(a) Based on Figure 24.23, explain why meta-dihydroxybenzene i...

Question

(a) Based on Figure 24.23, explain why meta-dihydroxybenzene is not oxidized to meta-quinone.

(b) If a meta-quinone is not produced, what would you expect the product of the oxidation of meta-dihydroxybenzene to be?

Chemical structure of meta-dihydroxybenzene with oxidizing agent notation and a red cross indicating no reaction.

Accepted Answer

Hi, everyone. Welcome back. Let's look at our next problem. It has two questions. Number one says, explain why the molecule below is not oxidized to a meta quinone. And then number two, what is the actual expected product of the oxidation? And our reactant is five ethyl benzene 13 dial. So this benzene ring with alcohol substituents at carbons one and three and an ayl substituent at carbon five. It's reacting with chromic acid and water. So first of all, what would a meta quinone be? That would be that benzene ring with two carbon groups meta to each other. So if both alcohol groups were converted to carbonel groups, but this is not going to be our product. It's like not because the second carbon group, there will be two carbon groups, but the second oxygen and the second carbon group is going to come from a water molecule, not from the second alcohol group. So one of the alcohol groups will get oxidized to a ketone. But the second carbonel group, it will be a result of water attacking a Carbocaine intermediate that will be generated and those positively charged carbons in the Carbocaine intermediates, there'll be several because we have this resonance structure going on. We'll be at the Ortho and prepositions not the metaphysic. So we'll take a look at those resonance structures to understand why this is. But the reason it's not oxidized to a meta quinone is again that, that second oxygen comes from water, not the alcohol group as a result of water attacking this Carbocaine intermediate being in the Ortho and paras positions. So now we need to look at what's the actual expected product. We'll look at the carbon canines that are formed. So we'll have that first alcohol group reacting to form this intermediate with a chromate molecule. And that chromate was going to leave generating a positively charged oxygen. So I've got my molecule here and now my, one of my alcohol groups, I've used the one on the top. I have this o bonded to cro three H the electrons bonding the oxygen to the chromic to the chromate. We'll go on to the chromium atom leaving behind this positively charged oxygen. As we know, oxygen is not happy with a positive charge. But fortunately being on his benzene ring, it has the potential for these resonance structures to stabilize it. So we have this double bond in the benzene ring adjacent to the carbon bonded to the oxygen. So those electrons from that double bond can jump up to form a double bond between the carbon and oxygen which will move the positive charge to that Ortho position. So there's our first resonance structure. So we've got a carbonel group on top, then a positive charge to the left of it and then just two double bonds left in the ring. Well, again, we can move this positive charge around, we can have the double bond that's on the carbon that has the second oxygen group can hop on a over to the next position just moving the positive charge now to the paras position from that carbonel group. And finally, we can move it one more time to generate another Ortho Carbocaine, moving that final double bond down and popping that positive charge up to the other Ortho position on the other side. And you can see we have these four resonance structures. Yeah, the most stable of these carbo CS of course, will be the three where the or that are actually carbo cion as opposed to the positively charged oxygen in the first resonance structure. Of course, we can regenerate by movie mats co double bond down. And so now, as we said, we're, we have this water, the water molecule can attack at any one of those three Carbocaine positions. So we'll have three possible expected products of the oxidation representing those three places where the water molecule can attack. So let's draw those. So you can see that we have the two carbonel groups and one alcohol group. In each case, we have two molecules where the two carbonel groups are uh Ortho to each other and one where their para to each other. And a quick side note, as we saw, we only reacted one of the alcohol groups with our chromic acid. It, this molecule that we were given was symmetrical. So it could have been either alcohol group, it doesn't matter the molecule symmetrical. So the end products would have been the same. So important little note um as to why we only had to demonstrate with one of those alcohol groups. So we shown explained why the molecule is not oxidized to meta quinone because we generate this carbo cion where the positive charge is in the Ortho and pers position where the water attacks to form the ketone. And then the actual expected products we've drawn are two Ortho carbonel groups and one para carbonate set of carbonate groups. See you in the next video.

(a) Based on Figure 24.23, explain why meta-dihydroxybenzene is not oxidized to meta-quinone.
(b) If a meta-quinone is not produced, what would you expect the product of the oxidation of meta-dihydroxybenzene to be?

Key Concepts

Dihydroxybenzene Structure

Oxidation Mechanism

Resonance and Stability

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