 So good morning and welcome back to the course on Classics in Total Synthesis Part 1. So in the last lecture we talked about synthetic utility of radical cyclization in total synthesis of trichonanes and where we talked about synthesis of 2 linear trichonanes namely Hirsutin and Capnallin, these 2 were reported by Dennis Courant and today we will continue our discussion on the applications of radical cyclization to 2 more trichonanes. One is an angular trichonane, other one is a propylene type, okay. So the first one, the angular trichonane which we will talk about today is called Silphi Perfolin. So this we already discussed by a different method but here what Courant has done is if you look at the linear synthesis where he has made a cyclopentene first then he tried to have a side chain, he tried to have a side chain with a radical and that will undergo 2-phi-exoradical cyclization. So for this angular trichonane again he wants to have only 1-phi-ampered ring that is cyclopentene, only the position of the 2 side chain differs. So if you look at the left hand side it remains same, okay. However, this side chain comes to here, okay. So now if you look at this carefully, the 5-exoradical cyclization onto this double bond followed by another 5-exoradical cyclization will give you the angular trichonane. So this was the key strategy which he wanted to use in the synthesis of Silphi Perfolin, okay. Let us see his retrosynthesis and his idea is to have this vinyl bromide, this vinyl bromide should in principle undergo 2 successive 5-exoradical cyclization whereas in the case of capnoline and Hirsutine which are linear angular trichonanes he has used 5-exoradical trich followed by 5-exoradical cyclization here it is 2 consecutive 5-exoradical cyclization. So this can be obtained from this particular cyclopentenone. So if you look at this carefully, first you can do 2 alkylation, okay. One with methyl, the other one with this allyl halide substituted allyl halide then followed by the Grignard addition of 4 carbon unit onto this ketone and hydrolysis will give this precursor, okay. So the synthesis actually started with 3 etoxycyclopentenone which is very easy to make from cyclopentene 1,3-dione by treatment with ethanol and HCl. One can prepare this in large quantity. Now you do the alkylation first with methyl iodide so you can introduce a methyl group at alpha carbon then again another alkylation with this substituted allyl bromide. So you could introduce 2 alkyl groups alpha to the carbonate, okay. Now what you need is you need to add a Grignard reagent. So the Grignard reagent is made from bromobutene. So the butenal magnesium bromide was added to this ketone in a 1,2 fashion to get this allylic tertiary alcohol. Now this upon protonation, okay, once you protonate you get this positive charge. Now the lone pair on this etoxy group will come and then pushes this water molecule out so you get the corresponding cyclopentene. So this is the key precursor for the radical cyclization. So it took this compound and one problem with this is this is the Michael accepter, okay. If you do a radical cyclization, first 5 exotrack radical cyclization will work however the second exoradical cyclization may not work. So that is why it is better to protect the carbonyl group. So the carbonyl group was protected as ethylene ketol by treating with ethylene glycol and acid. So then he carried out the key tandem radical cyclization. So the tandem radical cyclization worked well to give a mixture of two diastereomers. So this is the required isomer but he also got the unwanted isomer in the ratio 3 is to 1 in favour of the required isomer, okay. So now what is required is you have to remove the ketol as well as remove the ketone. So the ketol was removed using treatment with acid and the keto group was removed using Wolf-Kristener reduction. So that is how you could complete the totals in this is of sylphi portfolio that the epimer also was you know made where the ketol was removed and Wolf-Kristener reduction gave the corresponding epi sylphi portfolio. So in summary if you look at this synthesis, Dennis Courant who reported this synthesis in 1987 started with the cyclopentane 1,3-dione and used like in the case of Hirsutine as well as in Capitoline, he used the tandem 5-exotric radical cyclization to accomplish the total synthesis of sylphi portfolio. Overall it took about 7 linear steps with a yield of 6.4 percent. So now from linear to angular to we will go to propylene trichunates. How propylene trichunates can be synthesized using radical cyclization? So these are some of the propylene trichunates and the basic one is called modiphane and there are some oxygenated modiphanes, 13-acetoxy modiphane, modiphane epoxide and pulley carol. So this modiphane, first let us look at the modiphane. So you can call this as 3-3 propylene system, 3-carbon, 3-carbon, 3-carbon all 3-carbon atoms are joined together, so that is why this is