 So good morning and welcome back to the NPTEL course on Classics in Total Synthesis. In the last class we talked about the total synthesis of strychnine by Woodward. So we will continue our discussion on total synthesis of strychnine by two more groups. Today we talk about total synthesis of strychnine by Vireesh Ravel's group from University of Chicago and the second one is from Larry Overman from University of California, okay. So when you talk about total synthesis of strychnine, each group use some key reactions to construct the center 6-membered ring. Ravel's group use two key reactions, one is an intramolecular reaction, the second one is an intramolecular Dielsall reaction to construct this 6-membered ring, okay. So the 6-membered ring they constructed using intramolecular Dielsall reaction. But the first retrosynthesis is they started from isostrychnine, okay. So from isostrychnine if you know as I discussed in the Woodward synthesis, if you treat with base, so the double bond, the double bond will migrate and at the same time oxamychyl will take place which will give you strict. So the precursor for strychnine is normally isostrychnine or they can also use wheel and tumlege aldehyde. So in the case of Ravel's total synthesis he has used isostrychnine root. So how he made isostrychnine? So this is where the first key reaction, key disconnection he used was the HEC reaction an intramolecular HEC reaction. So that he envisaged from this starting material as you can see here you have the vinyl iodide. So palladium insertion can take place and then you have a double bond. So the intramolecular palladium catalyzed HEC coupling followed by migration of the double bond will give you isostrychnine, okay. So that was the first important disconnection Ravel has used. Next he thought this can be made from this intermediate, okay. So if you see this what he has to do is remove the ester and then attach this whole side chain, okay. Now if you look at this particular intermediate, you can see a cyclohexene unit, okay. So whenever you see a cyclohexene unit, one reaction which should come immediately is the Diels-Alder reaction, okay. So the Diels-Alder reaction so can be easily thought of to construct this compound. So how do you do if you see this diene, okay, the diene and the dienophile. This is the dienophile and this is the diene, okay. So now an intramolecular Diels-Alder reaction should give the tetracyclic compound. And how do you make this compound? If you have an amine, okay, and an aldehyde. So then you can couple to form an enamine, okay. So what you need is the whole unit should be an aldehyde, okay. Then aldehyde can react with this amine to form emine and that will isomerize to enamine, okay. That enamine can undergo Diels-Alder reaction. This can be made from this immunocyclopropane, okay, the immunocyclopropane. So we know if you have a double bond, then it is called vinyl cyclopropane. So vinyl cyclopropanes are known to undergo vinyl cyclopropane rearrangement to give cyclopentene, okay, vinyl cyclopropanes are known to undergo vinyl cyclopropane rearrangement upon heating to form cyclopentene. He thought about a similar reaction instead of the double bond, he wanted to use an emine, okay. So that way he can get dihydropyrone, okay. How this can be obtained? This can be obtained from this cyanide and you can treat with a 1-2-dibromo compound base and a 1-2-dibromo compound. You can easily introduce the cyclopropane unit here. Then you know some functional group transformation. One can easily get the starting material. So now let us see how he has really carried out the total synthesis. First let us see the starting materials, how he prepared. As I said first the phase transfer mediated, base catalyzed, the cyclopropanation took place. Then you have a cyanide and that cyanide should be converted into you know aldehyde and then subsequently emine. So first reduce that with dibol, so cyanide is reduced to aldehyde. Now you treat with benzylamine, okay. It forms the corresponding skiff base, okay, benzylamine and then the emine. This upon treatment with TMS chloride and sodium iodide, okay. So the nitrogen lone pair will attack the TMS, nitrogen lone pair will attack the TMS and the TMS chloride and TMS iodide, sodium iodide it will form, TMS iodide in situ, okay. So that iodide will come out. So now you will have this intermediate. Now what will happen? The iodide which came out that will attack the cyclopropane and it will open the cyclopropane and the double bond will neutralize the positive charge on nitrogen which will lead to the formation of this intermediate. So now the lone pair on nitrogen will attack carbon bearing the iodide, thus forming the five membered ring and the loss of TMS will give the first starting material which is required. However if you have looked at retro synthesis, you do not need benzyl group here. What you need is ester group that NCO2Me, okay and also this nitro group should be reduced to corresponding amine. So both can be done easily and first if you treat this N benzyl with chloromethyl formate, easily one can replace benzyl group with ester, okay corresponding CO2Me. Now the nitro group can be reduced by transfer hydrogenation method. So the ammonium formate in the presence of palladium charcoal in situ it will generate hydrogen, carbon dioxide and ammonia, okay. So that will reduce the nitro group to amine to get your