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: