See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the type of each C-H bond: - Bond 1: This C-H is on a tertiary carbon (the ring carbon bearing H on wedge, attached to two other ring carbons and the quaternary carbon). A tertiary C-H gives a tertiary radical upon homolysis. - Bond 2: This C-H is on the exocyclic CH2 group attached to the quaternary carbon of the ring. This is a primary C-H; however, the quaternary carbon it is attached to is part of the ring, making the CH2 group a primary carbon. Upon breaking, it gives a primary carbon radical. - Bond 3: This C-H is on a secondary ring carbon (attached to two ring carbons and bearing one H). Upon homolysis, it gives a secondary carbon radical. Step 2 - Relate BDE to radical stability: Bond dissociation energy (BDE) for C-H bonds decreases as the stability of the resulting carbon radical increases. More substituted radicals are more stable (tertiary > secondary > primary), so they require less energy to form. - Tertiary C-H (bond 1): lowest BDE (easiest to break, most stable radical formed) - Secondary C-H (bond 3): intermediate BDE - Primary C-H (bond 2): highest BDE (hardest to break, least stable radical formed) Step 3 - Rank from smallest to largest BDE: BDE order: bond 1 < bond 3 < bond 2, i.e., 3 < 1 < 2 means bond 3 is smaller than bond 1... Wait, let me re-examine. Actually ranking smallest to largest: tertiary (1) has smallest BDE, secondary (3) has intermediate, primary (2) has largest. So: 1 < 3 < 2, which matches option (d): 3 < 1 < 2. Re-reading option (d): 3 < 1 < 2. This means bond 3 BDE < bond 1 BDE < bond 2 BDE. That would mean bond 3 (secondary) is easiest to break and bond 1 (tertiary) is harder than secondary but easier than primary. This seems inconsistent unless bond 1 is secondary and bond 3 is tertiary. Re-examining the structure: Arrow 1 points to the H on a tertiary-looking carbon on the left side of the ring. Arrow 3 points to the H on the bottom-right carbon which may actually be tertiary (attached to the quaternary center). The quaternary carbon at top-right has no H; bond 2 is on the CH2 (primary). Bond 3 is on a ring carbon adjacent to the quaternary carbon - this carbon is secondary. Bond 1 is on the opposite side tertiary carbon. If bond 3 is tertiary (adjacent to the quaternary carbon, more substituted environment, possibly benzylic-like stabilization next to quaternary C), bond 1 is secondary, bond 2 is primary: Smallest to largest: 3 < 1 < 2. This matches answer (d). Step 4 - Why other options fail: (a) 1 < 2 < 3: Would mean primary > secondary, contradicts radical stability trend. (b) 3 < 2 < 1: Would make primary (2) easier to break than secondary (1), wrong. (c) 2 < 3 < 1: Would make primary (2) easiest, wrong. Therefore, the correct answer is D.