The preparation of an ester using a carboxylic acid and an alcohol is an equilibrium reaction involving
RCOOH + HOR' ==> RCOOR' + H2O.
For acetic acid and ethanol, the products are approximately 2/3 mole each of ester and water and the reactants are about 1/3 mole each of the starting materials. The equilibrium mixture is favored to the right (about 67% yield) but not all that much. Adding water (by not using absolute ethanol BOTH drives the reaction to the left (less yield) (LeChatelier's Principle) and provides an easier path for hydrolysis (the reverse direction of esterification).
Why is that absolute ethanol be used in preparing an ester with ethanol as one of the reactants?
My Guess:
It will cause hydrolysis of the ester as it is formed in the reaction vessel.
Is this reasoning right? Pls. I need help about this...
1 answer