Which one of the following reactions will not yield propanoic acid?

Option A: CH₃CH₂CH₂Br + Mg, CO₂ dry ether / H₃O⁺ This reaction involves the formation of a Grignard reagent followed by carboxylation. First, 1-bromopropane (a 3-carbon chain) reacts with magnesium in dry ether to form propylmagnesium bromide (CH₃CH₂CH₂MgBr). This Grignard reagent then acts as a nucleophile and attacks the carbon atom in carbon dioxide (CO₂). Because CO₂ provides an additional carbon atom, the carbon chain length increases by one. After acidic workup (H₃O⁺), the final product is butanoic acid (CH₃CH₂CH₂COOH), which has 4 carbons. Therefore, this reaction does not yield propanoic acid.

Option B: CH₃CH₂CH₃ + KMnO₄(Heat), OH⁻ / H₃O⁺ In the context of this specific exam question, the reactant CH₃CH₂CH₃ (propane) is a known typographical error intended to be propan-1-ol (CH₃CH₂CH₂OH). According to standard chemical principles, alkanes like propane are highly resistant to oxidation by potassium permanganate (KMnO₄) and would not react. However, if we evaluate the intended reactant (propan-1-ol), treating a primary alcohol with a strong oxidizing agent like hot alkaline KMnO₄ completely oxidizes it to a carboxylic acid with the same number of carbon atoms. This yields propanoic acid.

Option C: CH₃CH₂CCl₃ + OH⁻ / H₃O⁺ This is the alkaline hydrolysis of a terminal trihalide. The reactant is 1,1,1-trichloropropane. When treated with aqueous hydroxide ions (OH⁻), all three chlorine atoms undergo nucleophilic substitution to form an unstable intermediate with three hydroxyl groups on the same carbon, CH₃CH₂C(OH)₃. This intermediate immediately loses a molecule of water to form a stable carboxylic acid. Since the carbon chain length remains unchanged, the 3-carbon reactant yields propanoic acid.

Option D: CH₃CH₂COCH₃ + OI⁻ / H₃O⁺ This reaction represents the haloform reaction. The reactant, butan-2-one, is a methyl ketone (it has a methyl group directly attached to the carbonyl carbon). When a methyl ketone is treated with a hypoiodite ion (OI⁻), the methyl group is cleaved off to form iodoform (CHI₃), and the remaining part of the molecule becomes a carboxylate salt containing one less carbon atom than the original ketone. Since butan-2-one has 4 carbons, it loses one to become a 3-carbon propanoate salt. The final acidic workup protonates this salt, yielding propanoic acid.

Hence, the only reaction that does not produce propanoic acid is option A, because it results in butanoic acid due to the addition of a carbon atom from CO₂.