 Good morning everyone and welcome back to the course on Classics in Total Synthesis part 1. So we have been discussing about you know 3-mumber ring, 4-mumber ring, 5-mumber ring, 6-mumber ring based natural products and today we will talk about very interesting alkaloids. These alkaloids are very very interesting natural products. So we will be discussing many synthesis of alkaloids and the first synthesis of alkaloid today we are going to discuss about perhydroestrionicotoxin, the name itself looks little longer but you can break it like this, perhydroestrionicotoxin, perhydroestrionicotoxin okay. So there are 3 natural products isolated from the same source, the first one is called estrionicotoxin okay. So here you can see there are 2 side chains, there are 2 side chains okay having an enine moiety okay, there are 2 side chains having an enine moiety but the core structure is a spirofused system okay, 1-6-mumber ring fused with a piperidine ring okay, a 6-mumber ring fused with a piperidine ring with 2 side chains having an enine moiety okay. The second natural product is the one where the triple bond is reduced to a double bond okay, the triple bond is reduced to the double bond while the double bond is fully reduced. So this is called octahydroestrionicotoxin okay. The third natural product is fully reduced natural product called perhydroestrionicotoxin, it is completely reduced, the side chain is completely reduced. So this was isolated from the venum of a columbian frog called tendrobatus estrionicus. This was isolated and reported way back in 1971 and 73 by Vidkop and coworkers okay and in fact they asked Professor E.J. Coray to see whether this can be synthesized by his laboratory okay and immediately E.J. Coray's group took upon the total synthesis of this molecule and then they reported the first total synthesis of perhydroestrionicotoxin and his synthesis involved 2 important reactions, one is pinnacle, pinnacle and rearrangement, other one is Barton reaction. So before we actually go into the details of the total synthesis of perhydroestrionicotoxin by E.J. Coray, we will briefly discuss about or briefly recall the pinnacle, pinnacle and rearrangement and Barton reaction I am sure all of you would have gone through these 2 reactions but it is a brief recall, it is very important to know before we actually proceed to look at the total synthesis of perhydroestrionicotoxin reported by E.J. Coray which involves these 2 key reactions. So what is pinnacle, pinnacle and rearrangement? So if you have a 1-2 diol okay if you have 1-2 diol and upon treatment with acid this undergoes a facile rearrangement to a ketone called pinnacle. So the whole process is called pinnacle to pinnacle because it goes from pinnacle to pinnacle. How does it happen? So when you have a diol one of the alcohol is protonated. So once it is protonated automatically it becomes a good leaving group. So once it goes then it generates a carbocation. Though this carbocation is stable because it is a tertiary carbocation it is stable but the presence of a hydroxyl group, presence of a hydroxyl group adjacent to that makes it less stable because the lone pair on the oxygen now can push one of the alkyl groups or aryl groups to migrate so that you will get this intermediate where now the positive charge which is formed here is stabilized by the oxygen okay. So this is more stable than the tertiary carbocation. So this is the driving force for the migration of an alkyl or aryl group from the adjacent carbon with respect to the carbocation form okay. So one can ask why this migration should take place when already the carbocation is tertiary and there is no ring strain but still the migration takes place because the stability of the carbocation by the lone pair on the oxygen okay that actually helps the migration of an alkyl or aryl group to facilitate this pinnacle and pinnacolon rearrangement okay. So there are many examples I will just give few examples at least 2 or 3. So when you have a diol like this, this upon pinnacle, pinnacolon rearrangement first protonation takes place and followed by the leaving of water to generate the carbocation. Now the lone pair pushes this C-C bond okay. So it is a symmetrical does not matter this C-C bond to migrate or in other words ring enlargement takes place because of the migration of the C-C bond ring enlargement takes place. So what you get is a spiro system okay, the spiro 4-5 system you get okay. Then one can also see if we have unsymmetrical diols which alcohol will be protonated first. So if you have an unsymmetrical alcohol obviously it will protonate the alcohol which will give more stable carbocation. So now if you look at this example there are 2 alcohols. So between