Robinson Annulation

by Kiley Lynch

The Robinson Annulation is one of the longer mechanisms you’ll encounter. It’s a way of taking a ketone and an enone (a.k.a. “alpha,beta unsaturated ketone”) into a six membered ring with a new double bond.

When you’re asked to give out the mechanism of a reaction like this, I’d advise starting in one place.

Wait for it….. I know I sound like a broken record…. “what bonds are formed, and what bonds are broken?” I say this to my students over a dozen times per day. I am not kidding.

In this instance I ask this question because it gives you a Road Map to what your mechanism should look like.

Let’s look closely at the Robinson. I’ve numbered the carbons to make things more clear.

The first point should be obvious. All the bonds that form and break should be accounted for in the mechanism! 

This is a good thing. It means that you can rule out certain things. Having constraints helps narrow your focus.
Don’t make things more complicated than they actually are.

Let’s narrow things down a bit more, and analyze the bonds formed/broken one by one. Let’s pay special attention to the electron density on each. You should ask – is this atom electron-rich or electron poor? Or is there something about the reaction conditions that could make it more electron-rich or electron poor?

 

The first clue to the mechanism is the fact that we are adding base. Remember, “the conjugate base is a better nucleophile”*. That means that we should consider positions that could be deprotonated here as potential nucleophiles. Especially enolates.

OK. Let’s break this Robinson down.

Bonds Forming

  • C3-C4  – C3 is next to ketone (could be from enolate – nucleophilic). C4 is on beta position (electrophilic).
  • C2-C7 – C2 is electrophilic (carbonyl carbon) C-7 is next to ketone (potentially from enolate)
  • C2-C7 (pi bond). All C-C Pi bonds come from elimination reactions of some kind. We will need a good leaving group on one of these carbons.
  • C5-H – This is adjacent to a ketone: could have come from an enolate.

 

Bonds Breaking

  • C3-H Breaking the C-H bond adjacent to the ketone means we are forming an enolate under these conditions (basic).
  • C7-H Adjacent to a ketone. Also implies enolate formation.
  • C2-O (Pi) – all C-O (pi) bonds are broken through (1,2)-addition reactions. Therefore, a nucleophile must have added to this carbon.
  • C2-O (single) – here, we broke a C-O bond. This carbon becomes part of a double bond, so this is the leaving group in the elimination reaction that formed C2-C7 (pi).

We can get all of this information from just examining the bonds!  The bonds that form and the bonds that break give you clues about the mechanism.

Here’s some clues, made more clear:

  • Every time you break a C-O (pi) bond, it had to happen through addition of a nucleophile to a carbonyl.
  • Every time you see a proton adjacent to a carbonyl missing, it is a VERY strong indicator that an enolate formed there.
  • Every time you see a C-C (pi) bond being formed, it is the result of an elimination reaction (usually a condensation, where you’re losing water)
  • Every time you see a C-C (pi) bond being broken, it is likely the result of a conjugate addition (1,4-addition) especially when it’s adjacent to a carbonyl (C=O).

See how far you can get with this mechanism, based on just that information.

I’ll go through the mechanism tomorrow.

Thanks for reading ! James

 *P.S. The conjugate base is a better nucleophile. Also, the conjugate acid is also a better electrophile.