 So good morning and I will come back to NPTEL lecture series on classics in total synthesis. We have been discussing about total synthesis of various alkaloids. So we have been talking about total synthesis of alkaloids and in the last lecture we talked about the total synthesis of resurpin and today we will talk about synthesis of eohembine which has close similarity with resurpin. So if you look at the structure of eohembine and it has the same core structure of resurpin but it has acyl group here with trimethoxy aryl group attached to the carbonyl and you also have a OME group in resurpin. So this natural product was isolated way back in 1880 and it was a major alkaloid isolated from Corinander eohembi and this correct structure was reported by Wittkopp in 1943. You can see it took about 63 years to assign the correct structure of eohembi and it took another few years to complete the first total synthesis which was reported by Van Tommelen in 1958 and the key reactions involved in the total synthesis of eohembine was Dielsall reaction. So the Dielsall reaction was used to construct the E ring and followed by dihydroxylation, cleavage and cyclization which is almost like Woodward's total synthesis of resurpin. So now let us see how Van Tommelen approached the total synthesis of Tommelen and let us see his retrosynthetic approach. So he thought first he can cyclize this amide and this carbonyl group followed by cyclization at this carbonyl, he tries to form the C D ring using these 3 functional groups. If you look at this di aldehyde, this di aldehyde, the di aldehyde can be obtained by cleaving the cyclohexene. If you do a host analysis or if you do dihydroxylation followed by cleavage, you can get this di aldehyde and once you see this cyclohexene, wherever you see a cyclohexene, one reaction which should come to your mind immediately is Dielsall reaction. So that is what he proposed, a intermolecular Dielsall reaction between butadiene and benzo quinone, parabenzo quinone as the key step and that is the first step in the total synthesis reported by Van Tommelen. Now let us see how he did or how he accomplished the total synthesis of Van Tommelen, the combined starting from butadiene. So the first step as he proposed in the retro synthesis was the Dielsall reaction between butadiene and parabenzo quinone. So in benzene, when you reflex, you get this bicyclic compound. Now one can selectively reduce the e-n-dio, the double bond of the e-n-dio by treating with zinc and acetic acid without touching the isolated double bond. So this double bond is in conjugation with two carbonate group which can be easily reduced selectively by treating with zinc and acetic acid. Now this is symmetrical diketone, okay. So next is carrying out Darsen's reactions. You know Darsen's reaction is to homologate. So here the homologation was done with alpha-chloroethyl acetate. You generate an anion here that attacks the carbonyl group and this O- comes and expels the chloride to form the epoxide, okay that is the first step in Darsen's reaction. So now simple hydrolysis will give the carboxylic acid because it has to undergo decarboxylation followed by opening of the epoxide to get the corresponding aldehyde, okay. So the hydrolysis, the saponification was done with the aqueous sodium hydroxide and followed by opening and protection of the resultant aldehyde, okay. It is a homologation to get the aldehyde and the protection with diethylene glycol, okay. So this was done in the presence of diethylene glycol to get the corresponding aldehyde. So homologation was done from carbonyl group to CH2CHO. Now the aldehyde was easily oxidized at room temperature with silver oxide to get the corresponding carboxylic acid and that carboxylic acid was converted into corresponding acid chloride followed by treatment with tryptamine, okay. So it forms basically the corresponding amide, okay. Once you see this, now the next step is to cleave this double bond, okay. Once you cleave this double bond you get a dialdicate. So this was done in two-step protocol. First you carry out the dihydroxylation to get the diol and then you reduce this ketone, okay. You reduce this ketone under hydrogenation condition, okay. That is little bit tricky. So you can see that normally one would use sodium borohydride or lithium aluminium hydride but you also have the amide if you use a LAH that can reduce the amide but they have used simple hydrogenation condition to reduce the ketone to get the corresponding triol, okay. So you have 1, 2, 3 alcohols in the products, okay. But one can selectively cleave 1 to diol with sodium borohydride can cleave this 1 to diol, okay. And this structure also one can redraw this way, okay. This is for the sake of cyclization, okay. Once you cleave this, this aldehyde has to cyclize with the amide, okay. So you rotate, rotate this bond, okay. You rotate this bond, okay. Now this diol upon cleavage with sodium borohydride you get this aldehyde and that aldehyde immediately you can see it forms an aminol, okay. Immediately it forms an aminol with this particular aldehyde, okay. And this untreatment under the same condition, okay, it undergoes cyclization, okay. It undergoes cyclization to get these products. Here during sodium borohydride reaction so many reactions are happening. One it forms the dioldehyde, then this amide amine adds to this aldehyde to form this aminol. Then you can see the indolmite coming and attacking here, okay. That leads to the formation of the third ring, C ring, okay. But the problem is the stereochemistry at this ring junction, okay. Here the hydrogen