Question

Mechanism of epoxidation of trans-but-2-ene to meso-butane-2,3-diol.

Propose mechanisms for the epoxidation and ring-opening steps of the epoxidation and hydrolysis of trans-but-2-ene shown above. Predict the product of the same reaction with cis-but-2-ene.

Accepted Answer

Hey everyone and welcome back into this video. We are going to solve the following problem. A proposing mechanisms for iP oxidation and asset capitalized ring opening of trans hex three in and be predict the final product of the reaction of csX three in under the same conditions, let's start with a it suggests that our starting material is trans hex freeing the structure is already given. So you may just redraw it right now. This is our trans hex tree and of course we can show these implicit hydrogen atoms and the first step is an iP oxidation. Stop keep in mind that whenever we want to perform iP oxidation, we want to use a proxy asset. Right? And the proxy asset that we are going to use for the purposes of this problem is parasitic acid. First. See the gas, it has a formula of CH three C 03 H. Let's see how that looks like. Now, if we draw it below our structure in particular, we are going to expand that oxygen hydrogen bond, then there's another oxygen, our carbonell group. Bandits another oxygen. And there's a method group. So that's parasitic asset, right? We C 00. H. That makes sense. So now let's see what's happening here. It's very important to understand that this oxygen in particular is partially positive and this one is partially negative because electrons are withdrawn by this auction and the carbon group. So even though it's unusual to see such a scenario is actually partially positive here and because it's partially positive, there might be an attack by the pipe electrons over here onto that oxygen. But if we're forming a bond, the adjacent bond must be broken. So we're going to take that bond and transfer these electrons into a new pi bond. But because once again we're forming a new bond here, this pi bond must be broken. And the way to break it is to actually capture this hydrogen from here. Right? And now the bonding electrons from here will be transferred into the formation of the Depok side. So as you can see what's happening is that we're capturing that oxygen and then we're also transferring the electrons into a new carbon oxygen bond. So we're going to take it and we're going to essentially from our epoxy. Let's redraw it a bit more careful. So if we draw that arrow pointing on to auction, then the final arrow is just pointing down towards the carbons form that carbon auction bond. So, if you see here we are forming a rip oxide. Right? Don't worry about the intermediate reform here. That will be another asset. But that's still fine. We're only interested in the alc in itself. So let's concentrate on that. Now. The box that we are getting here must have the trans arrangement. We may start by drawing the the three member ring. Right? And essentially what we want to understand, we want to place our group's appropriately. So one ethnic group will be pointing up and another one will be pointing down because we start with a transaction. So that be transit box side. So you can say that this is our CH two CH groups. So let's just say CH three H two C. Let's draw it more appropriately, just the way we wrote it before. So let's say H two C H three C two. Clearly show the bonding. Now, similarly here we have CH two C3. Those are our ethyl groups and now the two hydrants are also on opposite sides. So one of them is pointing down hydrogen trans arrangement. Or basically in this case that B and C. To this hydrogen. And we got out of the box side, we're going to think about it about our next step. And the next step is the acid catalyzed ring opening. And here we are going to use reaction number two. We're going to use a strong asset A three or plus because it's an acid catalyzed ring opening. We're going to use the lone parent auction within the box it and we're going to protein ate it using hydro knee. Um so that would be our hydro knee. Um with a with a formal positive charge. There's an attack to protein ate the box side followed by the loss of water were forming our protein ated intermediate. So let's draw that the pertinent and intermediate would just have oxygen being protein ated with hydrants. So there will be formal positive charge and oxygen with the original groups. So that's hydrogen. let's use abbreviation here. Okay, that'd be ethel there will be an ethyl group over here and hydrogen. So that's how protein ated intermediate. And now we want to think about the nuclear attack step. The water molecule formed will under acidic conditions it will attack a more substituted position but the two positions are equally substituted meaning we may attack any of the two. But actually we will just think about the attack at one of the two positions and we will be able to draw it in and humor right? We will just have an opposite configuration at the other center. So if we have an attack here that's followed by the loss of the leaving group and you will