How to prepare acid anhydrides from carboxylic acids. Created by Jay.
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- At6:30, you did not mention mechanism of dehydration. Could you elaborate? It is important to show mechanism for first-time learners, even if it is simple for the instructor.(6 votes)
- Dehydration is typically acid- or base-catalyzed. Heat is used in order to break the bonds required to remove the hydroxyl groups, oxygen, and/or hydrogen. In the case that you were talking about, a the following mechanism could be proposed:
Both of the oxygens in the hydroxyl groups have a slightly negative charge, because they have a greater electron-withdrawing capability. Either one of them would attract the hydrogen from the neighboring hydroxyl group (call this the water-forming group). From there, the water forming group would leave, and the electrons that covalently bonded the remaining oxygen to the hydrogen would help to maintain the bond between the remaining oxygen and the carbon.
As I said, this is a mechanism that I proposed. I'm not sure if it is correct, but it makes sense to me.(3 votes)
- Is it important to know each step of the reaction for the MCAT? or general details about the reaction is sufficient?(2 votes)
- Just knowing the general details should be sufficient. As long as you know what the products formed are, you should be fine. They're going to test general knowledge and the application of that knowledge to novel situations so knowing how to apply what you know versus every single minute detail is more important.(5 votes)
- What would be the name of the compound formed at06:50? I'm talking about IUPAC name, just to be clear. Thanks :^)(1 vote)
- The preferred IUPAC name is 2-benzofuran-1,3-dione, but almost everyone calls it by its common name, phthalic anhydride.(3 votes)
- At1:40, why isn't it the Cl- that leaves the acyl chloride? Why do the electrons get moved onto the oxygen? Is it because O is more electronegative than Cl?(1 vote)
- In2:00,why don't we kick those electron off the chlorine to form the chloride anion?(1 vote)
- Is that not what happens by about the2:30mark?
Even though we show mechanisms in multiple steps that is for our clarity only, the reaction likely takes place in one very rapid step.(1 vote)
- why would the OH from NaOH attach to the H in terminal OH of the alkyl group? in earlier lectures; it could have attached to the alpha hydrogen or could it not have E2 to the carboneal carbon(1 vote)
Voiceover: Here's another carboxylic acid derivative so, this is an acid anhydride over here on the right. And we can form those from carboxylic acids. So if we start with the carboxylic acid, and our first step, add a base, like sodium hydroxide, and our second step, add an acyl chloride, then we'd form our acid anhydride as our product. Now if you think about a mechanism, sodium hydroxide's a base. The hydroxide anion is gonna take this proton, leaving these electrons behind on the oxygen. There are already two lone pairs of electrons on the oxygen to start with, so if we go ahead and draw the product, we would form a carboxylate anion. So, this oxygen right here would have three lone pairs of electrons on it, like that. And so if we follow some of those electrons, this would have a -1 formal charge, and if we put these electrons in magenta, those electrons come off onto our oxygen to form our carboxylate anion. And that's gonna function as our nucleophile. So in the second step, we add our acyl chloride. I'm just gonna go ahead and draw in our acyl chloride. It's going to be our electrophile, and let's think about why here. We have the oxygen withdrawing electron density from this carbon, because oxygen is more electronegative than carbon, so we have that. And then we also have this chlorine doing it as well. Chlorine is more electronegative than carbon as well, so we have these two things withdrawing electron density, and so this carbon is definitely electrophilic, right here. And so we have a nucleophile that's going to attack our electrophile, so our nucleophile attacks our electrophile, these electrons kick off onto our oxygen, and we can go ahead and show the result of our nucleophilic attack. So we would now have this carbon double-bonded to this oxygen, and then this oxygen is now bonded to this carbon, and then we have an oxygen up here, with three lone pairs of electrons, -1 formal charge. We still have this carbon bonded to a chlorine, and we still have an R prime group here, like that. So following some electrons, let's go ahead and put in these lone pairs here, the electrons in magenta form the bond between the oxygen and the carbon, and then we could say that these electrons in here moved off onto our oxygen, to give that oxygen a -1 formal charge. When we think about the next step, we know that the chloride anion is an excellent leaving group. So, if these electrons in blue move in here to reform our double bond, then these electrons would kick off onto the chlorine to form the chloride anion, which we know is a stable on its own, as a leaving group. And so, that forms our acid anhydride, so let's go ahead and draw in the final structure here. So we would have our oxygen, with lone pairs of electrons on our oxygen. We would have this bond. We reformed our carbonyl like that, and then the chloride anion was our leaving group, so now we have an R prime group, and so we've formed our acid anhydride. And so we could make these R groups the same, or we could make them different, and so this is a good way of forming a mixed anhydride, as well as one that is symmetrical. So let's look at an example. So let's take acetic acid and we're gonna do two different reactions with acetic acid. So the first thing we're gonna do, is add thionyl chloride to acetic acid, and we've seen that the addition of thionyl chloride converts a carboxylic acid into an acyl halide. So let's go ahead and show the conversion of that carboxylic acid into the corresponding acyl halide. And then we could take that acetic acid, and in a separate reaction, we could add sodium hydroxide, and the hydroxide takes this proton, which leaves these electrons behind on the oxygen. So let's scoot up a little bit, so we can see the mechanism that we talked about before. And so this would form sodium acetate, draw in our lone pairs of electrons on this oxygen, -1 formal charge, and a plus like that. So this oxygen already had two lone pairs of electrons. And so now you have the situation that we talked about up here. Your carboxylate anion functions as a nucleophile, attacks your electrophilic carbon on your acyl chloride. So we could show these electrons attacking this carbon, these electrons kick off onto the oxygen, and then when those electrons move back in, to reform your double bond, these electrons would kick off onto your chlorine, and then so we can go ahead and draw our product. So just thinking about what happens in this mechanism, we can go ahead and draw our products, which would be a symmetical anhydride, so we would have our oxygen right here, and then we would have our group over here. So thinking about the R groups this time, so this R group is a methyl group, and then we could think about this oxygen being this oxygen, and then this portion of the acyl chloride is this portion for our final product for our acid anhydride. And so this is acetic anhydride, which is the one that's used most commonly in an undergraduate lab. And so this is a nice way of preparing acid anhydrides. Let's look at another way to form acetic anhydride. You could start with two carboxylic acids, and this would be acetic acid and acetic acid, so the same carboxylic acid, and apply high heat, and this time you think about a dehydration reaction, so you could think about losing an OH from one, and a hydrogen from the other, to form water, so you could think about losing a water here, and your dehydration reaction. And then you can stick those portions of the molecule together, so you could take this portion, and then this portion, and put them together, and you can see that that is once again acetic anhydride, so let's go ahead and draw that. So we would form acetic anhydride here by dehydration. So this way of doing it is not always the best way, it works for acetic acid, but it doesn't work for most carboxylic acids. Here's one more case where it can work, if you have a situation like this. This is phthalic acid, so it's a dicarboxylic acid, And if you apply heat to it, you don't need as high of a heat as you need for the previous reaction, this heat is higher than this one, but you can once again form an anhydride. So if you think about a dehydration, losing OH from one, and H from the other, and then we can go ahead and draw the product here, so we would form our benzene ring, and then we would form our anhydride like this. So once again, loss of water. So the name of this acid anhydride would be phthalic anhydride because it's derived from phthalic acid. So this is a good way to form five- or six-membered rings. In this case, we have a five-membered ring. We have a carbon, an oxygen, a carbon, a carbon, and a carbon, so we have a five-membered ring this time. It also works for six-membered rings.