See image — Alcohols Phenols and Ethers Chemistry Question

💡 Solution & Explanation

Concept: When a carboxylic acid bearing a carbocation-stabilizing group is dissolved in concentrated H2SO4, it can undergo decarboxylation via formation of a stable carbocation intermediate. Step 1: Ph3CCO2H (triphenylacetic acid) is dissolved in conc. H2SO4. The acid is protonated to give Ph3CCO2H2+, and then the C-C bond between the triphenylmethyl group and the carbonyl undergoes heterolytic cleavage. This generates the highly stable trityl carbocation (Ph3C+) and CO2 (lost as gas) along with water. Step 2: The trityl carbocation Ph3C+ is exceptionally stable due to delocalization of the positive charge over three phenyl rings (extended conjugation and resonance). This drives the ionization readily in strongly acidic media. Step 3: When the resulting solution (containing Ph3C+ and HSO4-) is poured into methanol (CH3OH), the nucleophilic methanol attacks the electrophilic trityl carbocation: Ph3C+ + CH3OH -> Ph3C-OCH3 + H+. Step 4: The product X is therefore Ph3C-OCH3 (methoxytriphenylmethane / trityl methyl ether), which is option (c). Why other options fail: - Option (a) Ph3C-C(=O)-O-CH3: This would be the simple esterification product of Ph3CCO2H with methanol, but decarboxylation occurs first in conc. H2SO4, so CO2 is lost and the ester is not formed. - Option (b) Ph2CH-C(=O)-O-CH3: This implies a 1,2-phenyl shift (rearrangement) and re-esterification, which does not occur under these conditions. - Option (d) Ph3C-CH3: This would require a methyl carbanion or hydride, not available from methanol acting as a nucleophile. Therefore, the correct answer is C.