Mechanism of Reaction Between Butyraldehyde and HCl

What are the steps involved in the reaction between butyraldehyde and HCl to form 2-ethyl-3-hydroxy-hexanal?

Starting material: butyraldehyde

Solvent: ethyl ether

Reagent: HCl

Product: 2-ethyl-3-hydroxy-hexanal

The mechanism for the reaction between butyraldehyde and HCl in ethyl ether solvent to form 2-ethyl-3-hydroxy-hexanal involves several steps:

1. Protonation of the Carbonyl Oxygen:

In the presence of HCl, the carbonyl oxygen of butyraldehyde is protonated. This protonation makes the carbonyl carbon more electrophilic.

2. Nucleophilic Addition:

The nucleophilic ethyl ether attacks the electrophilic carbonyl carbon, leading to the formation of a tetrahedral intermediate.

3. Deprotonation:

The unstable tetrahedral intermediate undergoes deprotonation to restore stability. A chloride ion acts as a base and abstracts a proton from the tetrahedral intermediate, resulting in the formation of an alcohol.

4. Tautomerization:

The alcohol formed undergoes tautomerization, converting it into its keto form by migrating a hydrogen atom from the hydroxyl group to the adjacent carbon.

Overall, the reaction involves the protonation of the carbonyl oxygen, nucleophilic addition, deprotonation, and tautomerization, ultimately resulting in the formation of 2-ethyl-3-hydroxy-hexanal as the product.