1, 1, 2, 3, 4, 5, 6, 7 there are 7 rings, there are 7 rings one can imagine you know I am talking about when the molecule was isolated in 1918. So making a molecule of this complexity with 7 ring is not a joke and more importantly even the structural elucidation took considerably long time. It was isolated in 1918 but the correct structure was proposed by Woodward, correct structure was proposed by Woodward after 3 decades, okay it took 30 years to propose the correct structure of strychnine but interestingly if you look at how Woodward proposed the correct structure of strychnine was based on using uv, you know one should know those days NMR was not there, X-ray was not there. So one has to depend on 2 important techniques, one is degradation, okay. Those days strychnine was available in large quantity, okay one could isolate strychnine in large quantity so you keep on doing degradation, okay until you reach a known compound, okay. That is how you know, okay when you do a degradation you can expect what reaction it can undergo and what are the products you got, based on that you can work back, work back and then assign the structure. So those days degradation played a very, very important role in assigning the structure nevertheless the final proof for structural assignment of any natural product comes only in the form of synthesis, okay it was those days and you can imagine there were 400 research papers in the 30 years, 400 research papers have been published just to talk about only the structural elicitation of strychnine, okay. So partial structural elicitation, 400 research papers I do not think any other natural product would have got that much attraction particularly about structural assignment, okay that much complex this molecule was. And finally whatever structure Woodward proposed in 1948 based on degradation and UV was confirmed later by X-ray then you can imagine the intuition, the knowledge of various organic reactions by Woodward really stood very tall, okay exactly he proposed the correct structure without even looking at other techniques. So that tells volumes about his knowledge of organic chemistry and when you talk about synthetic challenges as I mentioned already there are 7 rings, okay so making 7 rings those days is not a joke it was very, very complex problem and one has to deal with it, okay. And if you look at the number of stereocenters, if you look at the number of stereocenters there are 6 stereocenters, okay in this molecule and among the 6 stereocenters 5 stereocenters are in this ring, okay 5 stereocenters are attached to one ring and next task is all the 5 stereocenters of this ring are contiguous, okay. So this makes much more challenging because you have to introduce 5 stereocenters in one ring and these 5 stereocenters are contiguous, okay. And another interesting aspect of this molecule is this 7-numbered ring this is a unique 7-numbered ring which was not heard or not seen in other natural products the first time they have seen such unique 7-numbered ring in an alkaloid, okay. Then there is a spirocenter so is it possible to locate a spirocenter in this molecule you can make easily you can see the 7 rings but can you look at the spirocenter there is one spirocenter that is this there is one spirocenter. So these are some of the synthetic challenges one could foresee before starting working on synthesis of strychnine, okay. After proposing the correct structure of strychnine based on various degradation studies Woodward thought definitely this is the molecule one should make, okay. So the more the challenge better for synthetic chemists to work on that because they used to love challenge, okay. And that time he made a very very famous statement and that statement is if we cannot make it if we cannot make strychnine then we will take it, okay. You know if you take strychnine what will happen, okay. That is how you took this as a challenge and started working on strychnine and as you know during these days the retrosynthetic concept was not there. So that is why I am not going to talk about retrosynthesis of strychnine by Woodward. Nevertheless we will see how he could go ahead and then make strychnine in reasonably good quantity, okay. So before that as I said the structure of strychnine was arrived based on combination of spectral data and degradation studies. The degradation studies of strychnine when you degrade strychnine there are two important substructure, okay. One was isostrychnine. So isostrychnine and strychnine if you look at isostrychnine can be converted into strychnine. How if you migrate this double bond? If you migrate this double bond you become alpha beta unsaturated system followed by oxamical addition will give strychnine. So that was the first degraded product from strychnine. The second degraded product was called wheel and gum leech aldehyde. If you look at this you can see that this is a lactol, isn't it? This is a lactol. If you do a stabilized wittic reaction, if you do a stabilized wittic reaction, okay. This lactol means it is aldehyde okay and alcohol. The aldehyde will undergo this stabilized wittic reaction to get alpha beta unsaturated ester then it can form an amide and then it can undergo oxamical addition to give strychnine. So in one step one can convert wheel and gum leech aldehyde to strychnine, okay. So if you look at subsequent total synthesis of strychnine, okay subsequent total synthesis of strychnine, most of the synthetic groups either used this key intermediate