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.

Week 13 of 52: Epoxidation of ethene

I am still behind, but I have resolved to catch up. This project will not die until I want it to. And I don't want it to. Not yet, anyway. I am so dedicated that I am starting a new post while sitting in my classroom. Class starts in fifteen minutes or so as I am typing this sentence, so I won't finish it yet. Shut up. Obviously I don't have an organic chemistry textbook in front of me, which has been my traditional method of searching for and selecting reactions to post here. And because of that, this time, we get a reaction not from the textbook at all. Pretty cool, huh?

This week's reaction, or the reaction for whichever week I'm on now, is another epoxide synthesis. Rather than working on an entire functional group (like the halohydrins from last week or last post or whatever), this one is specific to a single molecule: ethene. Being limited thusly is detrimental to a reaction's usefulness, assuming we want to construct a toolbox of reactions. However, this is still an important reaction because oxirane, the epoxide produced from it, is used extensively in industry. The reaction goes something like this...

7H₂C=CH₂ + 6O₂ → 6C₂H₄O + 2CO₂ + 2H₂O

In case it wasn't clear, the product (other than water and carbon dioxide) is oxirane, the smallest and simplest of the epoxides.

This reaction is mediated by a silver catalyst. Have I explained catalysts before? No? Too bad. Anyway, even though this doesn't work for larger alkenes, it's still important because oxirane is an important precursor in the manufacture of a lot of other things, most prominent among them being ethane-1,2-diol (ethylene glycol).

Week 12 of 52: Preparation of epoxides from halohydrins

I know I'm behind. Shut up. I'm also very busy. But I have a reaction for you. Learn it. Or else. Now, as you may have ascertained from the title already, this reaction is all about the preparation of epoxides. I introduced epoxides in the previous post, and of course you still remember them and love them. That's good. One way to make them is to use a halohydrin. "What's that?" you ask. Shut up and I'll tell you.

Halohydrins are themselves prepared from alkenes. But shut up. They look like this...
The first step of this reaction is a simple acid-base reaction in which a proton is stripped from that alcohol group. You've seen this before. It is not new. It is familiar. You are comfortable with it. What happens next is pretty cool: an intramolecular substitution reaction. The negatively charged oxygen forms a bond to the nearby carbon. The halogen is, of course, a leaving group. The end result is an epoxide: the oxygen is attached to both carbons, forming a strained ring.

Week 11 of 52: Breaking epoxides with nucleophiles

I think I missed a week and am behind on this. Whatever. I have no sympathy for you. Here is a weird reaction I arbitrarily chose. Learn it.

Epoxides contain a strained ring. It looks like this...
Look at those bond angles. Actually, don't. I mean, I just made that picture in a few seconds. It's not like it's accurate at all. But epoxide rings are strained. They're just waiting to pop open at any second if you give them reason to. Maybe. Actually, I made that up too. You probably shouldn't take this post too seriously. Just so you know.

One way for that ring to open is for a nucleophile to attack one of those α-carbons. Assuming that the reaction takes place in an aqueous environment, this leaves an alkoxide on the α-carbon that was not attacked by the nucleophile, which is protonated by the surrounding water.

The result of this reaction is that each of the carbons from the epoxide now has a different functional group attached to it. One has the nucleophile (whatever that was) and the other has an alcohol. This is probably useful for something. Ugh, sorry. That sounded lame. This isn't working. I need to change the way I do these posts. I need to make them better. This one sucks. I'm so sorry.

Week 10 of 52: Preparation of alkoxide salts from alcohols

Last week, I showed the Williamson ether synthesis, a nucleophilic substitution reaction in which an alkoxide ion attacks an alkyl halide. I noted that the preparation of the alkoxide itself was another reaction and that I would save this reaction for later. Well, it's later now. I think. Anyway, if you actually bothered to read the title, which really isn't all that impressive of an accomplishment, you would knot that I am indeed using alkoxide preparation as this week's reaction.

This is an acid-base reaction. I hope you remember how those work, because I shan't be reviewing it. Go back and find those posts yourself. Or don't. Whichever. Even if you are a bit familiar with acid-base reactions, you might find this one slightly peculiar. You might notice that an alcohol is not normally a good acid. That's why this time, we're using a super-strong base. Sodium hydride is one of my favorite bases ever and it's a fairly standard one for this. Probably. I think. So we'll use that as our example. Behold, a reaction:

CH₃CH₂O—H + NaH → CH₃CH₂ONa + H₂

See how awesome sodium hydride is?

Week 9 of 52: Williamson ether synthesis

I'm sure that even you managed to deduce from the title that this post will cover a method of synthesizing ethers that is, for some reason, named after "Williamson." Good job. No, not really. I mean, no it wasn't really a "good job" that you figured this much out. Of course I mean that this is about the synthesis of ethers. "Williamson" turns out to be the person who invented this. Or discovered it. Whatever, I don't care which. You suspected all along that this was the case, but you couldn't be sure until I told you just now. Anyway, the Williamson in question was Alexander Williamson. He came up with this back in 1850, so you can safely assume that he is now dead.

Really, this is a simple S_N2 reaction, which is my way of saying that I won't be spending a great deal of time on this. But don't conclude that this is some minor, throwaway reaction I am lazily posting to keep up my weekly quota. You'd be wrong about that. Well, you'd be wrong about part of it anyway. This reaction is, to this day, the main way ethers are manufactured. Ethers are important for industrial stuff probably. I mean, I assume they are.

The substrate for this reaction is an alkyl halide. Yes, again. Why not? What's wrong with alkyl halides. I heard that you like them a lot. And you should. As you already realize, the halide acts as a leaving group here. But this time, the nucleophile is an alkoxide ion. Alkoxides are of the form R—O^-. They are typically prepared as salts. I could describe how, but it occurs to me that I can use that reaction to fill in another week, so you'll just have to wait. So cruel, I know.

Anyway, the alkoxide attacks, bumping the halide off and attaching to the α-carbon. So we get an ether. R—O—R'. Also, if the ether is unsymmetrical, we could potentially have either side be the alkyl halide or the alkoxide, but one of the two possible configurations is more efficient. If I revisit this topic in the future, you must remind me to explain that.