See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Concept: Radical stability increases with (1) greater substitution (more alkyl groups stabilize via hyperconjugation/induction) and (2) resonance delocalization (allylic radicals are stabilized by conjugation with a π system). Step 1 – Identify the nature of each radical: (a) Secondary cyclopentyl radical — the radical is on a ring CH carbon (secondary, no resonance). (b) Tertiary cyclopentyl radical — the radical is on a ring carbon bearing one methyl group and two ring carbons (tertiary, no resonance). (c) The radical is on the carbon that sits between the cyclopentane ring and the vinyl group (CH2=CH–). This carbon bears the ring (one substituent), the vinyl group, and a hydrogen, making it a secondary allylic radical — stabilized by resonance with the vinyl π bond. (d) The radical is directly on the cyclopentane ring carbon that bears the vinyl substituent (CH2=CH–). This is a tertiary allylic radical — the ring carbon has two ring C–C bonds plus the vinyl group, so it is tertiary AND allylic, providing both substitution and resonance stabilization. Step 2 – Rank by stability: • (d) Tertiary allylic radical: MOST stable (rank 1) — resonance + tertiary substitution. • (c) Secondary allylic radical: second most stable (rank 2) — resonance but only secondary. • (b) Tertiary non-allylic radical: third most stable (rank 3) — no resonance but tertiary substitution. • (a) Secondary non-allylic radical: least stable (rank 4) — no resonance, only secondary. Step 3 – Why other orderings fail: • (b) cannot outrank (c) because allylic resonance (delocalization over two carbons via the π system) provides more stabilization than the incremental gain of going from secondary to tertiary without resonance. • (a) is clearly the least stable: secondary with no resonance. Therefore, the correct answer is {"a": 4, "b": 3, "c": 2, "d": 1}.