 OK. So let's force this acetal to go, though. Usually, when you're going to do acetals, you want to put a little drop of acid in there, because it catalyzes them to form. It makes it form real fast. So let's work with our diol, our 1, 2 diol, and acetone here. And we'll just write a little drop of acid in a different way than you've seen it before. Let's just write that. So you can write that, what we've been saying, like H2SO4, H2O. You could also write H3O plus. Of course, all of those are the same thing. H3PO4, cat, it'll say sometimes. So I actually don't prefer to write just H plus, because honestly, that ion doesn't really exist, H3O plus, and even bigger ions than that. But sometimes in the book, you'll see it as just H plus. So I don't want you to get confused, OK? So anyways, what's the first step that's going to happen in this reaction? The third rule of organic chemistry, maybe. Acid-basic reactions happen faster than any other reaction. So when I ask and you say, I know, tell me, OK? Just tell me. That's all you've got to do. Let's just, I'm going to erase that. We should have a proper, let me see what that will do. Oh, I put it in front of me. So you could think of it doing that, OK? Or honestly, probably what's going to happen is this is going to deprotonate the acid first, and then it's going to transfer it to that, OK? Because this is more basic than that, OK? But it doesn't matter, OK? Because eventually, you have to get that proton on that ketone oxygen, is that OK? So when I do that. And in fact, let's just erase this, because that doesn't have anything to do with that first step. Now we have our dial. So now it's going to happen. There's a little bit of a tank. Those electrons will go under it, push that double bond up. So this thing is a good electrophile, right? This is like that super electrophile that we're talking about. So just like we're saying, the next step is going to be nucleophile electrophiles. And those are the arrows you want to make. Very good. I like it. That's what I want. You know it, tell me. So let's draw our product of that. So remember, what is the hybridization of that carbon there? That's SB2. SB2. And the next one's going to be SB3, right? So if we had two different substituents on this carbonyl carbon, there's going to be a stereocenter being formed, OK? So you've got to watch out for that. But of course, in these general reaction equations. Create that tetrahedral entropy. The other thing you always want to keep in mind, of course, here, we didn't write our water molecule. You know, as organic chemists, we don't always like the balance equation. But you always want to keep in mind that if you've got a positive on one side, you want to keep that positive going. So the next step is going to be what? Can anybody guess what the next step is? It's going to be y, because we've got an acid there. Deprotonation. It's going to be a two-step deprotonation. It's not this one. It's not going to deprotonate. It's too hard for it to reach around, OK? So we're going to have to have water deprotonating here. And I'm just going to erase like this arc here. Let me show the whole work. So that step, OK? So I don't have to write the whole thing again. Now that, is everybody OK with that, OK? So now what's going to happen is one of these, it'll keep going back and forth, back and forth, back and forth, until eventually this one gets it, OK? So it's going to deprotonate like that. And in fact, that's one thing we should be showing here, is that all of these are equilibrium arrows. And there's a driving force of this reaction. We'll talk about that later. So of course, that's also an all-acid base reaction. So a good leading group, right? We've created a good leading group. I go down, and under the water, we create those. So I'm going to come back up into that leading group. It'll be. And now we have a point in turn. So, OK, one second. Let's see if, is everybody OK with that step? So the next step is the lone perelectrons from this oxygen, is that what you're saying? Lone perelectrons from this oxygen are going to collapse down on here, right here, OK? Yeah, it's a very important step. Because it's going to kick out that leading group. Is it easier for this thing? Is it easier for me to close my eye or to make my foot kick my eye closed? Do you know? Yeah, if I'm real talented and I don't have an eyelid, you know? It'd definitely be funny. So the closest proximity thing is going to happen first. So that's pushing the reaction, actually, because that's a stable molecule, right? And then this thing, the water's not going to be able to come back in time before this thing closes back. OK, so let's draw what we get next. I could see some of you are like, what's the water molecule? Is it easier for that to hit back or that to hit back? Which one? This one or this one? Why? Because it's attached to it, you know? It's easier to hit yourself than for somebody else to hit you. So what's going to happen? Is this oxygen is going to come back and do a nucleophilic attack on that electrophilic center? Notice this looks a lot like what it looked like up here, right, that plus charge. What's going to happen to the other one when you kick up? Like that. So that's the driving force of the reaction is losing that water molecule. I'm going to have to erase these steps over here. OK, is that cool? It's cool. Again, you can watch this video when I focus this. So when we do that, what do we got? We got that, that, that. Can't see that. Label your carbons and oxygens, right? So we can start on one end or the other. Which end do you want to start at? We'll start there. One, two, three, four, five, right? So the oxygen with the proton is number one, right? And the ketone carbon is number five. One, two, three, four, five. Is that what you did? Yes. And then what's the last step? You could say it louder. Deep protonating the oxygen. The last one. And in fact, that's going to be an equilibrium step until you extract it. And you get, what is this kind of compound called, guys? What are you making? Acetone. An acetone, yeah. That was the little bling to say that you got the right answer. Is there any questions on that one? Questions, questions? I guess the last, and we are missing. Remember, it's catalytic in acid. So that was a good question, Eve.