 Good morning and I will come back to our NPTEL lecture series on classics in total synthesis part 1. So we have been discussing total synthesis of trichunanes and in the last week, we talked about total synthesis of trichunanes using photochemical reaction, for example pattern of bookie reaction as a key reaction to form trichunanes reported by Viresha Ravels group. So we will continue our discussion on total synthesis of trichunanes where photochemical reactions have been used as the key reaction, okay. So today we will talk about maybe 4 or 5 total synthesis and the first one which we will discuss is the total synthesis of silphi, perfol, 16, 5 ohm and this synthesis, particular synthesis was reported by Demuth and Hinskan and interestingly what they have used is oxodipy-methane rearrangement induced by photochemical condition and overall if you look at the whole synthesis though the number of steps are little bit more, the strategy of using oxodipy-methane rearrangement to get the trichunane is a unique one and the starting material that is the diene which is required for the asymmetric Dielsall reaction was started from chiral you know which we will discuss. From the retrosynthetic point of view, this natural product can be obtained from this particular compound, you can see what one has to do is 2 methyl groups as well as the double bond should be introduced that can be done by first introducing a double bond here then you add a methyl group followed by quench with methyl iodide then introduce a double bond. So one can do it in 3 to 4 steps and this ketone can be obtained from another ketone and why this ketone was introduced you will know that is because for the photochemical oxodipy-methane rearrangement this ketone is required and before that this ketone should be reduced and protected and this particular ketone can be obtained from this tetracyclic compound. See now you can see a cyclopropane so this is formed through the oxodipy-methane rearrangement and reductive opening of this cyclopropane followed by quenching with methyl iodide you will get this intermediate okay and the key reaction here is the oxodipy-methane rearrangement of this tricyclic compound and when you look at this tricyclic compound as you know it can be obtained from 4 plus 2 cycloaddition reaction wherein this is the diene okay. This diene can be obtained from commercially available compound as well as one can make this in large quantity from Hages-Paris ketone and that is nothing but if you have this instead of alcohol if you have ketone that is Hages-Paris ketone and that can be obtained from 2 methyl cyclopentane 1,3-dione using L-polline as the catalyst one can do a Michael addition followed by Alder reaction which we normally call it as a Robinson annihilation sequence okay. So the starting material for this is 2 methyl cyclopentane 1,3-dione which is commercially available. So before we go into the details of the total synthesis reported by Marty Debout we will discuss the key reaction which is oxodipy-methane rearrangement. So what is that? So if you have a system like this that is you know you can call it as allyl ketone. Now under photochemical condition first this carbonyl will form a di radical same thing will happen with the double bond. So that can lead to the formation of this type of cyclopropane assume that this is CH2 okay. So this right di radical what will happen next it will open up. So how it will open up? So they will open up like this and it will lead to the formation of original ketone. Ketone we started with you will get the ketone back as well the resultant di radical now you can see here this is a radical this is a radical they will combine to form the cyclopropane assume that this is CH2 okay. So basically if you do an oxodipy-methane rearrangement you can see the formation of cyclopropane okay. And this one can do on bicyclic system, tricyclic system and here is an example where you have bicyclic system so that will give like this tetraradical and then what will happen this tetraradical immediately you can see it will form a cyclopropane okay. Then this radical on the oxygen will come back and then open this cyclopropane to give this bicyclic di radical this bicyclic di radical one can easily redraw like this okay you can see here this is 1, this is 2, this is 4, this is 5. So that is the 5 m board ring and the remaining is 6 m board ring. Now these 2 di radical will combine to form the tricyclic compound in that the third ring is a cyclopropane so this is the oxodipy-methane rearrangement product okay this can be easily obtained from any bicyclic compound having a carbonyl and double bond at appropriate place. When you do that one can also expect another product which is normally minor product that is arising out of 1, 3 acyl shift that is when you do a photochemical reaction on this ketone first NaRis type 1 cleavage will take place where you have you generate this di radical followed by migration of the double bond okay the double bond will migrate and you will get the corresponding ally radical okay what will happen now this will migrate or here you can write like this and these di radical will combine to form cyclobutanol okay. So one can see using oxodipy-methane rearrangement you can get a cyclopropane or if you use 1, 3 acyl shift then you will get a cyclobutanol okay 6 ombre ring fused with cyclobutanol. Now let us see how De Booth synthesized this natural product he started with 2 methyl cyclopentane 1, 3 diome which we saw during the retrosynthetic analysis then using the Robinson annihilation sequence first Michael addition with methyl vinyl ketone followed by treatment with S proline and dehydration you get a hadges-parish ketone okay. You have an enone and a ketone and the ketone can be selectively reduced with 0.25 equivalence of sodium borough hydride in ethanol at 0 degrees to get this aleric or this alcohol and once you have this alcohol then you can protect this alcohol as mometer okay and if you treat with LDA TMS chloride if you treat with LDA TMS chloride so it will generate anion here and then