the first starting material, okay. So this is fragment A, we can keep it. Then you need an aldehyde to form an amine, isn't it? So for that you started with this iodol alcohol, this is normally prepared from the corresponding triple bond and then CH2OH by treatment with LIH and iodine, okay. So this is a standard reaction, one can easily prepare from like this compound with LIH and iodine, okay. Then you do radical reaction, so upon treatment with tributyltin hydrate, vinyl radical is formed that vinyl radical will add to alpha, beta and such a register here it is methylacrylate to get this compound, okay. Now as I said next step is the oxidation of the allylic alcohol to get aldehyde that is what is required to form the amine, okay. Now we prepared both the starting materials, the next step is to combine these two. So when you combine these two, so first it will form an amine skip base, then it will undergo because you have extra double bond, it will undergo isomerization to generate this diene. Once a diene is there and already dienophile is present, then if you heat it, it can undergo intramolecular Diels-Hall reaction. Before that this amino group is protected by treating with chloromethyl formate, then you heat it, you get the cyclohexane ring. So if you look at carefully not only you get cyclohexane ring but there are two rings, okay 1, 2, there are two rings which are formed using the Diels-Hall reaction, okay. So now you see from 7 membered ring because diene has 7 membered ring, now how many rings are formed? 4 rings. So you need 3 more rings, so how 3 more rings are constructed? You can see one can easily connect these two, okay if you easily connect these two that will form one more ring, that is the fifth ring, fifth ring. So for that you treat with base and heat it. First when you treat with base and heat it, this protecting group will go, okay. Now the n- which is formed intramolecularly will attack the ester and form the corresponding lactate, okay. So now the fourth ring also is formed, okay at the same time that n CO2Me also here, n CO2Me also is removed. Now for the HEC reaction you have to attach the side sign at nH, okay. So this is also you know very easy to make, okay. So treat with sodium hydride, you pick up this proton then quench with this bromide. You can see the HEC precursor is ready now. Once you have this then treat with palladium acetate and tetrabital ammonium chloride in the presence of inorganic bases like potassium carbonate it undergoes intramolecular HEC reaction to give this compound. So 1, 2, 3, 4, 5, 6, 6 rings are already formed, okay. So now only the seventh ring that is a seventh number ring should be formed. So that is why very easy. So remove the protecting group, remove the protecting group. So any fluoride source for example if you use tetrabital ammonium fluoride, so TBS group can be removed. Once you remove that that is isostichnene and we all know that if you have isostichnene, if you treat with base, so the double bond will migrate and oxamical will take place to give stichnene. So this is one of the you know shortest and clever synthesis of stichnene reported by Viresh Ravel in 1994. So now we will move to the third total synthesis, first was Woodward which we discussed. The second one which we discussed was Viresh Ravel, it is not in the chronological order but we whatever we discussed there are two, one by Woodward, second one by Viresh Ravel and third one is by Obermann. Why Obermann synthesis is important because this is the first asymmetric total synthesis of stichnene reported in the literature. And what are the key reactions he used? One key reaction which he has used was azar cope manage reaction, I will come to that what is the azar cope manage reaction later. And overall he took about 20 steps, 20 steps to complete the total synthesis and with an impressive 3% overall yield considering the complexity of this molecule 3% overall yield is a very very good yielding total synthesis. Let us see how he has done the retrosynthesis, the earlier two retrosynthesis we saw went through isostichnene, Woodward as well as Viresh Ravel, whereas Laurie Obermann's retrosynthesis went through wheel and game cum leach all day. So now if you have wheel and game cum leach all day, if you treat with melonic acid, if you treat with melonic acid and heat it, this all day is converted into stichnene, okay in one step you can do that. So either you have to go synthesize isostichnene or wheel and game cum leach all day to synthesize stichnene in one step. So how this wheel and game cum leach all day is made or thought to be made by Laurie Obermann. So he thought if you look at this, this is nothing but an aldehyde, is not it? Nothing but an aldehyde. So he thought ideal precursor should be an ester, more stable compared to aldehyde, it should be more stable. So the precursor should be an ester and an alcohol. Once you reduce ester, alcohol will immediately cyclize to form the lactate. So this is a logical precursor for wheel and game cum leach all day. So the next step if you see this, so if you have a ketone, okay when you have a ketone, one can easily introduce a ester next to that, okay, one can easily introduce a ester next to that. Now this NH2 here the protected one upon removing the protecting group you will have NH2 that can form imine with the ketone, okay. Then it can undergo isomerization to form enamine, okay. So