Question:
The reaction of propene with HBr in presence of peroxide proceeds through the intermediate
The reaction of propene with HBr in presence of peroxide proceeds through the intermediate
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Peroxide effect → Anti-Markovnikov addition → radical mechanism → primary radical intermediate.
Updated On: May 1, 2026
- H$_3$C–$\dot{C}$H–CH$_3$
- H$_3$C–$\dot{C}$H–CH$_2$Br
- H$_3$C–CHBr–$\dot{C}$H$_2$
- H$_3$C–CH$_2$–$\dot{C}$H$_2$
- None of the above
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The Correct Option is D
Solution and Explanation
Concept:
In presence of peroxide, HBr adds via free radical mechanism (Anti-Markovnikov addition).
Step 1: Initiation
Peroxide generates free radicals.
Step 2: Propagation step
Br$\cdot$ attacks propene: \[ CH_3-CH=CH_2 \rightarrow CH_3-CH_2-\dot{C}H_2 \] Primary radical is formed because it leads to anti-Markovnikov product.
Step 3: Radical reacts with HBr
\[ CH_3-CH_2-\dot{C}H_2 + HBr \rightarrow CH_3-CH_2-CH_2Br \]
Step 4: Identification of intermediate
The intermediate radical is: \[ CH_3-CH_2-\dot{C}H_2 \] Final Conclusion: \[ \boxed{(D)} \]
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