Week 25 of 52: Acetylide as a nucleophile

In Week 18, I covered the use of terminal alkynes as weak acids. At the end of that post, I casually remarked that the conjugate base of such an acid, an acetylide ion, can itself be used in reactions. Since that post went up, you've been waiting in agony for a post about a reaction using an acetylide ion. Your wait is finally over.

Perhaps I erred in the earlier post, actually. I mean, it is accurate that terminal alkynes are very weak as acids, but just leaving it at that seems sort of pointless. Oh look, we took some very strong base and we protonated it. Lots of things can do that. For Week 18, what I should have done was emphasize that the point of that reaction would be to prepare acetylides. We can then use the acetylides we just prepared in nucleophilic substitution reactions.

Overall, this reaction is a fairly straightforward example of the S_N2 reaction, which was the first post in this whole series. I did write about that reaction first because it was important. After mentioning nucleophilic substitution so much in these subsequent posts, I think it should start to become clear just why it's important.

The textbook I've been using for most of this takes the opportunity to use this reaction as a starting point for discussion of multistep synthesis, which we've only seen a little so far on this blog, and retrosynthetic analysis. The whole point of learning these reactions in the first place is to understand how to convert something into something else. It's transmutation. I hope to have time in the near future to create a reaction map for use here, charting all the functional groups I've covered and all the reactions with them that have been shown so far. If you've been paying attention, and I know you haven't, this should give you an idea of the big picture.

Week 22 of 52: Halogenation of alkynes

I already covered addition of halogen to alkenes back in some previous week. I don't know. Look it up. I showed an alkene being turned into a vicinal dihalide. Alkynes are sort of like alkenes, but different. So this reaction is sort of like that one, but different. Everything is the same as everything else, only different. Amazing.

This reaction actually has two different products. That's because the halogen (either chlorine or bromine) gets added once, forming a trans dihalide. What that means is that there's still a double bond and that each carbon on the double bond gains a new bond to a halogen. Fine, here, it's like this...
Yes, the mechanism is complicated and you're freaked out and frustrated and you hate me right now, but shut up. I drew this piece of crap just for you, so be grateful. The first step here is addition, which forms that thing in the middle: a bridged halonium ion. This step isn't very fast, but the next step, a nucleophilic substitution, is.

But that's not all. This week is special for some reason, so you get two reactions instead of just one. No really. You see, the trans dichloride that is the product of this reaction will also react with chlorine. So given enough chlorine and time, the process repeats and we get a second addition reaction, just like before, but this time the initial substrate is a trans dichloride instead of an alkyne. The end product is a tetrachloride (or tetrabromide if this had been done with bromine).

So you do get two reactions, but really it's just up to two iterations of the same reaction. Still pretty cool, though.

Week 18 of 52: Terminal alkyne as an acid

Last time I showed how to make an alkyne. So now of course you will want to make an alkyne into something else. Well, this reaction is only for terminal alkynes. If the triple bond is in the middle of a chain, like this R—C≡C—R, then it won't work. But when the triple bond is between the last two carbons of a chain (also the first two, because you can count either way), then it can act as an acid with the hydrogen at the end (it looks like this: R—C≡C—H) leaving and reacting with a base.

Terminal alkynes are weak acids. Very weak, actually. There's this big fancy chemistry explanation for why this is the case, but it might seem pretty intuitive to conclude that this reaction would require a very strong base, which is the case. And when I say strong here, hydroxide isn't strong enough. Amide is though, and of course there's the awesome hydride.

Once the terminal alkyne is deprotonated, it can act as a nucleophile known as an acetylide anion. This ion can then be used to react with an electrophile.