Determine the structures of compounds A through G, including s...

Question

Determine the structures of compounds A through G, including stereochemistry where appropriate.

Accepted Answer

Welcome back, everyone draw the structures of the missing compounds. A through C. In the following synthesis, the starting material is cyclo heine, which first will reacts with perox benzo acid to produce a, a reacts with Meho magnesium bromide followed by the acidic work cup to produce b then be reacts with sulfuric acid in the presence of heat to produce C and CIA reacts with hydrogen gas in the presence of platinum to produce methyl cyclohexane. So let's start identifying the unknowns. The very first reaction that we have corresponds to an oxidation reaction with oxy acids. So if the starting material is an alkene and our region is a oxy acid, that means a must be an oxide, we're going to draw a six number. And if we break the pie bond, we can control our three membered ring. Right now, we can draw the three number drain, pointing up or down. It doesn't really matter because we can always flip it. And instead of having two dashes, we will get two wedges, it's perfectly fine because the structure is a miso structure. So well done. We got our first product which is a now, let's think about the formation of structure B. And to do that, we are going to essentially redraw the GNI peroxide to analyze it. And we have a green yard region and, and the first step, we are using methyl magnesium bromide and uh we are going to redraw it. So that will be methyl group bonnet to magnesium bromide. We have to recall that the regent is basic because we have a strongly polarized magnesium carbon bond towards carbon because magnesium the same metal. So in basic conditions, we can have a ring opening by an attack at the less substituted carbon. But because the two carbons are equally substituted, we have two possibilities which are equally likely. So the first one in blue will give us one of our products. We are going to draw a six mum ring now because we have a backside attack, we will have an inversion of configuration. So that means the metal group will be pointing up and then at the adjacent carbon, we will have our alcoy group pointing down because nothing changes regarding its configuration. It didn't participate in the reaction when we think about the adjacent carbon atom. And now, of course, the second step is the protonation step. So we can immediately state that after the protonation step, we, we are just going to form an alcohol. But before we do, so, let's consider the second possibility in green if the green region is used in attacking the other carbon then we will form an in ant humor to the previous structure. So let's go ahead and try a six membered ring, then we will get our Elcock side. So it will be pointing down because nothing changes regarding its configuration or specifically the configuration of that carbon. And we had a backside attack. So we can just draw a metal group bonded to the first carbon atom. Now, if we flip that structure, we will notice that we get a pair of in an tumors. So let's go ahead and flip it so that we, if we flip it, we understand that the metal group will be pointing down opposite to the configuration of the first structure and the alcoy group will be pointing up. OK. Well done. So we get a pair of in an tumors. And now, of course, the second step was the protonation step. So we are going to get two alcohol groups for each. So let's go ahead and try a six member grain for one of our products and then a six member grain for the other product. For the first one, we can say that the muscle group is pointing up and then the alcohol group at the adjacent position will be pointing down or the other product we get it's in anti. So we will have a metal group winding down and an alcohol group pointing up. Well done. So we now understand how the two structures for B are obtained Now, our goal is to go back to the reaction sequence and think about the formation of C. Now in C, the two alcohols from B will react with sulfuric acid in the presence of heat. Let's analyze it. First of all, we have to recall that sulfuric acid in the presence of heat whenever we have an alcohol indicates that we have and acid catalyzed dehydration of an alcohol. So let's not worry about stereo chemistry because we will form an al king. And in the general case, let's just draw a metal group and an alcohol group on solid lines. So first of all, when we are proton it with sulfuric acid, we get our Oxonian ion. Let's say that we are using sulfuric acid which acts as a proton donor. And then because we have a perfect leaving group, it will leave forming a secondary carbo ion. And now, of course, we have to think about the stabilization of that carbo car, which is indeed possible because we have a muscle group and an implicit hygiene atom. We will observe a one comma two hydride shift to form a more stable tertiary carbo Canion. So let's go ahead and draw a more stable tertiary carbo caron formed, which essentially tells us that we are indeed dealing with an E one reaction. So we have two equivalent beta hydrogens because the structure is symmetrical, we can choose any of them. And now the water molecule formed will act as a base to remove one of these hydrogens in an elimination reaction to form the required structure C, which will be our L king. In this case, we get a six membered ring and a double bond as well as the original metal group. That's how we get structure C, we are dealing with a basic acid catalyzed dehydration. So we can now easily draw structure C which corresponds to our six member ring, a metal group and a double bond. So that would be one methyl cycle. He one in, that's our product for C. And now let's see if there's anything else left. The final reaction tells us that if C is treated with hygiene in the presence of platinum, we must produce methyl cyclohexane, which makes sense because we are basically performing hydrogenation of an alkene. So the two hydrogen atoms will be used to form an alkin out of an alkin, right? To essentially get rid of that unsaturated. Well, then we got all of our products and we verified that indeed answer C must be correct because it is consistent with the outcome of the final product shown. Thank you for watching.

Determine the structures of compounds A through G, including stereochemistry where appropriate.

Key Concepts

Grignard Reagents

Acid-Catalyzed Reactions

Stereochemistry

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