Question

m–chlorobenzaldehyde on treatment with 50\(\%\) KOH solution yields

• A.

  

• B.

  

• C.

  

• D.

  

Answer 1

The Correct Option is B

Solution: When m-chlorobenzaldehyde is treated with 50% KOH solution, it undergoes a reaction known as the Cannizzaro reaction, which occurs in aldehydes that do not have alpha hydrogens.

Cannizzaro Reaction: In the Cannizzaro reaction, an aldehyde is converted into a carboxylic acid and an alcohol in the presence of a strong base, like KOH. The reaction mechanism involves the disproportionation of the aldehyde, leading to the formation of the corresponding carboxylate and alcohol.

Reaction Process: For m-chlorobenzaldehyde, the reaction can be summarized as follows:

2 m-chlorobenzaldehyde + 50% KOH → m-chlorobenzoate + m-chlorobenzyl alcohol.

Final Products: The final products of the reaction are m-chlorobenzoate ion and m-chlorobenzyl alcohol.

Thus, the correct product obtained from the reaction is: chlorobenzoate and m-chlorobenzyl alcohol.

Answer 2

The question asks to identify the products formed when m-chlorobenzaldehyde is treated with a 50% KOH solution.

Concept Used:

This reaction is a classic example of the Cannizzaro reaction. The conditions for a Cannizzaro reaction are:

1. An aldehyde that lacks an alpha-hydrogen atom.
2. Treatment with a concentrated strong base (like 50% KOH or NaOH).

The Cannizzaro reaction is a disproportionation (or self-redox) reaction. In this reaction, two molecules of the aldehyde react: one molecule is oxidized to a carboxylic acid (which exists as its salt in the basic medium), and the other molecule is reduced to a primary alcohol. The general form of the reaction is:

\[ \text{2 R-CHO} + \text{OH}^- \longrightarrow \text{R-COO}^- + \text{R-CH}_2\text{OH} \]

Step-by-Step Solution:

Step 1: Analyze the reactant and reagents.

The reactant is m-chlorobenzaldehyde. Its structure has an aldehyde group (-CHO) attached to a benzene ring. The carbon atom of the aldehyde group is bonded directly to a carbon atom of the benzene ring, which does not have any hydrogen atoms attached to it. Therefore, m-chlorobenzaldehyde has no alpha-hydrogen. The reagent is 50% KOH, which is a concentrated strong base. These are the characteristic conditions for the Cannizzaro reaction.

Step 2: Determine the oxidation product.

One molecule of m-chlorobenzaldehyde undergoes oxidation. The aldehyde group (-CHO) is oxidized to a carboxylic acid group (-COOH). However, since the reaction occurs in a strong basic medium (KOH), the carboxylic acid formed (m-chlorobenzoic acid) immediately reacts with KOH to form its corresponding salt, potassium m-chlorobenzoate. The product shown in the options is the m-chlorobenzoate anion (COO⁻).

\[ \text{m-Cl-C}_6\text{H}_4\text{-CHO} \xrightarrow{[\text{Oxidation}]} \text{m-Cl-C}_6\text{H}_4\text{-COO}^- \quad (\text{m-chlorobenzoate ion}) \]

Step 3: Determine the reduction product.

The second molecule of m-chlorobenzaldehyde undergoes reduction. The aldehyde group (-CHO) is reduced to a primary alcohol group (-CH₂OH). This results in the formation of m-chlorobenzyl alcohol.

\[ \text{m-Cl-C}_6\text{H}_4\text{-CHO} \xrightarrow{[\text{Reduction}]} \text{m-Cl-C}_6\text{H}_4\text{-CH}_2\text{OH} \quad (\text{m-chlorobenzyl alcohol}) \]

Step 4: Compare the products with the given options.

The two products of the Cannizzaro reaction are the m-chlorobenzoate ion and m-chlorobenzyl alcohol. Let's examine the options:

• (b) Shows the m-chlorobenzoate ion and m-chlorobenzyl alcohol. This matches our derived products.
• (a) Shows a diol formed by coupling, which is incorrect.
• (c) and (d) Show substitution of the chloro group with a hydroxyl group. Aromatic nucleophilic substitution of chlorine on the benzene ring does not typically occur under these conditions. The Cannizzaro reaction at the aldehyde group is the major pathway.

Therefore, the products obtained are the m-chlorobenzoate ion and m-chlorobenzyl alcohol, which corresponds to option (b).