Oxidation of Phenols to Quinones - Online Tutor, Practice Problems & Exam Prep

Oxidation of Phenols to Quinones Concept 1

Hey, everyone. So in this video, let's talk about the oxidation of Phenol into Quinone. Now, Phenol can be oxidized into Quinone in the presence of the strong oxidizing dichromate, which is Cr2O7-2. Now, Quinone itself is a conjugated 6-membered ring that possesses carbonyls that are either 1,2 or 1,4 to each other. Now, quinone is also named as cyclohexadienedione.

So cyclo is a ring, hexa is 6 carbons, diene is 2 double bonds, and dione is 2 carbonyls. Now, if we compare the oxidation of a tertiary alcohol to phenol, remember with tertiary alcohols, there is no alpha hydrogen on this carbon. This is the alpha carbon that possesses the OH. There's no hydrogen on it, so no oxidation can take place, so we would write no reaction. Phenol, though, is uniquely different.

We can oxidize it, creating our quinone. Here we're talking about 1,4-quinone, which means our carbonyls will be 1,4 to each other. And what was the other one? And remember, quinone is a diene as well, so we'd have 2 double bonds. This product here is called 1,4-benzoquinone or simply 1,4-quinone.

Just remember that tertiary alcohols do not undergo oxidation, but the unique ability of phenol exists where it can be oxidized. Here, it creates our quinone, which again is a 6-membered ring that's conjugated because it's double, single, double. We have 2 carbonyls and 2 double bonds.

Oxidation of Phenols to Quinones Concept 2

Now we know that phenol can be oxidized to quinone. Quinone itself can undergo reduction to yield either catechol or hydroquinone. Reduction can be accomplished by either using Tin(II) Chloride over hydrochloric acid or catalytic hydrogenation. Remember, catalytic hydrogenation is just H2 over a metal catalyst.

Typically, nickel, palladium, or platinum. One of these ones. So just choose one of the metal catalysts on the bottom. Now, here, if we take a look, we have 1,2-quinone because remember, quinone, the carbonyls can either align themselves 1,2 or 1,4 to each other. If we were to reduce this, it's going to give us catechol, which is our benzene ring, and we have our two OH groups next to each other.

So this would be catechol. And then we could oxidize this back into our quinone. Remember, phenol can be oxidized into a quinone here. And so think of catechol as a biphenyl. It can be oxidized to a quinone as well.

Now over here, we have 1,4-benzoquinone or 1,4-quinone. It can be reduced to hydroquinone. Hydroquinone is when the OH groups orient themselves 1,4 to each other on the benzene ring. And again, this is a biphenyl as well, so you can oxidize it right back to our quinone. Now one thing we should notice here is that both our phenol and our hydroquinone can be oxidized to 1,4-quinone.

So that's something they share in common. So just remember, these are essentially oxidation-reduction reactions where we're able to go from one to the other via either reduction or oxidation.

Oxidation of Phenols to Quinones Example 1

Hey, everyone. So in this example question, it says, beginning from phenol, determine the chemical steps needed to prepare the following compound. Alright. So, here we're starting with phenol and we're trying to make this diether. Well, if we think about reactions where we can go from an alcohol to an ether, we should think about the Williamson ether synthesis where we would use a strong base to deprotonate our OH group into a phenoxide ion in this case and then do an SN₂ reaction with an alkyl halide.

Now that could help us to make one ether, but here we have two ethers. So that tells me I need to find a way to get two OH groups onto this benzene that are ortho to each other. And we know that we can do that from what we've learned so far. But what we're going to do here is first oxidize my phenol into a quinone. So remember, we can oxidize it through the use of our strong oxidizing agent in the form of dichromate.

Right. So that'll oxidize this. Here goes our quinone, our ortho quinone. We can then reduce this. We could use tin(II) chloride over hydrochloric acid, and that changes it into hydroquinone.

Now that I have two OH groups, I can do Williamson ether synthesis twice. So what I could do here is, first, I can use a strong base. Let's say we use KOH, and then followed by an alkyl halide, we need two carbons. So the OH− would deprotonate this OH or the other one, making it O−, and then we do an SN₂ reaction hitting this carbon and kicking out the Br.

And we get our first ether. And then since we have another OH, we just do it again. Here, I decide to break them up into steps to show how each individual OH group is converted into an ether group. You could also say KOH two equivalents and then an alkyl halide two equivalents to say that it happens twice so that both OH groups are converted into ethers. So, doing that yet again gives us as our final answer, our diether.

So here goes the two ether groups that I've created. Right. So again, we're able to do this by remembering that a phenol can be oxidized into quinone, and then from there, reducing the quinone to hydroquinone, and then using a reaction we've seen in the past by changing an OH group into an ether group by way of the Williamson ether synthesis reaction.

Determine the products made during each of the following steps. 

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Beginning from phenol, determine the chemical steps needed to prepare the following compound. 

Hey, everyone. So here it says, starting from phenol, determine the chemical steps needed to prepare the following compound. Alright. So here we're starting out with phenol, and somehow we need to create this monstrous, this large structure. Now, we're going to say that this 6-membered ring is this 6-membered ring, which means I added these other 2 6-membered rings through some type of process. Now the key giveaway of what we did here is we created 6-membered rings, this one and this one, and at the end, we still have a pi bond present. This indicates to me that a Diels-Alder reaction happened in some way. Now, the way we have to think about it is we basically have to work backward here. This pi bond, you can move it up or down. So if I move this pi bond here, then this bond here moves here and this bond here moves here. That would mean that on this side I had this structure. And then the same thing over here. I can move it either way. Let's say move it down. Move this here, I move this here. Which means on this side, I had this. And right in the middle, I had quinone. Right? So this is what must have occurred. I changed phenol into quinone, and then I used 2 moles of these conjugated dienes in order to create these 2 6-membered rings, 1 on each side for our final product. So let's just walk us through it. So remember, phenol is susceptible to oxidation if we use a strong oxidizing agent. So, we're going to use sodium dichromate over H2SO4. That's going to oxidize this into quinone. And then for Diels-Alder, you would just say you're going to use 2 moles of this or 2 equivalents. To 2 equivalents of that. Right? And that's how we got our final answers here. So that's all we had to do. This problem seemed much more difficult than it really was. It was really looking at our final product. You're making 6-membered rings. There aren't many reactions that we know will make a 6-membered ring like that with an alkene still present. We know that Diels-Alder is a potential pathway. Remember that quinone itself can serve as a dienophile, so it all makes sense to go by this method. Right? So, again, we just had to oxidize our phenol into quinone, and from there, do Diels-Alder twice to get our final product at the end.