Question

In the presence of ${{H}_{2}}O$ concentrated sulphuric acid, acetic acid reacts with ethyl alcohol to produce:
(A) Aldehyde
(B) Alcohol
(C) Ester
(D) Carboxylic acid

Answer 1

The following reaction is slow and reversible. Sulphuric acid protonates ethanoic acid and the proton becomes attached to one of the lone pairs on the oxygen which is double-bonded to the carbon. The transfer of the proton to the oxygen gives it a positive charge.

Complete step by step answer:
Esterification is the process of combining an organic acid (RCOOH) with an alcohol (ROH) to form an ester (RCOOR) and water; or a chemical reaction resulting in the formation of at least one ester product. Ester is obtained by an esterification reaction of an alcohol and a carboxylic acid.
The reaction of the condition described in the given question is as follows:
$C{{H}_{3}}COOH~+C{{H}_{3}}C{{H}_{2}}OH\xrightarrow{Conc{{H}_{2}}S{{O}_{4}}}~~\text{ }~~C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}+{{H}_{2}}O$

On the reaction of acetic acid and ethanol in presence of sulphuric acid, there is formation of ethyl ethanoate which is an ester, and water.
Sulphuric acid acts as a catalyst in the following esterification reaction. During esterification, oxygen or alcohol is found in the main product that is nothing but ester. This indicates that H of alcohol is eliminated as a water molecule.
So, the correct answer is “Option C”.

Note: Ethyl acetate (also known as ethyl ethanoate) is an organic ester compound with a molecular formula of${{C}_{4}}{{H}_{8}}{{O}_{2}}$. It is a colourless liquid with a fruity characteristic odour that is commonly recognized in glues and nail polish remover.