JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Alcohols Phenols and Ethers

The alcohol wanted carries a phenyl, a methyl, an ethyl and the $-\mathrm{OH}$ all on one and the same carbon, so it is a tertiary alcohol and must be assembled from a ketone and a Grignard reagent. Butan-2-one, $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{COC}{\mathrm{H}}_3$, with $\mathrm{PhMgX}$ delivers the phenyl onto the carbonyl carbon that already carries methyl and ethyl; propiophenone, $\mathrm{PhCOC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3$, with $\mathrm{C}{\mathrm{H}}_3\mathrm{MgX}$, and acetophenone, $\mathrm{PhCOC}{\mathrm{H}}_3$, with $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{MgX}$, each build the very same tertiary carbon, so all three of these routes succeed. Formaldehyde, $\mathrm{HCHO}$, brings no carbon substituent of its own, so whatever Grignard adds to it can only give a primary alcohol of the form $\mathrm{R}-\mathrm{C}{\mathrm{H}}_2-\mathrm{OH}$; that route yields $\mathrm{PhCH}(\mathrm{C}{\mathrm{H}}_3)\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{OH}$, a primary alcohol, and therefore cannot furnish the tertiary target.