Cleavage of Ethers

Ethers are almost completely unreactive to reactions. They can only undergo one useful reaction- they can break apart in the presence of strong acid. 

Concept: How to predict the products of Ether Cleavage. 

Video Transcript

Now I'm going to talk about all the different reactions that ethers undergo. It turns out that you guys are in luck because there's only one. Ethers are actually extremely unreactive. They're very stable. 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 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.
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 a negative and a positive, that's very difficult to do with ethers because they're very strong bonds. The only thing we can do is that we can cleave them in the presence of strong acid. The acid must be super strong. It can only be HI or HBr. What that means is HCl, HF are not strong enough to make an ether cleave.
Let me go ahead and show you guys this mechanism. Let's go ahead and work with HI since it's the strongest one. Basically what happens here is that the ether protonates. 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. 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, H, positive. 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. That mechanism should be what? Do you guys know? Since it's a primary alcohol, it should be an SN2 mechanism. I would do a backside attack once again.
What I'm going to wind up getting at the very end is I'm going to get another equivalent of alkyl iodide plus water. What this is going to create at the end is not one, but two equivalents of the same alkyl halide. The answer for this question is that what I should put is two times that. That is my actual answer because I'm splitting up the ether into two different sections. I'm cleaving it and I'm getting alkyl halides as a result.
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 three-membered rings. This is a five-membered ring, so we would just 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. 

Example: Predict the product of the following reaction.