When piperidine undergoes the series of reactions shown here, ...

Question

When piperidine undergoes the series of reactions shown here, 1,4-pentadiene is obtained as the product. When the four different methyl-substituted piperidines undergo the same series of reactions, each forms a different diene: 1,5-hexadiene; 1,4-pentadiene; 2-methyl-1,4-pentadiene; and 3-methyl-1,4-pentadiene. Which methyl-substituted piperidine forms which diene?

Diagram showing the reaction of piperidine to form 1,4-pentadiene, with steps and products labeled.

Accepted Answer

All right. Hello everyone. So this question says as shown below, 13 beta diane is synthesized from polity. If a methyl substituted palain undergoes the same reactions, it produces 13 beta diene, two methyl, 13 bene and one for PTA diet, identify the methyl substituted poullain that is needed for the synthesis of each night. So even though we are given a reaction scheme of the Hoffman elimination, let's go ahead and walk through the reactions themselves using our methyl or methylated derivatives of pity. So recall first and foremost, that Hoffman elimination is an E two reaction in which a quaternary ammonium ion undergoes E two using a strong base. So recall that we have three methylated derivatives of P impossible. Our first is one methyl pro in which our methyl group is attached directly to the nitrogen of the ring. So what's happening during this reaction? Well, first and foremost, we have to make nitrogen a ordinary ammonium ion. So the way that this happens is by using an excess of in this case, methyl bromide with K two co three. And the second step, we add silver oxide and water. And so what this does is install an additional methyl group onto nitrogen directly and make our ordinary ammonium isle. So not only do we have our ordinary ammonium ion, we also have hydroxide here as a counter ion. So at this point now that we have our ordinary ammonium salt, the next thing that needs to happen is that the ring needs to open. And so this is done in the presence of heat. So as a result of this opening that I talked about earlier, we end up with a tertiary amine attached to the carbon chain. In this case, the carbons from the palain ring itself, but nitrogen still has two methyl groups attached to it to make it a tertiary, I mean at this point. And so the opening of the ring actually results in a double bond, meaning that our intermediate right now is nn dimethyl butte three in one amine. So this is where we get to the next part of the reaction, which is the actual elimination step. So let's clarify, right. The leaving group in question is the amino group itself. And so the alpha carbon is the carbon from the original palatine ring that's attached directly to the amino group. The beta carbon is the carbon next to the alpha carbon. And so what's happening here is that the proton gets removed from the beta carbon and the leaving group, the amine is attached to the alpha carbon. This creates a double bond between alpha and beta. And it's also worth mentioning that if there's more than one beta carbon possible or present, I should say, then there are going to be major and minor products. But it just so happens in this particular case that we do not have any other beta cars. So really we end up with one product. So here it's a multi step process. The first thing that happens is once again the addition of excess methyl bromide with K two co three because the idea is to go ahead and regenerate our ordinary ammonium salt. So second step is the addition of 20 with some water. And the third step is heat to eliminate everything right. And so at this point reacting with one methyl Perine is going to result in our product being 13 beauty dy in and try methyl amid. So just to re summarize one methyl Peridin results in 13 beta dyne. So here I went ahead and wrote that down on the screen to highlight that during our final answer later. So now what happens when you add methyl groups to the carbon chain of pala in itself? What I mean by that is, for example, our next methylated derivative which is two methyl Perine, meaning the methyl group is attached to the carbon adjacent to the nitrogen inside of the ring. So the bottom line is that the reaction is going to stay the same. But the difference is that now the carbon chain is going to be longer. So the first step or the first sequence of steps I should say is similar in the sense that we have to make a ordinary ammonium ion first and then open the ring. But the difference now is that we have this amino group attaching to a five carbon chain because once again, this methyl group is not attached to nitrogen. So we have a five carbon chain when considering two methyl pool. The principle stays the same though. We have the alpha carbon attaching to the amino group and we have the beta carbon right next to it. So the process really is going to repeat itself, right? We have to reform the ordinary ammonium ion and then we have to eliminate it. So this time around, we still have only one product because there's only one beta carbon. So this time around we end up with a five carbon diane. So that's one for P tody, but we still end up with tri methyl amine all the same. So here just to go ahead and clarify that we would get one for Penta dyne from to Mer. So now let's consider our third and final starting material. So our third and final methylated derivative is going to be three methyl pool. So we had the methyl group two positions away from the nitrogen inside of the ring, but the reaction is going to start in the same exact way. So we have formation of the ordinary ammonium ion followed by opening of the ring. So we have our amino group at the very end attaching to a four carbon change from the original ring. But the difference this time is the presence of a methyl group, two positions away from the amino group itself. So this time around what's interesting is that not only do we have the same alpha carbon attached to the amino group, we have our methyl group attaching to the beta carbon, but there's only one beta carbon still just to clarify. So process repeats itself, right? We have reformation of the quaternary ammonium ion followed by elimination and the presence of a new double bond between the alpha and beta carbons. So we have a four carbon diane with a methyl substituent on carbon two. So that's two methyl, 13 beta diane and try me for me. So here I would write that we get two methyl, 13 be dying from three methyl purity and there you have it. So just to go ahead and re summarize right. 13, Beter Diene is synthesized from one methyl Perola, 14 pine is synthesized from two MethylPro ladine and two methyl, 13 Beter diane is synthesized from three methyl purity. And so with that being said, thank you so very much for watching and I hope you found this helpful

Key Concepts

Piperidine Structure and Reactivity

Diene Formation Mechanism

Substituent Effects on Reactivity

Watch next