See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: In free radical chlorination, the reactivity of C–H bonds depends on two factors: (1) stability of the radical formed at that carbon, and (2) inductive/electronic effects of substituents. Fluorine is a strongly electron-withdrawing group (EWG) by induction, which destabilizes adjacent radicals by withdrawing electron density. Step 1: Identify the positions. - Position a: CH3 (terminal methyl, primary C–H, 3 H's, farthest from F) - Position b: CH2 adjacent to CH3 (internal primary/secondary, 2 H's, beta to F... actually this is the second carbon) - Position c: CH2 next to CH2F (secondary, 2 H's, beta to F) - Position d: CH2 directly attached to F (2 H's, alpha to F) Step 2: Assess radical stability at each position. - All interior carbons (b and c) are secondary carbons, so they form secondary radicals, which are more stable than primary radicals (at a or d). - Position d (alpha to F): The fluorine directly attached to the adjacent carbon strongly destabilizes the radical at d through inductive electron withdrawal, making d the least reactive. - Position a (terminal CH3, primary): forms a primary radical, less stable than secondary but not destabilized by F (F is four carbons away, negligible inductive effect). - Position c (beta to F): secondary carbon, but mildly destabilized by the inductive effect of F (two carbons away). - Position b (gamma to F): secondary carbon, very little inductive destabilization from F (three carbons away). Step 3: Rank reactivities. - b (secondary, gamma to F, least inductive destabilization among secondary): highest reactivity - c (secondary, beta to F, slightly more destabilized than b): second - a (primary, far from F, minimal inductive effect): third - d (alpha to F, strongly destabilized by F): lowest reactivity Order: b > c > a > d Why other options fail: - (a) a > b > c > d: Ignores that b and c are secondary carbons and more reactive than primary a. - (b) b > c > d > a: Places d above a, but d is alpha to F and strongly deactivated, so d should be least reactive. - (d) c > b > a > d: Reverses b and c; b is farther from F and thus less destabilized, making b more reactive than c. Therefore, the correct answer is C.