 Okay, so let's change this question up a little bit It says draw the products of the following de Kling condensation. Let's also draw the mechanism of that same reaction What do you say? Okay, okay so the first thing we're going to do is we've got the ester here, so remember put all your lump hairs first Act with sodium methoxide there, and then the second step is going to be H3O plus I guess let's write out what the product would be first Okay, so with that So the product is going to be this and condensation product like you would imagine it to be But let's do the mechanism So you know where we're going Okay, so Just like always whenever we're using that sodium methoxide. That's a strong base Remember and then we have acidic protons here on the alpha-carbon Okay, so those are the most acidic protons So the other thing we can imagine is doing is attacking the carbonyl carbon going up and back down kicking off That is leaving group, but that's the same thing as that so if it did it it wouldn't Change anything of the starting material. Okay, that makes sense too, right? So let's look at those two protons there those the acidic ones obviously, right? So we're going to Have those deprotonated That's it's a massive base reaction Get it deprotonated this one over here as well. Okay, I just picked that one because It's over on the left side and I right from right to left. Okay, the other one could have been deprotonated you get the same product Remember a deekman condensation. That's just a Claisen condensation. That's intramolecular. Okay, so Whenever it says deekman condensation look for an intramolecular reaction That's going to happen a cyclization. Okay Specifically a cyclization that'll be a five carbon cyclic ring or a six carbon because those are the most stable any other one is very difficult to form Anything smaller than five are bigger than six So what we're going to do remember we're going to be attacking this carbon here Okay, so we're going to count one two three four five Okay, that's the one that's going to be attacking there. Okay, so let's just show the mechanism That's going to come down like that That's going to go over And attack like that and that's going to make those electron go down like that Okay So now let's count what are carbons. Okay, so Okay, so what do we got one two three Four and see where the electrons are coming from that fifth carbon there, okay Five so we want to make a five-membered ring a cycle pin came right it might be easy for you Easier for you if you've labeled the carbons here one two three four five and erase them in a second So Attached to carbon one Right, you're gonna have the O minus now because look the electrons bounced out But you also have attached to carbon one the O ethyl group, right? Okay, so let's draw that now Does that make sense what we've done there? Yes? Okay? Wait So carbon one carbon one. Okay, so it makes sense because this thing's still attached to it Okay, okay, so carbon one is there carbon twos or anything attached to it. No, no carbon three carbon four carbon five Well, what is attached to it? Well one thing in particular. What is it? Okay? You got to say something. We're on the we're on camera. Everybody's waiting for your Answer. What is it? Why don't we say how many carbons we got to put on there? How many carbons are here before we hit the oxygen? One and then what do we put? Okay, the carbonyl and what else is attached to The O that thing right like a thought secret You could have said the ester to So that's what you get right? So this SP3 center is not stable. Okay, and remember what our react our product was Okay, so we're going to try to make that beta Ketone with the acid right so beta acid ketone or beta keto acid So we're going to knock that thing off like that There's our five-membered ring with the ketones Okay So I guess I'll just draw it over here carbon one. There's now a ketone Right and carbon two. There's nothing three. There's nothing for there's nothing and we knocked that other thing off the oxide group off of carbon one So it's out there by itself And on carbon five We've still got that ester So we've done that Right Okay, so now second step H3O plus So one thing we can think of H3O plus is going to do You have proton A the alkoxide So we're going to make ethanol as a byproduct of this reaction And more acid So that'll do that. We'll put another H3O plus Prodenate those things you're going to do those electrons there the carbonyl electrons are going to deprotonate like that I'll erase the top up there. So it's not so cluttered. Does that make sense now? Okay So we've got that and we've got water Water is nucleophilic to that electrophilic carbon Okay It's going to attack there Knock those electrons up like that when we do this and deprotonate that water. It's going to happen Is this thing here from this reaction here That's going to protonate the Alkoxy functionality giving us the H there plus charge You can see of small molecule Right that has a positive charge on it. It'll be a good leaving group. Hopefully you can see that's ethanol right there Okay, that's going to be a really good leaving group. So what's going to happen is these electrons are going to come down And knock that good leaving group out reforming the carbonyl And that's going to deprotonate giving us the final product deep condensation product beta keto acid Any questions on that one? Still a lot of steps, but Okay