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 21 of 52: Oxidation of an alkylborane

I will properly introduce oxidation reactions at some point. Or perhaps not. I don't know. Anyway, this is one of them, although this is not a proper introduction to them. The product from the previous reaction, an alkylborane, is oxidized here, yielding an alcohol. Really, this kind of like a second addition reaction (although it isn't one). Like the hydroboration reaction, this one is seemingly simple, but has some caveats. However, this time the textbook mostly glosses over those caveats, so this post will be pretty brief.

The reagent used here is hydroxide in hydrogen peroxide. The bond to boron is replaced by a bond to hydroxide, yielding an alcohol. And that's it. Well, not really, but I'm leaving it at that, so there.

In summation, the hydroboration-oxidation sequence takes us from an alkene to an alcohol, with the hydroxyl group bonded to the less substituted alkene carbon.

Week 20 of 52: Hydroboration

This one is an addition reaction and the product is the reactant for next week's reaction (which will actually not be in a different week at all, because I am catching up). I was going to do both reactions in a single post because these reactions don't take a long time to explain, they go together (one immediately follows the other), and the textbook does list them together before listing them separately. However, they are different reactions and I am going to err on the side of caution and split this across two posts. The first one (this one) is hydroboration of an alkene. The second post (the one right after this one) is oxidation of an alkylborane.

If you were really all that smart, you'd have gathered from the final sentence of the preceding paragraph that the product of this reaction is an alkylborane. This reaction is really quite simple...

Alkene + BH₃ → Alkylborane

Of course, that's only helpful if you know what an alkylborane is. Well, it's an alkane with a boron group of some sort attached to one of the carbons. Since this is an addition reaction and borane is adding across the double bond, one of the alkene carbons gets a hydrogen and the other alkene carbon gets a BH₂ group.

This is very simple, but the actual reaction is much more complicated, so perhaps it's best that I do this as its own post and note some of the ways this reaction is not as simple as it first appears...

1. BH₃ is a highly reactive gas and tends to form dimers (a molecule of borane will react with another molecule of borane to form diborane, B₂H₆).
2. Because borane and diborane are so reactive, they are impractical for this reaction. This problem is addressed by combining borane with a Lewis base to form a more stable complex. Tetrahydrofuran (pictured below) is apparently a favorite for this. I haven't yet talked about coordination complexes on this blog, so you have pretty much no idea what I'm talking about. So sorry.
3. The alkylborane formed through this reaction can still have the borane group react with other molecules of the original alkene (or any other alkenes that happen to be lying around). So the real product is a trialkylborane. The boron atom has three bonds, each one to a carbon that used to be an alkene carbon. The other three former alkene carbons have a hydrogen bound to them.
4. This reaction is regioselective. That's another topic that I suppose I've neglected on this blog. I guess I suck at this. In this case, what I mean is that the boron atom ends up on the less substituted carbon (if there is one). If the alkene is symmetrical, this doesn't matter. Otherwise, it has important implications on exactly what the product will look like.

Tetrahydrofuran:

Week 19 of 52: Halohydrin formation

I noted that epoxides could be formed from halohydrins, but still have not described how to form halohydrins in the first place. I aim to correct this oversight now. That's why this post is about halohydrin formation. If you'd been paying attention, which you haven't, you'd have seen this in the title.

The short version of this story is that exposing an alkene to a halogen and water yields a halohydrin. So the double bond in an alkene (C=C) becomes a single bond and one of the carbons gains a bond to a halogen and the other carbon gains a bond to a hydroxyl group. I just described the same thing twice and you still want a picture? Fine. I live to serve...
Making that picture just took valuable time that could have been spent playing Oblivion. I hope you're happy.

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.

Week 17 of 52: Alkyne synthesis by two successive dehydrohalogenations

I've already covered the E2 mechanism by which an alkyl halide can be converted to an alkene. With a particularly strong base, a dihalide like the one shown in the previous post can undergo an E2 reaction twice, yielding an alkyne.
Like that! Or something. Note the use of sodium amide. I put it there because this reaction requires a very strong base. There's an explanation for this, but having just read over it, I find it beyond the scope of what I'm doing here (I definitely haven't introduced the concepts needed to understand it). So we'll just leave it at that. This reaction requires a very strong base.

Week 16 of 52: Halogenation

It's getting to me that I'm obviously rusty on this stuff. I don't like it. I see the phrase, "forming a vicinal dihalide" and I think to myself that I have no idea what a "vicinal dihalide" is. Have I ever even seen the word "vicinal" before? No matter, I just figured it out because of my magnificent intellect. A vicinal dihalide must be one in which the two halogens are bonded to adjacent carbons. A dihalide in which the halogens were bonded to carbons farther from each other would be some other sort of dihalide, presumably. I guess. As you can see, I'm not an expert. I'm just pretending to be one. Because pretending is fun.

This reaction is pretty simply though. Alkene + halogen yields vicinal dihalide. Wow, that is simple. Fine, here's a picture...
That's pretty good, if I do say so myself. Anyway, this reaction is normally only done with chlorine or bromine. Addition of iodine is often too slow to be practical and addition of fluorine is apparently explosive. Fun. Oh, and then there's this part about how dichlorides and dibromides formed this way are themselves used as reactants for the synthesis of alkynes. It looks like I have my next post all figured out...