Ace Your Next Organic Chemistry Exam.

With these Downloadable PDF Study Guides

Our Study Guides

The Robinson Annulation

Description: The Robinson annulation is a method for forming six membered rings. It is a sequential combination of the Michael reaction and the Aldol condensation.

Notes: Although here shown using basic conditions (NaOH as base) the reaction can also be done under acidic conditions. “Heat” is sometimes mentioned under the reaction arrow to denote that conditions are used that will favor elimination.


Notes: Examples 1 and 2 show the exact same reaction, although example 1 is under basic conditions, and example 2 employs acidic conditions. Substituted ketones can also be used (Example 3), as can substituted α-β unsaturated ketones (enones). A number of bases can be used in this reaction; usually the choice is hydroxide or an alkoxide base. Heat helps to drive the reaction forward towards elimination.


The Robinson annulation is made up of two steps: a Michael reaction and then an Aldol condensation. The first process (the Michael reaction) begins with deprotonation of the ketone to form an enolate (Step 1, arrows A and B). The enolate then performs a conjugate addition (or “1,4 addition”) on the α-β unsaturated ketone, forming a new enolate (Step 2, arrows C and D). The enolate is then protonated by solvent (Step 3, arrows E and F) to give the Michael addition product.

The next “phase” of the Robinson annulation reaction is an aldol condensation. This begins with deprotonation of the ketone (Step 1, arrows A and B) to give an enolate, which then adds to the carbonyl carbon six atoms away (Step 2, arrows C and D). After protonation of the resulting alkoxide (RO-) (Step 3, arrows E and F) an equivalent of base then removes a proton from the ketone, making an enolate (Step 4, arrows G and H). The enolate then displaces the hydroxide group, forming a new π bond (Step 5, arrows I and J) to give the final Robinson annulation product.

Notes: After the deprotonation step (Step 1 of the Michael) only one resonance form is depicted for simplicity. After Step 2 of the Michael reaction, notice how we have an enolate on C-5. Why doesn’t it add to the carbonyl carbon on C-2 at this point? Because it would lead to the formation of a 4-membered ring, which would have lots of ring strain! That’s why it’s necessary to perform the subsequent protonation at C-5 and deprotonation at C-7 to form the enolate that can then form the six-membered ring.

It’s also possible to combine steps 4 and 5 of the aldol reaction mechanism (as drawn) to show deprotonation happening with elimination of hydroxide ion. It’s also reasonable to draw the resonance form of the enolate ion.