it will form an enolate that enolate will become enol TMS and this is the diene now he did deal solve reaction with mellic anhydride to form this tetracyclic comp okay. So this is the intermediate and this intermediate you can see this is the transient state okay this is the transient state and this gives this tetracyclic comp. Now this anhydride if you look at this anhydride should be converted into double bond so he used a electrolysis method like hydrolysis followed by di-ticarboxylation to give the double bond and this is the key precursor for the oxodipy-methane rearrangement. So taking this compound and then shine light it undergoes oxodipy-methane rearrangement and as I mentioned earlier about how oxodipy-methane rearrangement takes place first it will form this tetraradical and followed by formation of the cyclo propane and opening of this three membered ring you will get another di-radical and whereas this will be a carbonyl group and once you have this di-radical and you can number it you know for better understanding always better you number the starting material okay give numbering. So the six membered ring you can give numbering and then see where the radicals are in the starting material and when you redraw the structure for better understanding you can give the number and then you can easily make out where these two radicals are. Once these two radicals you can write and that will form the corresponding cyclo propane as I said there is a possibility of forming 1, 3 acyl shift also of course the yield is only 5% and the main product is only the required one that is the cyclo propane formation. So this tetracyclic compound next he reductively open okay so lithium tri di isopropyl amine and TMS chloride. So what happens this will be cleaved this cyclo propane will be cleaved and then it will form it will form the corresponding enol TMS okay this will form the corresponding enol TMS because you are using lithium diisopropyl amine and conging with TMS chloride because that enolate once it is generated you trap that enolate with the corresponding TMS chloride. Once you have that that is the first step and the second step the enol TMS ether can be cleaved with any fluoride source. So that is what they have done they have done with this benzyl trimethyl ammonium fluoride to cleave the enol TMS to generate the O- and the O- you conge with methyl iodide okay you conge with methyl iodide you stereo selectively introduce the methyl group okay. So now if you count the number of carbon atoms you will see here in this ring you have 6 here you have 2, 8 and then here you have 3 and then here you have 1 okay. Next step is to remove the carbonyl group so you do not want the carbonyl group okay this can be done in many ways but what they have done is they have protected the ketone group as diethiol derivative okay. Then you remove the man group okay you remove the man group is using titanium tetrachloride you get the hydroxyl group at this point you can remove the titanium using a combination of titanium tetrachloride Lewis acid and reducing agent so they use titanium tetrachloride and lithium aluminum hydride to get the corresponding CH2. So the deoxygenation of carbonyl group was done in 2 step first you protected a diethion then remove that with titanium tetrachloride and LAH. Now what you need to do you have to oxidize the alcohol to ketone okay then you have to introduce the 2 methyl groups and then double bond. So here you use some expensive reagent so first you treated with LDA TMS chloride which you know it will form the corresponding enol TMS. Then for introducing the double bond you use GDQ and this exotic reagent which looks like exotic reagent but it is not it is a reagent derived from trifluoroacetamide okay take trifluoroacetamide and then you treat with trimethylsilyl chloride you get this. So this is the reagent you know you can get a double bond or you can introduce a double bond next to the ketone. Next you have to introduce 2 more methyl groups one here and one here but at the same time the double bond should be kept intact. So how it can be done? First you add the 1, 4 addition on the double bond with lithium dimethylcuprate and quench the enolate with methyl iodide to get the 2 methyl groups and then repeat the same process repeat the same process. So that is you introduce a double bond via enol TMS followed by oxidation to get the natural product okay. So that way if you look at this synthesis the key step in the synthesis was the oxa diphyomethane rearrangement and they started from commercially available 2 methyl cyclopentane 1, 3 dione and then they used Robinson enolation as a key step to make the next starting material that is agisperage ketone and overall this total synthesis was accomplished in 11 longest linear steps with an overall yield of 1.6 percent yield. So now we will move to another very short total synthesis reported by Wehera okay. Here again as I said we will be discussing only the photochemical reaction which has been used as the key reaction in the synthesis of trichonates and what he has used in this particular synthesis is photochemical 1, 3 acetylcide. The earlier synthesis which we saw where oxa diphyomethane rearrangement as the key reaction and there also he got 1, 3 acetylcide product as a minor product. In this particular case Wehera has used photochemical 1, 3 acetylcide as the key step and that is a main reaction to get the key starting material or key intermediate. So let us see his retro synthesis and when you look at this capnoline as I said whenever you have a functional group or if you do not have many functional groups you can introduce a functional group okay. So here you have a functional group a double bond but that may not be sufficient sometimes when you do a proper retrosynthetic analysis. So it is better to introduce another functional group which can facilitate the retro synthesis to get a simpler starting material. So that is how what Wehera has done he has introduced a carbonyl group at the middle ring. The idea is if you introduce a carbonyl group then one