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:

a.

Accepted Answer

All right. Hello everyone. So for this question, let's go ahead and draw the expected product from the reaction of the molecule shown below with hydroxide and water. Ignore stereo chemistry. Here are starting material is cyclo dode canine 15 Dion, which notably is a di ketone reacting with hydroxide. That is if you recall a strong base based on this combination of reagents, the ketone is going to react with the base to ultimately produce an Enola. So lets talk a little bit about EOL late chemistry. More specifically the Aldo condensation because we are dealing with two ketones when it comes to a ketone, right, we have two R groups on either side of the car bottle. Now, the two carbon atoms that are directly adjacent to the car bono itself are known as the alpha positions. So when an alpha proton is subsequently removed, and there's a negative charge on the alpha carbon, that species is known as the EOL late that's resonant stabilized. And so the Enola can attack another key toone in a nucleic addition reaction, which means that the carbon atom of the EOL late with a negative charge can directly attack the that pushes the pi electrons of the carbot upwards towards oxygen. And it therefore combines these two carbon chains. And so eventually, we end up with a beta hydroxy compound where the second carbel is converted into an oh group. This is after the addition of some acid, but we can take this a step further if heat is added to the system. So if heat is added to the system, then the hydroxy group of the beta position can be eliminated. And that's going to result in an alpha beta unsaturated compound. But in this particular case, there is no mention of heat in the reaction conditions. So we're assumed or rather it's assumed that we can stop after the condensation step with the beta hydroxy product. And so from here, the reaction involving the given starting material is going to proceed in much the same way. But the difference is that it's going to be an intra molecular reaction because the two car bono groups are in the same compound in the same parent chain. So now let's analyze the starting material to understand which alpha position is going to form the Inuit. Now, we have two ketone functional groups, which means that there are a total of four possible alpha positions that can be detonated. Now, there is a symmetry to the molecule which means that the alpha positions I've labeled here as one and two are equivalent to each other and the ones labeled three and four are equivalent to each other as well. So here, because this is an intra molecular reaction, the product is going to have two ring structures. So the idea is to compare the possible alpha positions and understand which rings will be formed depending on which one of them becomes the ulate. Well, let's go ahead and count. And for this demonstration, I'm going to focus my attention on the alpha positions labeled one and three towards the left side of the structure. So the first possibility is to detonate the carbon labeled number one and make that position the EOL late. That's going to result in a new six member ring after the innate attacks. The second car bono, this means that the remaining carbons on the other side will be joined together into an eight member ring. So let's compare the other possibility. The other possibility involves the formation of a four member ring and a 10 member ring. Now that is not necessarily favorable, right? Because the six member and the eight member rings would be the more stable products because they experience less strain. That's what we're going to see here. We're going to see the formation of the more stable rings. So let's go ahead and draw that. And I'm going to go ahead and re label my carbons here for each ring. Lets start with the six meri. So now that we have the six member ring, I am going to go ahead and label my carbons just to better keep track of all my substituents. So on position number two goes my cart Bonnel and on position number six is the beta hydroxy group. So now the bond between carbons one and six of the six member ring is going to be the bond between carbons one and eight in the eight member ring. All right. So I did have to briefly pause the recording just to make sure that I have the right amount of carbons. But here we have an eight member ring that has no additional substituent. And so there you have it. Here is our final outdo condensation product and just to clarify this is Aldo condensation, but it's disregarding the dehydration step. And so with that being said, thank you so very much for watching. And I hope you found this helpful.

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

Key Concepts

Acid-Base Reactions

Functional Groups

Reaction Mechanisms

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