See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Step 1: Identify the structure. The compound is a six-membered ring containing one oxygen atom and one NH group. Attached to the ring carbon between the two carbonyl groups is an acetyl (CH3CO-) substituent. Step 2: Identify functional groups present. - Ketone: The acetyl group (CH3-C(=O)-) attached to the ring carbon is a ketone because the carbonyl carbon is bonded to two carbon atoms (the methyl and the ring carbon). Present. - Ester: The ring contains a -C(=O)-O- linkage (lactone), where a carbonyl is bonded to the ring oxygen. This is a cyclic ester (lactone). Present. - Amide: The ring contains a -C(=O)-NH- linkage (lactam), where a carbonyl is bonded to the ring nitrogen. This is a cyclic amide. Present. - Ether: An ether requires a C-O-C linkage where neither carbon bears a carbonyl directly making it an ester. The oxygen in the ring is part of the ester (-C(=O)-O-CH2-), so the O is part of the ester/lactone functionality. There is no simple ether (C-O-C without an adjacent carbonyl on the same oxygen) present in the molecule. Step 3: Why ether is not present. In the ring, the oxygen atom is flanked on one side by a carbonyl carbon (making it an ester oxygen) and on the other side by a -CH2- group. Because this oxygen is directly part of the ester linkage, it is classified as an ester, not an ether. No separate C-O-C ether group exists anywhere in the molecule. Step 4: Why other options are incorrect. Ketone (a) is present via the acetyl substituent. Ester (b) is present via the lactone in the ring. Amide (c) is present via the lactam in the ring. Therefore, the correct answer is D.