One last (weird) reaction to show you with respect to elimination reactions. Can you see what’s weird about it?
How did that double bond get over there? Normally when elimination occurs, we remove a hydrogen from the carbon adjacent to the leaving group. But here, something extra has taken place.
Let’s look at all the bonds that form and the bonds that break so we can track down exactly what happens:
Notice how it differs from a typical elimination reaction? Sure, we’re forming C-C (π), and breaking C-H and C-OH, but we have an extra C-H that forms and an extra C-H that breaks.
This is a sure sign of a rearrangement step!
So what’s going on here?
Well, we start by protonating the alcohol. This allows for water to leave in the next step, which is going to form a carbocation. Here’s the thing: the carbocation is secondary, and we’re adjacent to a tertiary carbon. So if the hydrogen (and its pair of electrons) were to migrate from C3 in our example to C-2, we’d now have a tertiary carbocation, which is more stable. Then, a base (water in this example) could remove C-H, forming the more substituted alkene (the Zaitsev product in this case). And that’s how the alkene ends up there.
OK. So that’s one mystery solved.
You might remember that these types of rearrangements can occur in SN1 reactions too. And if you read that post, you might recall that in addition to shifts of hydrogen (“hydride”, because there’s a pair of electrons attached) we can also have alkyl shifts. Here’s a final example. Note – I’ve also made a video of this, you can watch it here.
This pretty much does it for elimination reactions.
In the next series of posts, let’s go though one of the biggest questions students struggle with. Okay, now that we’ve gone through substitution and elimination reactions, HOW DO WE DECIDE WHICH ONE IS GOING TO OCCUR IN EACH SITUATION?
Great question. That’s next.