Epoxidation - Online Tutor, Practice Problems & Exam Prep

Epoxidation is an addition reaction that introduces an oxygen atom to a double bond, forming an epoxide, a cyclic 3-membered ether. This reaction typically utilizes peroxy acids, such as mCPBA, to achieve the transformation. The mechanism involves nucleophilic attack by the double bond on the peroxy acid, leading to a transition state with partial bonds. Additionally, epoxides can be synthesized from halohydrins through an intramolecular SN2 reaction, where a base deprotonates the alcohol, facilitating the formation of the epoxide ring. Understanding these pathways is crucial for further organic synthesis applications.

General properties of epoxidation.

So now I want to talk about another addition reaction, and this one only adds oxygen by itself to a double bond to form a completely new functional group called an epoxide. Needless to say, the name of this reaction is called epoxidation. So before we can even get started, I kind of wanted to define what an epoxide is because some of you guys might not know. Alright? And an epoxide is a functional group that's just made of a cyclic 3-membered ether. So what that means is that remember that the definition of an ether was R-O-R, that was an ether. Okay. Well, an epoxide is just going to be a cyclic ether. So what that means is it's an O with 2 R groups on both sides, but they're attached to each other. So this would be an epoxide. Okay? Epoxidation, needless to say, is going to add that one oxygen to the double bond to make it into that 3-membered ring. Okay? So how do we add epoxides to double bonds? Well, we do it by using a type of molecule called a peroxy acid. Peroxy acids are the molecules that are used to make them, and this is the general formula of a peroxy acid. What you're going to notice is that it actually looks a lot like a carboxylic acid. Remember that carboxylic acid? Oops. Remember that carboxylic acid looks like this: OH. So really, it's the same thing as a carboxylic acid except it has one more O. So remember that the definition of a carboxylic acid is CO2H. That's the condensed formula. Well, for a peroxy acid, it's going to be RCO3H. So what you're going to notice is that it's really the same exact thing except I've just added one more oxygen. So it's CO3H. Alright, so that's the first thing. Now, you could use any peroxy acid you want to make an epoxide, but the common ones that are used are MCPBA and MMPP. These are 2 reagents that you don't need to know exactly what they look like as long as you can recognize that these are types of peroxy acids. Okay? The only thing that changes is the R group, but the COOH is the same. Alright?

The mechanism of how peroxy acids make epoxides.

So the mechanism for this guy is actually kind of crazy. A lot of professors don't need you to know it, but some do. So I'm just going to tell you anyway, just in case. Also, because I want you guys to be the smartest kids in the class, so might as well learn it. Okay? So where do you think the first arrow is going to come from? That part is easy. The first arrow comes from the double bond. That's my nucleophile. And it turns out that it's going to attack the very last oxygen on my peroxy acid. So that part is easy. Now everything else is a lot of arrows because when I make that bond, I have to break a bond because that oxygen would break its octet if I just added a new bond to it. So what's going to happen is the electrons from this bond are going to make a double bond here. Does that make sense so far? So basically, if I make a bond, I have to break a bond. But now if I make a double bond here, that means that this carbon, this carbonyl carbon would break its octet. So if I make that bond, then I have to break this bond and move those electrons up to the oxygen. Okay? But now that I have these electrons up to the oxygen, this oxygen is going to have a negative charge. That negative charge is going to be attracted to the hydrogen on the other side. So then the electrons from this oxygen are going to grab that hydrogen. But does hydrogen like to have 2 bonds? No, it doesn't. So what that means is that the electrons from here are finally going to go back and attack the double bond. So you're going to have 5 arrows in total. I know that explanation kind of sounded crazy. That's probably the best way that we can think about it. Think about how the double bond is starting it and then there's kind of just like a flow, a transfer of electrons through the whole molecule. Now what you can imagine is that this is going to make a crazy looking transition state. So like I said, 90% of professors are not going to require you to draw this. But, just in case, you might have that one guy that needs you to do it, so we're going to go ahead and do it. So we're going to have to draw everything small because there's a lot of pieces here. What the transition state is going to look like is that all these bonds that are being created and destroyed have to be drawn as partial bonds. So it's going to be partial, partial, then a partial bond to oxygen, then a partial, a straight, a single bond, but then a partial double bond, then a single bond to oxygen, single bond to R with a partial double bond. Then this has a partial bond to hydrogen and then this has a partial bond to this oxygen. Okay. Isn't that crazy? So that is what the transition state is going to look like. Basically, partial bonds in every single reaction, every single place that we were making or breaking a bond is going to be a partial bond. At the end, what we're just going to get is that we're going to get an epoxide, which just looks like this, oxygen with 2 bonds. And then on the other side, we wind up actually getting a carboxylic acid. Okay? So I mean, I kind of drew that wrong. It would be, an oxygen up here and then a double bond oxygen here and then an R, which is a carboxylic acid. Okay, so those are the 2 products. The one we really care about is the epoxide. Okay, because that's what we're going to use in the following steps. Okay? So this is the number one way to make epoxides. You're going to see this over and over again, in future chapters even, you're going to see this.