or isostrychnine as a key intermediate. So they come up to this and from here it is no, okay. What are the key disconnections? I will not go into the complete retro synthesis. What are the key disconnections of some of the synthesis? There are many synthesis. I will talk about only four total synthesis of strychnine and today I will talk about Woodward's total synthesis. So Woodward's idea was first to construct the spiro system and the spiro system how we construct is first you make this imenium ion okay, this imenium ion then use the lone pair on the nitrogen of indole ring that will come and then neutralize the positive charge on the imenium. So that is how you construct the spiro as well as the searing, okay. Now Obermann, Obermann almost followed similar trend that is the imenium ion but what he did was the spiro system was constructed by a reaction called managed reaction, okay. So this enol, so now this will come and this will attack. So the managed reaction was the key reaction in Obermann's total synthesis of strychnine and Magnus. Magnus also almost followed similar method of Woodward's, okay. The form imenium and the indole, indole double bond attacks and neutralizes the positive charge on the nitrogen. One synthesis which was completely different than these three approaches and was well received was Vireche Raoult's. So Vireche Raoult constructed this particular ring, okay, this is a six-numbered ring. So always when you see a six-numbered ring what one reaction we should come to everybody's mind is Diels-Alder reaction, is not it? Normally when you talk about six-numbered ring, two reactions will come to your mind. One is Rabin's annihilation sequence, the other one is Diels-Alder reaction. So he cleverly used Diels-Alder reaction as the key reaction to construct this CD, this ring, okay, okay. So these are some of the key disconnections used by synthetic chemists across the globe for the synthesis of strychnine. Now let us, as I mentioned, let us discuss total synthesis of strychnine by RB Woodward today. So he started with fissure indole synthesis, first he started with phenylhydrazine and treated with this ketone. This two upon treatment with polyposphoric acid form this indole, okay. So now what he has to do is he has to functionalize the carbon number 3, okay, he has to functionalize carbon number 3. Now what he did? He used a managed reaction, okay. So this is obtained from formaldehyde, dimethylamine, okay and HCl, okay, this is you know standard intermediate for managed reaction. So he introduced first the CH2NMe2 here, CH2NMe2, okay, afterwards he added methyl iodide. So that means already you have a tertiary amine, the tertiary amine was quaternized, okay, this tertiary amine was quaternized. So now once you have quaternized amine, it can be a good leaving group, it is a good leaving group. So if you treat with any nuclear file, the NME3 is a good leaving group, it can go. So what he did? He took this compound and treated with sodium cyanide, okay. When you use sodium cyanide, it underwent SN2 displacement to get the corresponding cyanide, okay. So basically in 3 steps he could introduce this functional group that is CH2CM. Next one has to reduce the cyanide, so that was easily done by treating with LAH that is lithium aluminium hydride. So he could get CH2CH2MH2 in good yield. The subsequent step was the cyclization step, okay, the key step to make the spiro system. The first step, first step was just make the amine, just make the amine. You have aldehyde and you have aramine, you make the amine. Then you treat with tosyl chloride. What will happen with tosyl chloride? When you treat with tosyl chloride, the amine nitrogen will be tosylated, amine nitrogen will be tosylated, okay. So that is what happened. You can see first amine is formed, then N tosylation took place, then followed by cyclization. All this happened when you treat with tosyl chloride to get the spiro system, okay. So now you can see you have made the spiro system and you introduced one chiral centre and in the process this became amine, okay. So once it becomes amine, you have to reduce the amine, isn't it? The amine was reduced to the corresponding amine by sodium borohydride and that resultant NH, the indole NH was acetylated to get the corresponding N acetyl group. So now you can see there are 3 chiral centres, 1, 2, 3, were fixed using this reaction, okay. All are relative, okay, all are relative and they are not absolute, okay. Next step, he did OS analysis. Guess what would have happened? When you do OS analysis of this, what would have happened? Are there double bonds? No, you have 2 aromatic rings, isn't it? You have 2 aromatic rings. More substituted double bonds will be cleaved under OS analysis condition, okay. So if you look at this aromatic ring and this aromatic ring, this is the double bond, okay. This is the double bond which is tetrasubstituted, isn't it? This is the double bond which is tetrasubstituted. Now if you cleave this particular double bond selectively, if you cleave this particular double bond selectively, what we will get is the corresponding dieth, okay. And the terminal, since you had OME, this will become ester, okay, so diester, okay. So this is a very, very important and clever idea of using aromatic ring, okay. To generate the bis ester, using selective OS analysis, that was one of the classical thinking, okay. Then you treat with HCl methanol, okay. HCl methanol, first it removes the acetate, first it removes the