these 2 alcohol, this alcohol will be protonated because that will give a carbocation which is more stable than the other carbocation. This carbocation will be stabilized by 2 phenyl groups okay. So that is why that will be more stable and then this bond will migrate so leading to the formation of 1 and this is the predominant or major products and not the other one okay. So the migrating group preference is important but at the same time first the formation of most stable carbocation is the real driving force for the migration of the next one okay. The second reaction which is used by EJ Coray in the synthesis of perhydro nicotoxin is Barton reaction. The last 2 decades or more one has witnessed large number of publications on CH activation and functionalization but Barton has reported this reaction long time ago where angular methyl group angular methyl group in steroids can be easily functionalized by this reaction. What is this reaction? So if you have an alcohol like this it is very important that alcohol should be axial okay axial. Now if you treat with NOCl and pyridine okay so that OH will become ONO okay OH will become ONO. Now this I am shining with light it pumps oxygen radical okay and NO also comes out. Now if you look at this carefully through a 6 membered transient state this oxygen radical can pick up this hydrogen. If that happens then you will get OH and the CH3 now will become CH2 radical okay. The CH2 radical immediately will combine with the NO radical which came out when this ONO bond got cleaved you got oxygen radical and NO radical. So now what will happen the CH2 radical will combine with the NO radical to form CH2NO. The CH2NO immediately tautomerize the CH2NO immediately tautomerize to give the corresponding oxide okay. So what you have seen now in the whole process the angular methyl group which is really very difficult to functionalize has been functionalized by this reaction. So this reaction was reported by Barton so normally this is called Barton's reaction. Once you have this oxide one can hydrolyze the oxide to get the aldehyde or if you have an oxide you know another famous rearrangement can be considered. So Wegman rearrangement can happen so that way the CH3 will be converted into amine okay amide followed by amine. So many things can be done having the oxide moiety at angular methyl group. So these are the two reactions which you should remember when we talk about totals synthesis of perhydro histrionicotoxin reported by E.J. Corre. So this is the structure of perhydro histrionicotoxin then the first step the first retrosynthetic step was to remove the 5 carbon units okay. But before I actually talk about retrosynthesis if I tell that this compound this natural product was made from cyclopentenol can you believe this compound was synthesized from cyclopentenol. Cyclopentenol is the commercially available starting material and that was the starting material for the synthesis of perhydro histrionicotoxin. It may be difficult to believe for the simple reason that if you look at the natural product there is no 5-mombard ring is there any 5-mombard ring? No you have 2 6-mombard rings one is normal cyclohexane other one is 5-mombard ring derivative okay they are fused in a spiro fashion and I am climbing that the starting material is cyclopentenol. So that is the power of retrosynthesis if you logically think and logically write a proper retrosynthesis that can lead to a very very simple commercially available inexpensive starting material and also the whole reaction sequence will be very simple and straightforward okay. So now the first retrosynthesis was the addition of this 5 carbon unit addition of this 5 carbon unit to this imine okay. If you have an imine okay if you have an imine the equatorial addition equatorial addition of this 5 carbon Grignard or lithium that will give you the natural product and afterwards you have to protect the remove the protecting group of course when you do this the hydroxyl should be protected. So that is the first 2 retrosynthetic steps. Now once you have this imine this imine can be obtained from the lacto okay this imine can be obtained from the lacto in 2 steps okay. So whenever you have a lactum or whenever you have an amide there are many reactions you can think of. One you can have a carboxylic acid and amine and intramolecular coupling reaction will give you the corresponding amide here it is cyclic so it is a lactum. Then one can also think about Beckman rearrangement is not it one can also think about Beckman rearrangement where you have an oxyde that oxyde can undergo Beckman rearrangement. Again the position of the nitrogen depends on the regiochemistry of the oxyde. So for example if you have this oxyde okay