What stereoisomers are obtained from hydroboration–oxidation o...

Question

What stereoisomers are obtained from hydroboration–oxidation of the following compounds? Assign an R or S configuration to each asymmetric center.

a. cyclohexene

b. 1-ethylcyclohexene

Accepted Answer

Welcome back, everyone identify the products of the high tation oxidation reaction of the following compounds. Number one, cyclops, number 21 isopropyl cyclops, one in label each chiral center sr or S and we can begin this problem with the first structure cyclopse. We want to draw a five member grain and add a double bond. According to the problem, we are performing a hydration oxidation reaction. It requires us to use Bora in the presence of detroy fluorine, followed by the addition of an aqueous solution of hydroxide and peroxide. OK. So we have to recall that this reaction produces the anti marco of product we're going to form and alcohol. And we also have to understand that the alcohol group gets bonded to the less substituted carbon. But in this case, both carbon atoms are equally substituted and the structure is symmetrical, meaning there's only one possible product, we want to add hydrogen to the more substituted carbon. In this case, it doesn't matter which one. And we want to add the alcohol group to the less substituted one. So if we draw both of our possibilities, we get the same product. Now, the question is, do we have any chiral centers present. Well, the answer is no, we don't have any chiral centers present. Even though this reaction follows the sin addition principle. Since we are not forming a chiral center, there's no necessity to label any R or S configurations because the rank produces two symmetrical substituents. And therefore, we can move on to the second reaction. In this case, it states a one isopropyl cycle band one in. So let's draw a five member ring. Let's add a double bond. And now it states that carbon number one has an isopropyl substi, we are performing the same reaction. Number one, let's add bahrain in the presence of tetra hydro urine. And number two, I trioxide of water and peroxide. Those are the regions needed. And in this case, since the structure is not symmetrical, let's remember what we said before. We want to produce the anti mark cogni of products. So we're going to break the pi bond, we're going to add hydrogen to the more substituted carbon and the hydroxyl group to the less substituted carbon. So in this case, we are going to form two chiral centers. And let's recall that this reaction follows the sin addition principle, meaning hydrogen and hydroxyl will be added on the same face off the five member ring. So let's break the pie bond. And let's think about the possibilities. Let's suppose that hydrogen and hydroxyl are both added from the top base, which means that we're going to draw a wedge or hydroxyl hydrogen is going to be pointing up on a watch. But because it's implicit, we can simply draw a dashed bond or the isopropyl group enough for the second product, we can draw a five member ring. And we want to draw the second possibility. What if we add the two subscriptions from the bottom phase hydroxyl will be pointing down hydrogen will be pointing down, meaning the isopropyl group must be pointing up. And since we are producing two chiral centers, since their two carbon atoms with or different substituents, we need to assign the R and S configurations. Let's start with the first carbon that is bonded to the alcohol group. We want to identify priorities. Carbon is bonded to oxygen that has the greatest atomic number, right? Meaning oxygen gets where one, then we will have an implicit hygiene atom which gets priority number four because it has the lowest atomic number in between the two carbon atoms, the more substituted one gets a higher priority. So on the right, we get priority number two. On the left, see each two group gets priority number three. According to the Khan Ingle Pre Law rules, we can assign the rotation directly because priority number four is pointing down. So we go from 1 to 2 to 3, we get a counterclockwise rotation. And that means the configuration would be S OK. So we are done with the configuration of the first carbon atom, let's move on to the second one. Now, what substituents do we have? Well, essentially we have an implicit hygen pointing up it gets priority number four. And we also have we different carbon atoms. First of all, there is a one tertiary carbon atom, one secondary and also one secondary that is bonded to oxygen. So apparently the carbon atom that is bonded to oxygen will get the highest priority because oxygen has the greatest atomic number, the tertia carbon atom goes next and the secondary one gets priority number three, right? Because an increase in the degree of carbon essentially allows us to get a higher priority, right? Unless we have a hetero atom, a hetero atom such as oxygen will get a higher priority because it has a greater atomic number than carbon. And in this case, if we trace our path 1 to 2 to 3, we get a counterclockwise rotation, but we have to flip it because priority number four is pointing up. So the actual rotation will be clockwise. Meaning the configuration is R for the second structure. We don't need to do anything. We have the alcohol group pointing down instead of up, meaning we're flipping the configuration to R and the is isopropyl group is pointing up instead of down, meaning we're flipping the configuration to us and that would be it, we have our answers. So let's label them. Thank you for watching.

What stereoisomers are obtained from hydroboration–oxidation of the following compounds? Assign an R or S configuration to each asymmetric center.
a. cyclohexene
b. 1-ethylcyclohexene

Key Concepts

Stereoisomerism

Hydroboration–Oxidation

R and S Configuration

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