Move electrons using arrows to predict and label products of t...

Question

Move electrons using arrows to predict and label products of the following reaction:

Two chemical structures shown: ethoxide ion and acetic acid, with a plus sign between them indicating a reaction.

Accepted Answer

Alright. So let's go ahead and get started. So the first thing is I have to figure out which one's the acid and which one's the base. Is there an obvious acid here? Actually, there is. Okay? If you remember, the name of this functional group is carboxylic acid. Okay? And whenever you have a carboxylic acid, you know that's gonna be your acid. Okay? But let's say you forgot that. Let's say you did not. I know a lot of you guys are just like, oh, that's a carboxylic acid. Let's say you forgot that. Well then how could you know what that is? Well Well you could think about the other one. The other one has a negative charge. The negative charge must be the base. Right? Because remember that I said that the negative charge is always a base. Okay? That also means it's a nucleophile. You have to be thinking about that in the same way. Nucleophile and base, same thing. K? Then that means that this one if that one's my base, this one must be my acid, even if you didn't remember that was carboxylic acid. So I know that my arrow is gonna start from where? My arrow has to start from the o. So I'm gonna start from the negative charge. So I'm gonna start going in this direction, but now what I don't know is where I'm gonna go from there. Okay? So what I'm looking for here is any hydrogens that have a strong dipole on them. Okay? How many hydrogens are on that carboxylic acid? Actually, there's 4. Okay? There's one here, but then there's also 3 hydrogens here. Okay. Which one do I pick to pull off? Okay. Well, you could try to pick off one of these h's. Okay. The only problem is that if you think about it, what's attached to that H? A carbon. Is there a strong dipole between carbon and H? No. There isn't. In fact, that's that's a covalent bond, so there's not gonna be any partial charges. So what that means is that there's no reason that it would attack that h. It would actually just as likely attack that carbon and that's not gonna give me anything. So what that means is that all these h's are off limits because they have no dipole, no partial charges. Okay? So now look at this last one, the o. Does that one have a dipole? Yes it does, and that dipole is pulling towards this o here. So what that means is I have a partial negative and I have a partial positive and it looks like we have a winner. Okay? So I'm going to draw my first arrow from this negative to that h, and that means that my negative is going to attach to the attach to the positive. Okay? Now I ask myself, are we done? No. We're not done because hydrogen only wants to have 1 bond, not 2. So if I make a bond, then I have to break a bond. And that bond would be gonna delete my dipole. Right there. I'm gonna take those 2 electrons and give them to the o. Now let's go ahead and draw our conjugates. So what that means is that now I'm going to have an o that has a new single bond. K. That new single bond came from the make and it's gonna have a hydrogen on it. K. Then I'm also gonna have my carboxylic acid, but now that carboxylic acid is gonna be missing an H and it's gonna have an extra lone pair. It had the 2 lone pairs from before and now it has the extra lone pair that came from that bond breaking. Okay? Now we have to do formal charges. Is this o happy the way it is? Actually, yeah. That one doesn't get a formal charge. How about this one? No. This one isn't happy. It has a negative. K. So notice that this is an acid base reaction because I just exchanged a negative charge and an H with each other. K? So now we're almost done with this question. The only thing you need to do is tell me which one which the conjugate acid and which one is the conjugate base. So what do you guys think? Well this one was the base, so that means this must be my conjugate acid. K. Then the carboxylic acid was my acid so that means that this must now be my conjugate base. Interesting. Right? I hope that this is getting a little bit more comfortable for you guys. It's a big step. I acknowledge that. But I hope that this is making sense. We're always gonna follow the same pattern. So finding the acid and the base, have the base attack the acid, draw arrows until all the octets are fine, and then draw your conjugates. Okay? So I hope that made sense. Let's move on to our next topic.

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