Question

Identify the correct sequence of reactions and products for the following scheme.

• A. \( P_1 = \text{Phenol}, P_2 = \text{Salicylaldehyde}, P_3 = \text{Cyclohexanol}, P_4 = \text{Phenylacetate} \)
• B. \( P_1 = \text{Benzene}, P_2 = \text{Salicylic acid}, P_3 = \text{Cyclohexanol}, P_4 = \text{Phenylacetate} \)
• C. \( P_1 = \text{Phenol}, P_2 = \text{Salicylic acid}, P_3 = \text{Anisole}, P_4 = \text{Aspirin} \)
• D. \( P_1 = \text{Phenol}, P_2 = \text{Salicylic acid}, P_3 = \text{Anisole}, P_4 = \text{Phenylacetate} \)
• E. \( P_1 = \text{Cyclohexanol}, P_2 = \text{Salicylic acid}, P_3 = \text{Phenol}, P_4 = \text{Aspirin} \)

Hint:

In reactions involving phenol and sodium hydroxide, typical reactions include the Kolbe reaction (formation of salicylic acid) and ether formation via methylation.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This question requires the identification of products from four different named reactions starting from or involving phenols. We need to recognize each reaction and predict its outcome.
Step 2: Detailed Explanation of Each Reaction:
Reaction (i): Sodium benzenesulfonate to P\(_1\)
Starting material: Sodium benzenesulfonate (C\(_6\)H\(_5\)SO\(_3\)Na).
Reagents: 1. NaOH (fusion), 2. H\(^+\) (acidification).
This is the Dow process for the preparation of phenol. Fusing sodium benzenesulfonate with sodium hydroxide at high temperature produces sodium phenoxide (C\(_6\)H\(_5\)ONa). Subsequent acidification protonates the phenoxide ion to yield phenol.
\[ \text{C}_6\text{H}_5\text{SO}_3\text{Na} \xrightarrow{\text{NaOH, }\Delta} \text{C}_6\text{H}_5\text{ONa} \xrightarrow{\text{H}^+} \text{C}_6\text{H}_5\text{OH} \] So, P\(_1\) is Phenol.
Reaction (ii): Phenol to P\(_2\)
Starting material: Phenol (from step i).
Reagents: 1. CO\(_2\)/NaOH, 2. H\(^+\).
This is the Kolbe-Schmitt reaction. Phenol is first converted to sodium phenoxide with NaOH. The phenoxide ion, being highly activated, undergoes electrophilic substitution with the weak electrophile CO\(_2\). The carboxylation occurs primarily at the ortho position. Subsequent acidification gives salicylic acid (o-hydroxybenzoic acid).
\[ \text{C}_6\text{H}_5\text{OH} \xrightarrow{\text{1. NaOH, CO}_2, \Delta, P \text{ 2. H}^+} o\text{-HOC}_6\text{H}_4\text{COOH} \] So, P\(_2\) is Salicylic acid.
Reaction (iii): Phenol to P\(_3\)
Starting material: Phenol.
Reagents: NaOH, CH\(_3\)Cl.
This is the Williamson ether synthesis. Phenol, being acidic, reacts with a base (NaOH) to form sodium phenoxide. The phenoxide ion then acts as a nucleophile and attacks the alkyl halide (CH\(_3\)Cl) in an S\(_N\)2 reaction to form an ether.
\[ \text{C}_6\text{H}_5\text{OH} + \text{NaOH} \rightarrow \text{C}_6\text{H}_5\text{ONa} \xrightarrow{\text{CH}_3\text{Cl}} \text{C}_6\text{H}_5\text{OCH}_3 \] The product is methoxybenzene, commonly known as Anisole. So, P\(_3\) is Anisole.
Reaction (iv): Salicylic acid to P\(_4\)
Starting material: Salicylic acid (P\(_2\)).
Reagents: (CH\(_3\)CO)\(_2\)O, H\(^+\) (acetic anhydride in acid catalyst).
This is the acetylation of the phenolic hydroxyl group of salicylic acid. Acetic anhydride is an acetylating agent. The -OH group acts as a nucleophile and attacks the acetyl group, forming an ester. The carboxylic acid group does not react under these conditions.
The product is acetylsalicylic acid, which is the chemical name for Aspirin.
So, P\(_4\) is Aspirin.
Step 3: Matching Products with Options:
P\(_1\) = Phenol
P\(_2\) = Salicylic acid
P\(_3\) = Anisole
P\(_4\) = Aspirin
This set of products matches option (C).
Step 4: Final Answer:
The correct identification of the products is P\(_1\) = Phenol, P\(_2\) = Salicylic acid, P\(_3\) = Anisole, P\(_4\) = Aspirin. This corresponds to option (C).