See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the hybridization and carbon type for each C-H bond: - Bond x: The carbon bearing H(x) is a secondary sp3 carbon (it has an ethyl group on one side and a CH2 group on the other, plus two H atoms). This is a secondary (2°) aliphatic C-H bond. - Bond y: The carbon bearing H(y) is also a secondary sp3 carbon (CH2 group flanked by the x-carbon and the vinylic carbon). This is also a 2° aliphatic C-H bond, but it is allylic because it is adjacent to the C=C double bond. Allylic C-H bonds have lower BDE because the resulting radical is stabilized by resonance with the double bond. - Bond z: The carbon bearing H(z) is an sp2 vinylic carbon (part of the C=C double bond). Vinylic C-H bonds have the highest BDE among these because the resulting vinylic radical is in an sp2 orbital, which has higher s-character than sp3, making it less stable (higher energy radical), thus requiring more energy to break. Step 2 - Rank BDE based on radical stability (lower radical stability = higher BDE): - Vinylic radical (from z): least stable radical (sp2 carbon, no resonance stabilization) → highest BDE. - Secondary aliphatic radical (from x): moderate stability → intermediate BDE. - Allylic secondary radical (from y): most stable radical (resonance-stabilized) → lowest BDE. Step 3 - Decreasing order of BDE: z > x > y Step 4 - Why other options fail: - (a) y > x > z: Incorrect; this would imply allylic C-H has the highest BDE, which contradicts the stabilization of the allylic radical. - (c) z > y > x: Incorrect; this would place allylic C-H (y) above secondary C-H (x), but allylic radicals are more stable than secondary radicals, so BDE of y < BDE of x. - (d) y > z > x: Incorrect; placing allylic above vinylic contradicts the high BDE of vinylic bonds. Therefore, the correct answer is B.