Animated Solution for Chemistry - Organic Chemistry: $p$-cresol reacts with chloroform in alkaline medium to give the compound $A$ which adds hydrogen cyanide to form the compound $B$. The latter, on acidic hydrolysis gives chiral carboxylic acid. The structure of the carboxylic acid is

Organic synthesis often feels like a puzzle where each reagent is a clue leading to the final picture. In this problem, we embark on a three-step journey starting from $p$-cresol, transforming it through classic named reactions to yield a complex chiral carboxylic acid.

Our starting material is $p$-cresol (4-methylphenol). The first step involves treating it with chloroform ($CHC{l}_3$) in an alkaline medium ($KOH$). This specific combination of reagents should immediately ring a bell: it's the Reimer-Tiemann reaction.

The Reimer-Tiemann reaction is a hallmark method for the ortho-formylation of phenols. The strong base reacts with chloroform to generate dichlorocarbene ($:CC{l}_2$), a highly reactive electrophile.

Now, we must determine where this electrophile will attack the aromatic ring. The hydroxyl group ($-OH$) is strongly activating and directs incoming groups to the ortho and para positions. However, in $p$-cresol, the para position is already occupied by a methyl group ($-C{H}_3$). Therefore, the formyl group ($-CHO$) has no choice but to attach at the ortho position. This yields Compound A: 2-hydroxy-5-methylbenzaldehyde.

With our newly formed aldehyde (Compound A) in hand, we introduce hydrogen cyanide ($HCN$). The carbonyl carbon of the aldehyde is electrophilic, making it a prime target for nucleophilic attack.

The cyanide ion ($C{N}^{-}$) acts as the nucleophile, attacking the carbonyl carbon and pushing the pi electrons onto the oxygen atom. The negatively charged oxygen then picks up a proton to form a hydroxyl group. This transformation converts the aldehyde group ($-CHO$) into a cyanohydrin group ($-CH(OH)CN$). We have now successfully synthesized Compound B.

The final leg of our journey involves the acidic hydrolysis of Compound B. When a nitrile group ($-C\equiv N$) is subjected to aqueous acid ($H_3{O}^{+}$) and heat, it undergoes complete hydrolysis.

The nitrogen atom is eventually lost as an ammonium ion ($N{H}_4^{+}$), and the carbon atom of the nitrile is converted into a carboxylic acid group ($-COOH$).

Let's examine our final product: 2-hydroxy-2-(2-hydroxy-5-methylphenyl)acetic acid. The carbon atom that was originally part of the aldehyde group is now bonded to four distinct groups: a hydroxyl group ($-OH$), a carboxylic acid group ($-COOH$), a hydrogen atom ($-H$), and the substituted aromatic ring. Because it is attached to four different groups, this carbon is a chiral center, perfectly aligning with the problem's description of a "chiral carboxylic acid."

Comparing our derived structure with the given options, it is evident that it matches option (c).