Using any necessary reagents, show how you would accomplish th...

Question

Using any necessary reagents, show how you would accomplish the following syntheses.

(e)

(f)

Accepted Answer

Hi, everyone. Welcome back. Let's look at our next problem. It says, show how the following syntheses can be accomplished. And we have two reactions shown here. Box A and box B both start with the same initial reactant. A ring containing five carbon atoms and a nitrogen and the nitrogen has an ethyl group attached to it. So in both cases, we start with a tertiary mean because the nitrogen has the two carbons in the ring attached to it. And then the ethyl group in box A, our product is an a mean oxide, an oxygen atom has been added to that nitrogen. So this reaction to box A is an oxidation of a tertiary. I mean in box B, our final product the ring has been broken open and we now have a chain, the nitrogen has an O H group, this ethel group. And now instead of the ring, it's a five carbon chain with AC C double bond at the end. So we know that AC N bond was broken since the ring was opened at the nitrogen and a new C C double bond was formed. Well, when we have that nitrogen there, we know, and we, we've added, should say, excuse me, we've added an oxygen, we've oxidized the amine since we have that O H group. When we see that combination, the oxidized amine C N bond, broken new C C double bond that tells us that we had a cope elimination reaction. So we can see when we look at this, then that our box A is actually the first step along the way to the reaction in box B because we oxidize the amine that takes place in the cop elimination as that reaction is a self elimination. So let's show how we can do these reactions. Well, box A is really straightforward. It is simple and easy to oxidize tertiary means. And so all we need to do is add hydrogen peroxide and that reaction will take place. So I've just written H2O two under the reaction arrow because that will generate this molecule. It has now a positively charged nitrogen and we've added a negatively charged oxygen atom to the nitrogen very straightforward. Now that molecule, so I'm going to dry it underneath the product that was generated in box A is what's going to take part in the cob elimination. So when we look at this and try and predict our cope elimination, it could actually happen in two directions because there's a potential for two different alpha carbons. When we look at our nitrogen, we could use the carbon next to the nitrogen in within the ring where the ring gets broken open or you could use the carbon in the ay group, that's a substituent on the nitrogen. However, in this case, we're given the product we're looking for and that involves breaking the ring open. So we don't have to decide which way to go we're told in our problem which way the reaction happened. So I know that the bond that will break is that carbon nitrogen bond in the ring. So I'm gonna draw a little squiggly line just to remind us that that's what we're breaking across that carbon nitrogen bond. I'm also going to number the carbons in the chain because that helps me when I'm drawing the product to make sure I've got them all in this case. Again, it's, it's pre drawn for me, but it's just a good habit I find to get into so on the carbon that will still be bound to the nitrogen. I'm going to call that number one and then just go around the ring and number the five carbons to keep straight. So if I'm breaking the C N bond uh between carbon five and the nitrogen, then I know carbon four is my beta carbon and that's where the proton is going to come from in my coal elimination reaction. So I'm going to draw hydrogen atom on that carbon just to keep track of everything going on. Now, my negatively charged oxygen atom is going to take the proton from that carbon, the electrons that were in that bond come to form the new C C double bond and the electrons that were in the bond between the carbon and the nitrogen now go on to the nitrogen, breaking that carbon nitrogen bond. So our end product will be what is shown. And all that we have to do to this, we added hydrogen peroxide to generate this ami oxide as we saw it in box A. So all that we have to do in the second step is heat. It, this is a reaction that will take place just with a little heat added, we're increasing entropy. And so we only have to heat. So now to get generate the product again, it's shown in the box. But just to show how you would draw that out if you didn't have it given to you, I always like to start with my nitrogen. Oops I lost a bond there. The nitrogen now has an O H because the oxygen picked up the proton and it has its eyl group and then we've opened up the ring. So we have carbon one and then all the way around to five carbons on that chain coming from the nitrogen and carbons four and five now have a double bond at the very end of that chain. So we've generated the product in box B. So we know that for the synthesis to take place, we needed to add hydrogen peroxide and heat. And that gave us first the amine oxide and then we went through the cope elimination, opened it up and got that chain with a new C C double bond. So that's how these reactions took place. We'll see you in the next video.

Using any necessary reagents, show how you would accomplish the following syntheses.
(e)
(f)

Key Concepts

Cope Elimination

Amines and Their Reactivity

Reaction Mechanisms

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