Question

The major products \(P\) and \(Q\) of the following reactions respectively are: \[ \text{I.}\quad n\text{-Pentyl bromide}\xrightarrow{Zn/H^+}P \] \[ \text{II.}\quad n\text{-Pentyl bromide}\xrightarrow[\text{Dry Ether}]{Na}Q \]

• A. Pentane ; Decane
• B. Pent-1-ene ; Decane
• C. Pentane ; Pentane
• D. Pentane ; 1-Decene

Hint:

Alkyl halides on reduction with \(Zn/H^+\) give alkanes, while alkyl halides with sodium in dry ether undergo Wurtz reaction to form higher alkanes.

Answer 1

The Correct Option is A

Step 1: Identify the first reaction.
In the first reaction, \(n\)-pentyl bromide reacts with \[ Zn/H^+ \] This reagent reduces alkyl halides to alkanes.
The bromine atom is replaced by hydrogen.
Thus, \[ CH_3CH_2CH_2CH_2CH_2Br \xrightarrow{Zn/H^+} CH_3CH_2CH_2CH_2CH_3 \] Therefore, \[ P=\text{Pentane} \]

Step 2: Identify the second reaction.
In the second reaction, \(n\)-pentyl bromide reacts with sodium in dry ether.
This is Wurtz reaction.
In Wurtz reaction, two alkyl halide molecules couple together in the presence of sodium and dry ether.
The general reaction is \[ 2R-X+2Na \rightarrow R-R+2NaX \]

Step 3: Apply Wurtz reaction to \(n\)-pentyl bromide.
Here, \[ R=C_5H_{11} \] So, \[ 2C_5H_{11}Br+2Na \rightarrow C_{10}H_{22}+2NaBr \] The product \[ C_{10}H_{22} \] is decane.
Therefore, \[ Q=\text{Decane} \]

Step 4: Final conclusion.
Hence, \[ P=\text{Pentane} \] and \[ Q=\text{Decane} \] Therefore, the correct answer is \[ \boxed{\text{Pentane ; Decane}} \] Hence, the correct option is \[ \boxed{(1)} \]