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Elimination Reactions

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.