 Welcome back to classics in total synthesis lecture series and we have been talking about total synthesis of natural products having 6-phombed ring. So we spoke about total synthesis of lungifoling in the last lecture. So today we will continue our discussion on total synthesis of few more natural products. Particularly today we talk about total synthesis of one interesting natural product called carpano. So this carpano was isolated from the bark of carpano tree and as you can see from the structure it has 3 6-phombed rings 1, 2, 3 and 1 aromatic ring and of course you have 2 5-phombed rings okay. So it poses a enough synthetic challenge and it also has 5 contiguous asymmetric centers okay 1, 2, 3, 4, 5 there are 5 contiguous asymmetric centers. But there was very interesting total synthesis reported by Chapman and this involved 2 key reactions one is intramolecular Diels-Aubre reaction, so other one is oxidative coupling reaction. So this was reported in 1971 and when you look at this molecule you can see at least there are 2 double bonds in a cyclohexene ring okay. So whenever you have a 6-phombed ring with a double bond then one reaction which should come to your mind is Diels-Aubre reaction. See for example if you look at this molecule you can see there is one double bond here that one can think of being constructed using a Diels-Aubre reaction there is another one here so 1, 2 this also in principle can be thought of being made through Diels-Aubre reaction there is one more double bond okay which you cannot see properly as a part of cyclohexene but this is part of an aromatic ring but that also can be thought of as a double bond which could be formed as a result of Diels-Aubre reaction. But here in this case it should be formed as a result of hetero Diels-Aubre reaction. So now let us see intramolecular Diels-Aubre reaction could be successfully used in this total synthesis. For example if you have to make this double bond using intramolecular Diels-Aubre reaction okay what should be the precursor? Likewise if you have to make this double bond using intramolecular Diels-Aubre reaction what should be the precursor and the third option is this okay. Now let us see each double bond and how they can be made using the key intramolecular Diels-Aubre reaction or what should be the precursor for making these 3? Then we will choose the right one of course it should be followed by oxidative phenolic coupling this is what Chapman has proposed first option A if you want this double bond to be made using intramolecular Diels-Aubre reaction what should be the precursor? The precursor should be this one and what you can see here you have a diene in that diene one of the double bond is part of ketene as you know ketene is unstable and second difficulty is you have a cyclic allene a 6 numbered cyclic allene almost very difficult to make. So this disconnection leads to highly unstable precursor okay so option A could be ruled out now let us see option B. Option B you want to make this double bond using intramolecular Diels-Aubre reaction so then what should be the precursor? Your precursor should be this okay it is still possible but maybe the precursor to this triene may not be that easy okay. So let us look at option C and then see whether that will be easy to make so for example if you use or if you think about this double bond being made by intramolecular Diels-Aubre reaction then what you expect is this precursor. So now if you look at this precursor one can also further visualize first of all it can undergo intramolecular Diels-Aubre reaction and during that process this ring gets aromatics okay so that is the driving force. Second as you have seen here if you break this one that can lead to a symmetrical alcohol symmetrical phenol okay this symmetrical phenol can undergo oxidative phenolic coupling that will give you this intermediate which spontaneously can undergo an intramolecular Diels-Aubre reaction to form Carpano. So what should be the starting material? The starting material is this Carpano. So before we actually go into the details of the total synthesis of Carpano by Chapman let us briefly discuss the two key reactions which are being used to synthesize Carpano. The first one is the intramolecular Diels-Aubre reaction see normally when you talk about Diels-Aubre reaction there are two types of Diels-Aubre reaction one is intermolecular Diels-Aubre reaction see this intermolecular Diels-Aubre reaction is the most common one you will see and there are many intramolecular Diels-Aubre reaction which can give more rings okay when you talk about intramolecular Diels-Aubre reaction they are mentioned as IMDA intramolecular Diels-Aubre reaction say they write IMDA and again this IMDA is of two types one is type 1 the other one is type 2. So what is type 1? Type 1 intramolecular Diels-Aubre reaction means you have the dienophile attached to carbon number 1 of the diene. So if you look at this diene if you look at this diene you start numbering from here 1 2 3 4. Now the dienophile if this is attached to carbon number 1 if this is attached to carbon number 1 then this intermolecular Diels-Aubre reaction is called IMDA type 1 okay because it is attached to carbon number 1. So that normally gives a fused bicyclic system okay that normally upon intramolecular Diels-Aubre reaction gives fused bicyclic system whereas the type 2 where your dienophile is attached to carbon number 2 okay you can see 1 2 3 4. Now your dienophile is attached to carbon number 2 this preferably and most likely gives bridged bicyclic system okay this gives bridged bicyclic system and if you look at this to form the bridged bicyclic system you should have minimum 7 membered ring. So this will be always 6 membered ring is not it 4 plus 2 cycloaddition will give 6 membered ring but the ring size of the other ring which is going to be formed should be minimum 7 membered ring then only type 2 IMDA Diels-Aubre reaction will work. Now the reaction which we thought about or which Chapman used in the synthesis of carbonone is type 1 IMDA reaction because that gives fused bicyclic system. The other key reaction which he used is penacol coupling which I will come to that later. So for IMDA type 1 I am giving couple of examples so you can see it is type 1 because the whole dienophile unit is attached to carbon number 1 that undergoes intramolecular Diels-Aubre reaction to give this fused bicyclic system and this is another complicated system but that also you can see the dienophile is attached to carbon number 1. So this also should give the fused system whereas type 2 is where you have the dienophile attached to carbon