Question

A carboxylic acid is formed when an $\mathrm{\alpha }$-haloketone reacts with hydroxide ion. This reaction is called a Favorskii reaction. Propose a mechanism for the following Favorskii reaction.

Short answer

Mechanism of given Favorskii reaction

Step-by-step solution

Step-by-Step Solution Step 1: Enolization of a carbonyl compound

The hydrogen atoms on the alpha-carbon of a carbonyl have an acidic character due to the electron-withdrawing nature of the carbonyl group. Therefore, a strong base can abstract the alpha hydrogen of a carbonyl carbon producing a carbanion.

The carbanion thus formed undergoes tautomerization to form an enol in an acidic medium and enolate in a basic medium. An enol is a compound containing a hydroxyl group and a double bond on the same carbon.

The enol and enolate contain a nucleophilic carbon atom and therefore, can react with an electrophilic carbon-containing compound forming a carbon-carbon bond.

Favorskii reaction

Favorskii reaction occurs in an alpha haloketone compound with alpha hydrogen. It is a rearrangement reaction initiated by enolization. The hydrogen from the alpha carbon other than the carbon containing ketone is abstracted.

The mechanism involves the formation of a ring, and the final product of the reaction is a carboxylate ion in the aqueous medium. A base, generally a hydroxide ion, catalyzes the enolization reaction.

Mechanism of the given reaction

The hydroxide ion in the solution abstracts the alpha hydrogen from alpha-haloketone to form a carbanion. The carbanion formed attacks the halogen containing carbon, causing the removal of the halide.

The intermediate ring thus formed has an electrophilic carbonyl carbon which is attacked by another hydroxide ion from the solution. The adduct thus formed undergoes rearrangement to form a carboxylic acid containing a carbanion.

The carbanion abstracts the acid proton to form a carboxylate ion in the final step.

The reaction mechanism can be described as:

Mechanism of given Favorskii reaction