See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Bromine addition to alkenes proceeds via anti addition through a bromonium ion intermediate, and the stereochemical outcome depends on the geometry of the starting alkene. Step 1 - Draw cis-2-butene: In cis-2-butene, both methyl groups are on the same side of the double bond. The carbons involved are C2 and C3, each bearing one methyl and one hydrogen. Step 2 - Bromonium ion formation: Br2 attacks the pi bond from one face, forming a cyclic bromonium ion. The two faces of cis-2-butene are enantiotopic, so attack occurs with equal probability from either face. Step 3 - Nucleophilic opening by Br-: The bromide ion attacks anti to the bromonium ion. Because the double bond is cis, anti addition places the two bromine atoms on opposite faces relative to the carbon skeleton. Step 4 - Determine the product stereochemistry: Anti addition to cis-2-butene gives (2R,3S)-2,3-dibromobutane and (2S,3R)-2,3-dibromobutane in equal amounts. These two structures are non-superimposable mirror images (enantiomers), so the product is a racemic mixture. Step 5 - Check why other options fail: - (a) Achiral: incorrect; the individual enantiomers are chiral. - (c) Meso: incorrect; meso would result from anti addition to trans-2-butene, not cis-2-butene. - (d) Optically active: incorrect; because equal amounts of both enantiomers are formed, the mixture is racemic and shows no net optical rotation. Therefore, the correct answer is B.