The saponification of ethyl acetate and sodium hydroxide

The saponification of ethyl acetate and sodium hydroxide

The saponification of ethyl acetate and sodium hydroxide

At the beginning of this reaction, ethyl acetate ($CH_3COOC_2H_5 $) is mixed with sodium hydroxide ($NaOH $). In the case of sodium hydroxide, a strong electrolyte also quickly dissociates into sodium ions ($Na ^ + $) and hydroxyl ions ($OH ^ - $) in water. Hydroxide ions are strongly nucleophilic, while the carbonyl carbon of ethyl acetate is partially positive, and the two are attracted.

Hydroxide ions attack the carbonyl carbon of ethyl acetate and form a tetrahedral intermediate. This intermediate is unstable, and then the ethoxy group ($OC_2H_5 $) leaves to form acetate ion ($CH_3COO ^ - $) and ethanol ($C_2H_5OH $). The acetate ion combines with sodium ion to form sodium acetate ($CH_3COONa $). The chemical reaction equation is as follows:

$CH_3COOC_2H_5 + NaOH\ longrightarrow CH_3COONa + C_2H_5OH $

This reaction is an irreversible second-order reaction, and the reaction rate is related to the concentration of ethyl acetate and sodium hydroxide. During the reaction, the conductivity of the solution changes with the type and concentration of ions. At first, the concentration of sodium hydroxide in the solution is high, and the conductivity is large. As the reaction progresses, sodium hydroxide gradually decreases, and sodium acetate gradually emerges. Because the mobility of acetate ions is less than that of hydroxide ions, the conductivity of the solution gradually decreases.

This saponification reaction has important applications in chemical, pharmaceutical and other fields. Understanding its mechanism and controlling its rate can lead to efficient production and high-quality products. Scholars should study it in detail to explore the wonders of chemistry and apply it to the world.