A tertiary amine reacts with hydrogen peroxide to form a terti...

Question

A tertiary amine reacts with hydrogen peroxide to form a tertiary amine oxide.

Tertiary amine oxides undergo a reaction similar to the Hofmann elimination reaction (Section 10.10), called a Cope elimination. In this reaction, a tertiary amine oxide, rather than a quaternary ammonium ion, undergoes elimination. A strong base is not needed for a Cope elimination because the amine oxide acts as its own base.

Tertiary amine oxide undergoes Cope elimination to form an alkene and a secondary amine with hydroxyl group.

Does the Cope elimination have an alkene-like transition state or a carbanion-like transition state?

Accepted Answer

All right. Hi, everyone. So this question says that the following reaction undergoes cope elimination. A reaction similar to the Hoffman elimination. A strong base is not needed in this reaction because the amide oxide acts as its own base. Determine if the transition state of the given reaction is alkane like or carbon ion. Like option A says all keen like option B says carbon ion like option C says either or or option D says neither. So the reaction in question is or rather involves NBN 23 tri Meho button amine reacting with hydrogen peroxide in the presence of heat. Now recall that the Cope elimination is a reaction that will oxidize the starting a mean, right. So after the oxidation process using hydrogen peroxide, we end up with an intermediate N oxide. Now, when that an oxide is subjected to heat, then the oxide itself behaves as a base in an intra molecular elimination to produce an all keen and a hydroxyl amine like the ones depicted on the screen here. So to understand the transition state, right, we have to understand how the reaction proceeds. So let's go ahead and start out with the mechanism So on the bottom of the screen here, I'm going to go ahead and redraw our starting material so that we can use it to go through with this transformation. So heres my nitrogen and here are my methyl groups good. So now let's go ahead and proceed with the actual mechanism. The first step being the interaction with hydrogen peroxide. So here is hydrogen peroxide. And so first off right, the pair of electrons on our nitrogen in our amine is going to go ahead and interact with one of the oxygens in hydrogen peroxide, which will therefore break the bond between the two hydroxy groups. Now, as a result, right, he is our nitrogen. And at this point of the reaction, a hydroxy group will attach itself to our nitrogen, but I cannot proceed without drawing the rest of the molecule. So let me go ahead and do that. There we go. But like I said, previously, the hydroxy group or one of the hydroxy groups has at this point attached itself to the nitrogen of our starting amine, which will therefore give our nitrogen a formal charge of plus one. So at this point, at this point, we're going to reintroduce the hydroxide ion that formed as a result of the bond between the hydroxy groups and hydrogen peroxide breaking apart because a tear that hydroxides, remove the hydrogen from our hydroxy group to subsequently yield our N oxide. So the N oxide is going to look very similar to the compound in the step that we just went over, except for the fact that our oxygen is going to have a negative charge. So here are my carbons just to make sure I don't forget them. And like I said previously, our oxygen is going to have a negative charge, but but my nitrogen is still going to have a positive charge because that oxygen is still attached. Now, it's at this point that the intra molecular elimination reaction takes place. Now, in order to form the four carbon, all keen depicted in our final product, right, the beetle group of the amine here has to break off. So in order for the beetle group to break off, we have to eliminate a hydrogen from carbon to because here, once my negatively charged oxygen eliminates this proton, we end up with a double bond between carbons one and two of the side chain of the beetle group, which will therefore break the bond between carbon carbon, one and our nitrogen. So we end up with our double bond and we also give back a pair of electrons to nitrogen, making it neutral once more. So it's at this point that we can go ahead and discuss the transition state and the transition state must be depicted in brackets to showcase that this is a temporary structure. So the transition state at this point is going to depict a few things, right? It must depict the breakage of the bond between carbon and nitrogen. It must depict the breakage of the bond between carbon and hydrogen and the formation of not only the pi bond of the alkene but also the bond between oxygen and hydrogen. So here is my beetle group right on the bottom and I'm going to draw a dash line over carbons, one and two or in between carbons, one and two. The showcase that there is a pi bond in the process of being formed. Now, our carbon two was my hydrogen and that hydrogen is now connecting with oxygen and my oxygen here is still attached to nitrogen, which used to be attached to carbon one of the beetle group. So the bottom line is that you must draw dash lines to showcase which bonds are being broken and also which ones are in the process of being formed. So here, right, in the very beginning of the step, my oxygen had a negative charge but because a new bond is in the process of forming oxygen is now going to have a partial negative charge, whereas nitrogen is going to have a partial positive charge because it had a fully positive charge in the previous step. And a bond is in the process of breaking. And so here at this point, the transition state collapses right, the transition state breaks, so to speak, and we end up with one beauty and our amine that has a neutral hydroxy group attached here are my two metal groups and here is the complete structure of N 23 dimethyl butty n methyl hydroxylamine. So now let's talk about the transition state. Now, the hydrogen that was being removed, right, was the beta carbon of our beal side chain because the new bond, the pi bond was being formed in between the alpha and the beta carbons. Now, according to the mechanism itself, the hydrogen on the beta position is being removed. So as the pie bond is being formed, there would actually be a partial negative charge on the beta carbon. Why? Because hydrogen is being removed and electrons are to a certain extent being added during the formation of the Piedmont. So that gives the beta carbon a partial negative charge, which means it's a transition state would be more reminiscent of a Caronia, which means that our answer at long last right is going to be option b in the multiple choice, it's going to be carbon ion like because the beta carbon adopts a partial negative charge during the actual in intermolecular elimination step and there you have it. So with that being said, thank you so very much for watching and I hope you found this helpful.

Key Concepts

Cope Elimination

Transition States

Amine Oxides

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