Animated Solution for Chemistry - Organic Chemistry: Phenol on treatment with ${\text{CO}}_2$ in the presence of NaOH followed by acidification produces compound X as the major product. X on treatment with $({\text{CH}}_3\text{CO}{)}_2\text{O}$ in the presence of catalytic amount of ${\text{H}}_2{\text{SO}}_4$ produces:

Welcome to a classic organic chemistry problem that beautifully bridges the gap between textbook reactions and real-world pharmaceuticals. We are presented with a two-step reaction sequence starting with phenol, and our goal is to identify the final product. Let's break down the chemistry step by step.

Kolbe's Reaction

In the first step, phenol is treated with carbon dioxide ($C{O}_2$) and sodium hydroxide (NaOH), followed by acidification ($H^{+}$). This specific combination of reagents is the hallmark of Kolbe's reaction (or the Kolbe-Schmitt reaction).

The reaction begins with the sodium hydroxide deprotonating the phenol to form a phenoxide ion. The phenoxide ion is highly activated towards electrophilic aromatic substitution because the negatively charged oxygen strongly donates electron density into the benzene ring.

When exposed to carbon dioxide—a weak electrophile—the phenoxide ion attacks it. Interestingly, this attack happens almost exclusively at the ortho position. This regioselectivity is driven by the chelation effect of the sodium ion, which coordinates with both the phenoxide oxygen and the oxygen of the incoming $C{O}_2$ molecule, guiding it to the adjacent carbon.

Following the electrophilic attack and subsequent rearomatization, we get a carboxylate salt. The final acidification step protonates this salt, yielding ortho-hydroxybenzoic acid, widely known as salicylic acid. This is our intermediate compound, X.

Selective Acetylation

Now we move to the second part of the sequence. We take our intermediate, salicylic acid, and treat it with acetic anhydride ($(C{H}_{3}CO{)}_{2}O$) in the presence of a catalytic amount of sulfuric acid ($H_{2}S{O}_4$).

Salicylic acid has two functional groups that could potentially react: a carboxylic acid group ($-COOH$) and a phenolic hydroxyl group ($-OH$). However, they are not equally reactive. The phenolic $-OH$ group is a much better nucleophile than the $-COOH$ group. The lone pairs on the oxygen of the carboxylic acid are delocalized through resonance with the adjacent carbonyl group, making them less available for nucleophilic attack.

Therefore, the phenolic $-OH$ group selectively attacks the electrophilic carbonyl carbon of the acetic anhydride. The acid catalyst ($H_{2}S{O}_4$) helps by protonating the acetic anhydride, making it an even stronger electrophile.

As a result of this acetylation, the hydrogen of the phenolic $-OH$ is replaced by an acetyl group ($-COC{H}_3$), converting it into an ester group ($-OCOC{H}_3$). The carboxylic acid group remains completely unaffected.

The final product is 2-acetoxybenzoic acid. You might not recognize it immediately by its IUPAC name, but you certainly know its trade name: Aspirin. It is one of the most widely used medications globally, acting as an analgesic (painkiller), antipyretic (fever reducer), and anti-inflammatory drug.

Looking at our options, the structure of Aspirin matches perfectly with option (a).

To deepen your understanding, consider what would happen if we changed the reagents in the second step. What if, instead of acetic anhydride, we treated salicylic acid with methanol ($C{H}_{3}OH$) and an acid catalyst?

In that scenario, the reaction would be a Fischer esterification. The carboxylic acid group ($-COOH$) would react with the alcohol to form an ester, while the phenolic $-OH$ would remain untouched. The product would be methyl salicylate, commonly known as oil of wintergreen, which is used in topical pain relief ointments.

This highlights a crucial lesson in organic synthesis: understanding the relative reactivity of different functional groups allows chemists to selectively modify molecules to create vastly different and useful compounds!