Question

The molecular formula of second homologue in the homologous series of monocarboxylic acid is

• A. \( \text{C}_2\text{H}_2\text{O}_2 \)
• B. \( \text{CH}_2\text{O} \)
• C. \( \text{C}_2\text{H}_4\text{O}_2 \)
• D. \( \text{C}_3\text{H}_6\text{O}_2 \)

Answer 1

The Correct Option is A

The second homologue in the homologous series of monocarboxylic acids is determined using the general formula \(\text{C}_n\text{H}_{2n+1}\text{COOH}\), where \(n\) is a positive integer.

Monocarboxylic acids, also known as aliphatic carboxylic acids, feature a carboxyl group (\(\text{COOH}\)) bonded to an alkyl chain. The series begins with formic acid (HCOOH), corresponding to \(n = 0\).

To identify the second homologue:

1. Formic acid: This is the first compound, with \(n = 0\), yielding \(\text{HCOOH}\) (or \(\text{CH}_2\text{O}_2\)).
2. Acetic acid: This is the second compound, derived by substituting \(n = 1\) into the general formula: \(\text{C}_1\text{H}_{2(1)+1}\text{COOH}\), which simplifies to \(\text{C}_2\text{H}_4\text{O}_2\).

Therefore, the molecular formula for acetic acid, the second homologue in the monocarboxylic acid series, is \(\text{C}_2\text{H}_4\text{O}_2\).

Correction: The initial answer presented, \(\text{C}_2\text{H}_2\text{O}_2\), was incorrect. Re-evaluation confirms that \(\text{C}_2\text{H}_4\text{O}_2\) accurately represents acetic acid, the second homologue, according to the established formula.

• Option 1 (\(\text{C}_2\text{H}_2\text{O}_2\)) does not adhere to the general formula for monocarboxylic acids.
• Option 2 (\(\text{CH}_2\text{O}\)) lacks the essential carboxyl group.
• Option 3 (\(\text{C}_2\text{H}_4\text{O}_2\)) is the correct molecular formula for acetic acid.
• Option 4 (\(\text{C}_3\text{H}_6\text{O}_2\)) represents propionic acid, the third homologue in the series.