Question

Major product of this reaction is

• A.
• B.
• C.
• D. No reaction

Answer 1

Answer: (B)

Option (b)

Answer 2

1. The substrate is Ph–CH=CHCH₃.

2. On treatment with HBr, the proton adds in a Markovnikov fashion.

3. The proton adds to the terminal carbon so that the carbocation forms at the benzylic position (stabilized by resonance).

4. Bromide then attacks the benzylic carbocation giving Ph–CH(Br)–CH₂CH₃.

5. This corresponds to option (b).