COMEDK 2025 Evening Solved Paper

Concept:Dehydrogenation of a secondary alcohol over copper at 573 K gives a ketone, which then reacts with methyl Grignard to form a tertiary alcohol.Explanation:Copper at 573 K acts as a dehydrogenation catalyst for alcohols.For a secondary alcohol, the product is a ketone.The only secondary alcohol among the options that yields a ketone capable of producing tert‑butyl alcohol upon Grignard reaction is propan‑2‑ol.Propan‑2‑ol ($\mathrm{CH}_3\mathrm{CHOHCH}_3$) is dehydrogenated to acetone ($\mathrm{CH}_3\mathrm{COCH}_3$):$$\mathrm{CH}_3\mathrm{CHOHCH}_3 \xrightarrow{\mathrm{Cu}/573\,\mathrm{K}} \mathrm{CH}_3\mathrm{COCH}_3 + \mathrm{H}_2$$Acetone then reacts with $\mathrm{CH}_3\mathrm{MgBr}$ (methyl Grignard reagent), followed by hydrolysis with $\mathrm{H}_2\mathrm{O/H}^+$ to form tert‑butyl alcohol ($(\mathrm{CH}_3)_3\mathrm{COH}$).$$\mathrm{CH}_3\mathrm{COCH}_3 + \mathrm{CH}_3\mathrm{MgBr} \longrightarrow (\mathrm{CH}_3)_3\mathrm{C\!-\!OMgBr}$$$$(\mathrm{CH}_3)_3\mathrm{C\!-\!OMgBr} + \mathrm{H}_2\mathrm{O/H}^+ \longrightarrow (\mathrm{CH}_3)_3\mathrm{C\!-\!OH} + \mathrm{MgBrOH}$$Thus $[\mathrm{X}]$ is propan‑2‑ol.Answer:Propan‑2‑ol (Option C).