See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Hydroboration-oxidation (BH3 then H2O2/OH-) proceeds via syn addition of H and OH across the double bond. Step 1: Identify the substrate. trans-2-butene has the two methyl groups on opposite sides of the double bond (E configuration). The two carbons of the double bond (C2 and C3) are equivalent. Step 2: Determine the stereooutcome of syn addition. Syn addition means both H and OH add to the same face of the double bond simultaneously. When BH3 adds to trans-2-butene, it can add from the top face or the bottom face with equal probability (both faces are enantiotopic). Step 3: Analyze the products. - Addition from the top face of trans-2-butene gives (2R,3R)-2,3-butanediol... wait, let us reconsider. In trans-2-butene, the two methyl groups are trans. Syn addition from one face gives one enantiomer of 2,3-butanediol, and syn addition from the other face gives the other enantiomer. - Specifically, syn addition to trans-2-butene produces (2R,3R)-butane-2,3-diol from one face and (2S,3S)-butane-2,3-diol from the other face, each with equal probability. Step 4: Determine the nature of the product mixture. The (2R,3R) and (2S,3S) enantiomers are produced in equal amounts, giving a racemic mixture. Neither enantiomer has an internal plane of symmetry (the meso form would be (2R,3S)), so the product is not meso. Step 5: Evaluate options. (a) Achiral - incorrect; the individual products are chiral. (b) Racemic - correct; equal amounts of (2R,3R) and (2S,3S) enantiomers are formed. (c) Meso - incorrect; meso-butane-2,3-diol (2R,3S) would result from anti addition to trans-2-butene, not syn addition. (d) Optically active - incorrect; the racemic mixture is optically inactive overall. Therefore, the correct answer is B.