in your amine should be alpha but what he got is beta. Meanwhile you can also see this hydroxyl group attacks this aldehyde to form a 5-ambered lactol, okay. In one step as I said many reactions are taking place. Now this lactol you treat with para toluene sulfonic acid and methanol so it forms the corresponding lactol methyl ether, lactol methyl ether. The lactol methyl ether if you reduce with LAH, the LAH is known to reduce the lactol, okay. LAH can reduce the lactum to corresponding tertiary amine. So you got the tertiary amine. Now you do the demethylation by treating with aqueous HCl to get the corresponding lactol. Then acetylate you get the corresponding acetate, okay. So basically one why they have to do more steps because if you have a lactol, okay. If you have a lactol, LAH will reduce the lactol also, okay. That is why the lactol should be protected, okay. Then your lactum should be reduced. Now the acetate you treat with or you carry out a pyrolysis that means you heat it at very high temperature and this can undergo elimination. As you know, acetates when you heat at very high temperature it can undergo cis elimination to introduce the double bond, okay. The pyrolysis gives the double bond. Now again you can cleave this double bond. This is a enol ether. So if you do, if you treat with osmotic oxide again you get the diol and the diol if you treat with sodium paravidate it gives one side aldehyde and other side OCHO, okay. What you need? Here you need OH, isn't it? And here you need ester, okay. So this can be easily done by treating with chromium trioxide methanol. So chromium trioxide methanol what happens? It oxidizes the aldehyde to corresponding carboxylic acid and since you are using methanol that gets esterified and also during the process this OCHO also gets hydrolyzed to get OH. So now if you look at this structure it has almost everything in place except this stereo center. This stereo center should be opposite. So that is one thing. Second if you want optically pure Yohembi, okay then you have to resolve, okay. So the resolution was done by treating with camphor sulfonic acid. So with camphor sulfonic acid first you got pseudo Yohembi. Why it is called pseudo Yohembi? As I said the stereo center is opposite at this carboxylic acid, okay. And it is known in the literature earlier when they did isolation of Yohembi. This stereo center can be inverted by treating with mercuric acetate, okay. So when you treat with mercuric acetate it forms this imenium ion, okay. When you treat this with mercuric acetate it forms this imenium acetate. The imenium acetate if you reduce it now then the hydrogen comes from alpha, okay. So that is what you did. So from pseudo Yohembi first you treat with mercuric acetate to get this imenium ion and which is in situ reduced with pellet platinum methanol to get the corresponding isomerization at this carbon to get the natural product Yohembi, okay. So that is how 1 time million completed the total synthesis of Yohembi and the key steps as I discussed was a Dielsall reaction followed by one pot cleavage of diol to dioldehyde and then cyclization on the top portion to get ABC ring and the southern hemisphere to form the lactob. There are two key reaction, one is Dielsall reaction, other one is sodium peroidate cleavage to form two rings, okay. The synthesis involved 20 longest linear steps and overall yield of 0.024% from Dielsall reaction attack, okay. Now we will move to the second synthesis. Here the synthesis is asymmetric one, earlier one which you, when we talked about total synthesis of 1 time million, it was resemic synthesis but they resolved, they resolved at pseudo Yohembi stage and then converted into a naturally occurring Yohembi. Here what we will do, we will discuss about Mamos total synthesis Yohembi, it is an asymmetric synthesis and he used an intramolecular Michael reaction, an intramolecular Michael reaction as the key reaction to form the D ring and that is where he introduced the chirality, okay. So let us see how he synthesized. Yohembi was known to be made from this 2, 3, C-Co Yohembi, that means this has been already converted into Yohembi by 2, 3 groups, Hilbert's torque and others, they have completed 2, 3, C-Co Yohembi to Yohembi. So his idea was to make the 2, 3, C-Co Yohembi, if you could do that, that completes the formal asymmetric synthesis of Yohembi, okay. So his idea is basically to make this bicyclic compound, okay, in optically active form. So for that he proposed this can be made from this bicyclic enone, okay, this bicyclic enone and this bicyclic enone can be made from this keto ester, okay, by intramolecular cyclization and this keto ester, if you look at this carefully, he wants to use an asymmetric Michael reaction, that means you generate an anion here and add to this alpha beta unsaturated ester, asymmetric Michael reaction, that is the key step, okay. Let us see how almost completed the asymmetric total synthesis of Yohembi. He started with protected benzylamine, okay, trifluoroacetyl benzylamine, then treated with sodium hydride and coged with this bromide. So this bromide can be obtained in one step from acrolein. So this is acrolein, okay, in one step one can make this bromide, okay, it is a protected compound. So basically you remove this hydrogen and coge with this bromide. So you get the corresponding N-alkylated compound, okay. Now you can deprotect or remove the acetyl using oxalic acid to generate the aldehyde. Once you have the aldehyde, then you do a stabilized Wittig reaction to get the corresponding trans alpha beta unsaturated ester. So