Question

If the Favorskii rearrangement of 2-chlorocyclohexanone is carried out using sodium ethoxide in ethanol, the product is ethyl cyclopentanecarboxylate. Propose a mechanism for this reaction.

Short answer

Question: Propose a mechanism for the Favorskii rearrangement of 2-chlorocyclohexanone using sodium ethoxide in ethanol, which results in the formation of ethyl cyclopentanecarboxylate. Answer: The mechanism involves the following steps: 1. Formation of the enolate ion by deprotonating 2-chlorocyclohexanone with sodium ethoxide (NaOEt). 2. Intramolecular nucleophilic substitution (S_N2) where the enolate oxygen displaces the chlorine atom, leading to the formation of an oxirane ring intermediate. 3. Ring-opening of the oxirane ring by ethoxide ion, resulting in an alkoxide ion intermediate. 4. Protonation of the alkoxide ion by ethanol, leading to the formation of ethyl cyclopentanecarboxylate.

Step-by-step solution

Formation of the enolate ion

First, the sodium ethoxide (NaOEt) reacts with the 2-chlorocyclohexanone to form the enolate ion. Sodium ethoxide is a strong base and deprotonates the α-hydrogen present in the 2-chlorocyclohexanone, resulting in the formation of an enolate ion and sodium chloride as a byproduct. $$ \text{2-chlorocyclohexanone} \ + \ \text{NaOEt} \longrightarrow \ \text{Enolate ion} \ + \ \text{NaCl} $$

Intramolecular nucleophilic substitution

The enolate ion formed in step 1 has a negative charge on the oxygen atom and is a nucleophile. This nucleophile will attack the electrophilic carbon that bears the chlorine atom. This step involves an intramolecular nucleophilic substitution reaction (S_N2) where the enolate oxygen displaces the chlorine atom, leading to the formation of an oxirane ring intermediate. $$ \text{Enolate ion} \longrightarrow \ \text{Oxirane ring} $$

Ring-opening of the oxirane ring

The oxirane ring intermediate is highly strained due to the three-membered ring. The ethoxide ion, which is still present in the reaction mixture, acts as a nucleophile again and attacks the electrophilic carbon of the oxirane ring. This results in the opening of the oxirane ring and the formation of an alkoxide ion intermediate. $$ \text{Oxirane ring} \ + \ \text{EtO}^- \longrightarrow \ \text{Alkoxide ion} $$

Protonation and formation of ethyl cyclopentanecarboxylate

Finally, the alkoxide ion formed in step 3 gets protonated by an ethanol molecule present in the reaction mixture. This results in the formation of the desired product, ethyl cyclopentanecarboxylate. $$ \text{Alkoxide ion} \ + \ \text{EtOH} \longrightarrow \ \text{Ethyl cyclopentanecarboxylate} $$ In summary, the proposed mechanism for the Favorskii rearrangement of 2-chlorocyclohexanone using sodium ethoxide in ethanol involves the formation of an enolate ion, intramolecular nucleophilic substitution to form an oxirane ring, ring-opening, and finally protonation to result in ethyl cyclopentanecarboxylate as the final product.