Week 26 of 52: Hydrogenation of alkenes

This is an addition reaction because something is added across a double bond. It's also a reduction reaction because the alkene carbons are losing bonds to each other and gaining bonds to hydrogen. Oxidation/reduction is a very important concept in chemistry, but the terms are really quite terrible and deserve their own post or series of posts. However, when I took organic chemistry, oxidation and reduction were simplified for the sake of our reactions: carbon gaining bonds to more electronegative elements is oxidation and carbon gaining bonds to less electronegative elements is reduction. That simplification doesn't work for inorganic chemistry at all, but for our purposes, just note that this is a reduction.

Like other addition reactions, this one follows the pattern of RC=CR' → RZC=CZR', where "Z" is the thing being added across the double bond. Or to represent it pictorially, since you still don't get it for some reason, here's the reaction with hydrogen...
As you can see, it's not our most confusing reaction. This is a simple way to turn an alkene into an alkane, or to remove C=C bonds in general. The catalyst here is a metal, often palladium. As far as I know, the metal catalyst is mixed into charcoal to maximize the surface area for the reaction, but I would imagine that there are other methods used in some cases. If this is done with palladium, a typical abbreviation for the catalyst is "Pd-C" (standing for palladium on carbon).


Also, π-bonds and rings are known as degrees of unsaturation. Each π-bond or ring in a molecule counts as one degree of unsaturation. So replacing the double bond with bonds to hydrogen is a way of "saturating" the molecule. You've probably encountered this concept with saturated and unsaturated fats. But I explain no further. Good day to you.

Week 23 of 52: Hydration

I was about to write a post about the addition of water to alkynes, but I just realized that apparently I've yet to write one on addition of water to alkenes. This oversight on my part is unforgivable and you should berate me for it. Too late, as by the time you read this I will have already corrected my error in the form of a new post, this very post, in fact.

Hydration of an alkene is a specific case of an addition reaction. Unlike with hydrohalogenation, water does not itself provide a strong acid to attack the alkene. So we use sulfuric acid. Problem solved! The product is, of course, an alcohol.

The lone hydrogen tends to add to the less substituted carbon. The hydroxyl group adds to the other carbon. This is in accordance with something called Markovnikov's rule. But I have not explained this. How negligent of me.

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 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...

Week 15 of 52: Hydrohalogenation

I was getting caught up. And then I stopped. I blame school. And myself. Mostly school. but I am not giving up. I missed May and June, but I will get caught up by September. And you will read it. We are making this happen. A lot. I'm not sure quite how, though. Aside from being busy with school, I'm also finding this project harder now. I've lost track of which reactions I've written about. I've forgotten a lot of reactions. This is not good. I haven't been taken chemistry and I haven't been focused on it. Enough whining.

Hydrohalogenation is a good word. I like it. Before I went on this stupid, two-month hiatus, I wrote about electrophilic addition. Hydrohalogenation is a specific case of electrophilic addition. This textbook says, "Hydrohalogenation is the addition of hydrogen halides to alkenes to form alkyl halides." And of course you remember that alkyl halides themselves can be used in substitution reactions. And there's even elimination! You could do an addition on an alkene to make an alkyl halide and an elimination on that alkyl halide to make it back into an alkene! It would be useless, but I think it would be fun.
That's an image I made. It depicts the reaction. Obviously.

Week 14 of 52: Electrophilic addition

The catching up continues furiously. Or maybe just aggressively. With a scowl-like expression at the very least. I don't feel like doing this right now, but I am forcing myself to, alright? I could force myself to do my actual schoolwork, but I'll do that later. Yes, I'm procrastinating on my schoolwork by writing a summary of a reaction. It's not that weird. There are weirder people. Plus, this hardly even counts because I'm padding it with nonsense like, well, pretty much this whole paragraph. So there's that...

An electrophilic addition reaction involves the breaking of a π-bond and the formation of two σ-bonds. For now, let's keep it simple and consider alkenes. These reactions also work on other molecules, like alkynes (hydrocarbons with at least one triple bond), but we'll move on to them later (or never).

And electrophile is sort of the opposite of a nucleophile. And you already know about nucleophiles because I already explained them. Remember?

Nucleophiles are attracted to positive charge. Remember: nucleii of atoms are positively charged.

Well, electrophiles are attracted to negative charge. And, as we all know, electrons are negatively charged. Alkanes consist of C—H σ-bonds and C—C σ-bonds. But in alkenes, there is at least one C=C bond (a π-bond). The double bond is stronger than the single C—C bonds are by themselves, but the π-bond portion of that double bond is significantly weaker and sort of more spread out. The electron density is more exposed to attack. And like nucleophiles, electrophiles attack.

I won't provide a list of common electrophiles right now. Maybe some other time (probably not). However, here's the general form of an electrophilic addition...
And there would be an electrophile in there somewhere, which would probably take up two of those new bonds that formed. You'll hopefully become more comfortable with this over the next month or so. I plan to post a few specific versions of addition reactions on alkenes, so perhaps May will be the month of addition reactions. Well, I'm actually still behind, so that doesn't really work. But shut up.