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 starting material and reagents.
Starting material: n-Pentyl bromide (CH3CH2CH2CH2CH2Br), a 5-carbon primary alkyl halide.
Reaction P uses: Zn/H+
Reaction Q uses: Na/Dry Ether (Wurtz reaction conditions)

Step 2: Analyse Reaction P -- n-Pentyl bromide + Zn/H+.
Zinc in the presence of a proton source (Zn/H+) acts as a reducing agent. The zinc metal donates electrons to the C-Br bond, cleaving it homolytically and replacing the bromine with hydrogen. This is a simple reduction (dehalogenation): \[ \text{CH}_3(\text{CH}_2)_4\text{Br} + \text{Zn/H}^+ \rightarrow \text{CH}_3(\text{CH}_2)_3\text{CH}_3 \] The product is n-pentane (5 carbon atoms). The carbon chain is preserved, only Br is replaced by H.

Step 3: Identify Product P.
Product P = n-Pentane (C5H12).

Step 4: Analyse Reaction Q -- n-Pentyl bromide + Na/Dry Ether (Wurtz Reaction).
The Wurtz reaction couples two alkyl halide molecules using sodium metal in dry ether. Sodium abstracts the halogen from each molecule, and the two resulting alkyl radicals (or carbanions) combine to form a new C-C bond: \[ 2\ \text{CH}_3(\text{CH}_2)_4\text{Br} + 2\text{Na} \rightarrow \text{CH}_3(\text{CH}_2)_8\text{CH}_3 + 2\text{NaBr} \]

Step 5: Identify Product Q.
Two n-pentyl groups (each with 5 carbons) join together to give n-decane (C10H22), a 10-carbon straight-chain alkane. This is a key method for synthesising higher alkanes with an even number of carbons.

Step 6: State the final answer.
Reaction P (Zn/H+, reduction) gives Pentane; Reaction Q (Wurtz reaction, Na/Dry Ether) gives Decane.
\[ \boxed{P = \text{Pentane (C}_5\text{H}_{12}\text{)},\ Q = \text{Decane (C}_{10}\text{H}_{22}\text{)}} \]