Fischer esterification of carboxylic acids. Created by Sal Khan.
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- So, in general: an carboxylic acid + alcohol = ester + water...?(51 votes)
- yes, but you do need an acid as a catalyzer.(48 votes)
- At the end, should it be spelled ethylhepatanoate or ethylheptanoate?(14 votes)
- It is called Ethylheptanoate.(18 votes)
- if an atom tend to have 8 electrons in outermost shell, why does S in H2S04 form so, so many bonds when it already has 6 valence electrons? I thought it might want only two more electrons?(6 votes)
- For period 3 elements, they have empty d-orbitals, so instead of limiting itself to a maximum of 4 bonds (eg. 4 sp3 orbitals for carbon), sulfur can form hybrid orbitals which can hold more than 4 bonds by incorporating the d-orbitals during hybridisation (eg. 6 sp3d2 orbitals in sulfuric acid). Sulfur and other group 3 and above elements can expand their octlet since they have energetically accessible d-orbitals.(14 votes)
- 7:01.... hang on, didn't Sal say that the ethanol was already protonated by the sulfuric acid? Therefore, how can we have an unprotonated ethanol now floating about, which can stage a nucleophilic attack on the carbonyl carbon? Sulfuric acid is so strong that I thought all the ethanol would have been protonated by now?(6 votes)
- Remember that sulfuric acid is a catalyst and is present in small amounts. It protonates every ethanol molecule it can until it is all used up. But ethanol is the solvent and is present in much larger quantities (the excess reactant, if you like), so there is lots of unprotonated ethanol still remaining in the mixture.(5 votes)
- Ok this may sound stupid, but why does the protonated oxygen located on top lose a proton while the other (right hand one) gains one and acts as a leaving group? wouldn't it be possible for the 2nd (top) oxygen to accept a proton as well and act as the 2nd leaving water group?(1 vote)
- Exactly. The two oxygen atoms are equivalent, so, half the time the top oxygen will be protonated to form water as a leaving group. The other half of the time the other oxygen atom will be protonated. Both pathways are equally likely.(9 votes)
- you made the ester so what happened to agua(4 votes)
- I am actually looking at an updated organic textbook right now and I believe13:53should be be two different steps. Deprotanation occurs after the leaving group leaves hope that helps all you guys understand the mechanism better but thanks Kahn for doing most of it right!(3 votes)
- No. Actually you are mistaken. You are a step premature....there must be a proton transfer between OH and OR before the leaving group leaves. He has the correct mechanism...I have no idea what you're specifically referring to.(2 votes)
- Why would the oxygen form a bond with hydrogen and become positive? Isn't it more stable if it stays neutral?(2 votes)
- I assume you're asking about what Sal shows around3:00or6:00?
Yes, but you have to consider what else is present in the reaction — in this case there is a strong acid (sulfuric acid), which will force a proton onto the oxygen.
In other words, protonated ethanol and protonated heptanoic acid are more stable than sulfuric acid.
pKa tables are an important reference for figuring out where protons will end up — for example:
Found a pKa table that includes protonated carboxylic acids here:
- Since this is in equilibrium, what would promote an Ester product? or on the other hand the Acid product?(1 vote)
- Right. A Fischer esterification is an equilibrium reaction.
RCOOH + R'OH ⇌ RCOOR' + H₂O
Think of Le Châtelier's Principle if you want to push the position of equilibrium in one direction or the other.
To favour ester formation, you could
• Use a large excess of the cheaper reactant (usually the alcohol).
• Remove the water as it is formed (look up "Dean-Stark apparatus").
To favour conversion of ester to acid,
• Use an excess of water.
