The following cyclization has been observed in the oxymercurat...

Question

The following cyclization has been observed in the oxymercuration-demercuration of this unsaturated alcohol. Propose a mechanism for this reaction.

Accepted Answer

Everyone. And welcome back in today's video, we are going to propose a plausible mechanism for the October curation, dimmer curation of this unsaturated alcohol. And if we look at this problem, we indeed understand that this must be an oximeter curation immigration reaction because number one is the addition of mercury to acid eight. So that's our oxy Merck Eurasian steppe. And number two is the treatment of the result an intermediate with sodium boron hydride. So that's our Diemer curation stop often called reduction because we are using our reducing agent sodium carbohydrate, right. So we can start by drawing that mechanism. Let's suppose that we change the perspective of our bonds a bit. We will start with the double bond pointing to the right and we will just rotate a bond so that the methyl group is pointing out that would be simpler for us to visualize the whole mechanism. Okay. So that's how we're starting with cereal. In the first step, we are adding Marie Curie to acid eight. So we're going to draw to acid age groups. There's a lone pair on mercury that's important to understand. And then the first step of this mechanism we have a high density, high electron density within the pi bon. And since mercury is a metal, its electoral positive. So the high electron density can easily attack mercury while mercury can attack back because it has that lone pair on it. And this is how we are forming our merc uranium iron, there will be a bridge formed over here, right. So in addition to that, we want to make sure that we have the loss of the leaving group, one of the estate groups is going to leave and this is how we are forming our bridge. So if we draw our bridge, we will get the following O H. Now, the pi bon has been used into the formation of that Marik uranium iron. So we are already drawing our ring and it still has an acid, a group bonded, said mercury will now have a formal positive charge because it has three bonds, right? And we call this Marik uranium iron. Okay. So we have formed that now we expect to have a nuclear attack. And if you want a hint, you can easily take a look at the product form. You understand that you're starting with a structure which forms a ring, right? So you're going to make a close structure out of an open one. And that means you already have a nuclear file which performs that auction has a lone pair. So it can actually form a ring very easily because you have to Electra Felix centers, meaning you have your nuclear file within the structure itself. So that would be an intra molecular attack, meaning you're taking this lumpy and oxygen and you have to choose which one of the two carbon atoms you would attack. Recall that because this is an a wage group, this is an alcohol, it will attack a more substituted position over here, right? We have to elect carbon atoms, one has two hydrogen atoms and then that one has to alcohol substitutions. So we're going to go with the one that is more substituted. And this is consistent with the product as well because we want to form a smaller ring, we are not forming 1/6 member drain. We are forming a five member ring. So the lump air attacks that carbon. And because we're forming a new bond, we are going to break the adjacent bond and transfer these bonding electrons onto Mary Korean in the form of a lone pair. So we are ready to draw our resulting intermediate and we will label our carbon atoms for simplicity, that'd be 123 or right. And if we only think about these carbon atoms, these are forming a five member of the ring where auction is number five. So we can just throw our five member ring. Let's suppose that auction is at the very top, There will be four carbon atoms. Okay. And opposition number four has two substitutions. One of them is a muscle group. And another one after you break that bond would be over here that B CH two. So we're going to draw methylene right CH to monitor mercury And Mercury responded to Acid eight. Okay. Perfect. But in addition to that, let's not forget that we still have this hygiene. So let's show the presence of that hygiene. Meaning auction now has a form of positive charge. Okay. And now we want to deep throat in it that we don't want to have that hydrogen bonded to auction. If you look at the structure, we want to make sure that auction has three bonds. Now it has three and specifically what we can do here, we can use the acetate that we formed in the first step acid. It can act as a base and the negative charge nasty. It can essentially capture that proton and the bonding electrons will be transferred onto oxygen to form our next structure. We can redraw it. So that would be oxygen. Okay. Well done. So now the last step we are done with oximeter creation, which was the longest step. Possibly. We now want to think about the gamer curation stop or reduction. Notice how the final step essentially implies that there will be two metal groups over here and we can already predict what's happening in the reduction. Stop. Keep in mind that reduction step implies that you're going to break mercury apart from your alcohol chain. And you're going to replace it with hydrogen and study. So, because this was your methylene groups each to, if you're breaking, if you're removing hydrogen, my approach is if you're removing mercury here, you're going to add hydrogen instead of it. So mercury acid, it will be removed and you're adding one hydrant over here. So methylene siege to become CH three and that's how you're getting your metal group. And we can show that the final product indeed has two metal groups at that position because once again, there was that original method group and the additional method group is obtained after the reduction step where methylene gets hydrogen when we break apart mercury. Okay. So that seems consistent with our product. This is indeed the product that we're seeing in our problem and we can essential label everything here is our final mechanism because the problem was asking us to provide a plausible mechanism. So this is our mechanism. We can conclude that the correct answer choice to the small pool choice question is D however, if you have any questions, don't have states, leave your comment down below in the comment section. And thank you for watching.

The following cyclization has been observed in the oxymercuration-demercuration of this unsaturated alcohol. Propose a mechanism for this reaction.

Key Concepts

Oxymercuration-Demercuration

Mechanism of Cyclization

Stereochemistry

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