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...

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).