 Okay, so let's do this reaction, the Canis-R reaction. Remember this is a disproportionation reaction. Whenever we see the Canis-R reaction, it'll be on aldehydes that have no alpha proton, right? Do you see Benzaldehyde? Does this have an alpha proton on it? No. No alpha proton, right? Because that would be here, right? And there's no proton on it. So if we're going to use a strong-base NOH or KOH, right, on an aldehyde that doesn't have an alpha proton, it'll do this type of redox reaction, okay? And so when I say redox, right, hopefully you can see, right, when I go from Benzaldehyde over here to Benzoic acid, what's happened here? Has it been oxidized or reduced? Oxidized. Oxidized. And then Benzalalcohol? Reduced. Reduced. So in other words, we're going to have two Benzaldehydes in this mechanism, okay? So has everybody written down what we have here? Yeah. So I'm going to erase it, and then we'll actually draw the mechanism, okay? So I'm just going to write OH- because the potassium is a what? Spectator. Spectator, right? So we don't even have to worry about it, really. What's going to happen, as you might imagine? The hydroxide is going to attack the electrophilic carbonyl carbate. That's going to cause these electrons to go up to the carbonyl oxygen. And this reaction, like we were just saying, really can go back and forth, right? Because the OH can be the leaving group back, right? But what happens in the Canizaro reaction, as you might have already figured out, something weird is happening here, okay? So like we were saying, most of the time, probably it's going to go back that way, with the leaving group leaving as the OH. But every once in a while, and this is how the Canizaro reaction goes forward, right? Is it's going to have the leaving group be this hydroxide here, okay? So in order to have that hydride react with another benzaldehyde, we need to write that benzaldehyde. So let's go ahead and do that. So what did I say? Those electrons are going to come down, and these electrons are going to go with that hydrogen as a hydride and attack that electrophilic carbonyl, okay? So when that happens, those electrons go back. Is everybody okay with what we've done there? So hopefully you can already see one of them is being oxidized, one of them is being reduced, okay? So that's where we get our forward arrow. So we've made benzoic acid here. Do y'all see that? Benzoic acid. And then here, we're going to make the conjugate base of the benzalde alcohol. Remember, potassium hydroxide, that's usually in a water solution, okay? So what will happen is two things, right? We still have potassium hydroxide in there. That's an acid. So what's going to happen? We're going to depropanate that. And well, like I said, we have water in there, and it's no problem for us to make more hydroxide. So this will depropanate that. So when we do that, I'll just put these two in the next one. Is that okay the way I've drawn those arrows, like? Yes. So we have essentially what would be the potassium benzoate here, right? Because if we carried along that spectator ion, that would be the counter ion to that. And we now made benzalde alcohol. So that was one of our products, right? And remember, the second step was H3O+. Okay? So when we add H3O+, like that, what's it going to do? Yeah, very good. It's going to propanate our carboxylic acid. So the two products of the reaction, this particular one, are benzoyl acid and benzalde alcohol. Okay? So you need two equivalents of the benzalde hide to do this. And you all see, I know we mentioned it before, but that it's a redox reaction, disproportionation reaction. It's got that weird hydride leaving group step. So watch out about that. Any other questions on this reaction? Questions? Okay. Good.