Nomenclature and reactions of carboxylic acid derivatives
Acid-catalyzed ester hydrolysis
Voiceover: In the video on Fischer esterification we saw that if we took a carboxylic and alcohol, in an acid-catalyzed reaction, we produced an ester, and we also produced water. Our goal in that video was to make more of our ester, so we shifted the equilibrium to the right, to make more of our product. In this video, we're talking about the reverse reaction; we're going to talk about ester hydrolysis. So if we increase the concentration of water, that would shift the equilibrium back to the left, and we would hydrolyze our ester, and turn it into our alcohol and our carboxylic acid. So, it's important to think about what bond we're going to break; you can see that we're going to break this bond, in here, and this oxygen, and this R prime group turn into our alcohol, and so we'll see that in our mechanism. So let's go ahead, and look at the details of the mechanism, where we're starting with an ester, and we're going to, first, think about what else is present. One solution, H two O, and H plus, would give us H three O plus, and I'm going to go ahead and draw in the hydronium ion, over here, so this is H three O plus. The first step of the mechanism is to protonate the carbon EEL oxygen, so this lone pair of electrons picks up a proton from hydronium and let's go ahead and show the protonated carbon EEL, so now we have our oxygen, it has been protonated, so it has a plus one formal charge, so let's show those electrons, so these electrons right here, in magenta, pick up this proton to form this bond right here, and this activates our carbon EEL, so we've seen this in earlier videos, so when you think about a resonance structure, that actually makes this carbon more positive, it's more electrophilic, and therefore, that carbon is going to react with a nucleophile in our next step, and our nucleophile here is water, so water's gonna function as a nucleophile, the nucleophile attacks our electrophile, and that pushes these electrons off, onto this oxygen, so when we show the bond now, between the oxygen and that carbon, let me go ahead and draw in the rest of this: this would be a plus one formal charge on this oxygen, and let me highlight those electrons here, so these electrons in blue, are going to form the bond between this carbon and this oxygen. We still have an oxygen over here, on the left, and now, that oxygen has two lone pairs of electrons, so let me go ahead and show those electrons here, so let me make them green, so these electrons right here, in green, move off onto here, and are now a lone pair on that oxygen, we still have an oxygen bonded to this carbon, and an R prime group. The next step, we need to deprotonate; we need to get rid of that plus one formal charge, and so, a molecule of water can come along, and this time, function as a base, so water's gonna function as a base, gonna take this proton, leaving these electrons behind, on that oxygen, so let's get some space down here, to show the deprotonation, so we would now have this carbon, with an OH on the left, and after we deprotonate, we're also gonna have an OH on the right, so let me go ahead and draw in that OH, on the right, so, showing those electrons, let's make those red, so these electrons in here, are gonna move off onto this oxygen, and then drawing in everything else, we have our R group on the left, and we have OR prime on the right, and I'm gonna go ahead and put in lone pairs of electrons on this oxygen, because in the next step in the mechanism, we're going to protonate that oxygen. So hydronium is present, so H three O plus, so I'm gonna go ahead and draw that in here, so H three O Plus, this oxygen is gonna pick up a proton from hydronium, leaving these electrons behind, and so we're gonna protonate that oxygen, and the reason why we protonate that oxygen is it turns it into a better leaving group, so let me go ahead and draw in everything, so we have our OH groups at the top, we have our R group on the left, we have our oxygen, which has been protonated now, so it's gonna have a plus one formal charge, so plus one formal charge on this oxygen, let's show those electrons, so let's make those blue, here, so I'm saying that this lone pair is gonna take this proton, forming this bond, right here, and if you look closely, you now have alcohol hiding as a leaving group, because in the next step of the mechanism, we're going to reform our carbon EEL, and alcohol is going to leave, so if these electrons move into here, to reform our carbon EEL, that's too many bonds to carbon, and so these electrons are gonna come off, onto the oxygen, and so here's the bond-breaking step, where the alcohol leaves, and so let's go ahead and draw our product, so we're going to form our carbon EEL, and this oxygen is going to have a plus one formal charge now, and we have an R group, and, over here, would be an OH, because the alcohol is going to leave, so I'm gonna go ahead, and draw in the alcohol here, so OR prime, with now two lone pairs of electrons, so let's follow some electrons, so these electrons in blue, are the same ones as blue up here, and let's make these electrons in here red, so these electrons in red are going to come off, onto this oxygen, so we lose our alcohol at this step, so loss of our