Explain why ethyne should be alkylated before, rather than aft...

Question

Explain why ethyne should be alkylated before, rather than after, nucleophilic addition.

Accepted Answer

Hey everyone and welcome back to another video for the given transformation. Why should Ethan be calculated before an equival edition? And not after we're given a transformation from Ethan to an alcohol. And there are four answer choices. A the calculation of oxygen will compete with the calculation of carbon. If Ethan is calculated before the nucleic edition B, the calculation of oxygen will compete with the calculation of carbon. If Ethan is calculated after the nucleic edition, C nucleic edition would not occur, but Ethan will be oxidized if Ethan is cul before the nucleic edition and D al population would not occur if Ethan is alate after the nucleic edition. So let's understand what's happening here and what we're trying to do, we are starting with Ethan, which is acetylene. That's the simplest possible alk kind. And we understand that according to the question, we want to calculate it before we perform the nucleic addition to a carbonyl compound to form an alcohol. Let's recall that the very first step is the usage of sodium amide one equivalent to detonate acetylene. And in that case, we are forming the ayite and ion that satellite and ion can be calculated if we simply use an alkyl group, let's say, are one bonded to bromine, right? Essentially, we are performing an S N two reaction so that we are creating a longer chain alkin during an essence reaction, the negatively charged carbon attacks the Electrophysics Center followed by the loss of the leaving group. So that's how we are bonding R 12 hour S B carbon. Now, in the next step, we want to form a new file. So once again, we're using sodium amide, even though one equivalent is sufficient, we don't need to control it because we only have one acidic hydrogen so that we are forming another a satellite and I am. And afterwards, this is what we want to have here. We want to perform the new edition. Let's suppose that we are choosing a carbonyl compound such as a ketone. Let's take an example, a key zone with R two and R three groups. So now that nucleic addition would essentially result in the attack at the Electrophysics Center, followed by the cleavage of the pi bond. And now we have to think about the tetrahedral intermediate forms, oxygen will have a negative charge. There would still be R two and R three groups bonded to that carbon. And now we are essentially introducing our S P hybridized carbon which is now bonded to the carbon that initially belonged to the carbon group. Now, the very last step is the protonation step and we are forming an alcohol. That's the overall transformation. As suggested by the problem, we simply can use bidi ion a very weak acid or more accurately speaking a relatively weak acid, right? Because we are trying to proton oxygen, we don't need to use a strong acid. So when we use the perineum ion, we are finally forming our alcohol a structure. And then let's think about an alternative scenario. OK. So we have formed our alcohol and now we are going to think about a different scenario. What if we first of all perform the nucleic edition? So we are going to take a satellite and we are going to treat it with our carbonyl compound. In this case, we have chosen a ketone with substitutions R two and R three. So thinking about the nucleic attack cleavage of the pie bond, we're going to get our tetrahedral intermediate just as we had it before. So we can actually even added, right? Because it'll be exactly the same. I'm sorry, not, not exactly the same, right? Because we're starting with a different satellite and ion. So now we are going to bond our S P carbon and on the terminal end, we still have an acidic hydrogen. So now if we deproteinate it, right, that's what we're trying to do in an alternative scenario. If we use sodium amide, we're going to detonate that because we still have an acidic hydrogen bonded to an S P carbon. So now we have an additional negative charge. And if we try to perform an calculation from here, we clearly see what's happening. Let's suppose that we are using R one bro. And now we have a competition. So oxygen is negatively charged and carbon is negatively charged. So we have a competition specifically, we can just re tool R one B R for each case. And we noticed that we can simply attack the electrolytic carbon by each of our nuclear files, right? And that means we are going to get too calculated atoms, one of them will be oxygen, another will be carbon. And in that case, we are going to get the following product, oxygen will be bonded to R one. Then we are going to get R two are three. And then let's draw the resultant final substituent and we are not even getting and actual alcohol. That's really a problem because we have a competition. So we clearly understand why the second suggested method doesn't work, right? We cannot perform nucleic addition at the very beginning and calculation later because we have a competition between oxygen and carbon. So going back to the answer choices at the very beginning, we can state that B is the correct answer. The calculation of oxygen will compete with the calculation of carbon. If Ethan is calculated after the nucleic addition, that's exactly what we saw. We definitely want to calculate it at the very beginning. And then we want to perform the nucleic edition to form an alcohol. Otherwise, we cannot even get that alcohol to, to that competition. So to conclude the correct answer choice to this question is A B. However, if you have any questions, don't hesitate to leave your comment down below in the comment section. And thank you for watching.

Explain why ethyne should be alkylated before, rather than after, nucleophilic addition.

Key Concepts

Nucleophilic Addition

Alkylation

Reactivity of Ethyne

Watch next