See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: This is an SN2 halide exchange reaction (Finkelstein-type) using iodide (I-) as the nucleophile in DMF (a polar aprotic solvent). In polar aprotic solvents like DMF, iodide is a good nucleophile and the reaction proceeds via SN2 mechanism, where iodide displaces chloride. Step 1: Identify the structure of the starting material. The compound is Cl-CH2-C(CH3)-CH2-CH2-Cl, which is 1-chloro-2-(chloromethyl)-2-methylbutane... more precisely, it is a molecule with two C-Cl bonds: one at a primary carbon (neopentyl-type: -CH2-Cl attached to the quaternary-like carbon C(CH3)) and one at a primary carbon at the end of the chain (-CH2-Cl at the terminal position). Step 2: Assess reactivity of each C-Cl bond toward SN2. The terminal primary C-Cl (the -CH2-CH2-Cl end) is a straightforward primary alkyl chloride with no branching adjacent to it, making it more accessible to the nucleophile. The other C-Cl (the -CH2-Cl directly attached to the C bearing CH3) is a neopentyl-type primary chloride, which is severely hindered due to the quaternary carbon (bearing CH3 and two other carbon groups) adjacent to the CH2Cl group. Neopentyl-type systems are extremely unreactive in SN2 reactions. Step 3: Iodide preferentially attacks the less hindered, more reactive primary C-Cl (the terminal -CH2-CH2-Cl), replacing Cl with I. This gives Cl-CH2-C(CH3)-CH2-CH2-I, which is option (b). Step 4: Why other options fail. - Option (a): I replaces the Cl on the neopentyl-type carbon (hindered), which is disfavored in SN2 due to steric hindrance. This is the minor product. - Option (c): Elimination product with two double bonds shown; not consistent with simple SN2 halide exchange under these mild conditions. - Option (d): Elimination product; also not the major pathway under these conditions with iodide as nucleophile in DMF. Therefore, the correct answer is B.