COMEDK 2025 Afternoon Solved Paper

Concept:The iodoform test ($\mathrm{I}_2 / \mathrm{NaOH}$) gives a yellow precipitate of $\mathrm{CHI}_3$ only for compounds containing a $\mathrm{CH}_3\mathrm{CO}$– group or a $\mathrm{CH}_3\mathrm{CHOH}$– group.Explanation:Option A ($\mathrm{C}_6\mathrm{H}_5\mathrm{OH}$, phenol) and option B ($\mathrm{C}_6\mathrm{H}_5\mathrm{CH}_2\mathrm{OH}$, benzyl alcohol) lack any $\mathrm{CH}_3$– group next to a carbonyl or hydroxyl carbon. Option C ($(\mathrm{CH}_3)_3\mathrm{C}\textrm{-}\mathrm{OH}$, tert-butanol) is a tertiary alcohol and cannot be oxidised to a methyl ketone. Option D ($\mathrm{C}_6\mathrm{H}_5\mathrm{CHOHCH}_3$, 1-phenylethanol) contains the $\mathrm{CH}_3\mathrm{CHOH}$– fragment. On oxidation with $\mathrm{I}_2/\mathrm{NaOH}$, it forms acetophenone ($\mathrm{C}_6\mathrm{H}_5\mathrm{COCH}_3$), which then undergoes the haloform reaction to produce a yellow precipitate of $\mathrm{CHI}_3$.Answer:Option D