Write a stepwise reaction sequence using proton transfer reactions to show how the following reaction occurs. (Hint: As a first step, use${}^{\mathbf{-}}\mathbf{\text{OH}}$ to remove a proton from the${\mathbf{\text{CH}}}_{\mathbf{2}}$ group between the C=O and C=C.)

The bases (weak or strong) abstract the acidic proton and create a stable carbanion. The carbanion undergoes delocalization with the alternate pi bonds and forms resonance structures.

Protonation occurs due to the presence of carbanion. The carbanion abstracts the proton from water and gives the final product.

In the given reaction, the${}^{-}\text{OH}$ is a weak base and abstracts the acidic proton (exits between the carbonyl (C=O) group and a double bond (C=C)) and forms a carbanion.

The carbonyl and alkene groups both have empty orbital, which attracts the electrons from the adjacent group, making hydrogen acidic. The carbanion has a lone pair of electrons which delocalizes and forms another carbanion.

The carbanion at the para position abstracts the proton from water and forms the final product.

Mechanism of the formation of the end product