Question

\( CH_3-CH_2-CH=CH_2 + HBr \xrightarrow{Peroxide} \) product. In this reaction, the main product is —

• A. \( CH_3-CH_2-CH_2-CH_2-Br \)
• B. \( CH_3-CH_2-CHBr-CH_3 \)
• C. Both are main products
• D. None of these

Hint:

Peroxide means anti-Markovnikov addition; bromine attaches to the terminal carbon.

Answer 1

The Correct Option is A

Step 1: When HBr adds to an unsymmetrical alkene in the presence of a peroxide, the addition follows the anti-Markovnikov rule. This is called the peroxide effect or Kharasch effect.
Step 2: The reaction goes by a free-radical mechanism. The bromine radical \( Br^{\bullet} \) adds first, and it adds to the double-bond carbon that gives the more stable free radical.
Step 3: In \( CH_3-CH_2-CH=CH_2 \), \( Br^{\bullet} \) adds to the terminal \( CH_2 \) carbon, generating a more stable secondary carbon radical. Hydrogen then adds to the terminal carbon.
Step 4: So bromine ends up on the terminal (less substituted) carbon, giving \( CH_3-CH_2-CH_2-CH_2-Br \) (1-bromobutane) as the main product.
Correct answer: (i). Option (ii) would be the Markovnikov product formed without peroxide.