JEE Chemistry 2024 Question with Solution

Given: m-Chlorobenzaldehyde is treated with $50\%$ KOH solution.

Find: The products formed in the reaction.

This is a Cannizzaro reaction because m-chlorobenzaldehyde is an aldehyde that does not have an $\alpha$-hydrogen. In concentrated base, two molecules of such an aldehyde undergo disproportionation: one molecule is oxidized to carboxylate and the other is reduced to alcohol.

The reaction is:

$$2\,\text{m-chlorobenzaldehyde} + \text{OH}^- \longrightarrow \text{m-chlorobenzoate}^- + \text{m-chlorobenzyl alcohol}$$

So, one molecule gives m-chlorobenzoate and the other gives m-chlorobenzyl alcohol.

Therefore, the correct option is B.

Given: m-Chlorobenzaldehyde reacts with concentrated KOH.

Find: Which products are obtained.

Step 1: Identify the reaction type.

The solution states that this is a Cannizzaro reaction. Such a reaction is shown by aldehydes that lack $\alpha$-hydrogen atoms.

Step 2: Apply the Cannizzaro principle.

In Cannizzaro reaction, two molecules of the aldehyde react. One is oxidized to the corresponding carboxylate ion, and the other is reduced to the corresponding primary alcohol.

The general form is:

$$2\,\text{R-CHO} + \text{OH}^- \longrightarrow \text{R-COO}^- + \text{R-CH}_2\text{OH}$$

Step 3: Substitute the given aldehyde.

For m-chlorobenzaldehyde:

$$\text{m-Cl-C}_6\text{H}_4\text{-CHO} \longrightarrow \text{m-Cl-C}_6\text{H}_4\text{-COO}^-$$

This is m-chlorobenzoate.

And the reduced product is:

$$\text{m-Cl-C}_6\text{H}_4\text{-CHO} \longrightarrow \text{m-Cl-C}_6\text{H}_4\text{-CH}_2\text{OH}$$

This is m-chlorobenzyl alcohol.

Step 4: Match with the options.

The pair m-chlorobenzoate and m-chlorobenzyl alcohol matches option B.

Therefore, the correct option is B.