See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Acid-catalyzed hydration of an alkene (Markovnikov addition of water) and stereochemical outcome. Step 1 - Identify the alkene: The starting material is but-2-ene (CH3-CH=CH-CH3). The question shows the product as CH3-CH2-CH(OH)-CH3, which is butan-2-ol. This product has one chiral center at C2 (the carbon bearing the OH group). Step 2 - Mechanism of acid-catalyzed hydration: H3O+ protonates the double bond to give a secondary carbocation at C2 (or C3, which are equivalent here due to symmetry of but-2-ene). Water then attacks the carbocation. The carbocation intermediate (sp2, planar) is achiral, and water can attack from either face with equal probability. Step 3 - Stereochemical outcome: Because the carbocation is planar and achiral, nucleophilic attack by water occurs equally from both faces, producing equal amounts of (R)-butan-2-ol and (S)-butan-2-ol. This equal mixture of two enantiomers is called a racemic mixture. Step 4 - Why other options fail: - (a) A mixture of diastereomers: Diastereomers require at least two stereocenters; butan-2-ol has only one chiral center, so the two stereoisomers formed are enantiomers, not diastereomers. - (b) Optically active: A racemic mixture is optically inactive because the optical rotations of the two enantiomers cancel each other out. - (c) Optically pure enantiomer: This would require a stereospecific reaction with a chiral reagent or asymmetric induction, which does not occur here. Therefore, the correct answer is D.