 Okay, so let's try this mechanism that was on the quiz. This is essentially the DCC coupling. So remember, the DCC is the coupling reagent. Remember, we talked about that in class, okay? So it's going to effectively connect the amine, benzal amine here with acetic acid. So you see here we've taken carboxylic acid and a primary amine and we've made this amide with chroma, okay? So you can also see it's a dehydrating reagent. We lose water, okay? So since it asks us to write the mechanism, we're going to have to erase all this. And unfortunately, if we don't know what DCC looks like, then we're kind of up a creep. We won't be able to do this one at all, okay? For me personally, I like to put them all into a bond line form and start from there, okay? This is a rather involved mechanism, so we're probably going to have to erase a couple of times on this small board, okay? So let's erase everything. So remember this one, okay? Because remember, you have to remember where you're going or you're never going to get there, okay? So I'm going to take acetic acid and actually we don't need the benzyl amine at this point in the reaction yet because the carboxylic acid is actually going to react with the DCC first, okay? So we've got acetic acid, pairs have often helped, okay? DCC, remember, dicyclohexyl carbo-diamide, okay? So if you remember the name, it kind of helps you to remember the structure, okay? So this stuff is good because it's got like these big alkyl groups on it and it can dissolve it into organic solvents, okay? So what happens here, the first step is going to be the electrons on the oxygen here. Remember they're nucleophilic. And this is a really electrophilic carbon here. It's got these two double bonds, right? Bonded to the two nitrogens next to it. So this sp carbon is very electrophilic so it's going to be attacked like that by that oxygen. That's going to make this bond knock up to the amp, okay? So remember, you've got to have the one arrow attacked before the other one gets knocked away. So does that make sense? That first step, okay? So remember, if you don't know what DCC looks like, you're not going to be able to do this reaction. Okay, so you've got to memorize that structure. Okay, so let me mess up the drawing, too. I'm just going to keep making this carbon. So it's this carbon. We also have a positive charge there. Remember, since there's zero overall charge here, we've got to have a zero overall charge here, okay? So then I'm going to just say a base comes in, okay? You can imagine that there's more than one of these types of intermediates within your solution. So this would be a potential base for this reaction or something later on in the reaction sequence. You'll see a few other things with negative charges in there that could potentially be bases within the reaction solution. So I'm just going to put a generic base, okay? So like that, like we often do. And that's going to deprotonate that proton there. Okay, so remember, this can't reach around, this negative charge here. We can't reach around and grab that proton, okay? So we're going to have some other base do it. And then some acid is going to protonate that proton in the next step, okay? So are we cool with that one? Okay, so when doing this reaction, you've got to remember you're going to need a lot of room because of how big these structures get. And now we have some acid in our solution like that. And this base is going to deprotonate that acid like that. Does that make sense? Yes. Okay, so I'm going to ask you because I'm going to have to go backwards on the board now. Did you already write down all of this stuff over here? Can I erase all this stuff over here? Give me a second. Okay, so I'll leave this last picture here. Yes, the last picture. Okay, but everything else I can erase? Okay. Can you erase that one as well? Oh, I'll need it. Okay, so it's just an acid-base reaction so we'll keep using the equilibrium arrows. So something like that. And plus the B. Okay, so now the benzolamine comes in. So the other reagent that we added in the starting materials, do you remember from a while back we just erased it at the beginning? Yes. So benzolamine. So let's just draw it because this is when it comes in. So again, I like to draw it to see all the important bonds are really being shown and the lone pair of electrons like that. So what you may see, remember is the carbon of a carbonyl group is very electrophilic, okay? Also what we've done here is we've made a really good leaving group. So this thing here that I'm circling, when this is attacked and it comes back down it's going to make this small molecule that's very stable unto itself. It's called dicyclohexyl urea. Okay. And you'll see the solid stuff come and precipitate out of the reaction. It's white and it gives you an indication as to when your reaction is done as well. But anyways, it's very, I mean, it says it's easy to separate. I personally have used it and it'll get all the way through your column and stuff. So it's not the easiest thing, but it is very, it's fairly insoluble into an organic solvent. So anyways, let's go ahead and attack. I kind of drew it on the wrong side so we're going to have to go all the way around. Okay. Again, once this thing comes back down it's going to make this really good leaving group. Okay, kicked off. Does it make sense to where we have gotten to right now? So far so yeah. Okay. So I'm going to do the next step if we're cool? Okay. Wonderful. So like we're saying, just like normal these electrons are going to come back and kick back down just like that reaction you've seen a million times, okay? So those are going to come back down and make that carbonyl group again. But instead of kicking this back off the amine it's going to kick the oxygen off. Here. So what's going to happen is kind of this cascade like this. Okay. Okay. And you could think of it as deprotonating here or there's these acids in here so we'll just put an acid. So remember probably won't deprotonate this one because it's too close. Okay. It's on the same molecule. Have you ever saw this other step on the other side of the board? Okay. So that's, that reaction there is the driving force of the mechanism. Okay. So that's where our forward arrow is. And that's the dicyclohexyl urea coming out and I'll show it to you. Which one is the dicyclohexyl urea? You'll see it actually in two seconds I'll draw it for you. Okay. That thing there. The byproduct. Okay. So try to get that from this step here. But it's not the important portion. This is the important portion. Okay. And we have some base in there. We're going to do the last step of the reaction here. Okay. And that's going to just deprotonate one of these protons. Give the electrons to the nitrogen there. It would make the amygdala. Okay. So we'll do it this way. The byproduct of the reaction is this one here. Does that make sense? Okay. So any more questions on the camera? Okay. Wonderful.