number 2. So you can see here 1, 2 the dienophile is attached to carbon number 2 so that will give you the bridged bicyclic system. This is a classical example where you form this tricyclic system using an intramolecular type 2 Diels-Aubre reaction again here in this example you see 1, 2. So carbon number 2 of the dienophile is attached so that gives the bridged bicyclic system. So now what you have to look at is the ring size being formed as a result of this IMDA type 2 reaction. Obviously 1 ring will be 6 membered ring because this intramolecular Diels-Aubre reaction 4 plus 2 should give 6 membered ring but you should calculate the ring size. You should see the ring size of the other ring which is formed. So now if you look at this, this is 1, 2, 3, 4, 5, 6, 7, 8. As I said minimum 7 membered ring should be formed then only IMDA type 2 is possible. So here is 8 membered ring and in this example you can see 1, 2, 3, 4, 5, 6, 7, 8. Again this is 8 membered ring and this 8 membered ring could be successfully formed using IMDA type 2 reaction. Normally 8 membered rings are little difficult to form but using this IMDA type 2 one can easily make the 8 membered ring. The second key reaction is the oxidative phenolic coupling. In fact many biosynthesis involves this type of oxidative coupling reaction and Derek Barton has proposed that many benzyl isoquinolenes are converted into morphine alkaloids through this oxidative phenolic coupling reaction. So what happens? You take a phenol so under this condition it forms this O radical. The O radical can tautomerize and then you can get this ketone and this allylic radical. Now these two can combine one possibility and other possibility is the same thing can combine. So like this coupling reaction leads to the formation of dimerization. So I will give one example where how this oxidative phenolic coupling took place and what are the products possible for example if you take 2 naphthol and then treat with ferric chloride as you know this is one of the key reactions which is done in ton scale to form binols. And if you use further oxidation with potassium ferricinide so then it can form this bond. This is again through radical and then coupling reaction and it can also form to the other side. So this type of oxidative phenolic coupling is known and Chapman has cleverly used a combination of oxidative phenolic coupling and intramolecular type I Diels-Alde reaction as key reactions to synthesize carpano. Let us see. The starting material where you need the double bond trans double bond was made from this phenol. The first step was allylation with potassium carbonate and allyl bromide. Introduce the allyl group. Now what you need is that double bond should be isomerized. So now it is a terminal double bond. The double bond should go inside that should be internal double bond as well as you know it should be at this carbon. Two things we have to do. One this double bond has to go inside and that should be at the adjacent carbon. So the transfer of this allyl group to here can be easily done using glycentry arrangement. So you just take it and heat it at very high temperature maybe about 200. So one can easily get this glycentry arrangement product. Now still the double bond is in the terminal position. So what you need? The double bond should go here. So that is done using potassium tertiary butoxide and DMSO. So you treat with base strong base. So like potassium tertiary butoxide can generate and isomerize the double bond and isomerization takes place to get the trans allyl group. So the starting material is ready. Next key step is the phenolic coupling. So this phenolic coupling was best done with palladium chloride. So the palladium chloride it forms this dimer and once this dimer is formed so it can undergo you know the diradical and then the diradical will come and like this it will come in the same way it will come and then it will dimerize while dimerizing these 2 methyl groups will be trans to each other. These 2 methyl groups will be trans to each other. So now you can see that this is C2 symmetric compound and if you rotate this CC bond, if you rotate this bond like this what will happen you will get this particular intermediate. When you look at this particular intermediate you can see as originally planned by Chapman you can see here a diene this is a hetrodiene. You have oxygen as one of the atoms. You have oxygen as one of the atoms of the diene. So this is a hetrodeal salt reaction where hetro atom is present in the diene and that undergoes an intramolecular type 1 hetrodeal salt reaction straight away to give carpone. So basically from commercially available starting material from commercially available starting material carpone was synthesized by Chapman in 1971 through what one can call it as biomimetic route. So very elegant biomimetic synthesis of carpone was reported and his synthesis basically involved 4 steps starting from the corresponding phenol and the key steps are oxidative phenolic coupling and intramolecular cycloaddition reaction. And the overall yield was close to 50% you know you can imagine getting this compound natural product in 50% is really outstanding accomplishment. The second synthesis of carpone I will discuss is about Marsomoto synthesis. He also used the same intermediate the only difference is in the case of Chapman he used palladium here he used a saline complex. So the saline complex and this gave that directly the natural product not only the oxidative coupling took place but also the diene salt reaction. So he started from this phenol which was already reported by Chapman. So now he used molecular oxygen in the presence of a catalyst that gave directly carpone in one step okay. So what he did he took oxygen and cobalt to saline complex and that gives the radical. So that radical migrates and you get this allyl radical that spontaneously undergoes dimerization. So it does not give the cis-sysomer it gives only the trans-sysomer and the trans-sysomer as soon as it is formed it undergoes intramolecular supra-facial 4 pi plus 2 pi cycloaddition reaction to eat carpone in one step from the starting material which was already reported by Chapman okay. So if you look at this synthesis this is really a very elegant synthesis and this was obtained in 90% in a single step from the intermediate reported by Chapman okay. So I will stop here and then I will discuss about one more natural product called Nevinolene in the next class which is very interesting and complex natural product and that was the key starting point for making several you know cholesterol lowering drugs. See you next week. Thank you.