Show a complete mechanism for this reaction.

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Alright, hello everyone. So this question is asking us to write the detailed mechanism by which the following reaction occurs. Ignore stereo chemistry. So our starting material is seven methyl amino hepton, three own reacting with methanol in the presence of HCL to produce two ethyl two. So the most important thing to understand about this reaction is that our product is cyclic, right? We have a starting material with a relatively long carbon chain and it's producing a ring structure by the end. So this is an intra molecular cy ization reaction. So here to go ahead and start off the mechanism, let me just go ahead and redraw the starting material. All right. So here another thing that I want to go ahead and point out is that nitrogen in the structure of the product is connecting to the carbon attaching to oxygen. So the amino nitrogen is somehow going to attack the carbon carbon and the carbon oxygen somehow acquires a methyl group to becoming meth boxy group in the structure of the final product. So all these things need to be kept in mind. No, the amino group itself is a weak nucleo because it's neutral So really, our first step is to somehow activate the car bal so that the nucleic attack that we need is more likely to happen. So this is where HCL comes into play. So here, one of the lone pairs on the carbon oxygen can go ahead and take a proton from HCL to produce a pronated carbo intermediate. So here's the Aito group and this time we have the carbon oxygen with an additional hydrogen and therefore a positive charge because of that extra bond. So really, it's a re form of this intermediate that's going to help us in the next steps of the mechanism. And this is because recall that a positive charge on an electron atom isn't favorable, right? So the pi bond between carbon and oxygen is actually going to donate its electrons to oxygen directly. And so as a result, we get a resonance form in which oxygen is neutral now as a hydroxy group, but we have a carbo cion. So now I'm just enclosing my residence forms here in brackets. Really, it's the second resonance form here that's going to be more relevant. And the reason for this is because it explains why nitrogen is going to attack the carbon, not the oxygen, right. So now the amino nitrogen can go ahead and establish a bond between itself and the carbo CDI. So now this is step two. So let me just go ahead and label we have step one. And now we have step two. And so at this point, we have a ring. So it's going to be a six member ring, taking into account the nitrogen, the carbo cion and the carbons in between them. So we have nitrogen with its methyl group and it's hydrogen attached to it, which means that it's going to have a positive charge while part of this ring. So next to it, we have the hydroxy group and the remaining two carbons from the parent chain. So it's an ethyl group. So here our next step is to just go ahead and remove that proton from nitrogen to make it neutral because at this point, we have an ammonium ion. So methanol which is present in solution can go ahead and do precisely that. So that gives a long pair of electrons back to nitrogen ultimately making the nitrogen neutral in step three. All right. So remember how earlier it was established that there is a meth oxy group president solution that's going to come from methanol. So in order for methanol to install a meth boxy group onto the ring, the hydroxy group that's currently present must leave, it must detach itself. This is because Methanol itself is once again a weak nucleo file. So it needs to be activated. And so a carbo cion is a strong enough electro to attract the nucleic attack of methanol. So this is where we proceed with step four. Write in step four, we recall that there is a proton methanol present in solution from our prior step. So the hydroxy group, specifically the oxygen can take the proton regenerating methanol. And so it becomes a proton molecule of water attached to our ring, which this time around is an excellent leaving group. So our leaving group is going to go ahead and detach itself so that oxygen can be neutral. And this is where we get our carbo cion, Our second carbo cion actually in step five. So in step five, we have two resonance forms to consider. The first is with the aforementioned carbo cion. But we do also have to take into consideration the fact that nitrogen has a long pair of electrons. So in the resonance form actually moved this over for the sake of space. But here, right, in the resonance form, nitrogen can donate its low pair of electrons to establish a double bond between itself. And the cruel cut out, this is ultimately going to transfer the positive charge onto nitrogen. So because these are resonance forms, I'm going to go ahead and enclose them in brackets. But for the sake of the mechanism, the relevant resonance form is the one with the carbo cion because the carbo cion is what methanol or nucleo is going to attack. So methanol attacks the carbel cion and that is going to mark step six. So after step six, we are almost at the formation of our final product. But the difference. However, is that this time around we have a pronated meth boxy group attaching to or R. But after a final dero nation step, we're going to get our final product. So another equivalent of methanol available in solution can go ahead and remove that proton, resulting in or neutral final product with a neutral metox group. And there you have it. So that concludes step seven and the mechanism as a whole. So if you watch this video all the way through, thank you so very much for doing so. And I hope you found this helpful.

Show a complete mechanism for this reaction.
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Key Concepts

Reaction Mechanism

Nucleophiles and Electrophiles

Catalysis

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