Design a synthesis for each of the following compounds using t...

Question

Design a synthesis for each of the following compounds using the given starting material:

b. Chemical structures of two compounds on the left leading to a product on the right, with arrows indicating synthesis.

Accepted Answer

Welcome back everyone to another video for the following compound, write the synthesis using the provider's starting material. The starting material is diethyl 35 dimethyl heptane di eight. And we want to produce 45 dimethyl two prop cyclohexane one on. In order to provide an efficient synthesis, we first of all want to understand that we're starting with a di ester and we're forming a ketone. And for that purpose, one of our steps will be the use of diman condensation. The reason is very simple. We have a symmetrical dire and in diman condensation, we manage to get a beta Quito estra. So it's not really the final product that we observe, right? It's far from that, but it will essentially be helpful for us in the development of that six membered ring. What we can modify in our reaction scheme is that first of all, we notice that we have those two methyl groups, right? So we can essentially add those two methyl groups to our reaction scheme. And this essentially it tells us that we will have them in our ring. OK. And now what we're going to do is just replace those or groups with the actual groups that we have, we noticed that we have our oxy groups. So let's simply add our oxy groups that will be oxygen with an ethyl group on its head. Now, we want to use a base and we want to use acidic conditions. So let's choose some strong base. Let's suppose that we choose meide ch 30 negative to perform the first step of our reaction. And for the second step, we want to use a strong acid. So let's suppose that we go with hydrochloric acid hcl, this allows us to produce the beta keto era. We can once again replace our r with the ethyl group that we proposed. And of course, after that, what we notice is that we want to replace that ester group with an alkyl chain. We want to get apy chain. So let's think about this stuff. We can essentially make use of a substitution reaction. Why not? We know that we have an implicit acidic hydrogen in between the two carbonyl groups? So that alpha hydrogen based on its high acidity can be removed using a base. Once again, number one, we can use a strong base such as metox. And when we remove that acidic hydrogen, we will be able to produce a resonance stabilized carbon that carbon can act as a Nile on a specific alkyl Halide. Because we want to add the propel chain with three carbon atoms. Our alkyl Halide of preference will be prop bromide 1 to 3 carbon atoms. And let's add bromine to position one. We can choose any good living group, we can go with chlorine. But for the purposes of this problem, let's suppose that we choose bromine. So the negatively charged carbon will attack the electro carbon followed by the loss of the living group. And this is where we will have our substitution. So what we're going to do is simply replace that acidic hydrogen with a propel chain. So let's remove hydrogen and let's draw the resultant propylene, one, two and three carbon atoms. OK. Well done. Now, the final step that we want to perform would be the elimination of this ester group. Right? We don't have, we don't want to have it and we essentially want to replace it with hydrogen bond, it's the ring. So now how do we do that? Well, essentially we can perform the acidic hydrolysis of the ether to form a carboxylic acid. So that would be our first step that be hydrolysis in which we would produce the carboxylic acid as well as in alcohol, we would produce ethanol. And then when our carboxylic acid is obtained, we can use decarboxylate, which will allow us to successfully remove coo h from our structure in the form of CO2, right, that hydrogen will be used in the carbon hydrogen bond formation with the rank, we can perform everything in one step. So first of all, we want to perform the hydrolysis step. And for that purpose, we want to use a solution of hydrochloric acid. So this allows us to form a carboxylic acid group. And of course, we can then heat our mixture right, for the decarboxylate reaction to take place. And essentially, it will remove our carboxylic acid group and it will be replaced with an implicit hydrogen atom. So let's go ahead and draw the structure that we obtain. Simply speaking, we're going to remove this part. We know that there will be an A H right during the acidic work up. And then when the mixture is heated, we're going to remove our carboxylic acid, replace it with hydrogen and we can adjust the bond angle or the prop group. So essentially we have 123 carbon atoms and now that we got our file structure, this is exactly what we wanted to synthesize. We can just label our synthesis scheme and conclude the video. That would be it for today. Thank you for watching and I hope to see you in the next one.

Design a synthesis for each of the following compounds using the given starting material:
b.

Key Concepts

Retrosynthetic Analysis

Functional Groups

Reaction Mechanisms

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