• (Better) Use an excess of NaOH or KOH. The strong base will react completely with the acid as it forms and drive the hydrolysis to completion. This is called a saponification reaction.(5 votes)
- what are other names for this type of reaction besides esterification?(2 votes)
- there is no other bc this is a general idea (using an c acid to make an ester)... the exact opp of this is ester hydrolysis(2 votes)
Let's think about what might happen if we had a solution of this carboxylic acid here. We might as well name it just to get some practice. We have one, two, three, four, five, six, seven carbons. So this is heptan-, and then we don't write heptane, because this is a carboxylic acid. It is heptanoic acid. So let's see what happens if we have heptanoic acid reacting with-- this is one, two carbons, and then it has an OH group, so this is ethanol. That's what the OH group does. It makes this an alcohol. And it's in the presence of a sulfuric acid catalyst. This right here is sulfuric acid, one of the stronger acids. I'll actually draw its structure, because I always find it frustrating when people just write just the formula here without the actual structure, because the structure actually shows you why it's so acidic. Sulfur has six valence electrons, just like oxygen. So it has a double bond to an oxygen, another double bond to an oxygen, and then it has a single bond to an OH group, and then it has another single bond to an OH group. And notice, it has one, two, three, four, five, six valence electrons. Now, the reason why this is such a strong acid, is that if either of these oxygens take the electron from this proton, so actually give away the proton to the solution, there's a ton of resonance structures here. And maybe I'll make a whole video on sulfuric acid, just to show you all of the resonance structures. But, in general, whenever you see a reaction when they say it's catalyzed by an acid, all you have to do is realize it's just going to make the surrounding solution a lot more acidic, just a ton more acidic. Maybe we're in a solution of ethanol, and if we are in a solution of ethanol, it'll just add protons to the ethanol itself. So you can imagine this guy right over here. Let me draw the sulfuric acid a little bit differently. Let me draw these oxygen-hydrogen bonds, so you have this oxygen and it is bonded to a hydrogen there. This is floating around the solution. You have your ethanol that looks like this, so two carbons and then bonded to an oxygen, and then that oxygen is bonded to a hydrogen. The oxygen has two lone pairs, just like that. And so this guy really is good at getting rid of the protons. So you have a situation where this electron can be taken back by this oxygen, and then it can actually be given here, and there's all these resonance structures. But it's just very good at taking that electron. And that can happen at the exact same time that one of the ethanols capture the hydrogen proton, at the exact same time that this oxygen captures that hydrogen proton. And if you just look at this part right here, this will just result in-- this part alone will just result in a ton of having these protonated ethanols flying around. Actually, let me draw it over here. I want to make sure I'm using my screen real estate properly. So this will result in a ton of these protonated ethanols. So one, two carbons, it has this original hydrogen over here. But now it has one electron in that original pair, and it gave the other electron to this hydrogen, so it gave the other electron to that hydrogen nucleus. Let me do it in that same color. So it took a proton. It gave away an electron, so it now has a positive charge. And now what was a sulfuric acid now has a negative charge over here. So if I were to draw it, it would now look like this. Plus sulfur, two double bonds, two oxygens. You still have this OH group. Actually, it could still donate. This is still acidic, but now this oxygen right now gained an electron. It now has a negative charge. Now, the whole reason I did this is I wanted to give you a tangible sense of what sulfuric acid looks like and why it's acidic. But really, you just have to kind of internalize that you're just going to have a bunch of hydrogen protons floating around. They could be attached to an ethanol. If this was a water solution, they could create hydronium, so you have a bunch of hydrogen protons floating around that will catalyze this reaction. They will be used to facilitate the reaction we're going to explore, and then they will be let go. So hopefully, this right here, by having this right here, if I start involving some of these protonated ethanols in our reaction, you won't view that as a huge stretch of the imagination, because they would have gotten protonated by the sulfuric acid. Or if I just actually grab protons in our reaction, that actually might make it a little bit simpler. So let's start with the actual reaction. So let me read redraw my heptanoic acid. So I have one, two, three, four, five, six, seven carbons, double bond to an oxygen, and then we have