Halohydrins to epoxides via intramolecular SN2.

Now, there's actually one more way that we can make epoxides, and that's by using halohydrins. Okay? Now, remember that in the addition section, addition reactions, halohydrin is one of the addition reactions that we can use. Okay? And it turns out that halohydrins are also good at making epoxides. How? Through an intramolecular SN2 reaction. Remember that an SN2 is just a backside attack. So here's the way it works: basically, remember that I've got a double bond, and that double bond is exposed to diatomic halogen and water. What's going to wind up happening is that I get a halohydrin. You guys should all be able to follow-up to this point. Notice that the stereochemistry is, once again, anti. Okay? Cool. Now, also because this didn't have it was perfectly symmetrical, it doesn't matter which side I put the OH and the halogen. You could have picked either one. Okay? But now here's the interesting part. Once I have a halohydrin, I can react that with any base I want. And if I react that with a base, what's going to happen is that the base is going to wind up deprotonating my alcohol. So what I'm going to wind up getting is a nucleophile on one side of the molecule and a leaving group on the other side. Okay, so what we've basically done is we've made an oxide. We've basically made a nucleophile out of the alcohol. So what's going to happen here is that we're going to get an intramolecular reaction where this O, this O does an attack on that carbon and kicks out the X. So what winds up happening is that we wind up forming a ring that looks like this. We might wind up forming that this ring stays the same, but now this O is attached here and attached there. Because this new bond that I'm drawing in blue right here is the one that was created by the backside attack here. And then plus, I would get my leaving group X negative. Okay? So these are two different ways to make epoxides. Your professor may teach just the epoxidation with peracids. He may teach halohydrins as well. Okay? I want you guys to be responsible for both because I've seen them often enough that it's just important for you to know both of them. Okay?

Here’s what students ask on this topic:

What is epoxidation in organic chemistry?

Epoxidation is an addition reaction in organic chemistry where an oxygen atom is added to a double bond, forming an epoxide. An epoxide is a cyclic 3-membered ether, characterized by an oxygen atom bonded to two carbon atoms that are also bonded to each other. This reaction typically uses peroxy acids, such as mCPBA (meta-chloroperoxybenzoic acid), to achieve the transformation. The mechanism involves the nucleophilic attack of the double bond on the peroxy acid, leading to the formation of the epoxide and a carboxylic acid as a byproduct.

What are the common reagents used in epoxidation reactions?

The common reagents used in epoxidation reactions are peroxy acids. Two widely used peroxy acids are mCPBA (meta-chloroperoxybenzoic acid) and MMPP (magnesium monoperoxyphthalate). These reagents have the general formula RCO₃H, where the peroxy acid group (CO₃H) is responsible for the epoxidation process. The peroxy acid reacts with the double bond in the substrate to form the epoxide and a carboxylic acid as a byproduct.

How does the mechanism of epoxidation with peroxy acids work?

The mechanism of epoxidation with peroxy acids involves several steps. First, the double bond in the substrate acts as a nucleophile and attacks the terminal oxygen of the peroxy acid. This initiates a series of electron transfers, resulting in the formation of a transition state with partial bonds. The transition state collapses to form the epoxide and a carboxylic acid. The key steps include the nucleophilic attack, formation of a cyclic transition state, and the final ring closure to yield the epoxide.

What is the role of halohydrins in the synthesis of epoxides?

Halohydrins can be used to synthesize epoxides through an intramolecular SN2 reaction. In this process, a halohydrin is treated with a base, which deprotonates the alcohol group, forming an alkoxide ion. This alkoxide ion acts as a nucleophile and attacks the carbon atom bonded to the halogen, resulting in the formation of the epoxide ring. This method provides an alternative route to epoxide synthesis, complementing the use of peroxy acids.

What is the difference between epoxidation using peroxy acids and halohydrins?

Epoxidation using peroxy acids involves the direct addition of an oxygen atom to a double bond, forming an epoxide and a carboxylic acid as a byproduct. Common peroxy acids include mCPBA and MMPP. In contrast, epoxidation using halohydrins involves an intramolecular SN2 reaction. A halohydrin is first formed by the addition of a halogen and water to a double bond. The halohydrin is then treated with a base, which deprotonates the alcohol, forming an alkoxide ion that attacks the carbon bonded to the halogen, resulting in the formation of the epoxide ring.