Week 3 of 52: Bimolecular elimination

The bimolecular elimination reaction is typically just referred to as the "E2" reaction. It has some things in common with the S_N2 reaction featured in the first week. Both are bimolecular: the reaction is driven by a collision of two molecules. Both involve the concept of a leaving group: an atom or group of atoms that can accept electron density. A classic example is an alkyl halide. In both the E2 reaction and the S_N2, something attacks the alkyl halide, and the halogen is broken away as an anion. I was being lazy for the first two weeks of this and didn't use any illustrations, so let's throw in a picture for this...
I say that's totally a step up from my previous use of pictures in this blog. Anyway, in the S_N2 reaction, which I covered in the first week, not this week, the "something" would be a nucleophile and it would attack the carbon that the halogen is bonded to, the α-carbon. The bond between the α-carbon and the halogen would break and a new bond would be formed between the nucleophile and the α-carbon. Here's a reaction mechanism. Oh, I'm leaving out the hydrogens bonded to carbon in these pictures because I want to. But realize that the α-carbon has two hydrogens attached to it that are just sitting there, not doing anything...
This week's reaction has some important differences. Firstly, the "something" is a base. Basicity and nucleophilicity are similar concepts and the same thing can behave in both ways. A nucleophile attacks a relatively exposed area of positive charge, the nucleus of the α-carbon. A base participates in a traditional acid/base reaction, reacting with a proton. No bond is formed between the base and the alkyl halide. Instead, a bond is broken. Here, I'll show you, but this time, I need to draw some hydrogens...
All of the electron density in the C-H bond at that β-carbon (it's called a β-carbon because it's adjacent to an α-carbon) is dumped onto the carbon. This simultaneously forms a double bond between the α-carbon and the β-carbon and drives the leaving group (the halogen) away. More importantly, I think making this image has placed me officially beyond all redemption.

Emphasizing this yet again, rather than providing new information about this reaction, because I know you can only handle so many new things at once: in nucleophilic substitution, the nucleophile replaces the leaving group, hence the name. In elimination, a π-bond (double bond) is formed between two carbons while the leaving group and a proton are removed, hence the name. Easy, right?