Chapter 8 discussed the synthesis of cholesterol, which procee...

Question

Chapter 8 discussed the synthesis of cholesterol, which proceeds by a cationic cyclization cascade. Without looking back, suggest a mechanism by which the following reaction occurs. [The carbons have been numbered for you.]

Chemical reaction diagram illustrating epoxide reactions with numbered carbon atoms and sulfuric acid as a reagent.

Accepted Answer

Hello and welcome. Our next problem says the following synthesis is carried out via a konic CYA cascade that says what is its possible reaction mechanism? Note the carmens are numbered for you. So the molecule that's reacting is an oxide, that oxide is an oxygen bound to carbons two and three at carbon six or excuse me at carbon two, we have a methyl group. Carbon six. We have a methyl group and a double bond to carbon seven. A carbon 10. We have a methyl group and a double bond to carbon 11 by carbon 14, a double bond to carbon 15 and then a methyl on 15. And it has a total of 16 carbons in the chain. Your reaction occurs with hydro typher acid H two. So four and we end up with an alcohol. So our oxide ring is open so that alcohol groups coming out from carbon three and carbon two, we have two methyl groups. So that's carbon one and then the methyl group and we have a series of three rings altogether a six carbon ring next to in the chair configuration, next to a six carbon ring in the boat configuration. Next to a six carbon ring in a chair configuration. We've got that methyl group on carbon six methyl group, on carbon 10 methyl group on carbon 12 or excuse me, two methyl groups on carbon 15, one representing carbon 16. And there's a positive charge on carbon 15. So our peroxide ring has opened up. We've generated a Carbocaine and we've had this cascade of reactions forming three adjacent rings. So with this acidic environment here, we could start with a protonation of the oxygen in the oxide ring, which will make it more likely to leave to in the form of a ring opening and we'll add our lone pairs to that oxygen. Now, we know that the lone pairs on the oxygen can grab a proton with the electrons from that h bond going back to the oxygen. And once our oxide ring is proton, it will be easily opened due to the weakness we introduced by proton it. So now we have this proton oxide ring and we can have a nucleic attack by the electrons in one of our pylons to open that oxide ring, which of these carbons are going to bond to each other. Well, we can have a double bond from carbon six to carbon seven. So carbon six could attach to carbon two, carbon six has a substituent and carbon two does as well. Whereas if we had carbon seven attaching carbon seven has no substituents. So that means that carbon six being the one to attack, pull it to a more stable carbo cat I am. So let's show that happening. We've got the electrons of the pi bond between carbon six and carbon seven comes around attaches to carbon two. And then the electrons from the bond between the oxygen and carbon two go on to that positively charged oxygen from the oxide ring. We've broken open the bond and we have that as an alcohol group and we formed a 1234, 56 carbon ring, which is a very stable arrangement. So that would be a sort of favored reaction to happen. So to draw the product, we know we have the six carbon ring, there's no longer a double bond between carbon six and carbon seven. And let's put in our substituents in the right place. We know we have the alcohol group attached to carbon too or excuse me, sorry, carbon three, the electrons attacked carbon two. So the ring broke open there. So the alcohol groups attached to carbon three. So we're going to designate carbon three as the sort of uh not the top carbon in the six carbon rink, but the next one to the left. So that will be an oh group. And we're going to label that as carbon three and then we have carbon chew up at the top of a ring with, as we recall, two methyl substituents, essentially, there's carbon one coming off of it. And then there was a methyl group coming off of it. And then we go around the ring numbering. We know that we have an attachment to carbon seven with those pi bonds from the double bond between six and seven are the pi electrons, excuse me are what form that bond with carbon two. So we label carbon seven the right of carbon two and then go around the circle six 543 fact three, which has the alcohol. And then from carbon seven, our chain continues, we have carbon 89 and we just continue taking that chain out till the end. So we take the rest of that chain out from carbon seven. From carbon 789, 10. Carbon 10 is a methyl group and then a double bond to 11, then two carbon 1213, 14, a double bond between 14 and 15 and then a methyl group on carbon 15. And last of all, we need to add in one really important part, recall it. Carbon six had a methyl group. So we need to put that there. So we're gonna erase the six to make room here. And then of course, carbon six also now has a positive charge since we broke that double bond between six and seven and used the electrons to make a bond. So now we have another Carbocaine that can be attacked in the same way with the pi electrons from a double bond. So we go down to our next double bond between 10 and 11. And right away, we see that carbon 10 has a methyl group. Carbon 11 doesn't. So we know which arrangement will make them a stable Carbocaine. If we have carbon 10 bonded to carbon six, there draw these electrons coming over two carbo carbon six. The carbo C I in there when that breaks open and forms a new ring, we'll have a positive charge at carbon 10. So that's a tertiary Carbocaine and more stable. So I started by drawing my chair configuration of that six carbon ring with carbon three down in the bottom left corner with its 08 substituent. Then we'll go around the ring number, bring carbon four up above then 56 on the right six has that methyl group. It now no longer has a positive charge because it has a bond to carbon 11. And then let's go back down and put on the methyl group on carbon one or the metal group. That was carbon one label, carbon two, carbon seven. So now we have a bond from carbon six to carbon 11 and then carbon 11 down to carbon 10 and then we get nine and eight and linking back up with carbon seven. So now we've made our second six member ring again, that's a stable arrangement. Carbon 10, I'm going rates the number 10 briefly. So we can draw on its methyl group and then it also now has a positive charge. That's where our carbon cion is. And then finally, from carbon 11, we carry on the rest of the chain up to carbon 12, 1314, double bond to 1515 as a methyl group and then 16. So now we've got our second ring. So you know what to do. It's kind of the same thing we've been doing all along. We bring the electrons from that double bond between carbons 14 and 15 over to attack the carbo cion. And then we just have to generate our very last structure, but we don't actually need to draw that because at this point, we've reached our product. As drawn up above, we have that third ring formed by the linking of carbon 15 to carbon 10. The positive charge will now move on to carbon 15 and we've formed another six member ring. So luckily, at least in this case, we have the structure of the product provided for us and you can see how we generated those three rings in a row there and we're left with a Carbocaine at the end. See you in the next video.

Chapter 8 discussed the synthesis of cholesterol, which proceeds by a cationic cyclization cascade. Without looking back, suggest a mechanism by which the following reaction occurs. [The carbons have been numbered for you.]

Key Concepts

Cationic Cyclization

Epoxide Reactions

Acid-Catalyzed Reactions

Watch next