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:

d. Chemical reaction diagram showing the synthesis of a compound from given starting materials, including structural formulas and arrows.

Accepted Answer

Welcome back everyone to another video propose a synthesis for the molecule shown below, starting with the given compound. The given a compound is meloni er and we are producing cyclohexane carboxylic acid. Now, we can perform the synthesis taking into account the acidic hydrogens at the alpha position relative to each of our era groups. Now, what we're going to do is just use the de protonation step, right? Because those hydrogens are significantly acidic and we are going to form our carbon which is re stabilized, allowing us to produce in late. So let's see how that actually works. First of all, we're just going to read through our starting material and specifically our Melan ester. And in this problem, we are going to learn efficient techniques behind the meloni ether synthesis. So we got our ether and now we know that we have two acidic agens at the alpha position. What we're going to do is first of all, remove one of them. So let's expand the first one. And for this purpose, we want to use a strong base. Let's suppose that we choose oxide. So let's choose a th that is our strong base and let's indicate the dep protonation stuff. So we're capturing our hydrogen, we can essentially show that we are forming our carbon ion. Since there is no necessity to show in a late, we can just add that negative charge. And now what we can do from here is think about the formation of that six membered ring. Let's assume that we are just going to form that six membered ring at the negative charge, right? One of our carbon atoms will belong to the negatively charged position. And essentially what we're going to do is we're going to choose an alkyl hide, which would eventually allow us to perform an intra molecular substitution reaction if we're forming a six member ring and we already have one of our carbon atoms, that means we want to choose a five member chain. So let's draw 12345 carbon atoms. And what we want to do, we want to make sure that we are adding two halogens on each end. So let's suppose that we choose bromine atoms, bromine at position one and bromine at position five. The reason why we're doing this is because we have two hydrogens and we can substitute each hydrogen with carbon. OK. So now let's see what's happening. We're going to take our alky hide. And as we know, we can perform a nucleic attack at the electro position, followed by the loss of the living group, meaning we're observing a basic substitution reaction let's go ahead and show the result on product. Bromine is a perfect leaving group. That's why we're choosing it as our halogen. And now we're going to bond that chain to the original negatively charged carbon. So now we're going to draw one, two, three or five carbon atoms and we know that the fifth one, it will essentially have that bromine attached to it, right. So we still have that bromine. Now, what we're going to do from here is just repeat those steps. So first of all, we're going to add our own base that be oxide, we will remove the remaining hydrogen form under the carbon, which will essentially a sac the other carbon that is bonded to bromine and it will allow us to form that six member grain. So let's throw the resultant carbon, we will simply remove that implicit hydrogen at our negative charge, perform the nuclear attack step well, by the loss of the leaving group. And this is where we can say, OK, here we are forming our six member grain. Let's count the number of carbon atoms. That'd be 123456, right? So what we can essentially do is simply draw a six number ring. And now what we notice is that this six membered ring, we'll actually have those two ester groups. So we're just going to throw them out. So one of them that would be our co oh and there's that Ethyl group similarly, on the left, we are going to do the same thing. We're going to add co double bond another oxygen and an ethyl group. Well, then, so now we got our rank. The question is, how do we arrive at that answer? Well, basically, since we have a dias first of all, we can perform an acidic hydrolysis. And we know that during the acidic hydrolysis, we can form a carboxylic acid and an alcohol. So first of all, if we add, let's say hydrochloric acid as our strong acid, we're going to convert that dier into a dic carboxylic acid, which essentially means that we're getting rid of our ethanol molecules and we're getting our dic carboxylic acid with the target molecule. And finally, we can perform the decarboxylate. Meaning if we heat the results on dic carboxylic acid, we're just going to remove one coo h in the form of carbon dioxide. And that hydrogen will be used in the protonation or specifically hydrogen will just replace the whole carboxylic a group. And this is how we can essentially get our product. So now let's summarize all of our steps in a simple synthesis. So we're going to summarize the whole synthesis as follows. Now, once again, we will just redraw that Melano caster. So we have our carbon, right? So basically those be our ethel groups. Now let's add that to oxygen. OK. So now we have our ST material just as we discussed in the first step of the reaction, we're going to use our strong base. And for that purpose, we're going to use oxide. And then we are going to add our alkyl halite one comma five dye substituted, in particular, as we said, we're going to add 1 to 3 or five carbon atoms and bond two bromine atoms on each end. So that's our one comma five dye substituted alkyl bromide. OK. So now what we're getting here is the intermediate that we discussed. So we're just going to essentially replace the alpha hydrogen with our alky chain. So that'd be 12345 carbon atoms and bromine. OK. Well done. And now what we want to do from here, just as we said, we want to make sure that we once again treat our structure with a base. And the reason why we're doing this is because we want to repeat that substitution reaction once again, we're going to use exactly the same you. So sign and we're going to say that here, we are going to form our rank exactly what we wanted. So we are going to add that structure. And now our final step is to make sure that we are performing the high tr isis of the ether as well as decarboxylate. So what we can say is that we can perform everything at once, right? If we use hydrochloric acid and we're going to add water because that's a solution of hydrochloric acid as well as heat, we will be able to simultaneously convert the ester into acid specifically, we will get the carboxylic acid. And then since it's heated, we will also obtain our decarboxylate, which will allow us to achieve the final product. So we're going to get rid of everything and just read through the final product that we got. So let's add that Kook Silas group and we indeed got cyclohexane carboxylic acid as our final product. Let's label our synthesis scheme. And we can say that we are ready to conclude the video. Once again, we chose to remove our alpha hydrogen using a oxide. We formed our ring using a one comma two. I'm sorry, one comma five that substituted bromide having formed our rank structure, we simply performed the acidic hydrolysis and decarboxylate to get our acid. Thank you for watching.

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

Key Concepts

Organic Synthesis

Reaction Mechanisms

Functional Groups

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