if you have this oxyde then this can possibly this can possibly give you this lactum how because if this is the leaving group then the bond which is opposite to that leaving group only migrates that will give you exactly the same lactum okay. So the key reaction so far if you look at is the Beckman rearrangement to get the 6 number ring okay 5 number to 6 number ring okay. Now normally how do you get oxyde normally how do you get oxyde all of us will immediately think okay oxyde it is very easy you can start from the corresponding ketone is not it you can start from the corresponding ketone if you have the ketone then treat with hydroxyl amine you will get the corresponding oxyde that is how normal people will think but as I said this synthesis was reported by the Nobel laureate E. J. Corey so he has some special disconnection for this particular step so instead what he thought was this oxyde should be prepared or can be prepared from this alcohol and there is no ketone okay this is where you should recall the reaction which I mentioned that is Barton reaction. So if you have this alcohol and then treat with NOCl pyridine and the photochemical condition the NO group will be transferred to this CH2 through the 6 number transition state then that NO will become oxyde and that oxyde you can treat with acid to get the corresponding Beckman rearrangement product. So this is a very very important reaction and people normally would not have thought about using Barton reaction to get the oxyde okay that is a very clever thinking and because of that because of that the whole synthesis become much much simpler okay because you do not have to introduce a ketone just CH2 is there just you do this Barton reaction. Now the next step if you look at you need to introduce a hydroxyl group okay this is a phymobacteric no problem but here you need to introduce a hydroxyl group and as well as this 4 carbon unit. So what is the relationship between these 2 they are 1, 2 and they are trans to each other 1, 2 and trans to each other how you can introduce this okay how you can introduce this see if you are thinking about alcohol you can start from the ketone reduce it we do not know whether it will be selective in that case your butyl group is in axial position the hydroxyl is in axial position both normally you do not expect is not it the normally all the groups will try to go to equatorial position here both butyl as well as hydroxyl groups are sterically hindered axial position. So another very interesting reaction is a query used to fix this 2 stereo centers when you do a hydroboration when you do a hydroboration so what you do see for example if you take a phenyl cyclohexene okay. Now if you do hydroboration and oxidation if you do hydroboration oxidation hydrogen and boron hydrogen and boron will come from the same side because it is a cis addition okay hydrogen and boron will come from the same side that means the boron which will essentially be converted into hydroxyl will be opposite to that of phenyl group correct will be opposite to that of phenyl group basically what you get is the alkyl or aryl group and the hydroxyl group trans to each other and when the hydroboration takes place the hydrogen and boron will come from the least hindered side. So that means if you look at this the hydrogen and boron will come from the least hindered equatorial side soda equatorial side so that your alkyl group will go to axial so the precursor is nothing but this alkyl if you take this alkyl do a hydroboration oxidation you will get this compound okay then how do you get this compound very simple if you have this ketone then add butyl lithium followed by dehydration you will get this is not it butyl lithium and dehydration and this compound and I talked about pyrnacol pyrnacol rearrangement I told you this is a spiro ketone can be obtained from this pyrnacol okay can be obtained from this pyrnacol now this pyrnacol is obtained from cyclopentanone. So this is what I said when you can think about proper retro synthesis and use some clever disconnection and use some mice reactions one can get a very simple starting material and the whole strategy will be a classical one this is one of the classical total synthesis reported in the literature by E.J. Korn okay now let us see whether the retro synthesis what he had proposed he could easily follow it in the lab to complete the total synthesis okay those who are interested they can go through these two references first he started with cyclopentanone then pyrnacol coupling got the pyrnacol this upon treatment with acrosulfuric acid you get the corresponding the pyrnacol pyrnacol rearrangement. So this is quite easy and straight forward and once you have that add n butyl