See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: The Wurtz reaction couples two alkyl halides using sodium metal to form an alkane. When two different alkyl halides are used, three possible coupling products can form: same+same for each halide, and cross-coupling between the two. Step 1: Identify the alkyl halides. - Ethyl iodide (C2H5I) provides ethyl groups (C2). - n-Propyl iodide (C3H7I) provides n-propyl groups (C3). Step 2: Enumerate all possible Wurtz coupling products. - Ethyl + Ethyl → butane (C4H10) ✓ - n-Propyl + n-Propyl → hexane (C6H14) ✓ - Ethyl + n-Propyl → pentane (C5H12) ✓ Step 3: Check which option is NOT obtainable. - Butane (C4): formed from two ethyl groups → obtained. - Hexane (C6): formed from two n-propyl groups → obtained. - Pentane (C5): formed from cross-coupling of ethyl and n-propyl → obtained. - Propane (C3): would require coupling of a methyl group and an ethyl group, or two groups totaling 3 carbons — neither ethyl iodide nor n-propyl iodide can produce propane via Wurtz coupling. There is no combination of C2+C2, C3+C3, or C2+C3 that gives C3. Step 4: Therefore propane cannot be obtained from this reaction. Therefore, the correct answer is B.