an OH group right over here. Now, the first step of this reaction, this oxygen right here, we have all these protons floating around, very acidic environment. We have sulfuric acid there just giving protons away to the ethanol or to other things. So this guy can grab a proton either directly from sulfuric acid or maybe from one of the protonated ethanols, either one. So we could just draw it like this. He just grabs a hydrogen proton. The hydrogen proton might have had an electron associated with it that would then go back to a sulfuric acid, but just to make things simple, I'll just say grabs a proton from something else. So once he grabs that proton, then it looks like this. So I'll draw my two, three, four, five, six, seven carbons, double bond to an oxygen, single bond to an OH. This oxygen had two lone pairs, but now one of the lone pairs is broken up because it gave an electron to this hydrogen proton right over there. Hydrogen proton, it was positively charged. It gained an electron. Now it is neutral. But this oxygen right over here gave away the electron, so it is now positive. Now, the next step, we have all this ethanol floating around. So let me introduce some ethanol, and I'll do this in a different color. So we have this ethanol floating around, and this is one, two carbons. This is our ethanol. You have two electron pairs on that oxygen. And then you could imagine, especially because this is now a good leaving group, you can imagine that this acts as a nucleophile. It would do a nucleophile attack on this carbonyl carbon right here. And so what you could imagine is that this electron attacks, or gets given to, this carbon right here on the carbonyl. And at the exact same time that this happens, this oxygen is positively charged. It wants an electron. It wants to get neutral again. It's already hogging these electrons. That's why you have a partial positive charge on this carbon. That's why this guy might be attracted to this. We've see this in Sn2 reactions many, many, many, many times already. So this electron will be taken up by this top oxygen over there. So after that happens, what does our newly formed molecule look like? Let me just scroll down and have some clear space. After that, our newly formed molecule will look like this. We have two, three, four, five, six, seven carbons to get to the carbonyl carbon, and now it will have a single bond to that oxygen up there. That oxygen had one lone pair already. It had this bond to this hydrogen that it nabbed from the solution, that proton that it nabbed from the solution, and now it has another pair. It had this electron that was participating in a double bond with this carbonyl carbon, but now it took the other side of that bond back. So now it has that electron and the other one that it took from the carbon, so it has two lone pairs again. It had a positive charge, but now it took back an electron. Now, it is neutral. Now the rest of the molecule, we have this OH group right over here. We have that OH group right over there. And now we have the actual ethanol that has attached itself, so it is no longer ethanol. It's attached itself to this larger molecule. So this oxygen right over here, it still has one lone pair, but the other lone pair has been broken up, and it has given an electron to this carbon. It is now bonded to the larger molecule, and then the rest of it, you have these one, two carbons right there, and then you have this hydrogen right over there. Remember, we have a bunch of protons floating around. Actually, I should make it clear. These are all reversible reactions, so I actually shouldn't even draw one-way arrows here. Let me scratch that out. Actually, delete that. A better thing to do, instead of drawing these one-way arrows is to show that the reaction could actually go in either direction. It's just as likely to go from here to here as it is from there to there. So let me draw that. These are kind of in equilibrium with each other. So you can imagine the next thing that could happen is another oxygen could grab a proton from the medium. And actually, before I do that, let me make sure this guy can lose a proton to the medium. Actually, this guy could take the proton from that guy, but I won't do it that way. So you could imagine a situation where this proton just jumps off. I should have pointed it out. This guy gave away an electron to this carbonyl carbon. So this right here has a positive charge. In general, the oxygen, very electronegative, it's going to be hogging the electrons of this proton, so it can take them away. So far, we gained a proton, and now we can give back a proton. So this electron right here can go back to this oxygen, making it neutral. And the proton can be picked up by anything. It could be picked up by another molecule of heptanoic acid through this first protonation we saw. It could be picked up by another molecule of ethanol. Let me draw it like that. Let me draw it as getting picked up by another molecule of ethanol. It just gets thrown back in the solution. So this guy gives an electron to the proton, but the end result is the hydrogen leaves. The electron goes back to the oxygen. It now has a neutral charge. And once again, these are all reversible reactions. It can go in either direction. But we're now over here, and let me redraw my heptanoic acid. Two, three, four, five, six, seven single bond now to the oxygen, single bond to this OH. And now we have this bond right over here to this now deprotonated, what was ethanol, so you have O, and then we have one, two carbons, just like that. And this guy has a hydrogen attached to it. And I won't even bother to draw this protonated ethanol. The protons are flying around everywhere. Now, the next step, this guy, this OH group, and especially the oxygen in it, he could grab a proton from the surrounding solution. So he's got these two lone pairs. He can grab a proton from one of these protonated ethanols from the sulfuric acid, or maybe from one of these other intermediate molecules, from anywhere. That's the whole point of having this acid catalyst, so he could donate an electron to a proton and then form a bond with it. I'll keep it in that color. And so if that happened, then the next step in our reaction-- and remember, these can all go in either direction. The next step in our reaction will look like this. You would have your two, three, four, five, six, seven carbons, single bond to an oxygen, and then you have your bond to this oxygen, which is bound to two carbons, one, two carbons, just like that. And this guy on top is bonded to an oxygen. He's got two lone pairs. And this guy over here grabbed a proton. He gave an electron to a hydrogen proton. He had two, but now he gave one of them to this proton. And so he gave it to that hydrogen. That hydrogen is now neutral. It gained an electron. This oxygen is now positive, because it gave away an electron. It still has this other lone pair over here. It is now positive, and frankly, it is now a good leaving group. So, in the next step, you can have someone else. Remember, other people need protons earlier in this reaction. So this proton might get lost, maybe by one of the other ethanol molecules, so let me draw that. Let me do a color I haven't used yet. I'll use orange. So maybe one of the other ethanol molecules, or one of the other intermediaries in this whole reaction. So ethanol, I'll just do ethanol because it's easier to draw. It might give an electron to just the nucleus. And then this guy can take the electron back. This guy could take that hydrogen's electron back and give it to this carbon that was the carbonyl carbon several steps ago. And then since he's got that electron, he can then give back this electron to this, what was an OH group, but now it's got another hydrogen there, so he can then give it back to him. And then the resulting products will look like this. And this is all in equilibrium. We now have two, three, four, five, six, seven carbons. Now, this guy has a double bond again. So you have an oxygen right there. It is now a double bond. I'll draw this newly formed double bond in magenta. This guy has left as water, so I can just draw that, so now you have this other OH group that is bonded to this hydrogen over here. He has now left as water. And now you have this other thing over here. You have what was that ethanol group has now attached itself. That ethanol has lost its hydrogen. It has attached itself to what was a carboxylic acid, so now it looks like this. It now is bonded to an oxygen and is bonded to one, two carbons, just like that. And this whole reaction that I showed you is called esterification, one of those words that I have trouble saying. This one in particular is called the Fischer esterification. And he won the Nobel Prize in 1902 generally for his work in organic chemistry. And the reason why it's called that is we started with the carboxylic acid. We started with a heptanoic acid, and now we have an ester. An ester is something that, instead of an OH group like you have in carboxylic acid, you have an OR. You have an oxygen with an actual alkyl group attached to it. And this ester right here, and we'll probably talk more about esters in future videos, this ester right here, just to give you a hint of how to name it, you first start with the group that's attached to this oxygen right here. There are two carbons over there, so we'll use ethyl to name that over there. And then we have the rest of this, and we already know that that is one, two, three, four, five, six, seven carbons, including the carbon in the carbonyl group. And so that is hepta-, heptan-. And you would be tempted to call it heptanoic acid, but it is no longer a carboxylic acid. It is now an ester. So you call it heptanoate. And this is what tells you that you are dealing with an ester. And this tells you what's on the other side of the oxygen in the ester. This is telling you how many carbons you have attached to the carbonyl chain of the actual ester. Anyway, hopefully, you found that useful. I just wanted introduce you to a well-known reaction mechanism of creating an ester out a carboxylic acid.