basically in 2 steps one can make this compound from here, okay. So now look at this. So this compound can be redrawn like this. This compound can be redrawn like this. I just leave it for 30 seconds for you to visualize. So if you see this 5 umber ring, so that 5 umber ring is here and then you have the aromatic unit and then you have the 6 umber and then 6 umber both are behind, okay. Just try to understand, okay. This is very important when you go from here to further retro synthesis, okay, okay. Now let us see how he made this. This is where as I said he used a very, very simple but efficient key disconnection. If you look at this intermediate, okay, alpha and beta. The beta position with respect to ketone you have a nitrogen. So you can call it as beta amino ketones. In literature when you want to make beta amino ketones the best reaction you will consider is managed reaction. All beta amino ketones can be easily made by managed reaction. So the precursor for this should be this imenium ion. So now what will happen? You can make a negative charge that is enolate then intramolecularly attack and neutralize the positive charge on the nitrogen of imenium salt, okay. Now this if you look at can be obtained by an azacope rearrangement. If you look at this 1-5 diene, 1, 2, 3, 4, 5. You have 1-5 diene. When you have 1-5 diene if you heat it it should undergo coprene arrangement. But one of the carbon is nitrogen. So it is called azacope rearrangement. So if you heat it you will get this product. This is called azacope rearrangement, okay. This is managed, okay. Now this azacope rearrangement how will you get it? So you have a nitrogen and an epoxide that nitrogen can come and open the epoxide okay. You will get this alcohol. On the nitrogen if you treat with formaldehyde you get the corresponding imenium, okay. So the precursor for this is this. Later this can be easily obtained from this alpha beta unsaturated ketone. So once you have alpha beta unsaturated ketone there are 2 double bonds. One is electron deficient, other one is electron rich. The electron deficient double bond can be easily epoxylized by alkaline peroxide and afterwards you can convert the ketone into double bond, okay. So in 2 steps you can get this intermediate. Now this can be obtained from a carbonyl still a coupling reaction. You have a vinyl staining and then aiodoaniline. Now if you do a still a coupling in the presence of carbon monoxide then you can get the insertion of carbon monoxide here to get corresponding alpha beta unsaturated ketone, okay. So this is commercially available. This is available in cylinder. Now next we have to see how this can be made, okay. So take this compound and obviously whenever you have vinyl staining, okay, whenever you have vinyl staining it is easy to make from corresponding alpha beta unsaturated ketone. So you reduce the double bond and then generate the enolate, okay. Reduce the double bond, you alpha beta unsaturated ketone, reduce the double bond and generate the enolate and that can be converted into tributyltin derivative. And this alpha beta unsaturated ketone can be obtained from this cyclopentene derivative, okay. So this is normally done using a palladium catalyzed reaction and this can be obtained from this mono protected alcohol and that is normally obtained from cyclopentadiene, okay. These are all well known reactions, okay. Now let us see how the synthesis of this compound natural product worked, okay. And before that I should explain, so slightly about azacope managed reaction, okay. So this is 1,5 diene, okay and you have a nitrogen. So you can call it as azacope, okay. Then that should give you this compound, okay. Now the iminium ion, if this is oxygen, okay, then this can come and neutralize the positive charge. So this is managed reaction. So a combination of co-prearrangement, that is azacope rearrangement followed by managed reaction can give you a 5-ampered compound, okay. This is what exactly he has cleverly used in the total synthesis of technique. Now let us see first how he made a static material. So cyclopentadiene monoepoxides are well known and can easily prepare using parasitic acid. And you treat with palladium tetracase compound and acetic acid as the nucleophile. So one can easily open this epoxide, okay, to get the resimic compound, okay. One side you have alcohol, other side you have acetate. Now make the second alcohol also acetate and this can be easily resolved using an enzyme and that will give you. So only one isomer you can isolate. So this is optically active and you get in 99% E, okay. Now the other portion that is the iodoniline portion. So you can take 2-iodoniline, okay and then protect with this dimethylurea in the presence of formaldehyde, okay, in the presence of formaldehyde. So this is the formaldehyde coming from, it forms basically it forms iminium ion and then nitrogen, a nitrogen attacks and it are going to go 1-4 addition, okay. So this starting material is prepared, the other starting material is prepared. So then what you do, first you protect this alcohol as carbamate because allylic carbamate is required for polydehyde catalyst alkylation. So once you have this, then treat with this beta keto ester, okay, this beta keto ester. So what will happen, your pi allyl complex will form and the nucleophile will attack here. So that will give you this compound, okay. This chiral center is not fixed and