alcohol gives us this, over here, so we're almost done with our reaction, so this is really close to a carboxylic acid, all we would have to do is deprotonate, so we could think about another molecule of water coming along, and acting as a base, so water acts as a base, takes this proton, leaves these electrons behind, and that, of course, gives us our carboxylic acid. So, if I just go ahead and draw in our R group, and then we have ROH, and let's follow those electrons, so let's make those red as well, so these electrons, right here, come off, and we get our carboxylic acid as our product, so there's your mechanism for acid-catalyzed ester hydrolysis, which produces a carboxylic acid, and it produces an alcohol, and when we look at some reactions in a second here, we're gonna think about this part, right here, where we lose our alcohol as one of our products, to also give us our carboxylic acid. So let's look at a reaction: So, over here on the left, we have methyl salicylate, or oil of wintergreen, and we're looking for an ester, because we have H two O and H plus, giving us hydronium in solution, and so this is the portion of the molecule that's going to react, it's going to hydrolyze that ester, and we know from the mechanism, this is the bond that's going to break in our mechanism, and so, we can go ahead and draw our products, so we're gonna break that bond, and the whole left portion, so let me go ahead and draw that, this whole left portion here is going to form our carboxylic acid, so let's go ahead and draw in our carboxylic acid, as one of our products, so we have our ring we have our carboxylic acid, so this OH on our carboxylic acid, this OH came from H two O in the mechanism, and then we have this other OH, which produces, of course, salicylic acid, and then we're thinking about our alcohol product, let me go ahead and use green for this, so this is going to leave, we can see that, after you protonate, you get loss of that as your alcohol, and so if we just add a proton onto this oxygen here, we can see that we would form methanol, as our other product, so if I go ahead and draw in the methanol here, we have our two products: we have salicylic acid, and we have methanol, so acid-catalyzed ester hydrolysis. This reaction is at equilibrium, technically, and so you could do things like push the equilibrium to the right, and if you remember from the Fischer esterification video, this is what we use to make our wintergreen; we used methanol, and we used salicylic acid to produce our wintergreen, and so everything depends on reactions conditions, in terms of shifting the equilibrium. Let's look at another practice problem here: So, once again, we have our ester, and we have water, and I drew the water in a little bit of a weird way, and I'll show you why in a second, so acid-catalyzed ester hydrolysis, once again, we think about what bond is gonna break; this bond is gonna break, and the left side is going to turn into our carboxylic acid, so we can go ahead and draw our carboxylic acid, so we would have this portion of the molecule, so our carboxylic acid, like that, and once again, this OH, let me go ahead and highlight it, so this OH right here, in our carboxylic acid came from our water molecule, so we could think about losing a proton off of water, and then we form our carboxylic acid on the right, our other product, let's be consistent and stick with green here, this is the portion that's gonna turn into our alcohol, so we think about adding a proton onto that oxygen, and, once again we have methanol, as our other product, so, acid-catalyzed ester hydrolysis. Now the reason I drew the water molecule in a weird way, is because, what would happen if, instead of this hydrogen, what would happen if we had an R group, an alkyl group? Well, that would change this hydrogen into an alkyl group, and then we would form an ester as our product, so we'd be starting with one ester, and we'd be turning it into another ester, and that is called, "transesterification," so pretty much the same reaction that we've been talking about, except we wouldn't get a carboxylic acid, as one of our products; we would get another ester. And so, let's go ahead and do this again, starting with the same ester, but this time, we're gonna use butanol, instead of water. So, once again, we can think about losing, we're gonna break that bond, and we're going to lose this proton, and then we could stick those portions of the molecule together. So, if we stick this together with this, we can see the identity of the other ester that will form, so let's go ahead and draw in our product, so we would have an oxygen here, and then we'd have four carbons: one, two, three, four; so these four carbons from butanol, which form our ester. Our other product, once again, we see this portion over here, that would give us methanol, so once again, we have methanol as our product, and even though everything is at equilibrium, methanol has a very low boiling point, so we could boil off the methanol, and shift the equilibrium to the right to make more of our product. So we can see, once gain, we're starting with an ester of methanol, and we're converting it into an ester of butanol, simply by changing the reaction conditions, in a transesterification reaction, so this is a pretty useful reaction in industry.