So far I’ve mentioned substitution reactions, where we’re forming and breaking a bond on the same carbon.
Today let’s talk about a new class of reactions, called elimination reactions
The first question to ask is just to focus on the bonds forming and breaking.
Every elimination reaction follows the same pattern.
- We’re breaking C-(leaving group) bond,
- We’re breaking C-H (note how these bonds are adjacent to each other!)
- We’re forming a C-C PI bond.
- Also, we form a new bond between the H and a base.
Notice how the C-C Pi bond forms where those two bonds broke.
Here’s a typical example – a very simple one.
Here, we start with an alkyl halide (or other compound with a good leaving group), and we add a base.
Our product here is an alkene.
Look at the bonds formed and broken.
- breaking C-Br (this is our good leaving group)
- breaking C-H (this H goes to the base)
- forming a C-C pi bond
- forming an O-H bond
EVERY elimination reaction will form a new Pi bond and break TWO separate bonds on carbon.
Tomorrow: more details on how these reactions actually work.
Thanks for reading! James
P.S. Last note: see that we’re also forming a salt here. It’s safe to ignore this, since we generally don’t consider it important, but I like to draw out balanced equations so it’s clear that no atoms are being created or destroyed.