See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: Alkyl halides can be prepared by several methods including: reaction of alcohols with hydrogen halides (using catalysts like ZnCl2), free radical halogenation of alkanes, and the Hunsdiecker reaction (reaction of silver carboxylate salts with halogens). We need to identify which reaction does NOT yield an alkyl halide. Option (a): CH3CH2OH + HCl/ZnCl2 → CH3CH2Cl + H2O. This is a standard reaction of an alcohol with HCl in the presence of ZnCl2 (Lucas reagent) to give ethyl chloride, an alkyl halide. This works. Option (b): CH3CH2CH2CH2 + Cl2/UV light → free radical chlorination of butane gives butyl chloride (alkyl halide). This works. Option (d): CH3COOAg + Br2/CCl4 → This is the Hunsdiecker reaction. Silver acetate reacts with Br2 to give CH3Br (methyl bromide, an alkyl halide) + CO2 + AgBr. This works and yields an alkyl halide. Option (c): C2H5OH + NaCl → This reaction does not proceed under normal conditions. NaCl is a very weak nucleophile and a strong base (Cl- from NaCl) would be needed to displace OH-, but OH- is a very poor leaving group. Unlike HCl (which protonates the OH to make a good leaving group, water), NaCl does not protonate the alcohol, so the reaction essentially does not occur to give an alkyl halide. No alkyl halide is produced from this reaction. Therefore, the correct answer is C.