See image — Alcohols Phenols and Ethers Chemistry Question

💡 Solution & Explanation

Concept: K2Cr2O7 (acidic dichromate) is an oxidizing agent. When a secondary alcohol (OH on a ring carbon) is oxidized, it becomes a ketone. Additionally, beta-keto acids (a carboxylic acid beta to a ketone, or more precisely an alpha-keto acid or beta-keto acid) readily undergo decarboxylation upon heating. Step 1: Identify the starting material. The structure shown is 1-hydroxycyclopentane-1-carboxylic acid (or 2-hydroxycyclopentane-1-carboxylic acid — the OH and CO2H are on adjacent carbons of the cyclopentane ring, making it an alpha-hydroxy acid). Step 2: Oxidation with K2Cr2O7/heat. The secondary alcohol (OH) is oxidized by K2Cr2O7 to a ketone, giving 2-oxocyclopentane-1-carboxylic acid (an alpha-keto acid / beta-keto acid on the ring), which corresponds to option (c). Step 3: Decarboxylation. The heat (delta) condition promotes decarboxylation of the resulting beta-keto acid (alpha-keto acid). Alpha- and beta-keto acids lose CO2 readily on heating. The CO2H group is lost as CO2, leaving cyclopentanone. Step 4: Final product. After oxidation of OH to C=O and decarboxylation of the CO2H group, the product is cyclopentanone — a five-membered ring with one ketone group. Why other options fail: (a) Cyclopentadiene would require elimination reactions, not oxidative decarboxylation. (c) 2-Oxocyclopentane-1-carboxylic acid is only the intermediate after oxidation, not the final product under heating conditions. (d) Cyclopentane-1-carboxylic acid would require reduction, not oxidation. Therefore, the correct answer is B.