Week 11 of 52: Breaking epoxides with nucleophiles

I think I missed a week and am behind on this. Whatever. I have no sympathy for you. Here is a weird reaction I arbitrarily chose. Learn it.

Epoxides contain a strained ring. It looks like this...
Look at those bond angles. Actually, don't. I mean, I just made that picture in a few seconds. It's not like it's accurate at all. But epoxide rings are strained. They're just waiting to pop open at any second if you give them reason to. Maybe. Actually, I made that up too. You probably shouldn't take this post too seriously. Just so you know.

One way for that ring to open is for a nucleophile to attack one of those α-carbons. Assuming that the reaction takes place in an aqueous environment, this leaves an alkoxide on the α-carbon that was not attacked by the nucleophile, which is protonated by the surrounding water.

The result of this reaction is that each of the carbons from the epoxide now has a different functional group attached to it. One has the nucleophile (whatever that was) and the other has an alcohol. This is probably useful for something. Ugh, sorry. That sounded lame. This isn't working. I need to change the way I do these posts. I need to make them better. This one sucks. I'm so sorry.

Week 10 of 52: Preparation of alkoxide salts from alcohols

Last week, I showed the Williamson ether synthesis, a nucleophilic substitution reaction in which an alkoxide ion attacks an alkyl halide. I noted that the preparation of the alkoxide itself was another reaction and that I would save this reaction for later. Well, it's later now. I think. Anyway, if you actually bothered to read the title, which really isn't all that impressive of an accomplishment, you would knot that I am indeed using alkoxide preparation as this week's reaction.

This is an acid-base reaction. I hope you remember how those work, because I shan't be reviewing it. Go back and find those posts yourself. Or don't. Whichever. Even if you are a bit familiar with acid-base reactions, you might find this one slightly peculiar. You might notice that an alcohol is not normally a good acid. That's why this time, we're using a super-strong base. Sodium hydride is one of my favorite bases ever and it's a fairly standard one for this. Probably. I think. So we'll use that as our example. Behold, a reaction:

CH₃CH₂O—H + NaH → CH₃CH₂ONa + H₂

See how awesome sodium hydride is?

Week 9 of 52: Williamson ether synthesis

I'm sure that even you managed to deduce from the title that this post will cover a method of synthesizing ethers that is, for some reason, named after "Williamson." Good job. No, not really. I mean, no it wasn't really a "good job" that you figured this much out. Of course I mean that this is about the synthesis of ethers. "Williamson" turns out to be the person who invented this. Or discovered it. Whatever, I don't care which. You suspected all along that this was the case, but you couldn't be sure until I told you just now. Anyway, the Williamson in question was Alexander Williamson. He came up with this back in 1850, so you can safely assume that he is now dead.

Really, this is a simple S_N2 reaction, which is my way of saying that I won't be spending a great deal of time on this. But don't conclude that this is some minor, throwaway reaction I am lazily posting to keep up my weekly quota. You'd be wrong about that. Well, you'd be wrong about part of it anyway. This reaction is, to this day, the main way ethers are manufactured. Ethers are important for industrial stuff probably. I mean, I assume they are.

The substrate for this reaction is an alkyl halide. Yes, again. Why not? What's wrong with alkyl halides. I heard that you like them a lot. And you should. As you already realize, the halide acts as a leaving group here. But this time, the nucleophile is an alkoxide ion. Alkoxides are of the form R—O^-. They are typically prepared as salts. I could describe how, but it occurs to me that I can use that reaction to fill in another week, so you'll just have to wait. So cruel, I know.

Anyway, the alkoxide attacks, bumping the halide off and attaching to the α-carbon. So we get an ether. R—O—R'. Also, if the ether is unsymmetrical, we could potentially have either side be the alkyl halide or the alkoxide, but one of the two possible configurations is more efficient. If I revisit this topic in the future, you must remind me to explain that.