Draw the product of the reaction of each of the following comp...

Question

Draw the product of the reaction of each of the following compounds with a base:

b.

Accepted Answer

All right. Hello everyone. So this question is asking us to provide the expected product from the reaction of the molecule shown below with hydroxide ion and water. Here we are to ignore stereo chemistry. And so our starting material is 24 oxo exel cyclopentolate. So notably, right, our starting material is in fact a di ketone and we're reacting this di ketone with a base in the presence of water. Now, the reason why it's important to mention that we have a ketone functional group present is because the alpha carbon of an aldehyde or a ketone can be derotated in the presence of a base to produce an EOL ion. And just to go ahead and clarify the alpha carbon refers to any carbon that is adjacent to a carbonel ketones have two alpha carbons and aldehydes only have one. And so here we have several different alpha carbons to choose from for this dep protonation reaction. And so from there, EOL lights, if you recall are nucleic, which means that the olla formed on one side of the molecule would then attack the carbo on the other side. This is going to go ahead and create a ring. So before we go ahead and talk about how to draw the product, given this specific starting compound, let's just review the reaction in question. So as mentioned before, the EOL late is formed by dep proton an alpha carbon adjacent to a carbo in an aldehyde or ketone. And so if that inno late ion is in close enough proximity to another car bottle, it's going to attack the electro pilic car bottle directly which displaces the pi electrons of the car bottle towards the car bottle oxygen. And so what this does is establish a new bond between the alpha carbon and the other car bottle. The other car bals oxygen is going to have a negative charge thereby forming an oxide ion. And so the reason why it's important to have some water in solution is to provide a source of protons to go ahead and potentate that oxide ion and ultimately produce and alcohol on that side of the product. And so from here, our new product is a beta hydroxy carbonel compound. We could in the presence of heat. Take this step a little bit further and proceed with a dehydration reaction, removing the hydroxy group from that beta position. This again, in the presence of some heat would establish a new pie bond between the alpha and beta positions relative to our car bal that remains in the structure of the final product. So this creates an alpha beta unsaturated carbonel compound. But because heat is not mentioned in this particular question, we're only going to focus on the creation of the beta hydroxy car bottle compound. The reason why I talked about condensation, which is the elimination step anyways or in other words, dehydration is just for your information. But here we're going to stop at the beta hydroxy product. And so with this in mind, let's actually talk about the product. And so to talk about the product, let's go ahead and label all of the possible alpha carbons that are available to us in this starting material. On the cyclopentolate ring are two alpha carbons labeled one and two respectively. And so on, the ocho group branching off of the cyclops node ring are two more labeled three and four. So right away, the first alpha position that can be eliminated in terms of which one is going to create the Enola is position number two. And the reason for this is because position number two is tertiary and because it's tertiary, if it were to create the EOL late, then there's going to be some steri hindrance associated with that tertiary position, attacking the other car bal. And so now the question is between carbons 13 and four, which one is going to produce the eno like now because this is a di ketone and we're talking about an intra molecular, our dog reaction, right. Our goal is to produce the most favorable ring possible recall that in the context of organic chemistry, five and six membered rings tend to be the most ideal in terms of stability. And so any alpha carbon that will create either a five or six membered ring is most suited for this reaction. So here, let's go ahead and focus on carbon number one or the alpha position labeled number one. Just for the moment when trying to examine the ring that this is going to form. The idea is to count the number of carbons between that alpha position and the car bal that it's going to attack, including both the alpha position and the car bono carbon. And so when you do this, you recognize that this is going to form a seven member ring, which is not the most ideal, right? It's a little too big. So let's compare this now to the alpha carbon on the very right side that I labeled number four. Previously, if I count the number of carbons I will form in this particular ring, we're also going to have seven carbons, which is not ideal, right? So now let's go ahead and compare this with the alpha carbon to the left of the second carbo counting the number of carbons in between this alpha position and the other car bal will produce a five carbon ring, which is our ideal here. And so with that being said, we've established which alpha carbon is going to attack which car bottle and therefore what the product is going to look like. So let's go ahead and draw that now first, because our product is going to form a five member ring. My recommendation is to go ahead and draw the new ring first. Not only do I recommend drawing out the new ring, I also recommend labeling your carbons one through five just to keep track of substituents branching off of it. And so from here, it's going to be easier to recognize that branching off of carbs four and five of the new ring is the ring of the cyclopia known in the starting material and so on carbon number five, which if you recall is the car bottle that's going to be attacked, there should be a hydroxy group as opposed to the car bottle. And then from here branching off of carbon. Number one is the car bottle on the very right side of the starting material and there you have it. So here is our final product for this reaction. And so if you watch this video all the way through, thank you so very much. And I hope you found this helpful.

Draw the product of the reaction of each of the following compounds with a base:
b.

Key Concepts

Acid-Base Reactions

Functional Groups

Reaction Mechanisms

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