lithium so n butyl lithium adds to the ketone and then followed by dehydration done by thionylchloride pridine to get the corresponding alkene okay. So now that you have the alkene next step is the hydroporation oxidation yes hydroporation oxidation with hydrogen peroxide and sodium hydroxide gave the corresponding secondary alcohol and the secondary alcohol as well as the butyl group they are anti to each other okay. Now this can be written like this this can be written like this it is a relative stereochemistry it is not asymmetric synthesis this is relative stereochemistry and that is why I have written this solid bond okay. Now once you have this you can clearly see you have introduced the spiro system and the butyl group and the hydroxyl group with correct stereochemistry spiro system butyl group hydroxyl group with correct stereochemistry. Now you have to use this handle the hydroxyl handle to introduce the oxide okay you have to introduce the oxide as planned. So for that what one has to do Barton reaction. So he took this alcohol and then treated with NOCl pridine then shine light. So first it forms the ONO so under photochemical condition the ONO bond breaks and then it gives oxygen radical and NO radical. Now the oxygen radical picks up the hydrogen radical from here through a 6-membered transient state okay through a 6-membered cyclic transient state okay. It picks up the hydrogen and that gives you the corresponding radical in the 5-membered tree okay. Once the hydrogen is picked up you get the corresponding radical here. Then the NO radical which went out will combine with this cyclopentyl radical to form the corresponding NO okay. It is like cotton paste if you look at this chemistry it is very nice interesting chemistry. You attach NO cut it then attach to other side cotton paste chemistry. Then once you have NO this NO immediately tautomerizes to corresponding oxide immediately tautomerizes to the corresponding oxide. So you have the oxide the next step is carry out the Beckmann rearrangement okay the Beckmann rearrangement is very simple so one can use acid. So what they have done is they have treated with para toluene sulfonyl chloride okay. So selectively they can tocellate the oxide OH then followed by migration of this bond gives you the lactam. You have the lactam the next two steps first treatment with LAH what will happen if you treat with LAH what will happen when you treat a lactam with LAH it will become the corresponding amine. The carbonyl group will be completely removed the lactam will become amine okay. Then you have the free hydroxyl the free hydroxyl will be protected by the TBDMS chloride. So what you get is the corresponding OTBDMS or one can write OTBS also a TBDMS can be written as TBS as well. So now what is left? So you need to introduce the 5 carbon unit at this carbon. You need to introduce a 5 carbon unit at this carbon and that 5 carbon unit also 5 carbon unit also has to come from the equatorial side. So for that as per the original plan as per the original plan you need to introduce a double bond okay you need to introduce a double bond you have NH and you need to introduce a double bond. How will you do? Yes you can brominate now if you treat with base potassium amylate N amylate potassium N amylate it undergoes elimination of HBR okay it undergoes elimination of HBR to introduce the double bond okay. Now once you have the amine next you have to add the 5 carbon unit you have to add the 5 carbon unit. What you have to do? The 5 carbon unit is corresponding N amyl lithium amyl is 5 carbon okay. So you add the 5 carbon unit so once you do that then that 5 carbon unit also you can see it will add from the less hindered equatorial side okay the 5 carbon unit will add from the less hindered equatorial side. After that the Tbuff is a fluoride source is used to remove the TBS group. Addition of N amyl lithium followed by removal of TBS group you get the natural product that is per hydro is trionicotoxin okay. So as I said this is one of the classical synthesis of an alkaloid where two very simple reactions but cleverly utilized one is pinocchiole pinocchiole rearrangement the other one is Barton reaction. The Barton reaction the use of Barton reaction is really ultimate thinking otherwise normally people think that the oxide can be made from ketoglue but here we cleverly use the Barton reaction from CH2 to introduce the oxide that once you have the oxide that paved the way for the ring expansion to get the lactam and that is how you could make the other 6-membratory okay. So overall in few steps E.J. Coray and his group could synthesize this per hydro histio nicotoxin starting from commercially available cyclopentyl okay. So thank you I will stop here.