Week 4 of 52: Unimolecular elimination

Apparently January will have been the month of reactions on alkyl halides by bases and nucleophiles. I didn't initially set out to do it that way, but in retrospect it does make sense. I have no idea what reaction I'll start February with. Oh well, it's not time for that yet. It's time for unimolecular elimination.

If you've been paying close attention (which you haven't), you'll probably have deduced that this has the "unimolecular" feature of the S_N1 reaction and the "elimination" feature of the E1 reaction. So you practically know what will happen just from the name, you clever reader, you. But just in case you aren't that bright, which let's face it, is pretty likely, I'll offer a brief explanation...

• This is a two-step reaction.
• In the first step, the bond between the α-carbon and the leaving group breaks. The leaving group leaves.
• The departure of the leaving group results in the formation of a carbocation.
• In the second step, a base removes a proton from a β-carbon.
• As in the E2 reaction, the electron pair from the broken C-H bond forms a π-bond between the α-carbon and the β-carbon.

Still confused? Let's try some pictures. One note: with the S_N2 and E2 reactions, I just assumed that we were dealing with primary alkyl halides. With the unimolecular reactions, a carbocation is formed, so these reactions will be fastest with tertiary alkyl halides. So I'll draw some more groups, just to show that. So, here's the first step...
That's the same first step for both the substitution reaction and the elimination reaction. It's the second steps that are completely different. Here's S_N1...
And you already know exactly what the E2 reaction will look like now, but I'm showing you anyway, just on the off-chance that you really are that incompetent...