See image — Alcohols Phenols and Ethers Chemistry Question

💡 Solution & Explanation

Concept: The product shown is a cyclic carbonate (1,3-dioxolan-2-one), which is a five-membered cyclic carbonate formed from ethylene glycol (HOCH2CH2OH) and a carbonyl source. The product has the structure where both oxygen atoms of ethylene glycol are connected to a single carbonyl carbon (C=O), forming a cyclic carbonate ring. Step 1: Identify the product. The product is ethylene carbonate, a five-membered cyclic carbonate with structure: CH2-O\ /C=O / CH2-O (a cyclic structure where both -OH groups of ethylene glycol have reacted with a carbonyl source). Step 2: Determine what carbonyl source (A) reacts with ethylene glycol (HOCH2CH2OH) in pyridine to give a cyclic carbonate. The cyclic carbonate contains the -O-C(=O)-O- linkage, meaning the carbonyl carbon in the product came from reactant A and it must have provided a C=O unit that reacted with two -OH groups. Step 3: Analyze the options: - (a) CH3-C(=O)-CH3 (acetone): A ketone does not react with diols under pyridine conditions to form a cyclic carbonate. It can form a cyclic acetal/ketal but that would give C=O replaced by C(OR)2 with no remaining C=O in the ring. - (b) COCl2 (phosgene): Phosgene is Cl-C(=O)-Cl. It reacts with ethylene glycol in the presence of pyridine (which acts as a base to neutralize HCl). The two -OH groups replace both Cl atoms, forming the cyclic carbonate CH2-O-C(=O)-O-CH2 (ethylene carbonate). This is a well-known synthesis of cyclic carbonates. - (c) CH3-C(=O)-Cl (acetyl chloride): This would give an ester linkage, not a carbonate. Reaction with one -OH gives an ester; it cannot form the cyclic carbonate with -O-C(=O)-O- linkage since there is no second leaving group on the same carbon. - (d) CH3-C(=O)-OEt (ethyl acetate): This is an ester that could undergo transesterification, but it would not produce a cyclic carbonate with a C=O between two oxygens. Step 4: Only phosgene (COCl2) has two leaving groups (two Cl atoms) on a single carbonyl carbon, allowing both hydroxyl groups of ethylene glycol to displace them, forming the cyclic carbonate product with pyridine acting as acid scavenger. Why other options fail: Acetone (a) forms acetals not carbonates; acetyl chloride (c) has only one reactive site and would give a monoester; ethyl acetate (d) would not readily form a cyclic carbonate under these conditions. Therefore, the correct answer is B.