See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Electrophilic addition of bromine (Br2) to alkenes proceeds via anti addition (trans addition) through a bromonium ion intermediate. The stereochemical outcome depends on the geometry of the starting alkene. Step 1: Identify the starting material. (E)-2-butenedioic acid is fumaric acid, which has the two carboxylic acid groups on opposite sides of the double bond (trans configuration). Step 2: Mechanism of bromination. Br2 adds to the double bond via a cyclic bromonium ion intermediate. The nucleophilic bromide ion then attacks from the back side (anti) relative to the bromonium ion. Step 3: Anti addition to fumaric acid. In fumaric acid (E-configuration), the two COOH groups are trans to each other. Anti addition of Br2 means the two bromine atoms add to opposite faces of the double bond. Step 4: Determine the product stereochemistry. When anti addition occurs on fumaric acid: - The bromonium ion forms on one face. - Br- attacks C3 from the opposite face. - Due to the trans geometry of fumaric acid, anti addition produces a single stereoisomer where the two Br atoms and the two COOH groups have specific spatial relationships. - The product is the meso compound: (2R, 3S)-2,3-dibromosuccinic acid. This is because the molecule has an internal plane of symmetry, making C2 R and C3 S (or equivalently C2 S and C3 R, which is the same meso compound). Step 5: Why other options fail. - Option (b) (2R,3R) and option (d) (2S,3S) are enantiomers of each other (the chiral pair), which would result from syn addition, not anti addition. - Option (c) a racemic mixture of (2R,3R) and (2S,3S) would also result from syn addition to fumaric acid, not anti addition. - Anti addition to fumaric acid specifically gives the meso compound (2R,3S), confirming option (a). Therefore, the correct answer is A.