Problem 8-54 describes a new method to perform ozonolysis reac...

Question

Problem 8-54 describes a new method to perform ozonolysis reactions that used pyridine (py) to generate the final aldehydes and ketones in a non-aqueous reaction medium. In a subsequent publication (J. Org. Chem., 2013, 78, 42), Professor Dussault (U. of Nebraska at Lincoln) described a “tandem” process in which two reactions are performed sequentially without having to isolate the intermediate aldehyde or ketone. Show the final product from each sequence. (Hint: The isolated products were from the larger part of the structure. Ignore stereochemistry.)
(c) Chemical structure showing ozonolysis reaction steps with reagents O3, pyridine, CH3MgBr, and H3O+ for product formation.

Chemical structure showing ozonolysis reaction steps with reagents O3, pyridine, CH3MgBr, and H3O+ for product formation.

(d)

(c) Chemical structure showing ozonolysis reaction steps with reagents O3, pyridine, CH3MgBr, and H3O+ for product formation.

(d) Chemical reaction scheme illustrating ozonolysis with pyridine, followed by Grignard reagent and acid workup steps.

Accepted Answer

All right. Hello everyone. So this question says that in a publication journal of Organic Chemistry 2013, 7842 Professor do so at the University of Nebraska at Lincoln described a tandem process that involves sequentially performing two reactions without isolating the intermediate aldehyde or ketone give the final product of the reaction shown below. Hint the isolated products were derived from the larger portion of the structure disregard stereo chemistry. And here we have two incomplete reaction schemes showcasing the starting material that notably is an Keen reacting with a series of reagents. Now, in this particular case, the reaction in question is the tandem application of carbon carbon bond forming reactions with reductive oz analysis. Now, the first step of both of these reaction sequences is the reaction of our starting all keen with ozone at a temperature of negative 78 °C recall that this is the first step of oz analysis. So if we take a generic alkene here, let's say that on, on one end, I have our groups one and two. And on the other end, I have an R group three and a hydrogen. The first step of the sequence is the first step of the oz analysis reaction with ozone at a temperature of negative 78 °C. This step is very important for the oxidative cleavage of all keys because at this point, the two carbons of the starting double bond will now be separated from each other to create an Ozon intermediate. So the ozona intermediate looks like a five member ring where we have one oxygen connecting to two carbons and an oxygen to oxygen bond as well. So the two carbons of the ring were the two carbons of the starting a king. So on the left side, here, I have our groups one and two. And on the right side, I have our group three and hydrogen. So at this point, we have our Ozon intermediate, right? And this is where our gringer reagent comes into play or becomes relevant because recall also that the unique thing about a green yard reagent is that it makes carbon nucleic. So here one equivalent of the green yard region is going to go ahead and attack both carbons from the starting all keen and collapse the Ozon night intermediate. That means that we're going to end up with two distinct products. So one of them comes from an a keen or the alkene carbon to the left side, which was connected to our groups one and two. But now it's also going to be connected to the r group of the gringer reagent let's call that R prime. However, because of the oxygen atoms in the Ozon intermediate, we're going to end up with two oxide ions. So are other alcide ion includes the all keen carbon to the right side here with our three A hydrogen and our prime from the green yard wage. Now, the last step is the protonation of the oxide ions to create two alcohols. However, in this case, we're only going to focus on one alcohol product that stemmed from the larger fraction or the larger fragment of the structure. So with this in mind, let's go ahead and begin with part one. All right. So here considering the structure of part one, if we consider the two sp two hybridized carbons of the starting A Ken, we can get a sense as to what fragments are going to be present in either or oxide isle. Because if I imagine a split going down the middle of the double wand, I can see that the fragment to the left involves a five carbon ring. Whereas the other fragment to the right side involves a meth boxy group because the five carbon ring is physically larger that's going to have or that's going to produce our larger product that is the ultimately isolate isolated product. So when drawing the products of this three step sequence, the first thing I'm going to do is draw out those fragments of the double bond separately from each other. So now on each carbon atom that came from the double bond itself. I'm going to add a hydroxy group and the R group of the green yard region, which happened to be a fennel group. So at this point, we have two alcohols as are products of the sequence. However, once again, because the five carbon ring was the larger fragment, that's going to be our major product here. So for part one, the major product is 34 dimethyl, one phenyl cyclen one all. So now let's do the same thing with part two. All right. So now for part two, once again, we can go ahead and imagine a split going down the double bond of our given starting material. This leaves us with one fragment involving a six member ring and another fragment involving only a methylene carbon thats a ch two. So because of this, our major product is going to be the alcohol that includes the larger fragment. It being the six member ring, right. So first, I'm going to draw out each fragment separately and then on each of the carbons from the starting all keen, I'm going to go ahead and add a hydroxy group and a fennel group that came from the Green Yard Regent. So therefore our larger and therefore our isolated product is going to be six methyl, three phenol bicycle, 311 hep 10, 30. And there you have it. If you watch this video all the way through thank you so very much. And I hope you found this helpful.

Key Concepts

Ozonolysis

Tandem Reactions

Non-Aqueous Reaction Medium

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