Give the structures of the products represented by letters in ...

Question

Give the structures of the products represented by letters in this synthesis.

Part 1:

Accepted Answer

Hey everyone and welcome back in today's video, we are going to solve the following problem provide the missing products of each reaction in the following synthesis. Looking at the synthesis, we understand that we are starting with an AL king which is cyclamen scene. So we may simply redraw it for our analysis, where we can see clearly what's happening in the synthesis. We throw our double bond and what we understand is that in the first reaction we are using em CPB A and M C P B A stands for meta chloral, paroxetine, stomach acid. In other words, this is an oxidizing agent and it's very famous whenever we are dealing with Elkins and we want to produce a pock side rings. So simply speaking, we are breaking this carbon carbon double bond and the pi bond is used in creating two carbon oxygen bonds. We are going to produce any Parkside ring. So we are going to draw our cycle plantain ring. And instead of having a double bond here, we are going to draw any peroxide ring. Right, so that's our first product. We have produced any Parkside and we may label this as our a our first product. Now in the second step, this ip oxide is treated with sodium hydroxide and water. Recall that when ip oxides react with sodium hydroxide or in particular the hydroxide itself and water. We have an SN two reaction. So that be SNC, which means that if the attack place at this position they are equal is substituted. So I suppose we're taking this as our position of the attack, we will have an inversion of configuration and therefore we will have an anti arrangement between the two groups that we form. One of them will be a wage because that's what we get during the attack. And the the other one at first will be the Alcock said an eye on but then it is protein ated. Right? So we will have another alcohol group produced. So knowing that we may simply say that we are producing two alcohol groups and anti arrangement. We may draw our ring. One of the groups that we produce will be pointing down And the other one will be pointing up. Now this is our product. Notice that we have two alcohol groups. This is a dial and the arrangement of the two alcohol groups. It ends here right there on different sides of the of the plane of that ring. So this is our product. B. Now the next reaction is addition of sodium hypochlorite and A O. C. L. Access and then we also have acetic acid. Now, this region in particular is used in an oxidation reaction. It's a good oxidizing agent which is able to oxidize secondary alcohols. And that's indeed what we have into ketones. So we are going to turn every alcohol group into a ketone. That's how simple it is. Let's troll our result on product. So if we add to ketone groups. There will be two carbonell groups. One over here and another one over here. That's our product. See, right, so let's label it and let's look at the next reaction now, see is treated with the Green yard region. That's what you may notice here in particular. We are using method magnesium bromide. So that's CH three MG B R in easter. Recall that whenever we're using the Green Yard reagent and the general form Green Yard reagents are written as our MG X. Where X might be chlorine or bromine. What we're doing is we're performing an attack by these electrons which are polarized significantly towards the alcohol group and we are attacking the carbonell group which breaks the pi bond and we're producing an Elk oxide and iron. But simultaneously we're adding that alcohol group to our carbon atom. So because this is a metal group, we are going to add a methyl group at this position and a metal group at this position, let's do that. We get our Alcock side. So there would be a muscle group at this position right now. And also a negatively charged oxygen. Similar thing happens over here we have a metal group and a negatively charged oxygen right now, that's why it says that in the next step, we want to use an asset, H 30 plus two protein, eight every oxygen and get the alcohol. So our product d will be a dial once again. So there will be a metal group at this position. An alcohol group right now, because we have a protein ation step of the elk oxide, a metal group and another Oh age. We can say that this is our product, the well done. And now if we move on we need to get product e using sulfuric acid and heat recall that whenever we are treating our compound with sulfuric acid and heat in particular because there's heat present at favorites elimination, right? And sulfuric acid is a strong asset. So we may say that this reaction is an acid catalyzed dehydration. We want to produce, produce the same product. So we have to understand which beta hygiene will we eliminate this one? The elimination of this beta hydrogen would lead us to the production of a less substituted elkin. So we are going to remove this beta hydrogen instead. And we will essentially form a double bond in this position after its removal followed by the loss of the leaving group. Because in particular we will have a protein in the form of a wage here. But actually if you think about the mechanism in general, we would first of all protein ate it have the loss of the leaving group to produce a tertiary carbon. Okay, Ryan And then again, we're just thinking about which made up hydrogen, will we remove. Right, so we have two possibilities this one or that one. If we remove this one, we would produce a dice substitute at alc in But if we remove this one we would then produce a tricep stupid one. So we are going with this one and because there are two alcohol groups, we are going to remove two beta high genes, one from here and another from here. Due to the reasons stated before we have to beta hydrants to remove because there there will be two protein native groups that will form to carbo Karen's right? So the product that we get is simply our rain to begin with. Now the metal groups are still bonded over here and here. But now after the removal of this beta hydrogen and form and the formation of a double bonded disposition, we already don't have ohh anymore because it was protein ated and it acted as a leaving group. So now we have a double bond at this position and similarly after the protein nation and the formation of the carbon khurana. Then the removal of this beta hygiene. We are forming a double bond at this position. So we may say that our final structure or our compound E essentially looks like this. We may now label this as our E. And we have all of our products. We may essentially state that this is our complete synthesis with A B, C D and E. Now the correct answer to this multiple choice question is a However, if you have any questions don't have states leave your comment down below in the comment section and thank you for watching

Give the structures of the products represented by letters in this synthesis.
Part 1:

Key Concepts

Epoxidation

Grignard Reaction

Stereochemistry and Diastereomers

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