See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: The Wurtz reaction involves treating an alkyl halide with sodium metal in dry ether to couple two alkyl groups and form a higher alkane. The reaction is: 2 R-X + 2 Na → R-R + 2 NaX. The Wurtz reaction works best when both alkyl halide molecules are identical (symmetric coupling), because using two different alkyl halides gives a mixture of three products (R-R, R-R', and R'-R'). Step 1 - Analyze each option: (a) Iso-butane (2-methylpropane): To form iso-butane by Wurtz reaction, one would need to couple two identical alkyl groups. Iso-butane has a branched structure (CH3-CH(CH3)-CH3), and attempting to form it via Wurtz reaction would require coupling of a 1-carbon and a 3-carbon fragment from different halides, giving a mixture. It cannot be cleanly prepared by Wurtz reaction. (b) n-Butane (CH3-CH2-CH2-CH3): n-Butane can be prepared by reacting two molecules of ethyl halide (CH3CH2X) with sodium. Since both halides are identical (ethyl halide), the Wurtz reaction gives a single clean product: CH3CH2-CH2CH3 = n-butane. This is the ideal symmetric Wurtz reaction. (c) n-Pentane (CH3-CH2-CH2-CH2-CH3): n-Pentane has 5 carbons. To form it symmetrically via Wurtz reaction, one would need two identical fragments of 2.5 carbons each, which is not possible with whole alkyl halides. Using methyl halide + butyl halide or ethyl halide + propyl halide would give mixtures. So n-pentane cannot be best prepared by Wurtz reaction. (d) Iso-pentane (2-methylbutane): Iso-pentane has a branched structure and cannot be prepared cleanly from a single type of alkyl halide via Wurtz reaction. Step 2 - Conclusion: n-Butane is uniquely suited for Wurtz reaction synthesis because it results from symmetric coupling of two ethyl halide molecules (2 CH3CH2Br + 2 Na → CH3CH2CH2CH3 + 2 NaBr), giving a single pure product with no mixtures. Why other options fail: Iso-butane, n-pentane, and iso-pentane all require either asymmetric coupling (different alkyl halides) or impossible half-carbon fragments, leading to product mixtures and poor yields. Therefore, the correct answer is B.