Ether Cleavage - Online Tutor, Practice Problems & Exam Prep

Ethers are highly stable and unreactive compounds, similar to alkanes, with the exception of cleavage reactions in the presence of strong acids like HI or HBr. The mechanism involves protonation of the ether, followed by nucleophilic attack leading to the formation of alkyl iodides and alcohols. The alcohol can further react, yielding two equivalents of alkyl halides. Understanding this cleavage mechanism is crucial for predicting products in reactions involving cyclic ethers, which can undergo similar transformations.

How to predict the products of Ether Cleavage.

Now I want to talk about all the different reactions that ethers undergo. And it turns out that you guys are in luck because there's only one. Okay? Ethers are actually extremely unreactive. They're very stable. So what that means is that there's almost nothing you have to know about ethers in terms of their reactivity. They're pretty much almost as unreactive as alkanes. But there is one reaction that they can be forced to do in the presence of a very strong acid and that is the cleavage of ethers. Like I said, I was just trying to scare you guys. It's actually not that bad. Just one reaction and that's it. So ethers are very combustible like alkanes. They have a lot of energy stored in those bonds. But in terms of breaking those bonds heterolytically, so that you get like a negative and a positive, that's very difficult to do with ethers because they're very strong bonds. So what we can do is that we can cleave them in the presence of strong acid. And the acid is super strong, it can only be HI or HBr. So what that means is HCl, HF are not strong enough to make an ether cleave. All right? So let me go ahead and show you guys this mechanism.

Let's go ahead and work with HI since it's the strongest one. And basically what happens here is that the ether protonates. Okay? So what I wind up getting is that my oxygen and my carbon, that's a very strong bond. But what I can do is I can protonate it first. What that's going to wind up giving me is an ether with a formal charge. Now that I have that formal charge and now that I have such a good nucleophile present, remember, I^- is a very strong nucleophile, I can do a backside attack. So now what I can do is I can take this I and I can attack the carbon chain and I can kick out the OH as a leaving group. What that's going to give me is it's going to give me one equivalent of an alkyl iodide, so I'm going to wind up getting ethyl iodide. And then I'm also going to get ethanol because I wind up getting the OH turns neutral again, so I wind up getting this plus this. Cool, right?

But it doesn't end there because now that I have the alkyl iodide, that's done. The alkyl halide is done reacting. But the alcohol can actually react again because what I can do now is I can take my alcohol and I can protonate it with another equivalent of HI. So what I'm going to do is this and that and what that's going to give me is that's going to give me OH⁺. And then I have an I^-. Now guys, if you have already studied the conversion of alcohols to good leaving groups, then this is actually that same exact reaction. All we're doing is we're converting alcohol to a good leaving group using HI. And that mechanism should be what? Do you guys know? Since it's a primary alcohol, it should be an SN2 mechanism. Equivalent of alkyl iodide plus water. Equivalent of alkyl iodide plus water. So what this is going to create at the end is not one, but 2 equivalents of the same alkyl halide. So the answer to this question is that what I should put is 2 times that. That is my actual answer because I'm splitting up the ether into 2 different sections. I'm cleaving it and I'm getting alkyl halides as a result. Okay? I hope that made sense.

So now what I want you guys to do is predict the whole mechanism for this cyclic ether. Now this is a cyclic ether, but it's not technically an epoxide because remember epoxides are 3 membered rings. This is a 5 membered ring. So we would call this a cyclic ether. I want you guys to use HBr to figure out what the end product is going to be. This is going to react twice just like the other reaction. So try your hardest to get this and then I'll show you guys the way. Go for it.

Predict the product of the following reaction.

All right. So let's start off with the first step. HBr. We know that the O is going to protonate. So what I'm going to wind up getting is a protonated ether with a formal charge. And I also get this awesome nucleophile Br-. So the next step needs to be a backside attack. I'm going to go ahead and pick one of the sides, doesn't matter which one, to attack and kick out the good leaving group, which in this case is going to be a neutral alcohol.

So what I'm going to wind up getting now is something that looks kind of funny. It's going to break open the ring. So I'm going to get a bromine up here, and then I'm going to get the same ring as before and then I'm going to get OH. Just to make sure, let's check that there's the same number of carbons. 1, 2, 3, 4. Yes, we drew it correctly. Now keep in mind that we don't ha

Here’s what students ask on this topic:

What is ether cleavage and why is it significant in organic chemistry?

Ether cleavage is a reaction where an ether is broken down into smaller molecules, typically in the presence of strong acids like HI or HBr. This reaction is significant because ethers are generally very stable and unreactive, similar to alkanes. The ability to cleave ethers allows chemists to transform these stable compounds into more reactive species, such as alkyl halides and alcohols, which can then participate in further chemical reactions. Understanding ether cleavage is crucial for predicting the products of reactions involving ethers, especially cyclic ethers.

What are the conditions required for ether cleavage to occur?

Ether cleavage requires the presence of very strong acids, specifically hydroiodic acid (HI) or hydrobromic acid (HBr). Weaker acids like hydrochloric acid (HCl) or hydrofluoric acid (HF) are not strong enough to induce ether cleavage. The reaction mechanism involves the protonation of the ether oxygen, followed by nucleophilic attack by the halide ion, leading to the formation of alkyl halides and alcohols. The alcohol can further react with another equivalent of the strong acid to produce additional alkyl halides.

What is the mechanism of ether cleavage using HI?

The mechanism of ether cleavage using HI involves several steps. First, the ether oxygen is protonated by HI, forming a positively charged oxonium ion. This makes the carbon-oxygen bond more susceptible to nucleophilic attack. The iodide ion (I^-), a strong nucleophile, then attacks the carbon atom, breaking the carbon-oxygen bond and forming an alkyl iodide and an alcohol. The alcohol can further react with another equivalent of HI, undergoing protonation and nucleophilic substitution to form a second equivalent of alkyl iodide and water.

How does the cleavage of cyclic ethers differ from acyclic ethers?

The cleavage of cyclic ethers follows a similar mechanism to acyclic ethers but involves the breaking of a ring structure. For example, a five-membered cyclic ether will undergo protonation and nucleophilic attack by HI or HBr, leading to the formation of linear products. The ring-opening process results in the formation of two equivalents of alkyl halides. The key difference is that the cyclic structure must be broken, which can sometimes lead to different stereochemical outcomes compared to acyclic ethers.

Why are HI and HBr the only acids strong enough to cleave ethers?

HI and HBr are the only acids strong enough to cleave ethers because they are highly acidic and can effectively protonate the ether oxygen, making the carbon-oxygen bond more susceptible to nucleophilic attack. Additionally, the iodide (I^-) and bromide (Br^-) ions are strong nucleophiles, capable of attacking the carbon atom and breaking the bond. Weaker acids like HCl and HF do not provide sufficient protonation or nucleophilic strength to induce ether cleavage.