Density indicates C₆H₁₂; terminal alkene yields single carboxylic acid via cleavage Steps:

• Molar mass M ≈ density × molar volume; at ~25°C, volume ≈24.4 dm³ mol⁻¹, so M = 3.50 × 24.4 ≈ 85 g mol⁻¹ (matches C₆H₁₂ at 84 g mol⁻¹).
• A (but-2-ene, C₄H₈) and B (2-methylbut-2-ene, C₅H₁₀) have lower M (56, 70), so densities ~2.3 and ~2.9 g dm⁻³; eliminated.
• Hot conc. acidified KMnO₄ cleaves alkene C=C: =CH₂ to CO₂, =CH-R to RCOOH, >C= to ketone.
• Hex-1-ene (D) is CH₂=CH(CH₂)₃CH₃; cleaves to CH₃(CH₂)₃COOH (one pentanoic acid) + CO₂. Why D is correct:
• Terminal =CH₂ oxidizes to CO₂ (not carboxylic), leaving single RCOOH, so only one type of carboxylic acid forms. Why the others are wrong:
• A: Density too low (M=56 g mol⁻¹).
• B: Density too low (M=70 g mol⁻¹); also yields ketone + acetic acid.
• C: Symmetric internal alkene yields two propanoic acid molecules.

Final answer: D