JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Chapter_pending

Acetophenone reacts with phosphorus pentachloride, which replaces the carbonyl oxygen by two chlorines to give the gem-dichloride $\mathrm{Ph}\text{-}\mathrm{CC}{\mathrm{l}}_2\text{-}\mathrm{C}{\mathrm{H}}_3$. Three equivalents of sodium amide carry out two successive eliminations and then remove the terminal alkyne proton, giving sodium phenylacetylide; acidification liberates phenylacetylene, $\mathrm{Ph}\text{-}\mathrm{C}\equiv \mathrm{C}\text{-}\mathrm{H}$. Diborane followed by alkaline hydrogen peroxide adds water across the triple bond in anti-Markovnikov fashion, giving an enol that tautomerises to the final aldehyde, $\mathrm{Ph}\text{-}\mathrm{C}{\mathrm{H}}_2\text{-}\mathrm{CHO}$. In this phenylacetaldehyde the six ring carbons and the carbonyl carbon are $s{p}^2$ while the methylene carbon is $s{p}^3$, so the number of $s{p}^2$ carbon atoms is seven.