no problem. One can reduce the ketone to get corresponding alcohol and at this point you know the ratio was 20 is to 1 and next is you have to introduce a double bond here. That means you should make this hydroxyl as a good living group, simple treatment with DCC, simple treatment with DCC gives the corresponding alpha, beta and saturated ester, okay. This alpha, beta and saturated ester when you reduce with DIBOL, it will do 2 things. One, the acetate also will be reductively cleaved, acetate also will be reductively cleaved and then ester will be reduced to get the primary alcohol. Now the primary alcohol, you protect it as TIPS eter, okay, primary alcohol is protected as TIPS eter and this allylic alcohol can be oxidized to get the corresponding ENO, okay. Any oxidation can do but they have done Jones oxidation. So once you have this alpha, beta and saturated ketone, then you do treat with L-selectride. So L-selectride has hydride in a 1, 4 fashion in this. Then that enolate is quenched with phenyl n triplet, okay. So this is, this will give you corresponding enol triplet, okay. This enol triplet can be exchanged with exomethyl di tin, okay. So if you couple with palladium and then O-triplet will be replaced with corresponding trimethyl tin compound, okay. So this is fragment A. So now you have fragment A and already we made fragment B. So combine these two using still-lay coupling in the presence of carbon monoxide, okay. So what it give is the expected alpha, beta and saturated ketone. Once you have this alpha, beta and saturated ketone, the next step is to introduce the epoxide. Of course, alpha, beta and saturated ketone, if you want to epoxylase in the presence of other double bond, then you have to treat with alkaline hydrogen peroxide. So you can use Titan B and then tertiary-butylhydroperoxide that gives the alpha epoxide predominantly, okay. So once you have this alpha epoxide, now the next step is to remove the TIPS group, okay. You have to remove the TIPS group, convert that into amine, okay. So the TIPS is removed, before removing TIPS, you have to convert the carbonyl into double bond, okay. That is normally one can do in one step with Wittig reagent. Then followed by removal of the TIPS, give, this should be OH, okay. First step, Wittig gives the double bond and the second step TIPS will remove the TIPS and you will get alcohol. Now the alcohol is mesylated and OH is converted into O mesylate, making it as a good leaving group. Then this amine attacks the mesylate and you introduce the NHCOCF3, okay. The next step is this NH, the lone pair should attack the epoxide and open the epoxide to get the corresponding alcohol. So that is done using, you know, simple base, this is done with sodium hydride and intramolecularly you can also draw this compound in a different way for better understanding. So now the N- will attack here and open the epoxide and you get the corresponding alcohol, okay. Now it is all set for the Asocope rearrangement, okay. What you should do? Only thing is this NH should be converted to N double bond, okay. So that is very easy if you treat with formaldehyde. So then you get the corresponding imenium and it does not stop there. So it undergoes the Asocope rearrangement and followed by managed reaction to give this, okay. So this is as planned, you could successfully use the Asocope and then the managed reaction, okay. Now we will redraw this. So once you redraw this, this is what you got, okay. So what is left is, now you have to introduce the ester here and remove this and then form the ename, okay. That is the next step. So you can treat this ketone with the LDA and then Mander's reagent, cyanomethyl you get the corresponding ester and the same time after you work up with the 5% HCl methanol then you can remove this and it forms NH2, the NH2 attacks the carbonyl and it forms the ename, okay. So now from this particular route already you can see 5 rings are formed, 5 rings are formed. One more ring to form to get the Vlan Kamlich Aldi, one more ring, okay. So what you should do? You should reduce this double bond, okay. So that is straight forward, you treat with zinc dust in methanol, you reduce the double bond but however the chiral center here is a mixture, okay. So that if you treat with sodium ethoxide and methanol then the alpha ester gets epimerized to get the required beta, okay. Now once you have the beta bond, next is to reduce that with dibol. So dibol ester is reduced to aldehyde as soon as the aldehyde is formed, alcohol attacks that gives you the corresponding lactol, isn't it? That gives you corresponding lactol. Then you treat with malonic acid and acetic anhydride, reflects it, it will directly give strychnine, okay. In one part you can take Vlan Kamlich Aldihyde and that will convert, that will be converted into strychnine in one step, okay. So if you look at the synthesis of strychnine by Overman, it involved a very, very simple and important key reaction that is acercope and manage in one part, acercope reaction followed by manage reaction to get the 5-ohm buttering with all the stereo centers in place, okay. That is the beauty of this particular synthesis. And second important thing is, this is the first asymmetric synthesis of strychnine reported in the literature, okay. So thank you with this I will stop here and then we will discuss a couple of more totals in this substrate in the next lecture. Thank you.