 Okay, so let's try this problem. Two-methyl, two-propanol, reacts with strong acid to give this alkene shown here, right? Two-methyl, one-propane, and water. So what kind of reaction are we seeing here? Elimination, right? And what do you think it would be, E1 or E2 reaction? E1. E1, why is that, do you think? Because it's a tertiary alcohol, that's very good, okay? Because it's a tertiary alcohol. And so what's going to happen, you know, when we propane that tertiary alcohol, we're going to get the leaving group leaving, and we'll get a tertiary carbocation. That's what it'll be, E1. Okay, so I just want to go over the mechanism for you all, right, with cocaine. So everybody hopefully have written all this stuff down. So whenever you're doing the mechanism, remember, draw the structures out. It's much easier to figure out what's going on, okay? So remember, I should have asked you, right? So strong acid gets deep-propanated by the base, right? Yeah, just like you were saying earlier, right? The base is the alcohol because it's the longer term. So hopefully everybody would do both of those errors for the first time. So now what we have is that protonated alcohol. It's the step where you're asking, is it any one or any two, okay? Since it's tertiary, that leaving group is going to leave by itself, okay, to make that tertiary carbocation, because that tertiary carbocation is very stable, okay? So we're going to do primary or something like that, secondary. You would think maybe secondary, maybe E1, maybe E2, primary, always E2, okay? So we got that carbocation that we were talking about, like that. And now we got that water molecule that was shown in the products. Okay, so remember that H3O plus was a catalyst for this reaction, okay? So we've got to reform that. So catalysts, remember, they participate in the reaction, but they have to be regenerated, okay? Catalysts doesn't get used up in the reaction. So that's specified one hydrogen there, but we all know there's three of them there. So what's going to happen is we're going to keep that we were looking for, right? So there's your water, there's your algae, and that's okay. So that thing can go and do another one of those same reactions, okay? So what's going to happen if you were thinking about setting this up in the lab, right? Excuse me. You'd have the alcohol, maybe it would be neat or in a solvent, and you'd only put, like, a drop or two of, you know, a strong acid in there, like sulfuric acid or something, and this reaction would occur. You wouldn't need a molar ratio of the acid if it's a catalyst, okay? Are there any questions on this reaction? Yeah. Any questions? Okay, cool. Good job.