See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: NaNH2 is a strong base used to eliminate HX from vicinal or geminal dihalides to form alkynes. The sequence 3NaNH2 / mineral oil / heat drives double dehydrohalogenation to give a terminal or internal alkyne, and NH4Cl workup protonates the sodium acetylide to give the free alkyne. Step-by-step analysis of each option: Option (a): C6H5 — CBr2 — CH3 (a geminal dibromo compound, 1,1-dibromo-1-phenylethane) - Treatment with 3 equivalents NaNH2 causes double dehydrohalogenation of the geminal dibromide. - First equivalent of NaNH2 eliminates one HBr to give C6H5 — CBr = CH2 (a vinyl bromide). - Second equivalent of NaNH2 eliminates another HBr to give C6H5 — C ≡ CH (phenylacetylene) as the sodium acetylide, C6H5 — C ≡ C^- Na^+. - Third equivalent ensures complete reaction / drives the equilibrium. - NH4Cl workup protonates the acetylide: C6H5 — C ≡ C^- Na^+ + NH4Cl → C6H5 — C ≡ CH + NaCl + NH3. - Final product: Ph — C ≡ CH ✓ Option (b): C6H5CH = CH2 (styrene) - Treatment with Br2/CCl4 gives 1,2-dibromide: C6H5 — CHBr — CH2Br (vicinal dibromide). - Treatment with 3NaNH2 / mineral oil / heat causes double dehydrohalogenation of the vicinal dibromide. - First NaNH2: eliminates HBr to give C6H5 — CBr = CH2 or C6H5 — CH = CHBr (vinyl bromide). - Second NaNH2: eliminates another HBr to give C6H5 — C ≡ CH as the sodium acetylide. - NH4Cl workup gives Ph — C ≡ CH ✓ Option (c): C6H5 — C(=O) — CH3 (acetophenone) - Treatment with PCl5 converts the carbonyl to a geminal dichloride: C6H5 — CCl2 — CH3. - Treatment with 3NaNH2 / mineral oil / heat causes double dehydrohalogenation analogous to option (a). - First NaNH2: C6H5 — CCl2 — CH3 → C6H5 — CCl = CH2 + NaCl + NH3. - Second NaNH2: C6H5 — CCl = CH2 → C6H5 — C ≡ C^- Na^+ + NaCl + NH3. - NH4Cl workup gives Ph — C ≡ CH ✓ Since all three reactions (a), (b), and (c) ultimately yield Ph — C ≡ CH as the final product, option (d) All is correct. Therefore, the correct answer is D.