called 3-3-3 propylene system and it was isolated from the Goldenrod plan and which was well known for its toxicity to cattle and sheep and from the synthetic point of view, particularly from Dennis Curran's radical cyclization strategy point of view, he wanted to extend the same you know tandem radical cyclization. So however in this synthesis he faced quite a bit of problems, one there are 3 contiguous quaternary center in modiphane, 1, 2 and 3. There are 3 contiguous quaternary centers, so it is not that easy to form 3 contiguous quaternary centers using tandem radical cyclization, this is a big challenge. Then you also have equal number of chiral centers, 1, 2, 3, 3 chiral centers are there and particularly the chiral center with a methyl group that is not that easy to fix. And the molecular architecture of modiphane is such a way that it does not allow the tandem radical cyclization to take place. So he has to do stepwise radical cyclization to achieve the total synthesis of modiphane. So let us see how he achieved the total synthesis of modiphane. So he thought about 4 different strategies okay, so let us not go into the details of how these 4 different strategies he wanted to use, he wanted to make first without this methyl group. So to check his strategy, so that is synthesis of death methyl modiphane. So he started with cyclohexane 1, 3-dione and then if you treat with NBS, you can introduce the bromine at this carbon, then you treat with methanol PTSA, so that will form the corresponding enol ether. So once you have this enol ether, then again you add this 4-bitenyl magnesium bromide. What we have done for the synthesis of sylphyperfoline followed by hydrolysis, you get the 3-homo allyl 2-bromocyclopentenol, 3-homo allyl 2-bromocyclopentenol. So now you reduce the enol with sodium boroyalate cerium chloride under Lucia reduction condition to get the corresponding allylic alcohol. So once you have this allylic alcohol, then treat with isobutric anhydride okay. So when you treat with isobutric anhydride, it forms the corresponding ester. The idea is once you have this ester, he wanted to carry out a glycine rearrangement, intramolecular glycine rearrangement. So for that he has to treat with LDA and TMS chloride which in situ generate this enol TMS followed by glycine rearrangement, island ester glycine rearrangement to give this carboxylic acid okay. So now if you see you have that dimethyl group, it is a quaternary carbon and there is another quaternary carbon. So two quaternary carbons are made. Now what you need to do is to introduce the third quaternary carbon followed by cyclization. So the carboxylic acid was esterified with diazomethane to get the corresponding ester. Then he tried to cyclize this okay using radical cyclization. You treat with tributyltrinhydride and AABN. So here when you want to do radical cyclization, you can use stoichiometric amount of tributyltrinhydride and catalytic amount of AABN or one can also use catalytic amount of tributyltrinhydride, catalytic amount of AABN. So what does it mean that is the tributyltrinhydride which is formed can be further reduced with sodium cyanoborohydride or sodium borohydride. So that is how one can use tributyltrinhydride in catalytic amount okay. So now this reaction works well as you can see here 3 is to 1 ratio of the cyclized product and then simple reduced product and here the stereochemistry what he got was 5 is to 1 in ratio where the required isomer alpha is the major isomer okay. So successfully he could carry out the first radical cyclization okay successfully he could carry out the first radical cyclization. Now what he has to do is he has to connect here these 2 carbon, connect these 2 carbon to form the third ring okay. So for that what he did he reduced the ester to alcohol. So once you have the alcohol you oxidize to aldehyde and then treatment with triphenyl phosphine and CBR4 will give you corresponding dibromolkene okay. So he wanted to use this dibromolkene as the radical precursor okay. So you can see you have 2 bromines okay and if it cyclizes here so that will give you the corresponding molecule. So when he did this reaction with 1.1 equivalent of tributyltin hydride and ABM. So between these 2 bromides we should know which one will form radical first obviously so this is more exposed or least ended is not it. So that will form the radical quickly. So that radical was formed and if this radical is formed then this cannot cyclize. So what it will happen it will take up the hydrogen and it will simply reduce and this also does not isomerize okay this does not isomerize. So he thought this will isomerize to the trans compound and the trans compound can undergo the 5xO trig radical cyclization but the isomerization did not take place with 1.1 equivalent of tributyltin hydride and ABM only the less hindered bromine was replaced okay that only was productively removed okay the exchange did not take place. So what he