Question

Heating of 2-chloro-1-phenyl butane with EtOK/EtOH gives ‘X’ as the major product. The reaction of ‘X’ with HBr gives ‘Y’ as the major product. The ‘Y’ is}

• A. 1-bromo-2-phenyl butane
• B. 1-bromo-1-phenyl butane
• C. 3-bromo-1-phenyl butane
• D. 1-phenyl-1-butene
• E. 2-phenyl but-1-ene

Hint:

Whenever an alkene is attached to a benzene ring, addition reactions often proceed through a benzylic carbocation, which is highly stable.

Answer 1

The Correct Option is B

2-chloro-1-phenyl butane undergoes dehydrohalogenation with alcoholic KOH or EtOK/EtOH. The major alkene formed is the more stable alkene: \[ \text{1-phenyl-1-butene} \] Now HBr adds to this alkene according to Markovnikov’s rule. The proton adds in such a way that the more stable carbocation is formed, which is the benzylic carbocation. Then bromide attacks that carbon. So the major product formed is: \[ \text{1-bromo-1-phenyl butane} \]
Hence, the correct answer is: \[ \boxed{(B)} \]