Epoxide Reactions - Online Tutor, Practice Problems & Exam Prep

Epoxides, due to their strained three-membered rings, undergo ring-opening reactions through acid or base catalysis. In acid-catalyzed reactions, protonation occurs first, favoring nucleophilic attack at the more substituted carbon, resulting in anti addition products. Conversely, base-catalyzed reactions involve direct nucleophilic attack at the least substituted position, leading to vicinal diols that are also anti. Understanding these mechanisms is crucial for distinguishing between different types of dihydroxylation reactions, particularly syn and anti configurations.

Acid-Catalyzed Epoxide Ring-Opening

Now that we have epoxides and know how to make them, it's important to know what we can do with them. And it turns out that ethers are highly unreactive. Ethers barely react with anything. But because the fact that epoxides are 3-membered rings, they're very, very highly strained. So what that means is that even though ethers as a functional group are not very reactive, epoxides are because they have so much potential energy locked up in those bonds. They're going to want to break open first chance that they get. So it turns out that we have 2 different ways that we can open the epoxide ring and they have to do with different agents that we use. So these are called the acid-catalyzed ring-opening reaction and the base-catalyzed ring-opening reaction. Let's start off with acid and then we'll work on to base. Okay?

In an acid-catalyzed reaction, notice first of all that I'm using the words acid-catalyzed. Okay? That actually already gives us a hint about the mechanism. Okay? Remember when I said whenever you have an acid-catalyzed mechanism, what's the first step going to be? Protonation. You always need to protonate first. Okay? If it says acid-catalyzed, you always need to protonate first. So what that means is that the very first step is that if I'm using an acid with an epoxide, the nucleophilic O is going to go ahead and grab the H from the acid because that H has a positive charge. So now what I'm going to get is a molecule that looks like this, where everything is still in the same exact place. Okay? Except that I've got a formal charge on this O. Okay? Because the O has too many bonds. Alright?

So now what happens? I've got a Cl^- and the Cl^- wants to break open the ring, but it's trying to decide which side is going to break. Is it going to break the most substituted, the tertiary, or the least substituted, the secondary? Okay. That's supposed to be a 3 by the way. Let me draw that again. The tertiary or the secondary? And the answer to this is that the chlorine or whatever anion we're using, even if it was just a neutral nucleophile, is going to be the most attracted to the side of the ring with the most positive character. Okay? So what we're looking at is the side of the ring that could stabilize that the best because that, I mean, I'm sorry, I just said carbocation. It's not a carbocation. It's just a cation. It's just a positive charge. But that positive charge can delocalize a little bit into those 2 different atoms. So the question is, which side is going to be the one that has the most positive character? The secondary or the tertiary? The answer is the tertiary. So I'm going to go ahead and attack here and if I make that bond, I have to break a bond, so I'm going to break the bond to the O. What this is going to give me is a new compound that looks like this. Let's say that this bond right her

Base-Catalyzed Epoxide Ring-Opening

So let's go ahead and check out base catalyzed now. Let me know if you have any questions about acid. Let's go on to base. Okay? So in a base catalyzed ring opening, what I have is that I'm starting off with a nucleophile right away and there's no protonation step because obviously I don't have an acid. Okay? So what that means is that all I have is an O^- or some kind of nucleophile looking for a way that it can break open this ring. Okay? Because the ring is highly strained. Well, this O^-, since there's no positive charge, it's not going to be attracted to the side that's more substituted. It's actually going to be attracted to the side that's the easiest to access. So for this one, I'm actually going to go ahead and look for the least substituted position. And that's the one that's going to be favored because that's the easiest one to form. So in this case, if my choice is between secondary and tertiary, I'm actually going to go for secondary. So I'm going to go ahead and attack that carbon, which means I'm going to break that bond and what that's going to give me is a new molecule that looks like this, where let's say if this is my red bond again and this is the red bond here, what I would have is, let's say that the O attached from the front. Let's say that my nucleophile attached from the front. That means that, yep, I would get my propyl group and I would get my hydrogen, which I don't have to draw. On the other side, what I would now get is an O^- on the back side. Okay? Now notice that once again, these products are going to be anti to each other because every time you break open a 3 membered ring, they're going to be anti because they're going to snap out of place and they're they're going to just go as far from each other as possible. Okay? And then I have 2 methyl groups. So I'm just going to write that there. 2 methyls. Okay? So that's the end of that step. Now typically we're going to assume that the O^- will protonate. Okay? So typically there will be a protonation step, sometimes your professor will add like over water or something like that. And if there was water, usually we just assume that the end product is protonated. So if it was protonated, what would it give us? What it would give us is actually going to be what we call a diol. Okay? So check this out. Look what I'm going to get at the end. What I'm going to get is I'm going to get an O going towards the back and an O going towards the front. What's the relationship between these alcohols? Well, first of all, we call this a diol because there's 2 alcohols. That's easy. Just diol. There's 2 of them. There's actually a special type of molecule that we'll study more in-depth. Okay. So we have a diol, but also the relationship would be vicinal. Okay. Remember that vicinal is the word that we use to say that they're next to each other. That's the same thing as saying 1,2. Okay. So these are vicinal diols, but on top of that these are anti because they are trans to each other. That means that it was an anti mechanism. So the entire idea behind this is that this is what we would call an anti vicinal dihydroxylation. That just means we're adding 2 alcohols. Okay? Why is this important? Well, this isn't the only way, this isn't the only type of base catalyzed ring opening you could do. In fact, we could have used any nucleophile. I could have used a nitrogen or another type of oxide or whatever. But this specific one where I use NaOH is very important because it's really one of the only ways to make these antivicinodiols. That means I have 2 alcohols facing opposite directions next to each other. This is important because there's another reaction that you're supposed to know called synvicinlodials. And synvicinlodials or dihydroxylation is a completely different mechanism. So these are going to go hand in hand and you're supposed to recognize both of them. Okay? If you haven't learned synvicinaldials already, you will learn it soon because I'll teach it to you. Okay? But anyway, I just want you guys to know that it's supposed to be you're supposed to be able to compare these two reactions to each other and you're supposed to be able to know when do you use an epoxide with NaOH and when do you do the syndial reaction. Alright? So I hope that these two mechanisms made sense. If I were to sum it up once again, it would be acid is attracted to the most substituted, base catalyzed is attracted to the least substituted. You always get an alcohol at the end regardless. And that's probably all you need to know. And you're always going to get anti products because you're breaking open a 3 membered ring. Alright? So I hope that made sense guys. Let's go ahead and move on.

