Question:
\(\text{H}_{3}\text{C}-\text{CH}=\text{CH}_{2}+\text{H}-\text{Br}\xrightarrow{(\text{C}_{6}\text{H}_{5}\text{CO})_{2}\text{O}_{2},\ \text{Benzoyl peroxide}}\text{H}_{3}\text{C}-\text{CH}_{2}-\text{CH}_{2}-\text{Br}\)
Which of the following rules is followed when the above reaction occurs?
\(\text{H}_{3}\text{C}-\text{CH}=\text{CH}_{2}+\text{H}-\text{Br}\xrightarrow{(\text{C}_{6}\text{H}_{5}\text{CO})_{2}\text{O}_{2},\ \text{Benzoyl peroxide}}\text{H}_{3}\text{C}-\text{CH}_{2}-\text{CH}_{2}-\text{Br}\)
Which of the following rules is followed when the above reaction occurs?
Which of the following rules is followed when the above reaction occurs?
Show Hint
Benzoyl peroxide makes HBr add through free radicals so that bromine goes to the carbon with more hydrogens.
Updated On: Nov 12, 2026
- Markovnikovs rule
- Anti-Markovnikovs rule
- Saytzeff rule
- Huckel rule
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Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
In the presence of benzoyl peroxide, HBr adds to propene by a free-radical chain mechanism (Kharasch effect).
Step 2: Mechanism
The Br radical attacks the terminal \(\text{CH}_{2}\) carbon and gives the more stable secondary radical \(\text{CH}_{3}\dot{\text{C}}\text{HCH}_{2}\text{Br}\). This radical then takes H from HBr.
Step 3: Result
The product is 1-bromopropane. Bromine has gone to the carbon with more hydrogens, the reverse of Markovnikov's rule.
Final Answer:
The reaction follows the anti-Markovnikov rule. \[ \boxed{\text{(B) }\text{Anti-Markovnikov's rule}} \]
In the presence of benzoyl peroxide, HBr adds to propene by a free-radical chain mechanism (Kharasch effect).
Step 2: Mechanism
The Br radical attacks the terminal \(\text{CH}_{2}\) carbon and gives the more stable secondary radical \(\text{CH}_{3}\dot{\text{C}}\text{HCH}_{2}\text{Br}\). This radical then takes H from HBr.
Step 3: Result
The product is 1-bromopropane. Bromine has gone to the carbon with more hydrogens, the reverse of Markovnikov's rule.
Final Answer:
The reaction follows the anti-Markovnikov rule. \[ \boxed{\text{(B) }\text{Anti-Markovnikov's rule}} \]
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