JEE Chemistry 2024 Question with Solution

Given: The question asks for the molecular formula of the second homologue of mono carboxylic acids.

Find: Which option matches the second member of this homologous series.

Monocarboxylic acids follow the general formula

$$\mathrm{C_nH_{2n+1}COOH}$$

where $n$ is a non-negative integer.

The first member is formic acid:

$$\mathrm{HCOOH = CH_2O_2}$$

This corresponds to the first homologue.

The second member is acetic acid. Putting $n = 1$ in the general formula:

$$\mathrm{C_1H_{2(1)+1}COOH = C_2H_4O_2}$$

Therefore, the molecular formula of the second homologue is $\mathrm{C_2H_4O_2}$.

So, the correct option is B.

Note: The provided the solution contains inconsistent option-label mapping, but its worked conclusion clearly identifies the correct formula as $\mathrm{C_2H_4O_2}$, which matches option B in the given options.

Given: Homologous series of mono carboxylic acids.

Find: The second homologue.

List the first few members:

1. Formic acid: $\mathrm{HCOOH}$ or $\mathrm{CH_2O_2}$
2. Acetic acid: $\mathrm{CH_3COOH}$
3. Propionic acid: $\mathrm{C_2H_5COOH}$

Now write the molecular formula of acetic acid:

$$\mathrm{CH_3COOH = C_2H_4O_2}$$

Hence, the second homologue is $\mathrm{C_2H_4O_2}$, so the correct option is B.