Here’s what students ask on this topic:

What is the mechanism of acid-catalyzed ring opening of epoxides?

In acid-catalyzed ring opening of epoxides, the first step is protonation of the epoxide oxygen, making it more electrophilic. The protonated epoxide then undergoes nucleophilic attack at the more substituted carbon, as this position can better stabilize the positive charge. This results in the formation of an anti product, where the nucleophile and the hydroxyl group are on opposite sides of the former epoxide ring. The general mechanism can be summarized as:

$R_1-C_2-O_1-C_3-R_2+H_{3}O+\to R_1-C_2-{\mathrm{OH}}_1-C_3-R_2$

How does base-catalyzed ring opening of epoxides differ from acid-catalyzed ring opening?

In base-catalyzed ring opening of epoxides, there is no protonation step. Instead, a nucleophile directly attacks the least substituted carbon of the epoxide, as this position is less sterically hindered. This results in the formation of anti products, where the nucleophile and the hydroxyl group are on opposite sides of the former epoxide ring. The general mechanism can be summarized as:

$R_1-C_2-O_1-C_3-R_2+{\mathrm{OH}}_{\mathrm{-}}\to R_1-C_2-{\mathrm{OH}}_1-C_3-R_2$

What are the products of acid-catalyzed and base-catalyzed ring opening of epoxides?

In acid-catalyzed ring opening of epoxides, the product is an anti addition where the nucleophile attaches to the more substituted carbon, and the hydroxyl group attaches to the less substituted carbon. This results in a compound with the nucleophile and hydroxyl group on opposite sides of the former epoxide ring.

In base-catalyzed ring opening, the nucleophile attacks the least substituted carbon, leading to the formation of vicinal diols that are also anti. This means the two hydroxyl groups are on adjacent carbons but on opposite sides of the ring. The general products can be summarized as:

Acid-catalyzed: $R_1-C_2-{\mathrm{OH}}_1-C_3-R_2$

Base-catalyzed: $R_1-C_2-{\mathrm{OH}}_1-C_3-{\mathrm{OH}}_2$

Why are epoxides more reactive than regular ethers?

Epoxides are more reactive than regular ethers due to the significant ring strain in their three-membered ring structure. This ring strain arises from the bond angles in the epoxide being much smaller than the ideal tetrahedral angle of 109.5°, leading to high potential energy. As a result, epoxides are more eager to undergo ring-opening reactions to relieve this strain. In contrast, regular ethers do not have this ring strain and are therefore much less reactive. The reactivity of epoxides can be exploited in both acid-catalyzed and base-catalyzed ring-opening reactions, making them versatile intermediates in organic synthesis.

What are vicinal diols and how are they formed from epoxides?

Vicinal diols are compounds with two hydroxyl groups on adjacent carbons. They can be formed from epoxides through base-catalyzed ring-opening reactions. In this process, a nucleophile attacks the least substituted carbon of the epoxide, leading to the formation of an anti product where the hydroxyl groups are on opposite sides of the former epoxide ring. The general reaction can be summarized as:

$R_1-C_2-O_1-C_3-R_2+{\mathrm{OH}}_{\mathrm{-}}\to R_1-C_2-{\mathrm{OH}}_1-C_3-{\mathrm{OH}}_2$