Hey, everyone. So in this question, now we're going to look at the mechanism for the Kolbe-Schmitt reaction. So step 1 involves phenol being deprotonated by sodium hydroxide. Alright. So here, let's draw our phenol.

We're going to deprotonate it with Sodium Hydroxide. So, here this is going to come in and deprotonate, grabbing this H. Oxygen holds on to the electrons. We're going to make our Phenoxide ion, which is resonance stabilized. Because here, once we make it, it could resonate into the Benzene ring.

So this could come here to make a double bond, kicking this bond here. And here, we're going to have another form that it could take. So now this is negatively charged. It could continue to resonate. So this moves here, taking this bond here.

So here goes a negative now. This can resonate here, causing this to resonate here. Okay. So this is negatively now. And then it could technically resonate back, kicking the double bond from the carbonyl back up to the oxygen, recreating benzene.

Here, we're not showing that last one, but it does exist. And in fact, we'll show that. So this could resonate here, keeping this bond up here. And so all the different ways we could show the resonance structures of the phenoxide ion. Now, here, phenoxide ion reacts with carbon dioxide through EAS or electrophilic aromatic substitution.

And we're going to say, here goes our Phenoxide ion here. And in fact, what we're going to do is we're going to just show one of the resonance structures that we had. Let's show this one. And there goes carbon dioxide which is C=O. So this decides to come and attach here, kicking 1 of these pi bonds to the oxygen.

So here goes. The bond connecting to it. Okay. So we've added it. So we've done electrophilic addition.

There's still an H here the whole time. And that's important because we're going to need that H in order to restore aromaticity. So the substitution part is going to come in. As remember, EAS itself is an addition-elimination type of mechanism. And then we're going to have there.

Alright. So then we're going to say here, in order to do this, we're going to use another mole of OH-, which is going to come in, deprotonate, grab this H, fall here to make a double bond, and kick this bond up. And here we're going to write down what that looks like down here. So we've recreated our benzene ring, but now we have this negative oxygen, which needs to be converted into a neutral form. And we also have this negative oxygen that is part of the carboxylic acid that got added.

All we have to do is protonate both of them. Alright. So we're going to protonate both of these with H+. Okay. So here we're going to say this can get protonated by this H+, and then another H+ can be used to protonate this oxygen that's negative.

And at the end, doing that gives us our phenol back and then our carboxylic acid that is ortho to it. Alright. So that's how we get our salicylic acid at the very end. So this is the general mechanism when it comes to the Kolbe-Schmitt reaction.