can do an intramolecular alkylation okay. So that is what he planned. So you can see you can if you make this as a good leaving group okay. If you make this as a good leaving group then you can generate anion and you can form the third fiber boundary is not it? So that was the idea. So for introducing the carbonyl group then you can introduce a double bond here. The reason for introducing the double bond is this four carbon unit can be added stereo selectively using a 1, 4 addition reaction. And now if you look at this you reduce this double bond and introduce a double bond here what for that is how he can use the photochemical reaction that is 1, 3 acyl shift to get this compound. Let us see how he made this compound in the DL synthesis and how from there he used the 1, 3 acyl shift to get this bicyclic compound. The total synthesis started with metacrysal methyl ether, metal ammonia reduction gave the diene and you treat with sodomyte when you treat with sodomyte then it will generate anion and the migration of the double bond will take place to give this type of diene or you can this will give this type of diene but this is the most stable diene. So you get this diene and this can undergo Diels-Aul reaction with alpha-chloroclonitrile alpha-chloroclonitrile is a synthetic equivalent for ketene, okay so ketene can normally cannot undergo 4 plus 2 cycloaddition reaction. So that is why indirectly you have to use some equivalent which can give ketene in the product. So alpha-chloroclonitrile is one of the best synthetic equivalents for ketene. So you do the Diels-Aul reaction with alpha-chloroclonitrile. Now if you treat with potassium hydroxide and DMSO that will hydrolyze this to ketone, okay try to write mechanism for this, okay so mechanism for the hydrolysis of chloronitrile adduct to the ketone it is a very interesting mechanism. Take this bicyclic ketone and then treat with potassium tertiary butoxide excess potassium tertiary butoxide and methyl iodide you can introduce 2 methyl groups next to the ket, okay. Then when you reduce this with sodium borohydride or lithium aluminium hydride you get a mixture of endo and exo alcohol. So this is endo and this is exo alcohol so almost you get 1 is to 1 ratio. These 2 will give 2 different products upon treatment with acids for example if you take this endo alcohol so first it will protonate the hydroxyl group and that will be a leaving group. Now the bond which is anti to the leaving group so this is the bond which is anti to the leaving group that is the OH bond and this is the bond which is anti to that, okay that will migrate and what will you get if you use Lewis acid is okay you will get this product. So this is nothing but if you look at you can write like this that is because the C this C C bond which is anti to the leaving group migrates. So if you do the same thing with exo alcohol now which bond is anti to that so this is the bond and this bond is anti to this. So that will migrate if you treat with acid and this will migrate so that will give you this bicycling ketone okay and this can be redrawn like this okay it is a 5 umber ring here and 6 umber ring here. So what you have to notice in this is if you use endo alcohol you get the corresponding alpha beta unsaturated ketone but if you use exo alcohol you get ketone and the double bond moves to the 5 umber ring other ring okay no problem and you can also write it like this and then now you shine light. So what it will happen first the Norris type 1 cleavage to generate the diradical followed by migration of the double bond. Now this C C bond can rotate is not it this C C bond can rotate and when you rotate you can see this diradical this diradical will combine to form the corresponding 5 umber ring and if you rotate this by 180 degree okay you can go through this and rotate it by 180 degree and you will get this couple okay hydrogenate the double bond as I said you have to shift the double bond here you do not want the double bond here you want the double bond here. So first you reduce the double bond then introduce the double bond. So this is done using a Suji trust method where you generate the enol carbonate enol allyl carbonate and followed by treatment with palladium acetate you get the double bond. So once you have the double bond next is to add the 4 carbon unit in a 1 4 fashion. So take this enol and add this 4 carbon unit you will get this product. Now as I said next you have to make this as a good leaving group generate anion and then form the third ring. So for that you have to remove the TBDPS get the alcohol converted to good leaving group. So TBDPS can be easily removed if you treat with acetic acid get the alcohol convert that into tosylate. So tosylate is a good leaving group. Now you take this compound treat with lithium hexamethyl disolxide okay so it will generate anion and intermolecularly attack the carbon bearing O tosylate in a sent to fashion and you introduce the third ring. So what is left now is to remove the ketone. So reduce the ketone to alcohol convert that into zandate sodium iodide carbon disulfide methyl iodide OH will be converted to zandate and the zandate can be reductively cleaved using tributatin iodide and AABN. So to summarize so if you look at the total synthesis of Wehera and Shinari Mamoto of Capnallin they started with a simple commercially available starting material that is Metacrysa methyl ether. So they did birch reduction conjugation real solid reaction to get the bicyclic 2 to 2 octinone then the key reaction that is 1 3 acyl shift was done under photochemical condition to get a 5 ombre ring and 6 ombre ring and later they did functional group transformation to achieve successfully the total synthesis of Capnallin. Overall they took about 13 longest linear steps however the overall yield is quite good about 3.7 percent yield. So with this we have discussed few more total synthesis based on photochemical reaction as the key reaction. We will continue our discussion on synthesis of few more trichunanes again using photochemical reaction in the next lecture. Thank you.