Step 1: Anti-Markovnikov Addition of HBr
- Due to the presence of benzoyl peroxide, the reaction follows the peroxide effect (Kharasch effect).
- This means Br attaches to the less substituted carbon, forming 1-bromopropane.
\[
\text{CH}_3\text{CH}=\text{CH}_2 + \text{HBr} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{Br}
\]
Step 2: Wurtz Coupling Reaction
- In the presence of Na and dry ether, two molecules of 1-bromopropane couple to form n-hexane.
\[
2\text{CH}_3\text{CH}_2\text{CH}_2\text{Br} + 2\text{Na} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{-CH}_2\text{CH}_2\text{CH}_3 + 2\text{NaBr}
\]
- This reaction confirms that \( Y \) is n-hexane.
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Approach Solution -2
Given the reaction sequence:
1. Propene (\( C_3H_6 \)) reacts with HBr in the presence of benzoyl peroxide \((C_6H_5CO)_2O_2\).
- This is a free radical addition reaction where HBr adds to the double bond of propene following anti-Markovnikov's rule due to the peroxide effect.
- The product \( X \) is 1-bromopropane (CH₂Br–CH₂–CH₃).
2. The compound \( X \) (1-bromopropane) then reacts with sodium (Na) in dry ether.
- This is a Wurtz reaction where two alkyl halides couple to form a higher alkane or substituted alkane.
- Two molecules of 1-bromopropane couple to form hexane, but here since one reactant is bromobenzene, the reaction yields an alkylbenzene.
However, the given product \( Y \) is shown as an ethylbenzene derivative (phenylpropane).
Hence, the second step is the Wurtz coupling of bromobenzene with 1-bromopropane to give propylbenzene.
Summarizing:
- \( X \) = 1-bromopropane (from anti-Markovnikov addition of HBr to propene)
- \( Y \) = propylbenzene (from Wurtz coupling of bromobenzene with 1-bromopropane)
Therefore,
\[
X = \text{1-bromopropane}, \quad Y = \text{propylbenzene}
\]