you have introduced the Michael acceptor now, okay, you can see that Michael acceptor has been introduced. Now what we need is you have to introduce the Michael donor. So for that, you hydrolyze the trifluoroacetyl group with potassium carbonate ethanol, you get the secondary amine. Now you do a Michael addition, another Michael addition, okay, this Michael addition with methyl vinyl ketone, you get the corresponding methyl ketone. So now the stage is set for the key intramolecular asymmetric Michael reaction. So now let us see which chiral reagent he has used for the key intramolecular Michael reaction. So he took this compound and then treated with alpha phenyl ethyl amine, alpha phenyl ethylamine. So that is the chiral amine. So alpha phenyl ethylamine is, you see, this is the one, okay. So he took one isomer that is the plus isomer of alpha phenyl ethylamine and then made the corresponding enamine, okay. He took alpha phenyl ethylamine and then treated with this ketone. So you have a ketone and then treat with primary amine, it can form imine that imine can undergo you know isomerization to form enamine. So this is the enamine. Now this enamine can undergo an intramolecular Michael reaction, okay. So you can see this can be redrawn in this confirmation. I will leave it for 30 seconds so that you can see how this can be redrawn like this, okay. So once you know how to redraw this and it is pretty easy how the chirality is transferred. Chirality that is this, it is like a chiral auxiliary, okay. Alpha phenyl ethylamine is a chiral auxiliary that is used successfully for the asymmetric Michael reaction, okay. So that is how the two chiral centers, okay you can see the two chiral centers are fixed using this asymmetric Michael reaction, okay. Next you do not want the benzyl group because you have to remove the benzyl group so that the N-h can be attached to the indole, okay. So hydrogenalysis will remove the benzyl group to get the Piperidine ring, substituted Piperidine ring. Now you protect the Piperidine, Piperidine with Boc and hydride to get corresponding N-boc then you try to cyclize these two, you need a 6-perma ring, okay. The Claisen reaction you do this on this keto ester to form the corresponding 1-3 diketo, corresponding 1-3 diketo. Once you have this 1-3 diketone, now if you treat with para-tolybine sulfonic acid and methanol it can form enol ether, this we have already discussed when you have 1-3 diketone and treat with para-tolybine sulfonic acid and alcohol, methanol, ethanol, isopropanol, butanol, so it will form the corresponding enol ether, okay. Since this is unsymmetrical you can get both isomers, however what he found out was keeping this reaction for long time, keeping this reaction for long time converts isomer A to B. So basically if you run this for long time initially you get a mixture of A and B, but if you keep it for long time the A is converted into B which is what he wants. So it took the enol ether, now you can reduce the ketone, okay, you reduce the ketone with di-ball to get the corresponding allylic alcohol, okay. Now if you treat with acid, para-tolybine sulfonic acid, so protonation will take place here, okay, then this lone pair will push the double bond and in the process you will get corresponding enol, okay. That enol, now you have acidic proton here, okay, you have acidic proton here, so that can be removed using LDA and quench with Mander's reagent, Mander's reagent is cyanomethyl formates, okay. So you can easily introduce CO-TME group at alpha composition, okay. Then you reduce the double bond with under standard hydrogenalysis condition, so you get the corresponding bicyclic compound, okay. Now what you need? You need to selectively reduce the ketone to alcohol, then you have to remove the bar group. So before removing the bar group, you have to protect the hydroxyl group, you protect the hydroxyl group as TBS and while protecting the alcohol as TBS group, the bar also got removed, then you alkylate the piperidine with this indole ethyl bromide, okay, indole ethyl bromide and now you can see you have the complete structure of your hembine except that this bond is not formed, okay, except that this bond is not formed, okay. But as I said, 2,3-C-Co-Euhembein has been already converted into Euhembein, okay, there are at least couple of reports where 2,3-C-Co-Euhembein has been converted into Euhembein. So if we can remove the TBS group, okay, so that will give you 2,3-C-Co-Euhembein. So from here the known steps are treatment with mercury acetate in the presence of 5% acetic acid and followed by sodium borohydrate reduction. So basically as you know it forms an iminium ion, okay, then the cyclization takes place again followed by iminium ion and then reduction with sodium borohydrate, the hydrate is delivered from the alpha side to get Euhembein, okay. So that is how he could accomplish the asymmetric formal total synthesis of Euhembein, okay. The key reaction in the asymmetric total synthesis of MOMOS is the asymmetric intramolecular Michael reaction. So if you look at the structure of Euhembein, only the D-ring has 2 chiral centers, okay. So the chiral centers were introduced by asymmetric Michael, intramolecular Michael reaction. Overall the synthesis was done in 13 longest linear steps and with impressive overall yield that is 25.6% yield. If you compare this yield with one-termilanes, this is significantly high, I would say it is 100 times higher than what one-termilanes reported, okay. Both are interesting synthesis and both had its unique key reactions to synthesize Euhembein, okay. So thank you.