Question:Show how you would use the Robinson annulation to synthesize the following compounds.

Robinson annulation is a chemical reaction that results in a ring formation. Robinson annulations reaction forms$\alpha ,\beta$ unsaturated cyclic ketones from methyl vinyl ketones andaldehyde or ketones.

Robinson annulation has two reactions: First, Michael reaction and then intramolecular aldol condensation to give a new six-membered ring containing an alpha, beta-unsaturated ketone.

Robinson annulation begins with a Micheal addition followed by aldol condensation:

• Essential condition present which removes$\alpha$ - proton from ketone.
• In Micheal’s reaction, the enolate formed attacks enone.
• Enolate protonates to give ketone.
• Base removes $\alpha$- proton.
• In an aldol reaction, the enolate attacks the ketone.
• Alkoxide protonates to provide alcohol.
• The command removes$\alpha$ -pro, ton and E2 elimination of hydroxide occur to give enone.

Steps of the process of Robinson annulations

As in the question, we are given the end product of Robinson Annulation instead of reactants so that we will work backwardly. We will first be seeing how the end product is synthesised from aldol condensation and then Michael’s reaction.

In Robinson, Annulation, dehydration occurs at last, but here hydration is done.

Hydration of product

In Robinson Annulation from cyclic aldol, a six-membered ring forms but here opposite occurs.

Cyclic aldol formation

Here Michael’s reaction is performed in the opposite direction.

Reverse Michael reaction

All steps together are: