What are the steps in a Reformatsky reaction?

A Reformatsky reaction involves the reaction of an alpha-haloester with a carbonyl compound in the presence of zinc.

The Reformatsky reaction is a type of organic reaction that was named after the Russian chemist Sergey Nikolaevich Reformatsky. This reaction is used to synthesise beta-hydroxy-esters from alpha-haloesters and carbonyl compounds. The reaction is facilitated by the presence of zinc, which is used to generate a zinc enolate intermediate.

The first step in the Reformatsky reaction is the formation of a zinc enolate. This is achieved by reacting an alpha-haloester with zinc metal. The zinc atom inserts itself between the alpha carbon and the halogen atom, resulting in the formation of a zinc enolate. This is a crucial step as the zinc enolate is the active species that reacts with the carbonyl compound in the next step.

The second step involves the reaction of the zinc enolate with a carbonyl compound. The carbonyl compound can be an aldehyde, ketone, or an ester. The zinc enolate attacks the carbonyl carbon of the carbonyl compound, leading to the formation of a new carbon-carbon bond. This results in the formation of a beta-hydroxy-ester.

The final step is the protonation of the oxygen atom of the newly formed beta-hydroxy-ester. This step is usually facilitated by the addition of an acid, which donates a proton to the oxygen atom. The end product of the Reformatsky reaction is a beta-hydroxy-ester.

In summary, the Reformatsky reaction involves three main steps: the formation of a zinc enolate from an alpha-haloester and zinc metal, the reaction of the zinc enolate with a carbonyl compound to form a beta-hydroxy-ester, and the protonation of the oxygen atom of the beta-hydroxy-ester to form the final product. This reaction is a useful tool in organic chemistry for the synthesis of beta-hydroxy-esters.