Question

What is the esterification reaction equation of benzyl alcohol and acetic acid?

Answer 1

When any alcohol reacts with carboxylic acid in the presence of an acid, then esters are formed. This process is called esterification. Esters are fruity smelling compounds having the formula $RCOO{{R}^{'}}$, where R is any alkyl group and ${{R}^{'}}$ can be different and same as R.

Complete answer:
Esterification reaction is the formation of esters by the reaction of carboxylic acids, or acid chlorides, or acetic anhydrides with alcohols, in the presence of an acid like sulphuric acid, ${{H}_{2}}S{{O}_{4}}$.
The esterification reaction involves the cleavage of the O – H bond of the alcohols, which is the result of the acidity the alcohols possess.
The reaction given here between benzyl alcohol and acetic acid, involves the protonation of the carboxylic acid, which generates an electrophile, on which the alcohol attacks. The intermediate will be formed, which has a leaving group as the water molecule, after the removal of the water molecule, the product formed is an ester. Here, the ester formed will be phenyl acetate.
The esterification reaction is as follows:
$(Ph-C{{H}_{2}}-OH)+(C{{H}_{3}}-COOH)\xrightarrow[acid\,catalyst]{{{H}_{2}}S{{O}_{4}}}(C{{H}_{3}}-COOPh)+{{H}_{2}}O$
Hence, from the reaction of benzyl alcohol, and acetic acid, phenyl acetate is formed.

Note:
The esterification reaction can also be carried out by replacing carboxylic acid with acid chloride or acetic anhydride. The reaction will be, $C{{H}_{3}}COCl+{{C}_{2}}{{H}_{5}}OH\xrightarrow{pyridine}C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}+HCl$ , and ${{(C{{H}_{3}}CO)}_{2}}O+{{C}_{2}}{{H}_{5}}OH\overset{{{H}^{+}}}{leftrightarrows}C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}+C{{H}_{3}}COOH$ respectively. These are autocatalysis reactions, where the acid is formed as one of the products that removes water. Pyridine is used to react with HCl so that the reaction may move in the forward direction.