thought so instead of using 1.1 equivalent of tributyltin hydride he thought he will use catalytic amount of tributyltin hydride and excess sodium cyanoboride okay and here the idea is first anyhow only this bromide is reduced okay. So let it reduce afterwards if you add excess sodium cyanoboride which will in turn generate more tributyltin hydride then this bromine will be replaced by radical and that can undergo cyclization. So with this he tried this reaction and as expected first the least hindered bromide was reductively removed then followed by further heating under the same condition the trans radical was formed and which underwent further cyclization to give death methyl modiculate. So if you look at the modiculant structure you need 1 extra methyl group here okay. So based on this he wanted to extend the same strategy to the synthesis of modiculate. So for that he started with this type methyl tin derivative okay. This time with the tin derivative what he did if you treat with LDA if you treat with LDA you can generate anion here okay that anion will come like this and then it will attack this cordon that is a primary halide so alkylation will take place at the primary halide and the secondary halide will not be affected. So that will give you this. This idea was now if you can generate radical here okay that radical can undergo cyclization 5XO and when it comes back the trimethyl tin radical will come out that way you have the double bond still intact. So with this he treated with tributythene hydride in the presence of ABN so he got exclusively this product. How did he get? The tributythene hydride first generates the tributythene radical okay and it is the exchanges with bromine so you get the secondary radical okay. So when I talked about radical cyclization the stereochemical outcome of radical cyclization particularly for 5XO radical cyclization can be done by drawing a chair like conformation. So exactly if you see this we have drawn a chair like conformation. So the first step is the addition of the radical to the double bond 5XO trig so that should give you this combo okay. I will leave it for few seconds so that now you can understand this. So this will undergo 5XO radical cyclization to give this radical. Now this radical will come like this and then eliminate the tributythene tin radical. So you can write like this yeah okay. So that is how the cyclization takes place and the methyl group which is alpha it is because here the methyl group was put in equatorial position. So that is how you know the chair like conformation will help in establishing the stereochemistry of the methyl group which is formed here okay. So once you have this bicyclic system the next step is you have to convert this ester convert this ester into a dimethyl group and then CH2. So for that first you hydrolyze the ester to carboxylic acid and then trip with acid chloride you get the corresponding acid chloride. So now what he did was he tried a saccharolyte reaction where the allyl TMS okay the allyl TMS was treated with this acid chloride in the presence of titanium tetrachloride to add this bromolyl group directly to the carbonyl group. So now if you look at this so you have the vinyl bromide which on treatment with tributythene hydrate should generate the radical and that radical should add here to form the third phibomb buttering otherwise that will give that is core structure of modified. So when you did this reaction yes the reaction worked well and then you could get the propylene structure. So now what is required is convert this carbonyl group. The carbonyl group should be converted into dimethyl group okay. So if you look at the radical cyclization first the radical cyclization took place followed by the migration of the double bond internal double bond instead of external double bond you get internal double bond that is because you have a ketone after radical cyclization because normally radical cyclization you do it at high temperature. So the double bond exocyclic double bond migrated to alpha beta unsaturated ketone. So what he did to introduce the two methyl groups first he treated with methyl lithium to get the corresponding tertiary alcohol. This tertiary alcohols are known if you treat with you know titanium tetrachloride and dimethyl zinc okay it is known well known in the literature that tertiary alcohol can be converted into quaternary by treatment with titanium tetrachloride and dimethyl zinc. So basically you know the Lewis acid, Lewis acid will coordinate with OH to make it as a good leaving group and dimethyl zinc will deliver the methyl group. One can also do the same thing by treating with dimethyl zinc titanium chloride in the presence of TMS bromide okay. So effectively what you are doing is you are converting the tertiary alcohol into quaternary. So this is what was the expected product but what he got was another product that can be easily explained when this OH goes as a leaving group