Preparation of esters via Fischer esterification
How to draw the mechanism for a Fischer esterification. Created by Jay.
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- This is a general question about the first part of the video dealing with a general Fischer esterification: there are so many alcohols hopping around, both protonating and deprotonating, and I find this extremely confusing! It seems that the alcohol is a reactant and a catalyst both. I'm not quite sure even what my question is, but all these different alcohols are confusing to keep track of. Why are there so many of them in the mechanism? or is it just the same alcohol hanging around and shuttling protons around?(7 votes)
- Protonation is necessary for shifting formal charges between molecules. This exchange of charge whether it is adding or losing a proton is how molecules combine and separate. Don't get overwhelmed by the multiple steps in this mechanism even though there are a lot. The reason why there are so many protonation reactions is because the carboxylic acid has to be altered in multiple ways at different points to yield an ester. If you get confused just think of the reaction as simply this: adding an R group to the hydroxyl group (OH) of a carboxylic acid via an oxygen bond.
To answer your second question: there is only one alcohol that is is used to protonate and deprotonate the the carboxylic acid and provide the R group that is to be attached.
I hope this helps.(7 votes)
- A question that has been haunting me for my entire Ochem career: Why is this mechanism taking the path it is taking? For instance, at1:20, the carbonyl carbon is positive enough for the nucleophilic attack, because of the carbonyl oxygen. But then at5:33, under almost the exact same conditions, the alcohol deprotonates the oxygen, instead of attacking the carbon.
Another example: At3:23, the hydroxyl group is protonated by ROH2, but why isn't the proton donating molecule just dehydrated instead? Why is the proton taking the time to leave its own ROH2 molecule to make another ROH2 molecule?(4 votes)
- These are all valid, good questions! The answer to them lies in how we look at organic reactions, and chemical reactions in general. Bear in mind that the aforementioned molecules are in a hectic state, frequently colliding and interacting. The possibilities you theorized in your questions do indeed occur; this is precisely why organic reactions almost never have 100% yields of the desired products. Mechanisms presented in lectures are not telling you exactly how each type of molecule in a solution interacts with the others, but rather highlights a certain pathway that produces the product of interest. Furthermore, these mechanisms are often simplified, and side reactions can also eventually give rise to the major products via more complicated pathways. Extending this concept to this particular reaction, remember that the steps are in equilibrium, meaning that although other products (such as the ones you mentioned) do form, they are less stable and easily revert back to their previous forms; given enough time, the reaction will favor the most stable product(s). The Fischer esterification is actually a good example of this: the reaction takes anywhere from 1-10 hours to go to completion! :)(2 votes)
- Can't you also shift the equilibrium to the right by adding excess carboxylic acid too?(3 votes)
- Yes you can, but it would typically be cheaper to add more alcohol or easier to remove water.(3 votes)
- If the salicylic acid was treated with h2so4 but no methanol would the formation of a ring (similar to the lactone reaction) be possible?(3 votes)
- ..2 years old, but I'd say it can't since there is only 1 carboxylic acid group.
Not only that, but the hydroxy group is coming straight from the benzene ring, so the final ring structure that you're thinking of would have small bond angles that are unlikely to form.(1 vote)
- From7:45on, a intermolecular Fischer esterification that forms lactone is shown. Is it possible that at the same time of forming lactone, an OH part of one molecule attack the COOH part of another molecule, and form a ester with two 5-carbon chain (and no ring) in it?(2 votes)
- yes, you can also get the two 5 carbon chain product and you could essentially form a polymer from it because the OH portion of one can keep attacking the COOH on the other end of the molecule. Then theres also the possibility of a ring forming at any point in that process as well.(2 votes)
- What are these R O O Rs ?(2 votes)
- ROOR is the peroxide functional group. Organic peroxides are organic compounds that contain the ROOR group.(2 votes)
- when making the lactone, the hydroxyl group attacks the carbonyl carbon....and then does the OH attached to the carbonyl group grab the proton from the hydroxyl on Carbon 5. And then from here it's a leaving group?(2 votes)
- That's as good a way as any of looking at it.
It's more probable that the proton transfers involve hydronium ions and water molecules in the reaction mixture, but that involves many more steps.
