 Okay, so let's try this one. It says draw the mechanism for the following clays and condensation And then it shows you these molecules up here a full benzoate and acetophenone are reacting with sodium ethoxide first and then the second step acid to make 1,3-diphenyl-propane, 1,3-dione Which is this dye ketone and then ethanol which is kind of the side product. This is the stuff We're really looking for that's the clays and condensation product, okay? So remember, what do we do when we're doing mechanisms? We're going to erase everything except for the molecules that are reacting in that initial step, okay? So when we look here, we've got sodium Ethoxide that's actually going to react with the alpha carbons, right? So When we look here, we look at this molecule and does it have any alpha carbons this molecule here? That have protons on them? No, right. What about this one? Yes. Yes. Do you know what an alpha carbon is? Okay, so that's something you need to know if you don't know what it is then how can you know, okay? So yes, you're absolutely right that this one does and this one doesn't so the first thing we're going to do Is erase everything that's not going to be right So you also have to remember where you're going, okay? So if you forget that you're making that dye ketone, it's going to be very difficult to Actually get there through your mechanism, so keep that in mind When you're drawing stuff, okay? You always want to keep looking back to where you're going So where's the alpha carbon with its proton? I'm just gonna write one of the protons you can write all three of them if you want so sodium ethoxide I like to put all of my Long pair electrons that we're going to have The deprotonation occur to make the enolate so remember that alpha proton is acidic We also have made so we've got that Okay, so we also had remember in our starting Reaction mixture ethyl benzoate So that's what we're going to use now So so it's got this electrophilic carbon right there And we've got the nucleophilic carbon here, of course remember the nucleophilic carbon, okay? Those electrons are going to come back down like that and then these electrons here from the double bond that are going to go And attack that electrophilic carbon like you might imagine them to Then that's going to make those electrons go up like that, okay? So do this step in two steps. Don't make those electrons come back down and knock the thing off Okay, because it is two steps So that's what we're going to see now. So Why don't we do it like this? Hopefully we can see down here But what do we have now so draw everything that those arrows are saying I want you guys to draw it first Or at least before I finish drawing, okay? So that what everybody drew is that what you were able to draw and you drew it too. Okay, wonderful So then it's just the last step So the electrons here are going to collapse down back down because of course they want to be SV2 hybridized That's going to kick off this good leaving group Remember we're in a solution of sodium methoxide Solusium and methoxide, right? That's what we're about to do isn't a big deal. There's ethoxides all over the place So normally you would think oh, that's a bad leaving group But in the case of being having a basic solution, it's not a bad leaving group And then that's going to be our forward arrow there And I'll just draw it of standing up of course have The sodium methoxide still out there, right? And then if you recall the second step was H3O plus, right? So why did we do that? It's because we want to protonate this methoxide here, and that's gonna This is the second step here, okay? So but the glazing Condensation product is that the dye Any questions on that one? Questions, got it. You want me to kill it?