Question

Find the total number of possible product(s) in the following reaction

$\xrightarrow[\text{CCl}_4]{\text{HBr}}$ Product(s)

Answer 1

2 (two enantiomeric products)

Answer 2

The starting alkene is a substituted cyclopentene having a CH₃ group on the sp² carbon. Under HBr addition (in CCl₄), the proton adds according to Markovnikov’s rule so that the more stable carbocation is generated on the carbon bearing the CH₃ group. Since the intermediate carbocation is planar, the bromide ion can attack from either face, giving rise to two enantiomers.

Core Explanation:

1. Mechanism: HBr adds to the double bond via a Markovnikov addition pathway.
2. Intermediate: A planar carbocation is formed at the substituted carbon.
3. Stereochemistry: Br⁻ attacks from above or below the plane, resulting in a racemic mixture of two enantiomers.