See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: To synthesize 1-butene (a 4-carbon terminal alkene) from ethanol (a 2-carbon compound), we need to extend the carbon chain by 2 carbons and introduce a terminal double bond. Step-by-step reasoning for option C: Step 1: Ethanol (CH3CH2OH) is treated with HBr and heat. This converts ethanol to bromoethane (CH3CH2Br) via an SN2-type substitution under acidic conditions. Step 2: Bromoethane (CH3CH2Br) is treated with NaC≡CH (sodium acetylide). Sodium acetylide acts as a nucleophile in an SN2 reaction, alkylating the bromoethane. The acetylide carbon attacks the electrophilic carbon of bromoethane, forming 1-butyne (CH3CH2C≡CH). This extends the carbon chain from 2 to 4 carbons with a terminal triple bond. Step 3: 1-Butyne is hydrogenated using H2 with Lindlar's catalyst (poisoned palladium). Lindlar's catalyst performs a syn-selective partial hydrogenation of the triple bond, converting the terminal alkyne to the terminal alkene: 1-butene (CH2=CHCH2CH3). This gives the cis (Z) addition, but for a terminal alkene like 1-butene, there is no geometric isomerism issue — we get 1-butene directly. Why other options fail: - Option (a): NaC≡CH reacts with ethanol? Ethanol is not an alkyl halide; sodium acetylide would simply deprotonate ethanol rather than alkylate it. No C-C bond formation occurs. Even if we consider ethanol reacting, the product would only be 4 carbons but the sequence skips conversion to a leaving group. Actually NaC≡CH + ethanol gives no useful alkylation — this sequence is invalid for chain extension. - Option (b): Same problem as (a) for step 1; additionally Na/NH3 gives a trans-alkene (Birch-type reduction), but the starting sequence is flawed. - Option (d): Step 1 gives bromoethane, step 2 uses KOC(CH3)3/DMSO which is an elimination to give ethylene (not useful for chain extension), step 3 then tries to react ethylene with NaC≡CH which is not a valid SN2 reaction on an alkene. This sequence does not logically produce 1-butene. Therefore, the correct answer is C.