Your mechanism is an easy way of remembering what happens.(1 vote)
- What happens with a tertiary alcohol? I've heard the reaction doesn't occur and it goes through elimination. If true, why?(2 votes)
- I can see two things that would favor an elimination over a esterification.
1) A tertiary alcohol is sterically hindered from attacking the carbonyl carbon in the acid.
2) A tertiary alcohol when protonated can form a tertiary carbocation, which is relatively stable. This will then allow an E1 type elimination reaction to produce an alkene.(1 vote)
- For any carboxylic acid, can we protect the carbonyl with like an acetal or thioacetal, leaving the hydroxyl group free to be manipulated? or does acetal formationn only work with aldehydes and ketones ?(1 vote)
- Acetal formation works only with aldehydes and ketones.(2 votes)
- At the outset of the video, where did the extra H on the R'-OH come from?(1 vote)
- You can get extra protons (H+) by providing an acid, usually from sulfuric acid (H2SO4). Although it is not explicitly stated it is assumed that there is an acid present that provides these protons but for the sake of simplicity he simply wrote (H+).
As for the alcohol (R'-OH) it is simply an alcohol of any kind that you can add; be it a primary, secondary, or tertiary alcohol. For example: If your alcohol is ethanol you will get ethyl acetate from this reaction. So if you add more ethanol you will get more ethyl acetate because adding alcohol shifts the equilibrium to the right as stated.(1 vote)
Voiceover: One way to make an ester is to use a Fischer esterification reaction. So if you start with the carboxylic acid, and you add an alcohol, and a source of protons, you're gonna form your ester, and you're also going to make water in this process. It's important to note that the oxygen in the R prime group come from your alcohol, and so we'll see that in the mechanism. Also, this reaction is at equilibrium, so if you want to make more of your ester product, you have to shift the equilibrium to the right, and so there are a couple ways to do that. One thing you could do would be to decrease the concentration of water, which would shift your equilibrium to the right to make more of your ester. You could also do something like increase the concentration of the alcohol, and that would shift the equilibrium to the right as well. Let's take a look at the mechanism to form esters. So, we start with our carboxylic acid, and we're going to protonate the oxygen, so a lone pair of electrons on the oxygen can pick up a proton, and leave these electrons behind, and so we can show that we have now protonated the oxygen right here, so it gets a +1 formal charge. Protonation of your carbonyl activates your carbonyl, it makes your carbon more electrophilic, so this carbon is now more electrophilic. So we saw that in some of the previous videos. And so if the carbon is more electrophilic, our next step that makes sense is going to be a nucleophilic attack. So, a molecule of our alcohol comes along, and you can think about a lone pair of electrons on the oxygen attacking this carbon right here, pushing these electrons off onto your oxygen. So let's go ahead and show the result of our nucleophilic attack. So we now have an oxygen up here. It had one lone pair on the left, now it has two lone pairs, bonded to a hydrogen, and then over here on the right is the bond that we just formed, the bond between carbon and oxygen. So let's show those electrons in magenta. So this lone pair on the oxygen forms the bond between carbon and oxygen, so there it is. Also on this oxygen, there's still a hydrogen. There's still an R prime group, and one lone pair of electrons, giving this oxygen a +1 formal charge. And there's still an OH bonded to our carbon like that. The next step is to get rid of this +1 formal charge on our oxygen, so another molecule of alcohol comes along, and this time acts as a base. So in the previous step, alcohol acted as a nucleophile. In this step it's going to act as a base, it's going to take this proton, and leave these electrons behind on the oxygen. So let's go ahead and get some more room down here. So we're going to deprotonate, so let's go ahead and show what we would make. So we have our carbon bonded to this top oxygen, two lone pairs of electrons on it, a hydrogen, an R group off to the left. I'm gonna draw this oxygen down here with lone pairs of electrons so we can show the next step. And then this oxygen right here now has two lone pairs of electrons on it, and it's still bonded to our R prime group. So let's show those electrons. The electrons in blue here move off onto the oxygen, so that's the deprotonation step. The next step is to protonate the OH at the bottom, so let's get some more, even more room to show that. So we're going to show this OH down at the bottom here being protonated, so I'm gonna draw in a source of protons right down here, so our protonated alcohol, +1 formal charge on our oxygen, R prime right here. So a lone pair of electrons on this oxygen can pick up a proton, leave these electrons behind, so we have a protonation step. And