Question

Why are acetyl chloride or acetic anhydride used instead of acetic acid in preparing esters of acetic acid?

Answer 1

When alcohol is added with acid it forms an ester with water as a by-product. The water so formed needs to be removed as fast as possible because it will further react with the form ester and the reaction will get reversed. To prevent this reversibility of reaction we add some other agent to the esterification reaction.

Complete answer:
When alcohol is added with a acid then the ester is formed and water is formed as by-product which can be represented as:
$C{H_3}OH{\text{ + }}HOOC - C{H_3}{\text{ }} \rightleftharpoons {\text{ C}}{{\text{H}}_3} - COO - C{H_3}{\text{ + }}{H_2}O$
The water so formed needs to be removed as soon as possible because it will further react with the formed ester and the reaction gets reversed and it will carry in reverse direction. Thus to prevent this we use acetyl chloride or acetic anhydride. Also for the reactivity of acetyl chloride or acetic anhydride is more than acetic acid. Hence we use acetyl chloride or acetic anhydride instead of acetic acid in preparing esters of acetic acid. This can be shown as:
$C{H_3}CO - Cl{\text{ + }}H - O{H_2}CC{H_3}{\text{ }} \to {\text{ C}}{{\text{H}}_3} - COO - C{H_2}C{H_3}{\text{ + }}HCl$
Here no such water is formed and the yield of ester is also better than using acetic acid.
$C{H_3}CO - OCOC{H_3}{\text{ + }}H - O{H_2}CC{H_3}{\text{ }} \to {\text{ C}}{{\text{H}}_3}CO - OC{H_2}C{H_3}{\text{ + HOCOC}}{{\text{H}}_3}$
Here acetic anhydride reacts with alcohol to form ester in a good amount.

Note:
By using acetyl chloride or acetic anhydride we can also produce a good yield of ester at low temperatures also. Acetyl chloride is a chloride substitute of acetic acid and acetic anhydride is formed when two moles of acetic acid are mixed and a water molecule is abstracted from it. Acetic anhydride is an anhydrous form of two moles of acetic acid.