Assessment 24.22 asked why meta-dihydroxybenzene could not be ...

Question

Assessment 24.22 asked why meta-dihydroxybenzene could not be oxidized to meta-quinone. The attempted oxidation instead gives rise to three different quinone products. Suggest a mechanism for the formation of each.

Chemical reaction showing meta-dihydroxybenzene oxidized by chromic acid, yielding three different quinone products.

Accepted Answer

All right. Hello everyone. So this question says that the oxidation of five methyl benzene 13 diol produces three products below. Instead of meta quino draw the mechanism for the synthesis of product. A All right. So here we have our starting material of five methyl benzene 13 dial which I redrew on the screen here towards the bottom for our mechanism. Later on. In any case, though our starting material is a derivative of phenol, right. So we have to recall a few things about the oxidation of phenol. Now recalled phenol is oxidized using a strong oxidizing agent to give a mixture of Ortho and peri. Now, the reason why a strong oxidizing agent is required is because the oxidation of phenol means that aromatic is lost. So an example of an oxidizing agent suitable for this reaction would be chromic acid. So that's H two co four. And so the first step of the mechanism is the formation of a chromate ester. So the chromium in ch in excuse me, in chromic acid is going to interact with one of the hydroxy oxygens in our phenol derivative here. So let's just say that it reacts with the one on the top. So I would redraw my ring, you have a metal group on the side, one hydroxy group that's unaffected. And so the hydroxy group on the top is going to lose its hydrogen atom because it's now going to establish a bond between itself and chromium. And then from there, chromium has two double bonds to two other oxygens and its own hydroxy group. It's worth mentioning also that the formation of this chroma ester produced water as a and so the next step is going to require the breaking of the chromate ester, right. So the bond between chromium and oxygen has to break. And this is actually going to reduce chromium from an oxidation state of positive six to an oxidation state of positive four. So as a result of the breakage here of the chromate Esther, what's going to happen? Well, what's going to happen is that we're actually going to end up with a cion where the positive charge is on the oxygen, right? Because chromium took those electrons with it. And so because we have the presence of this carbot on oxygen and the fact that we have pi bonds inside of the ring means that we are going to see some resonant structures. So what I'm going to do here is actually enclosed my first resonance structure here in brackets and move this to the side just to accommodate the other resonance structures here. All right. So from here, what's going to happen? Right. Well, first and foremost, a positive charge on an electron oxygen atom isn't favorable, right. So the pi bond of the ring that's closest to oxygen is going to establish a double bond between itself and oxygen. So that's going to create a car bottle with a carbo c inside of the ring. And from there, the carbo cion migrates. So now we have a carbo cion inside of the ring with another double bond moving towards it, thereby moving the location of the positive charge. So we have our car bottle still, right? And this time we have a tertiary carbo cion. So because there's one more pyon to consider inside of the ring, oh actually let me backtrack a little bit because there's no tertiary carbo cion. So I do apologize for that. Instead, the new carbo cion is going to be opposite of the carbon. That's what I meant. So in any case, we have one more pyon to migrate. So that moves towards our carbo cat I on here. And so that brings our carbo cion in between the car bono and the hydroxy group and that completes our resonance structure. So in addition to the resonance structures that we just went ahead and drew, we also have to account for the presence of chromium right, that's still present in solution. So we have chromium double bound to two oxygen still with one hydroxy group and an additional lone pair of electrons from when the chromate ester was broken. So now to form product, a specifically, we have to focus on one particular resonance form because lets consider the structure, right, recall it from here, we would have water attacking our carbo cion to ultimately produce a mixture of products. So when you consider the structure of product, a specifically, not only do we have one car bal at the very top of the structure from the hydroxy group of the corresponding starting material, but we also have a new one opposite to the first. And so what that means is that to make product a specifically water had to have attacked the residents form where the carbo cion is opposite to the carbo. So that's the one that we would focus on for this mechanism. So if we scroll down here and then have a molecule of water attack, that specific carbo CIA scroll down some more just for the sake of space. So from here, right, we no longer have a positive charge on carbon. Rather the oxygen of water is going to have a positive charge because of the fact that it has only one loan pair of electrons right now. And so the next step is to try and mitigate that right. The idea now is to go ahead and remove a proton from this proton water molecule. And so a second water molecule can go ahead and do that. So we have another equivalent of water depot in the first to make it a hydroxy group. This time, oops I meant to move this. But in any case, now we have a hydroxy group where the car bottle of our final product that is the second car bottle is supposed to be, right. So our final step, which we're not actually going to show the mechanism of here is another oxidation of that hydroxy group using chromic acid. So this gives us our two the car bottles and I may have to move this again, but there you have it. Now we have our completed mechanism. And so with that being said, if you watch this video all the way through, thank you so very much for doing so. And I hope you found this helpful.

Assessment 24.22 asked why meta-dihydroxybenzene could not be oxidized to meta-quinone. The attempted oxidation instead gives rise to three different quinone products. Suggest a mechanism for the formation of each.

Key Concepts

Dihydroxybenzene Structure

Oxidation Mechanism

Quinone Formation

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