Show how cyclohexanone might be converted to the following

Question

Show how cyclohexanone might be converted to the following δ-diketone (Hint: Stork).

Accepted Answer

Welcome back everyone to another video show how the following 15 day ket zone can be synthesized from cyclops. We're given the material that we want to synthesize and we want to think about its synthesis if we start with psycho penal. So when we consider the starting material to get cyclobenz, we simply want to eliminate our side chain. We want to remove it from the structure, we can forget about it for a bit and just read through the starting material. So now we have cyclops, what we're going to consider in this problem will be the Michael edition of enemys. So this is the idea that we're going to use in the synthesis specifically what we want to do at the beginning is synthesize an enemy. So we're going to start with our carbonyl compound. And let's remember that to synthesize an enemy, we can essentially use piperidine, we can draw our six member grain. So let's go ahead then to that and let's bond nitrogen as well as one hydrogen, right? We got the structure of piperidine. And let's remember that for the enemy information, we want to use acidic conditions. So we're also going to include hydro and essentially what we're going to do is just form our enemy. Let's see how that looks like. First of all, what we want to do is just eliminate our carbon. We want to introduce our carbon nitrogen bond. So let's bond our piper in ring to perform that bonding. Of course, we will not have to include that hydrogen. So we are just emitting that hydrogen and essentially introducing our newly formed carbon hydrogen bond. So now let's draw that six membered ring. And now let's remember that just as the name suggests, this is an enemy. So we need to add that pi bond within the five membered ring. OK. So now that we have our enemy informed, our next step is the Michael edition, right? Because we got our enemy, we have our micro donor with a high electron density. We have our pi electron density. And now we want to choose an appropriate Michael acceptor. Specifically, we want to notice that when we consider our side chain, the number of carbon atoms that we are adding would be equal to four, right? We have 1234. So we're going to take our side chain, we're just going to read through it. And let's remember that one of the best micro acceptors would be alpha, beta unsaturated ketones, right. So we can essentially add our pi bond between our alpha and beta positions. Specifically adding it there will allow us to get the side chain that we want, we will soon understand why. So this will be our micro acceptor because it essentially has an electro pilic beta carbon. Right. Essentially, we can check that by resonance if we shift our pi bond, that carbon atom will be electron deficient and we can shift it because we can break the adjacent pi bond we have that conjugation. So now what happens is that we are essentially going to use the lo pair on nitrogen, which will allow us to form a pi bond between nitrogen and carbon. We will perform an attack on the beta position. And when we do that, we will essentially shift our pi bond, break the adjacent pipe bond and get our next intermediate. So let's go ahead and show the result and intermediate, we can simply take our structure. Now, what we're going to do is simply add a double bond between nitrogen and carbon based on our arrows, we're breaking the pi bond within the ring and we are simply bonding our four member chain. So let's go ahead and do that. So that's one to 34 carbon atoms. We have our iny eight anion formed, right? Oxygen will have a negative charge. So there will be a pi bond. And now it is really important to also understand that Nel will have a form of positive charge. Now, finally, we can produce the required product in one single step and essentially what we want to do is just perform an acidic work up. The reason why acidic work up will form our final product immediately is because first of all, we will form an EO now we have an eno at anion when we form an EO it will undergo a keto eal totem organization. So that means if we think about the previously drawn intermediate, the acidic world cup will allow us to form a ketone out of an anal because the Keone form is more stable than an anal form. OK. So now after that to our eno converts into a K song. And finally, let's also remember that the stick work up will allow us to convert the I minium Ron back to the carbonyl group. In other words, we are just going to get rid of our rank as well as nitrogen and get our oxygen back. So this is how simple it is. We have achieved two steps by simply adding a strong asset. We have converted our innate into eno which undergoes keto eal to organization to produce our carbon group. And in addition to that acidic work up allows us to restore our original carbon group because the minium ion reacts with an acid to produce a ketone. That would be it. We can label our synthesis and include the video. Thank you for watching and I hope to see you in the next one.

Show how cyclohexanone might be converted to the following δ-diketone (Hint: Stork).

Key Concepts

Cyclohexanone

Stork Enamine Reaction

Diketones

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