Question

An ester of molecular formula \[{{C}_{4}}{{H}_{8}}{{O}_{2}}\] was produced by the reaction of an alcohol with a carboxylic acid.

	Alcohol	Acid
1	Methanol	Propanoic acid
2	Ethanol	Ethanoic acid
3	Propanol	Propanoic acid

Which of the following could be the alcohol and the acid?
a.2
b.1 and 2
c.1 and 3
d.1, 2 and 3

Answer 1

Hint: The given question states that the reactants are acid and carboxylic acid and the product is an ester. Start this question by drawing out the possible structures of ester with the molecular formula \[{{C}_{4}}{{H}_{8}}{{O}_{2}}\].

Complete step by step answer:
As we can see, the process occurring here is esterification, which is the reaction between alcohols and carboxylic acids to make esters. The general reaction is as follows –

According to the question, the compound has 4 carbons.
Therefore, the reaction can either be between methanol and propionic acid or ethanol or ethanoic acid to give \[{{C}_{4}}{{H}_{8}}{{O}_{2}}\].
Therefore, the answer is – option (b).

Additional Information: Other than this, esters are also formed by a reaction between acyl chlorides (acid chlorides) and alcohols, and between acid anhydrides and alcohols.

Note: Let us define esters, alcohols and carboxylic acids –
“Esters have a hydrocarbon group of some sort replacing the hydrogen in the -COOH group of a carboxylic acid. We shall just be looking at cases where it is replaced by an alkyl group, but it could equally well be an aryl group (one based on a benzene ring).”
“Carboxylic acids are compounds which contain a -COOH group. For the purposes of this page we shall just look at compounds where the -COOH group is attached either to a hydrogen atom or to an alkyl group.”
“Alcohols are compounds in which one or more hydrogen atoms in an alkane have been replaced by an -OH group.”