See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the starting material: The starting material is 2-acetylcyclohexan-1-one, which contains both a ketone (cyclohexanone) and a methyl ketone (-COCH3) group at C2. Step 2 - Reaction with I2/NaOH (iodoform reaction, step 1): The methyl ketone group (-COCH3) undergoes the iodoform reaction. I2/NaOH (haloform reaction) selectively reacts with methyl ketones (CH3CO-) to give iodoform (CHI3, yellow precipitate) and the corresponding carboxylate salt. The -COCH3 group is converted to -COO^- (carboxylate) with release of CHI3. This confirms yellow precipitate of CHI3 is formed in all options, so all options share that feature. Step 3 - After iodoform reaction: The product after step 1 (base conditions) is the carboxylate of 2-oxocyclohexane-1-carboxylic acid (i.e., sodium 2-oxocyclohexanecarboxylate). The -COCH3 has been cleaved to give -COO^- + CHI3. Step 4 - Step 2 (H+): Acidification converts the carboxylate salt to the free beta-keto acid: 2-oxocyclohexane-1-carboxylic acid. Step 5 - Step 3 (Delta, heating): The beta-keto acid undergoes decarboxylation upon heating. A beta-keto acid readily loses CO2 to give the corresponding ketone. Loss of CO2 from 2-oxocyclohexane-1-carboxylic acid gives cyclohexanone. Step 6 - Final products: CHI3 (yellow precipitate) + cyclohexanone. Step 7 - Why other options fail: - Option (a): gives 2-carboxycyclohexan-1-one — this would be the product before decarboxylation (step 3 not applied properly). - Option (b): gives 2-formylcyclohexan-1-one (CHO group) — incorrect, iodoform reaction does not give an aldehyde here. - Option (d): gives a dicarboxylic acid — incorrect, no such pathway exists here. Only option (c) correctly identifies cyclohexanone as the organic product after all three steps, along with CHI3. Therefore, the correct answer is C.