Question

In the following reactions, the major products E and F, respectively, are 
\[ \text{(i) NaOH/CO}_2,\; 125^\circ\text{C},\; 4–7\; \text{atm} \quad\quad \text{(ii) H}_3\text{O}^+ \quad\quad \longrightarrow E \quad\quad \text{(CH}_3\text{CO)}_2\text{O} \longrightarrow F \]

• A. A
• B. B
• C. C
• D. D

Hint:

Kolbe–Schmitt gives ortho-hydroxybenzoic acid from phenol; acetic anhydride converts phenolic OH to acetate (aspirin formation).

Answer 1

The Correct Option is D

Step 1: Reaction of phenol with NaOH/CO$_2$ (Kolbe–Schmitt reaction).
In the Kolbe–Schmitt reaction, phenoxide ion reacts with CO$_2$ at high temperature and pressure to form salicylic acid (o-hydroxybenzoic acid).
Electrophilic attack occurs predominantly at the ortho position due to the –O$^-$ directing group.
Thus, product E = o-hydroxybenzoic acid.
Step 2: Reaction of salicylic acid with acetic anhydride.
Acetic anhydride (\( (\text{CH}_3\text{CO})_2\text{O} \)) acetylates phenolic –OH groups.
In salicylic acid, only the phenolic OH reacts (the COOH group does not undergo acetylation under these mild conditions).
Therefore, the product is o-acetoxybenzoic acid (aspirin). Thus, product F = o-acetoxybenzoic acid.
Step 3: Matching E and F with options.
Option (A) correctly shows:
– E = salicylic acid (o-OH + CO$_2$H)
– F = acetylated product (OCOCH$_3$ + CO$_2$H).
Step 4: Conclusion.
Thus, the major products E and F correspond to option (A).