See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: The Wurtz reaction involves treating an alkyl halide with sodium metal in dry ether to produce an alkane. When CH3Br (methyl bromide) is used, the reaction can produce both methane and ethane in one step. Step 1: Identify what reaction can produce both methane and ethane simultaneously in one step. Step 2: The Wurtz reaction with CH3Br proceeds as follows: 2 CH3Br + 2Na → CH3-CH3 + 2NaBr (ethane formed by coupling) However, side reactions also occur: CH3Br can also react via other pathways, but more relevantly, a mixed Wurtz-type or the reaction of CH3Br with sodium can give both CH4 (via protonation/reduction side reaction) and C2H6. Actually, the key concept here is that CH3Br with Na can give ethane (Wurtz: 2CH3Br + 2Na → C2H6 + 2NaBr), and CH3Br can also undergo reduction or react with water traces to give CH4. More precisely, using CH3MgBr (Grignard) with water gives CH4, but the question asks about one-step preparation of BOTH methane and ethane. The correct interpretation: CH3Br treated with Na (Wurtz reaction) gives ethane as the main product, and also produces methane as a by-product in the same reaction mixture (one step). This makes CH3Br the only reactant listed that can yield both CH4 and C2H6 in a single reaction step. Why other options fail: (a) H2C=CH2 (ethylene): Hydrogenation gives only ethane, not methane. (b) CH3OH (methanol): Reactions give formaldehyde, CO, or CH4 depending on conditions, but not both CH4 and C2H6 in one step easily. (d) CH3-CH2-OH (ethanol): Reactions typically give ethylene or diethyl ether, not both methane and ethane in one step. Only CH3Br (option c) with sodium metal (Wurtz reaction) produces ethane as the primary product and methane as a side product, both in one step. Therefore, the correct answer is C.