When the following compound is hydrated in the presence of aci...

Question

When the following compound is hydrated in the presence of acid, the unreacted alkene is found to have retained the deuterium atoms. What does this tell you about the mechanism for hydration?

Chemical structure of a deuterated alkene with a phenyl group, illustrating acid-catalyzed hydration.

Accepted Answer

Welcome back, everyone. The un reacted alkin and the following compound retains the deuterium atoms after it was hydrated in the presence of acid. Which of the following claims regarding the mechanism for the hydration process is are true. We given the acid catalyze hydration of the alkin and therefore answer choices. Number one, the formation of the Carbocaine is the slow step when the alkin is hydrated in the presence of acid. Number two, the removal of the ethereum and the formation of the alkin make up the second step of the mechanism. Number three, the Carbocaine reacts with water in the fast step of the reaction. Thus, it does not have time to lose a proton or the ethereum to rebuild the alk. And number four, in the slow step of the reaction, the Carbocaine reacts with water. To answer this question, we have to consider possible pathways for our reaction to take place. We have our starting keen and first of all, we know that this must be an acid capitalized hydration. So we're going to recall the first step. In the first step, we are using the acid that would be sulfuric acid. We are proton the alkene according to the more cognitive stro hygienist added to the less substituted carbon. But in this case, if we consider the ethereum as an isotope of hydrogen, we can simply say that the two carbon atoms are equally substituted. We're going to add our hydrogen to the carbon atom that has the ethereum because forming a positive charge at the cilic carbon will be more stable due to the resonance effect. Now, when we form our carbo Canion, the water molecule can act as a nu phyle to form the carbon oxygen bond and to produce ox sodium. Meaning what we're going to do now is just introduce our carbon oxygen bond. So let's go ahead and do that. And finally, we perform the de protonation, stop using another molecule of water or basically the conjugate base of sulfuric acid that's also acceptable depending on how we look at this mechanism. And then we remove hydrogen, one of them to produce an alcohol. On the other hand, when we form our carbo Canion, we can observe a different scenario. We can actually reform that keen removing hydrogen or the ethereum atoms, right, simply performing an elimination reaction using another molecule of water, which can act as a base. So if we consider the blue pathway, we get an, we can also draw the resultant alkin here because we are simply removing the ethereum, the hydrogen atom is still present. So we are going to add hydrogen instead of the ethereum, since there's an implicit hydrogen, right? So now we understand the sequence of the reaction and our possibilities. And we simply want to understand which one is the slow step and which one is the fast step. Let's assume that the first step is the slow step, which is the formation of the car. If this is the slow step, then the next step, addition of water will be our fast step in which water will quickly attack. The carbo can formed to produce ox sodium and finally define alcohol. This means the carbo canon formed will quickly react with water in a nucleic attack manner. And there will be no time to form the Alcaine, right? And this is consistent with what we're seeing. If the first step is the slow step, then the alkin will not be formed because the carbo canon will exclude the react with water. And this means all of the reacted alkin will retain the deuterium atoms. On the other hand, if the first step is the first step, then there will be an equilibrium between the alkene and the carbo Canion. Specifically, the carbo Canion will have enough time to undergo elimination, right? Because the hydration would be a slow step, so that the carbo Canion would have enough time to undergo elimination and the alkin would be formed and the uteri atoms would not be reformed. So we understand that based on the context of the problem, the correct pathway is that the formation of the carbo canon is the slow step. And then we have a fast step in which there's a nucleic attack by the water molecules on the carbo Canion. Going back to the answer choices, we may simply conclude that the first option is correct. The formation of the carbo canon is the slow step when the alkin is hydrated in the presence of acid, right, that's because we assume the second step is fast and it is fast enough to not allow us have any time for the elimination. And similarly choice, number three is also correct. The Carbocaine reacts with water in the fast step of the reaction. Thus, it does not have time to lose a proton or the ether to rebuild the al this is the consequence of the first slow step because if the first step is slow, then the second step must be fast. And if it is fast, then the Carbo Canon does not have time to participate in an elimination. The second option is incorrect because we are not removing the ethereum. Just as explained by the context of the problem. We are not rebuilding an al since the carbo Canion is reactive and it quickly reacts with its water. And number four, in the slow step of the reaction, the carbo Canion reacts with its water. That's incorrect because as we concluded, it is the fast step, the slow step is the formation of the car and those would be our final answers options number one and we thank you for watching.

When the following compound is hydrated in the presence of acid, the unreacted alkene is found to have retained the deuterium atoms. What does this tell you about the mechanism for hydration?

Key Concepts

Markovnikov's Rule

Carbocation Stability

Deuterium Labeling

Watch next