acetate. Then what will happen? You have the NH, isn't it? This NH will attack this ester because if you rotate this CC bond, if you rotate this CC bond, it will come here. So what will happen? You get the corresponding lactam, you get the corresponding lactam. That is one step, one thing. Second thing is the double bond which is outside, now isomerized, okay, to this ring. So two reactions happen, one the N acetyl was cleaved followed by cyclization, the second the double bond isomerization to get this period boundary, okay. Now if you look at strychnine, out of 7 rings, 4 rings are constructed, 1, 2, 3, 4, okay. 3 more rings to be constructed. The fifth ring, okay, the fifth ring here was constructed using a Claisen reaction. You have COTME and COTET, one can generate anion here and attack. So that is what happened. And for that, before trying to do this base catalyst or base mediated the Claisen reaction, what happened? This tosyl group, the presence of tosyl group created trouble, okay. The tosyl group got eliminated and then sometimes it created double bond here, double bond here, it gave complex mixture. So what he thought was, first let us remove the tosyl group. So he removed the tosyl group with HI and red phosphorus. But when you try to remove the tosyl group with HI and red phosphorus, the esters also will get hydrolyzed, esters also will get hydrolyzed. So that is how you get the dicharboxylic acid. Then no problem, the NH was acetylated with acetic anhydride pyridine to get the diacid, okay. Diacid can be easily converted to the bis ester by diosamethane. The diester by diosamethane treatment, you get the diester. Now if you do the Claisen reaction, sodium ethoxide will generate anion and then attack the ester, so you will get the corresponding beta keto ester. But when you try to do that, the epimerization was taking place at this curve, epimerization was taking, you can see what was the stereochemistry here and what is the stereochemistry here. So the after epimerization, the cyclization takes place, okay, leading to the formation of beta keto ester, okay. So this is what you wanted, but this C-C bond formation that is the epimerization happened and then you got this combo, okay. No problem. Next what one should do is you have to remove the keto, you have to remove the keto, you do not want the keto, there you want CH2, okay. So the keto ester as you know, the keto ester can exist in enol form, okay. So once you have this in enol form, you can convert that into corresponding enol tosylate. So if you treat with tosyl chloride pyridine, it forms a corresponding tosylate, okay. So you want to make this combo, okay. So now once the tosylate is there, one can think of addition elimination reaction with sodium benzene thiolate, okay. So this will give you the corresponding SBN. So this will undergo addition and followed by when it comes back, the tosyl group will go, okay. So now SBN, as you know, SBN and the double bond can be reduced under hydrogenalysis condition. First Rane-Nickel will remove SBN, then simple hydrogen and palladium carbon will reduce the alpha-beta unsaturated ester to give this combo, okay. So now if you look at this closely, you have made 1, 2, 3, 4, 5, 5 rings are made, okay. Now you have to connect this ester to this, okay. You have to connect this ester to the amine. So first he hydrolyzed the ester to carboxylic acid with potassium hydroxide methanol and that time this particular carbon also underwent epimerization because when you draw a confirmation, you draw a possible confirmation of this molecule, after epimerization this carboxylic acid occupies equatorial position, okay. That is why it undergoes epimerization when you treat with potassium hydroxide methanol to get alpha carboxylic acid, okay. So you have the alpha carboxylic acid and in fact this is the compound. This is one of the compounds prepared in large quantity from strychnine by degradation, okay. This is one of the compounds prepared in large quantity from strychnine by degradation, okay. And this is also called relay compound. What is relay compound? Okay. What is relay approach? Okay. So before I go further, I will talk about what is relay approach. So those days when they work on total synthesis of complex molecules, okay. So they have to start from some simple commercially available sorting materials and then they go further. After 10, 15 steps, you see you will have 1 milligram or 2 milligram, okay. But interestingly, they will reach a very, very important intermediate. At intermediate, they might have isolated from the natural product through degradation, okay. Suppose if X is the natural product, through degradation after few steps, they get Y, okay. Now the same group or some other group works on the total synthesis of the natural product X and they follow a certain pathway and then reach Y. Y is the compound which from the natural product, they could degrade and get it in large quantity. Now what they do, they have already established a method for making Y from simple starting material, okay. They already established the method for making Y. Now what they will do instead of going back and then starting from the commercially available starting material to make Y, what they do, they take Y which was available from natural product by degradation because it is available in large quantity. By synthesis, they might have made only small quantity. They will take this degraded product again from the degraded product, they try to convert. They