notice that we are forming our product. Now if we draw products keep in mind that this is a backside attack meaning we have an inversion of configurations. So to draw our product properly, we are going to first of all start with this center so we have no age group pointing up. Nothing changes here. Right Haijun pointing up on a wedge group pointing down on a dash because it was a backside attack after the D protein nation stop this weight will be pointing down. And now the ethyl group is pointing up and the Haijun Adam sprinting down now we previously said that there might be another product form if we attack this position because it's also because the two electrolytic positions are equally substituted. But if you perform bond rotation over here, you will notice that this is a missile compound, it cannot happen in and humor. If you perform bond rotation, you will have a wage pointing up on a solid line, ethyl group pointing down right, and hydrogen pointing up. So there's an internal line of symmetry after you perform bond rotation. So it's a meso compound and we don't have any financial reform. So this is one and only product here. So we can say that for a We are done now, if we think about B, we are not asked to show the mechanism for me. Were just said that we are starting with CIS X three in So we are going to join this hex three in that will be our double bond and the two ethnic groups will be below the double bonds. We're going to draw these two ethnic groups. The structure is actually already given to us. So that's six hex three in And once again, if we think about the ip oxidation step, as we previously said, if this is a cis compound, then there will be he says a pox ID. So the two ethyl groups might be pointing up or down depending on the perspective. Right, let's just draw one of the perspectives, let's say that the two ethnic groups are pointing up. Keep in mind that once again, there are two hydrogen atoms that are implicit here. So those would be pointing down now in the step, we're using parasitic asset just to keep track of our work to form that Depok side, we need to use parasitic acid or M C P B. A. If you wish. That's fine. But let's just simplify that and use parasitic acid. So we're forming our Depok side. We said that we want to protein ate it right? And after the deep protein nation step, let's just say that this is number one. Number two is the protein nation step when we protein ate it reform. Oh, age with a positive charge. And then what are attacks any of the two positions? And now it might matter right, maybe we're not getting our meso compound anymore. So let's see what we're getting. If we perform an attack full by the ring opening, we're getting our first product, this is once again a backside attack. So we will have an inversion of configuration. So we're going to start with the second electra Phillip carbon atom. There's an A wage pointing up on a solid line, ethyl group pointing up on a wedge high chin pointing down now regarding this position, there will be an a wage to the backside attack is pointing down now the two remaining groups will be a wedge for NFL and a dash for hydrogen. And now this is where we do not have any internal symmetry. Right? If you perform bond rotation, ethyl group will will be pointing down instead of up after that bond rotation. So it's not a means of structure meaning we want to draw, it's an antimatter as well simply speaking what we want to do. We want to see what's happening if we have an attack at a different position so we try and attack at a different position over here, we may start drawing our next product. So we are going to do our central carbon carbon bond and starting with this position because there's a loss of the leaving group. So it's easier for us that would be just ohh pointing up ethel pointing up hye jin pointing down now just to the backside attack when we lose that hydrogen right? When we D protein eight, water that gets bonded, we get O. H. Ethyl group pointing up hae jin pointing it out. So now we get to products, these are a pair of swim a label, these is our products. Let's make it a bit clear to make sure that we get everything in frame. So Right, so let's just conclude our findings for a we started with trans hex three in and it gave us a missile compound after that ip oxidation and acid catalyzed ring. Opening up an ip oxide just to the fact that this is a miso compound, it's a single product, we don't have a pair of finance commerce but when we start with assist compound and use the same conditions, we produce a pair of finance farmers. Now the correct answer to the small cultures question is B. However, if you have any questions, don't hesitate to leave your comment down below in the comment section and thank you for watching.

Propose mechanisms for the epoxidation and ring-opening steps of the epoxidation and hydrolysis of trans-but-2-ene shown above. Predict the product of the same reaction with cis-but-2-ene.

Key Concepts

Epoxidation

Ring-Opening Mechanism

Stereochemistry of Alkenes

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