See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Free radical halogenation of butane (CH3-CH2-CH2-CH3) with Br2 under hv (light) involves abstraction of a hydrogen atom to generate a radical intermediate, followed by bromine capture. Step 1: Identify the sites of bromination. Butane has two types of hydrogens: primary (C1 and C4, 6H each) and secondary (C2 and C3, 4H each). Bromine radical is highly selective and preferentially abstracts secondary hydrogens. Thus, bromination predominantly occurs at C2 (or equivalently C3 by symmetry), giving 2-bromobutane as the major product. Step 2: Analyze the stereochemistry of the major product. 2-Bromobutane (CH3-CHBr-CH2-CH3) has one chiral center at C2. Free radical bromination is a non-stereospecific process — the planar radical intermediate at C2 can be attacked by Br2 from either face with equal probability. Step 3: Since attack occurs equally from both faces of the planar radical, the R and S enantiomers of 2-bromobutane are produced in equal amounts, giving a racemic mixture (50% R + 50% S). Step 4: Why other options fail: - (b) Meso: A meso compound requires at least two chiral centers with internal symmetry. 2-Bromobutane has only one chiral center, so meso is not applicable. - (c) Diastereomers: Diastereomers require at least two stereocenters with different configurations; not applicable here. - (d) Constitutional isomers: These would differ in connectivity, not stereochemistry; not the situation here since the major product is specifically 2-bromobutane. Therefore, the correct answer is A.