the methyl group can attack this carbon or the double bond can migrate the double bond can migrate the tertiary carbocation followed by methyl group attacking here. So he got a mixture of two methyl groups okay two gem dimethyl group here as well as here. So he thought okay this is not a good method so he wanted to introduce the gem dimethyl group first. So for that what he did you take ester and then convert into the tertiary alcohol. So now you see two gem dimethyl groups are introduced first itself at the right place. Now you have to homologate. So what he did he treated with TMS bromide to get the corresponding bromide. Now he wanted that vinyl group so he treated with titanium tetrachloride and then bromolil trimethyl silate. When he did this reaction he got again a mixture of two compounds and that again was obtained by rearrangement. So now when you use titanium tetrachloride this will go as a leaving group when it goes as a leaving group okay instead of this allyl TMS attacking this carbon either this bond of the five number ring or this bond can migrate. When the right hand side bond migrates you get this product when the left hand side migrates you get this bond what happens once this migrates then this allyl TMS this will attack the double bond and the double bond will come here to neutralize a positive charge. So ring expansion takes place okay. So instead of three five number ring fused you get one five number ring and one six number ring fused and you get a vinyl bromide also okay. So again you could not get what the precursor he was looking for to synthesize the modified okay. So what he went back to the ester hydrolyzed the ester and converted that into acid chloride. So now what he did he introduced a triple bond he added a triple bond to the acid chloride. So for that he followed Negishi's procedure. So peridium catalyst coupling reaction to introduce a triple bond. So his idea is later he wanted to convert this carbonyl into gem dimethyl group. Let us see how ester then TMS group was removed which is not required with fluoride source potassium fluoride TMS you can remove the TMS to get the alkynyl ketone okay. So once you have this alkynyl ketone now you have to introduce the radical precursor that is corresponding bromide or iodide vinyl bromide or vinyl iodide. So the treatment with TMS iodide so the TMS iodide it is again developed by you know Kishi. So it undergoes a 1, 4 addition to give transiodic compound okay the trans vinyl iodide okay. The trans vinyl iodide is required as you know for the radicalization to take place you need the trans iodic compound. So once you have this trans iodic compound then tributyten iodide reaction works very well to get the modifying that is the propylene skeleton. So once you have this propylene skeleton now what is required is converting this ketone into dimethyl group okay earlier he failed in converting this ketone into dimethyl group and when he started with dimethyl group then cyclization did not go okay. So how he did was when you have the ketone with the double bond that is cyclopentenome you add methyl grignard or methyl lithium to get the corresponding tertiary alcohol. Once you have this tertiary alcohol then next step is the oxidative transposition oxidative transposition of allylic tertiary alcohol. So this was reported by William Dobbin using PCC. So PCC that is pyridinium chlorochromate is well known for the oxidative transposition of allylic tertiary alcohol to corresponding enome. So here he used Jones reagent Jones reagent is nothing but chromium dioxide and sulfuric acid in acetone. So that oxidation gave the transpositioned enome once you have this enome then for the introduction of one more methyl group you normally use Gilman's reagent. So that way he could introduce now the gem dimethyl group okay. What is required is convert this into a methyl group. So treatment with vitic salt then you can get the corresponding exocyclic double bond but for modifying what you need is the endo double bond. So the exocyclic double bond on heating with para toluene sulfonic acid gave the modification. So he has successfully used not the tandem radical cyclization but two radical cyclization to form two five umber drinks he also started with one five umber drink and two more five umber drinks were added based on the radical cyclization. His synthesis started with cyclopentane 1,3-dion and he also used an intermediate cyclic vinyl stand in for the successful synthesis of modifying and overall this synthesis took little longer than the other two. So it took about 14 steps but the yield is quite good so 14 steps with 18% overall yield with complete control of stereochemistry is one of the key aspects of the total synthesis of modifying reported by correct. So it is a resemic synthesis it was one of the classical synthesis of typhonine okay with this I will stop and we will see more discussion on synthesis of typhonines in the next class. Thank you.