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.