See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: The starting material is cinnamaldehyde methyl ketone (PhCH=CH-CO-CH3, i.e., 1-phenyl-1-buten-3-one or more precisely trans-4-phenyl-3-buten-2-one, also known as benzalacetone). The target product is cinnamic acid (PhCH=CH-CO2H). The transformation requires converting the methyl ketone (CH3CO-) group into a carboxylic acid (-CO2H) while preserving the C=C double bond. Step 1 - Identify the functional group change: The -COCH3 group must be converted to -COOH. The carbon count decreases by one (the CH3 is lost), meaning the methyl group is oxidatively removed. Step 2 - Haloform reaction: I2/NaOH is the iodoform reaction. Methyl ketones (RCOCH3) react with I2/NaOH to give iodoform (CHI3) and the carboxylate salt (RCOO-Na+). Upon acidification with H+, the carboxylate gives the carboxylic acid RCOOH. In this case, R = PhCH=CH-, so the product is PhCH=CH-COOH (cinnamic acid). The C=C double bond is not affected by I2/NaOH under these mild basic conditions. Step 3 - Why other options fail: (a) KMnO4/Delta (hot KMnO4) is a strong oxidant that would oxidatively cleave the C=C double bond, giving benzoic acid and other fragments, not cinnamic acid. (c) H2/Pt would reduce the C=C double bond (hydrogenation), giving a saturated compound, not the desired product. (d) LiAlH4 is a reducing agent that would reduce the ketone to an alcohol, not oxidize it to a carboxylic acid. Therefore, the correct answer is B.