Salt formation yields highest molar mass product

Steps:

• Ethanoic acid is CH₃COOH (Mᵣ=60).
• React with each reagent to identify main organic product.
• Compute Mᵣ for each: A gives CH₃CH₂OH (46); B gives (CH₃COO)₂Mg (142); C gives CH₃COOK (98); D gives CH₃COOCH(CH₃)₂ (102).
• Select highest Mᵣ product.

Why B is correct:

• Magnesium hydride (MgH₂) reacts as: 2CH₃COOH + MgH₂ → (CH₃COO)₂Mg + 2H₂, forming magnesium ethanoate with Mᵣ=142 (C₄H₆MgO₄).

Why the others are wrong:

• A: LiAlH₄ reduces acid to ethanol (CH₃CH₂OH, Mᵣ=46).
• C: K₂CO₃ neutralizes to potassium ethanoate (CH₃COOK, Mᵣ=98).
• D: Propan-2-ol esterifies to isopropyl ethanoate (Mᵣ=102).

Final answer: B