the reason why this protonation is favored, is because this is going to create an excellent leaving group. So if you look closely, you're gonna see water there as our leaving group. So let's go ahead and show that. So now this oxygen down here has been protonated, so it has a +1 formal charge on this oxygen. So let's show those electrons. So these electrons right here on this oxygen pick up this proton, so forming this bond to that proton. So let's go ahead and draw in the rest of what we have. We have our oxygen with two lone pairs of electrons and our R prime group, like that. So in the next step, we just formed water as our leaving group in here, if you can see it, if these electrons in here were to come off on the oxygen, you can see that's water. So when these electrons on this top oxygen move in here to reform our double bond, that's when these electrons in here in blue are going to come off onto our oxygen, and water is an excellent leaving group. So let's draw what we have now. So once again everything's at equilibrium, so we're going to reform our double bonds, and this top oxygen now has a +1 formal charge. So let's show those electrons. Let's make 'em red here, so these electrons in red are going to move in here to reform our double bond. This carbon is still bonded to an R group on the left side, this carbon is bonded to an oxygen, and our R prime group over here, the oxygen has two lone pairs of electrons, and we just lost water. So let me go ahead and draw water down here. So, loss of H2O at this step. We're almost to our final product, because all we have to do is deprotonate right here, and we'll form our ester. So, we could show another molecule of alcohol coming along. So our prime, two lone pairs of electrons, taking this proton right here, leaving these electrons behind on our oxygen. So let's go ahead and draw the final structure of our ester. So we would have R, C double bond O, with two lone pairs of electrons. So let's show these electrons in here, move off onto our oxygen like that. And then our carbon is still bonded to this oxygen, and we have our R prime group. And so our end result is to form our ester and water. Ok, so that's a little bit of a long mechanism. Let's take a look at some reactions to form esters using the Fischer esterification reactions. So, let's start with this molecule over here on the left. So this is salicylic acid, and if we add methanol, and we use sulfuric acid as our source of protons, we're going to form an ester. And this is one of those famous labs that's always done in undergraduate organic chemistry. So if we think about the mechanism, remember that this oxygen on our alcohol, and in this case it's methyl, are going to ag. So we're going to lose this OH on our carboxylic acid, and we're going to put this oxygen and this methyl group on in place. So let's go ahead and draw the product. So we would have our benzene ring right here, and we would have our carbon double-bonded to our oxygen, and we would have the oxygen from the alcohol, from methanol, and then our methyl group like that, and then we still have our OH right here. The reason why this is one of those classic undergraduate labs, is this is wintergreen. So this is an incredible smell. It's always a lot of fun to do this in an undergraduate lab, because the lab smells great when you're done, so the synthesis of wintergreen. Alright, let's look at another Fischer esterification. This one is a little bit different. This one is an intramolecular Fischer esterification. So if we look at our starting molecule on the left. This time we have our carboxylic acid and our alcohol in the exact same molecule. And we have all these single bonds in here, which we know we can have some free rotation. So if we draw the molecule in a different conformation, so let's go ahead and do that, so we have our carboxylic acid up here, and let's count how many carbons we have, so let me use red for that, so we have one, we have carbon one, two, three, four, and five. So we have five carbons, so let's go ahead and draw them in, so there's carbon one, two, three, four, five, and then we have our OH. So let me go ahead and number those carbons. So this is carbon one, carbon two, carbon three, carbon four, and carbon five. And so in a different conformation, we can think about this oxygen attacking this carbonyl in the mechanism. So we know that we're going to lose this OH, we know we're gonna lose this Hydrogen, and so we can stick those together and think about our final product. So we're going to form an ester, but it's a different ester than what we've seen before. So we have our carbonyl right here, and then we have this oxygen. So this oxygen is a member of our ring now, and that came from this oxygen. So we can see our five carbons, one, two, three, four, five. We could think about losing water here. And so our product is called a lactone. So this is a lactone here. So, it's an ester that's in a ring. Here we have a six-membered ring, where oxygen is one of the members of the ring. So a pretty cool intramolecular Fischer esterification reaction to form a lactone.