try to achieve the synthesis of the target molecule X. So that is called relay approach, okay. So what Woodward has used here as relay approach? So sickening is available in plenty because from natural source. Now when you do potassium permanganate oxidation as you know this double bond gets cleaved, okay. This double bond gets cleaved. So one side it becomes ketone, other side it becomes carboxylic acid and not only that, not only that this becomes ketone, it also oxidizes the adjacent one, adjacent CH2 to get a keto, keto lactam, okay, it is a keto lactam. The next step what he did, so this is also called strychninonic acid. So then sodium amalgam, sodium amalgam selectively reduces this keto in the presence of two lactams. It will not touch the lactam. It will reduce only the ketone to get the corresponding alcohol. Now this alcohol upon treatment with base, what happens if you have hydrogen here, see this hydrogen is acidic. So base will pick up this proton and it undergoes elimination to give this alpha, beta and saturated lactam, okay. So this Woodward and his group made in large quantity. Okay, this is also called strychninolome, okay. Now if you treat with acetic anhydride, acetic anhydride pridine, so what you get? This OH will become OAC and base treatment also migrates a double bond, base treatment you can see base treatment migrates a double bond from alpha beta to beta gamma, okay. On this, now it treated with mercuric acetate, okay, mercuric acetate, so what happened? It introduced another double bond, okay, one more double bond introduced to get the six numbered lactam, six numbered lactam. What he has to do? He has to hydrolyze the acetate and then the resultant alcohol if you oxidize with chromium trioxide you get the corresponding keto lactam. This upon treatment with H2O2, H2O2 as you know it can undergo bare williger oxidation here, okay, bare williger oxidation followed by hydrolysis, this becomes the corresponding carboxylic acid, this becomes corresponding carboxylic acid and this undergoes decarboxylation and then you get corresponding NH. So this much happens in the first step that is hydrogen peroxide, bare williger oxidation and hydrolysis this becomes this becomes carboxylic acid and this becomes NH, that NH is acetylated to get this carboxylic acid. So now you know Woodward has come up to this stage starting from simple phenyl hydrogen he could come up to this stage. The same compound Woodward also got it by degradation from strychnine in large quantity. So now what he felt this compound made from strychnine in large quantity could be used further instead of starting from simple starting material. So and here also you can see that ester gets hydrolyzed to get the carboxylic acid. Then once you have that he treated with acetic anhydride. So basically he wants to decarboxylate this one, okay and introduce COCH3. So when he treated with acetic anhydride pyridine first the carboxylic acid was acetylated, carboxylic acid was acetylated COCOCH3. Since you are using pyridine so what happened the pyridine picks up this hydrogen, pyridine picks up this hydrogen then it attacks the acetyl carbonyl group. Basically when that happens you get a 4-mumbled ring, okay. The 4-mumbled ring if you see it can undergo elimination of carbon dioxide, 4-mumbled ring can undergo elimination of carbon dioxide and when it happens then you get the corresponding acetyl group. So this is what he wanted. He wanted the COCH3 that he did cleverly by treating with acetic anhydride and pyridine. Now again he has to treat with acetic anhydride pyridine to get the enacetate and aqueous HCl methanol, okay not only enacetate this also become enolacetate this COCH3 become enolacetate. Then if you hydrolyze with HCl methanol it becomes the COCH3 and also it becomes NH. So this becomes NH, selenium dioxide oxidation as you know if you have COCH3 selenium dioxide oxidation will give COCHO. So that is the first step it forms COCHO, now this NH intramolecularly will attack the CHO to form the aminol to form the aminol. Now the aminol again will get oxidized with selenium dioxide to get the ketone lactone. So now if you look at this carefully we have made 6 rings, okay 6 rings are done. So seventh ring that 7 ombre ring has to be done. So now you have lactone and ketone. So you can differentiate ketone from lactone. So you added sodium acetylide so that added to this ketone followed by reduction of the triple bond with LIH you got the double bond. Lithium aluminum hydride in addition to reducing the triple bond to double bond it also reduced the lactone to corresponding amin. This on treatment with HBr, okay allylic rearrangement took place and then you got isostrictive, okay. So HBr it forms allylic rearrangement to get the Br allylic bromide. This on treatment with H2SO for water you get the corresponding alcohol. What is this? This is nothing but isostrictive. And we all know isostrictive has been already converted into strychnine by base treatment. So he simply treated this isostrictive with potassium hydroxide methanol and it underwent first isomerization of the double bond to alpha, beta and such a system followed by oxa, Michael that is strictly. So this was considered as one of the most advanced and classical synthesis in 20th century. And this actually opened many, many synthetic methods and there are many total synthesis of strychnine reported since then. And we will discuss at least 3 more totals in the strychnine in the next class, okay. Thank you.