See image — Alcohols Phenols and Ethers Chemistry Question

💡 Solution & Explanation

Concept: Racemization of chiral alcohols under acidic conditions. Step 1: 2-octanol has a chiral center at C-2 (bearing OH, H, methyl, and a hexyl chain), making it optically active. Step 2: When treated with dilute acid (H+), the hydroxyl group at C-2 gets protonated to form a good leaving group (water). The resulting carbocation at C-2 is a planar, sp2 hybridized species. Step 3: A planar carbocation at a secondary carbon is achiral. Water (or any nucleophile) can attack from either face of the planar carbocation with equal probability, producing a racemic mixture (50% R and 50% S). Step 4: A racemic mixture has no net optical activity because the equal and opposite rotations cancel each other out. Thus, the compound rapidly loses its optical activity. Step 5: Why other options fail: - Dilute base (b): Base does not readily protonate or activate the OH group of a simple alcohol to form a carbocation; racemization under base is not a rapid, straightforward process for a simple secondary alcohol. - Light (c): Photochemical racemization is not a typical or rapid pathway for simple aliphatic alcohols like 2-octanol. - Humidity (d): Moisture alone does not cause racemization of a simple chiral alcohol. Therefore, the correct answer is A.