Starting with cyclohexane, how could the following compounds b...

Question

Starting with cyclohexane, how could the following compounds be prepared?

e. Structural formula of cyclohexanol with a methoxy group, illustrating the Grignard reaction process.

Accepted Answer

Hey, everyone. And welcome back to another video show how the following compound can be prepared. Starting with cyclopse. The given target molecule is three methyl cyclo and one in and we went to synthesize it from cyclopse. So let's draw cyclopse which is our five membered cyclic al came. Now, first of all, because we know that al canes and cycle alkanes generally are quite unreactive. We want to use one of the main reactions that allows us to add a halogen to our structure. Let's choose bromine for the purposes of this problem. And we know that according to the free radical mechanism, if we use bromine in the presence of light or heat, we will be able to replace one of the hydrogen atoms with bromine. And because all of the hydrogen atoms are equivalent due to the symmetry of the ring, we can state that any of the hydrogen atoms can be replaced with bromine. Now, if we have bromine in our structure, it can act as a good living group, right? We have to think carefully about the double bond of the target molecule. We are forming an alk. If there is a double bond we can actually easily format using an elimination reaction. Let's think about the beta hygienes that we have. Now their two equivalent beta hyen. Once again sees the line of symmetry passing through our compound because they are equivalent. It doesn't really matter what sort of base we use, whether it's a bulky base or a small base. But for the purposes to make sure that we are producing our elimination product as the major product, we're going to use a strong base, which is a poor nuclear file. And one of those bases would actually be TB oxide, right? Because it's a bulky strong base, which is also known to be a poor nuclear file. So what's happening is that when we think about our beta hydrogens, we are essentially going to remove one of them in an E two elimination manner. So we are going to capture the proton form a double bond followed by the loss of the leaving group. And that allows us to get cyclopse, we are forming our double bond. Now, from here, we have to carefully think about our next step. And if we compare cyclopse to three methyl cyclo one in, we notice that we're actually simply adding a muscle group at the ay position. And because we are considering the allylic position, we can think of the allylic bro in the allylic bro reaction, we want to use N BS in the presence of heat and peroxide. So this is our set of regions which is required to perform the ayro. And once again, because the structure has a line of symmetry, the two allylic positions are equivalent to each other. And we are going to replace one of the two equivalent hyen atoms. Actually, there are two. But once again, they're all equivalent, we're going to replace one of them with grooming. So if we draw our cyclo alk, we are just going to add bromine to the allylic carbon. And now from here, our last step is to actually replace Broin with the metal group. And for that purpose, we can remember the Gilman Regent which allows us to perform a successful conversion. The Gilman regent is an organ of metallic Regent specifically, let's remember that the general form of the Gilman Regent is R two, which represents that there are two alki groups that are bonded to copper that carries a negative charge and lithium which carries a positive charge and acts as a counter I and whenever we react the Gilman Regent, right? So we can label that the form of the Gilman regent is RT copper lithium. Whenever it reacts with an L kill Hali, such as R prime X, we are simply going to form our R R prime bond, right? So in this case, if we have bromine bonded to our cycle, and if we choose an appropriate agent, and in this case, because we need to have a methyl group, we are going to choose R to be the metal group. That means we are going to use C H three, which is the metal group, right? Since we want to incorporate it into our region, we're going to use two times of that copper lithium. And based on what we said, we are going to form our R R prime bond. So if R is the metal group and our prime corresponds to our cyclopse ring, we are going to bond them to get the final product. So let's draw the final product. That would be our five membered ring. We have our double bond and then we are going to add the metal group to our ring. And that's how we get our final product. So let's label our synthesis scheme and let's summarize our findings in the first step. We are going to use the free radical mechanism in which cyclo Alcaine is converted into kill bromide, right? We're using bromine in the presence of light. In the next step, we are using an E two elimination with a strong base, which is a poor nu file so that we can form our Cyclo alkin as the major product. Then we are performing the AY bro using N BS heat and peroxide. And after that, using the Gilman agent, we are able to replace bromine with the methyl group. That will be our final answer. And I would like to say, thank you for watching. I hope to see you in the next video.

Starting with cyclohexane, how could the following compounds be prepared?
e.

Key Concepts

Cyclohexane Structure and Reactivity

Substitution Reactions

Functional Group Transformation

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