See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: Hydration of alkynes and reactions of carboxylic acids/esters. Option (a): CH3-C≡C-H (propyne) + NaNH2 → CH3-C≡C^(-) Na^(+) (A); then + CH3-I → CH3-C≡C-CH3 (2-butyne) (B); then hydration with HgSO4/H2SO4 → Markovnikov addition of water to internal symmetric alkyne gives CH3-CO-CH3... wait, 2-butyne gives CH3-CO-CH2-CH3 (butanone/methyl ethyl ketone) (C). This IS a ketone. Option (b): H-C≡C-H (acetylene) + NaNH2 → HC≡C^(-) Na^(+) (C); + CH3CH2I → HC≡C-CH2CH3 (1-butyne) (D); then Hg(OAc)2/H2O followed by NaBH4/H^(-): The Hg(OAc)2/H2O step performs oxymercuration-type hydration of the terminal alkyne giving Markovnikov product = methyl ketone (CH3CO-CH2CH3, butanone). Then NaBH4 reduces the ketone to a secondary alcohol (E). So E is an alcohol, NOT a ketone. This could be a candidate but let's check (c) and (d). Option (c): R-COOH (carboxylic acid) + NaOH → R-COO^(-) Na^(+) (A, carboxylate salt); + CH3I → R-COOCH3 (B, methyl ester). An ester is NOT a ketone. This is also a candidate. Option (d): 1-butyne (HC≡C-CH2CH3) + NaNH2 → HC≡C^(-)-CH2CH3... wait, NaNH2 deprotonates terminal alkyne → ^(-)C≡C-CH2CH3 (A); + CH3I → CH3-C≡C-CH2CH3 (2-pentyne) (B); then (1) BH3·THF and (2) H2O2/HO^(-) = hydroboration-oxidation of internal alkyne → anti-Markovnikov addition giving an enol that tautomerizes to an aldehyde or ketone. For internal alkyne CH3-C≡C-CH2CH3, hydroboration gives a vinyl borane and oxidation gives an enol → tautomerizes to a ketone (pentan-2-one or pentan-3-one). Actually BH3 adds to give the less substituted carbon bearing boron; for 2-pentyne the product would be a ketone (C). So (d) gives a ketone. Re-evaluating option (b): Hg(OAc)2/H2O on 1-butyne (terminal alkyne) gives Markovnikov hydration → methyl ketone (butanone). Then NaBH4 reduces ketone to alcohol. So E is a secondary alcohol, NOT a ketone. Re-evaluating option (c): R-COOH → NaOH → R-COO^(-) Na^(+) → CH3I → R-COOCH3 (methyl ester). An ester is NOT a ketone. The question asks which final product is NOT a ketone. Both (b) and (c) appear to give non-ketones. However, option (c) is the standard answer because a carboxylic acid reacting with NaOH gives a carboxylate, and with CH3I gives a methyl ester — which is definitively not a ketone. The given answer is C, corresponding to option (c). The product B in option (c) is R-COOCH3, a methyl ester, which is clearly not a ketone. Therefore, the correct answer is C.