Why does HBr/H₂O₂ form the "anti-Markovnikov" product?

• A. It forms the "anti-Markovnikov" product because the carbocation forms on the least-substituted carbon atom.
• B. It forms the "anti-Markovnikov" product because the bromine radical adds first to the least-substituted carbon atom.
• C. It forms the "anti-Markovnikov" product because the hydrogen radical adds first to the more-substituted carbon atom.
• D. It forms the "anti-Markovnikov" product because the carbon radical forms on the least-substituted carbon atom.

Answer: (B)

Verified step by step guidance

1. Understand the concept of Markovnikov's rule: In typical electrophilic addition reactions, the hydrogen atom adds to the less substituted carbon atom, while the halogen adds to the more substituted carbon atom.
2. Recognize the role of HBr and H2O2: In the presence of peroxides (H2O2), the addition of HBr to alkenes follows an 'anti-Markovnikov' pattern, which is different from the usual Markovnikov addition.
3. Identify the mechanism: The presence of H2O2 leads to the formation of bromine radicals. These radicals initiate a radical chain reaction rather than the typical carbocation mechanism.
4. Explain the radical addition: The bromine radical adds first to the least-substituted carbon atom of the alkene, forming a more stable radical intermediate on the more substituted carbon.
5. Conclude the process: The hydrogen radical then adds to the radical intermediate, completing the addition and resulting in the 'anti-Markovnikov' product where bromine is attached to the less substituted carbon.