See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Addition of HBr to alkenes in the presence of peroxide proceeds via a free radical mechanism (anti-Markovnikov addition). The key is to determine the stereochemical outcome when HBr adds to cis-2-butene under radical conditions. Step 1: Identify the substrate. cis-2-butene has the two methyl groups on the same side of the double bond. The carbons of the double bond are C2 and C3. Step 2: Mechanism under peroxide conditions. Peroxides initiate free radical chain addition. A bromine radical adds to one carbon of the double bond (say C2 or C3), generating a secondary carbon radical at the adjacent carbon. The radical intermediate at C3 (or C2) is planar (sp2-like), meaning bromine can attack from either face with roughly equal probability. Step 3: Stereochemical analysis. Starting from cis-2-butene: - When Br• adds to C2, a radical at C3 is formed. This radical center is planar, so the second H (from HBr) can add to either face of C3. - Because the radical intermediate is planar at C3, the two possible products are (2R,3R)-2-bromobutane and (2S,3S)-2-bromobutane in equal amounts — OR (2R,3S) and (2S,3R) depending on the geometry. Step 4: Detailed stereo outcome. For cis-2-butene, after radical addition of Br to one face, the resulting radical at C3 is planar. The two methyl groups originally on the same side (cis). Attack on the radical from either face gives equal amounts of (2R,3R) and (2S,3S) products (the enantiomeric pair), which constitutes a racemic mixture. Step 5: Why other options fail. - (b) Diastereomer: Diastereomers would be a mixture of (2R,3S) and (2R,3R), which is not what is produced as the dominant outcome here. - (c) Meso: The meso compound (2R,3S)-2-bromobutane would require equal and opposite configurations at C2 and C3; while meso could form from trans-2-butene addition, for cis-2-butene radical addition the dominant product pair is the racemic (±) enantiomers. - (d) E and Z isomer: HBr addition is an addition reaction, not producing geometric isomers of a double bond product. The radical mechanism produces a planar radical intermediate, leading to equal amounts of (2R,3R) and (2S,3S)-2-bromobutane — a racemic mixture. Therefore, the correct answer is A.