See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: In radical halogenation with Br•, bromine radical is highly selective. It preferentially abstracts the hydrogen that gives the most stable (lowest energy) carbon radical. The stability order of radicals is: tertiary > secondary > primary, and radicals are further stabilized by adjacent heteroatom lone pairs (alpha to oxygen) or other stabilizing groups. Step 1: Identify each hydrogen and the radical that would be formed upon abstraction. - H^a: This hydrogen is on a carbon that also bears a CH3 group. Looking at the structure, H^a is on a carbon at the top of the ring that has a methyl group attached. The carbon bearing H^a appears to be a tertiary carbon (part of ring + methyl substituent + two ring carbons), giving a tertiary radical upon abstraction. - H^b: This hydrogen is on the carbon bearing the OH group (alpha to oxygen). Abstraction gives a radical alpha to -OH, which is stabilized by the adjacent oxygen lone pairs. This carbon is also a secondary/tertiary ring carbon alpha to OH. - H^c: This hydrogen is on an exocyclic CH2 group (primary carbon), giving a primary radical - least stable. - H^d: This hydrogen is on a secondary ring carbon, giving a secondary radical. Step 2: Determine which position gives the most stable radical. H^a is on a carbon that is tertiary (connected to: the ring carbon on one side, the ring carbon on the other side, and the CH3 group) - this gives a tertiary carbon radical. H^b is alpha to OH (secondary or tertiary ring carbon alpha to oxygen) - stabilized but likely secondary. H^a gives a tertiary radical, which is more stable than the secondary radical from H^d, more stable than the primary radical from H^c. Step 3: Compare H^a (tertiary radical) vs H^b (alpha to OH, secondary radical). A tertiary radical is generally more stable than a secondary radical alpha to oxygen in most cases, and Br• being selective will favor the most stable radical. The carbon bearing H^a has three carbon substituents (two ring carbons + one CH3), making the resulting radical tertiary. Step 4: Why other options fail. - H^c gives primary radical: least stable, least favored. - H^d gives secondary radical: moderately stable, but less than tertiary. - H^b gives a radical alpha to -OH (secondary), stabilized by oxygen, but tertiary carbon radical from H^a abstraction is more stable overall. Therefore, Br• most readily abstracts H^a to give the most stable tertiary radical. Therefore, the correct answer is A.