In the following molecule, which of the labeled hydrogens ($A$, $B$, $C$, or $D$) is most easily abstracted in a free radical bromination reaction? (Assume the hydrogens are: $A$ = methyl hydrogen, $B$ = primary hydrogen, $C$ = secondary hydrogen, $D$ = tertiary hydrogen.)

• A. Hydrogen $D$ (tertiary hydrogen)
• B. Hydrogen $B$ (primary hydrogen)
• C. Hydrogen $A$ (methyl hydrogen)
• D. Hydrogen $C$ (secondary hydrogen)

Answer: (A)

Verified step by step guidance

1. Understand that in free radical bromination, the rate-determining step is the abstraction of a hydrogen atom to form a carbon radical. The stability of this radical greatly influences which hydrogen is most easily abstracted.
2. Recall the order of radical stability: tertiary (3°) radicals are more stable than secondary (2°), which are more stable than primary (1°), and methyl radicals are the least stable. This is due to the inductive and hyperconjugation effects that stabilize the radical center.
3. Identify the types of hydrogens given: A is methyl (attached to a carbon with no other carbons), B is primary (attached to a carbon bonded to one other carbon), C is secondary (attached to a carbon bonded to two other carbons), and D is tertiary (attached to a carbon bonded to three other carbons).
4. Since the bromine radical will preferentially abstract the hydrogen that leads to the most stable radical intermediate, compare the radical stabilities formed by removing each hydrogen: the tertiary radical (from D) is the most stable.
5. Conclude that hydrogen D (tertiary hydrogen) is the most easily abstracted in free radical bromination because it forms the most stable radical intermediate.