JEE Main 2025 Apr 7 Shift 2, Chemistry Q25: Properties and Important Reactions of Aldehydes & Ketones

Treating acetophenone (Ph-CO-CH${}_3$) with PCl${}_5$ converts the carbonyl into a geminal dichloride: A = Ph-CCl${}_2$-CH${}_3$.

Reacting A with excess (3 equivalents) of sodium amide in liquid ammonia carries out a double dehydrohalogenation followed by deprotonation, converting the gem-dihalide into the terminal sodium alkynide: B = Ph-C$\equiv$C-Na.

Acidifying B simply protonates the alkynide to give the terminal alkyne: C = Ph-C$\equiv$CH (phenylacetylene).

Hydroboration of the alkyne with B${}_2$H${}_6$ followed by oxidation with alkaline H${}_2$O${}_2$ adds "H-OH" across the triple bond with anti-Markovnikov regiochemistry, initially giving an enol, Ph-CH=CH-OH, which immediately tautomerises to the more stable carbonyl form: D = Ph-CH${}_2$-CHO (2-phenylacetaldehyde).

Counting sp${}^2$ carbons in D: the six aromatic ring carbons of the phenyl group are all sp${}^2$, and the aldehyde carbon (C=O) is also sp${}^2$, while the -CH${}_2$- carbon is sp${}^3$. This gives $6+1=7$ sp${}^2$ hybridised carbons.