Step 1: Understanding the Concept:
When an unsymmetrical alkene (like propene, $\text{CH}_3\text{CH}=\text{CH}_2$) reacts with hydrogen bromide ($\text{HBr}$), the outcome depends entirely on the presence or absence of organic peroxides.
- In the absence of peroxides, the reaction proceeds via an electrophilic addition pathway governed by Markovnikov's rule, where the halogen adds to the more substituted carbon.
- In the presence of peroxides, the reaction switches mechanism entirely to a free-radical addition pathway. This is known as the Peroxide Effect or Kharasch Effect, which dictates an Anti-Markovnikov orientation.
Step 2: Detailed Explanation:
Let's trace the free-radical mechanism to see why the product forms this way:
1. Initiation: The organic peroxide ($\text{R-O-O-R}$) splits thermally or photolytically into alkoxy radicals ($\text{R-O}^\bullet$), which then abstract a hydrogen atom from $\text{HBr}$ to generate a reactive bromine radical ($\text{Br}^\bullet$).
2. Propagation Step 1 (Addition to Alkene): The bromine radical ($\text{Br}^\bullet$) attacks the double bond of propene ($\text{CH}_3\text{CH}=\text{CH}_2$). It can attach to either of the two carbon atoms:
- If $\text{Br}^\bullet$ adds to the central carbon ($\text{C}_2$), it creates a less stable primary carbon radical ($^\bullet\text{CH}_2\text{CHBrCH}_3$).
- If $\text{Br}^\bullet$ adds to the terminal carbon ($\text{C}_1$), it creates a highly stable secondary carbon radical ($\text{CH}_3\text{CH}^\bullet-\text{CH}_2\text{Br}$).
Because the secondary free radical is stabilized by hyperconjugation and inductive effects, it forms preferentially as the primary reaction intermediate pathway.
3. Propagation Step 2: The stable secondary carbon radical ($\text{CH}_3\text{CH}^\bullet-\text{CH}_2\text{Br}$) quickly abstracts a hydrogen atom from another $\text{HBr}$ molecule, yielding the final product: 1-Bromopropane ($\text{CH}_3\text{CH}_2\text{CH}_2\text{Br}$).
This yields the Anti-Markovnikov product, where bromine links to the terminal carbon atom containing more hydrogen atoms. This matches option (B).
Step 3: Final Answer:
The major product formed is $\text{CH}_3\text{CH}_2\text{CH}_2\text{Br}$.