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 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 24 of 52: Addition of water to alkynes

Last week's post show's how water adds across an alkene. An alkyne has the same kind of bond that an alkene does (alongside another bond, but more on that later or possibly never). So the reaction is similar for alkynes. However, the product of this reaction can potentially be something completely different (an alkene just becomes an alcohol).

My book points out that either a strong acid or a mercuric catalyst (I could explain what that means, but I don't feel like it) can facilitate a reaction in which water adds across one of the bonds, leaving a double bond and an alcohol on the same carbon. This is called an enol.
It's a hideous portmanteau of "alkene" and "alcohol." But nevermind that. What's important is that this sort of thing is unstable. So it does something super-cool. I am not kidding. This is tautomerization, and it's awesome.
The enol form of the molecule tautomerizes into the "keto" form (which could be either a ketone or an aldehyde). I won't spend any more time on tautomerization right now, despite how freaking cool it is.

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 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: