Acetate ion acts as base in E2 elimination

Steps:

• Identify reaction: Hex-2-ene forms via elimination of HI from 2-iodohexane using CH3COO⁻.
• Determine mechanism: E2 involves base abstracting β-proton, leading to C-I heterolysis and alkene formation.
• Classify CH3COO⁻ role: It accepts H⁺ from β-carbon, fitting Bronsted-Lowry base definition.
• Evaluate options: Match roles to mechanism details.

Why A is correct:

• Bronsted-Lowry base accepts proton; CH3COO⁻ + H⁺ → CH3COOH in E2 step.

Why the others are wrong:

• B: No change in oxidation state; elimination preserves carbon skeleton without oxidation.
• C: E2 breaks C-I heterolytically (electrons to I⁻), not homolytically (radical).
• D: No water involvement; it's elimination with acetate, not hydrolysis.

Final answer: A