Leaving group ability determines rate in SN2 for primary halides

Steps:

• Both are primary alkyl halides, so mechanism is SN2 with nucleophile OH⁻.
• Aqueous NaOH is polar protic but favors SN2 for primaries due to backside attack preference.
• Leaving group quality: I⁻ > Cl⁻, lowering activation energy for C–X bond break in SN2.
• Thus, 1-iodobutane reacts faster under identical conditions.

Why B is correct:

• Iodide's weaker C–I bond and better leaving group stabilize transition state in SN2, increasing rate constant per rate law: rate = k[RX][OH⁻].

Why the others are wrong:

• A: Chlorine's poor leaving group makes SN1 negligible for primary; rate too slow.
• C: Chlorobutane follows SN2 but slower due to worse leaving group than iodide.
• D: Primaries rarely follow SN1 without stabilization; iodide still prefers SN2 in this setup.

Final answer: B