See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: When a ketone reacts with an alcohol under acid catalysis (TsOH) in the presence of molecular sieves (to remove water), it forms an acetal (or ketal). Molecular sieves drive the equilibrium toward acetal formation by absorbing the water produced. Step 1: Identify the reactants. The ketone is acetone (propan-2-one, CH3COCH3) and the alcohol is isopropanol ((CH3)2CHOH). Step 2: Mechanism of acetal formation. Under acid catalysis, the carbonyl oxygen of acetone is protonated, making the carbonyl carbon more electrophilic. One molecule of isopropanol attacks to form a hemiketal. Loss of water (facilitated by molecular sieves) gives an oxocarbenium ion. A second molecule of isopropanol attacks to give the acetal (ketal): (CH3)2C(OCH(CH3)2)2, i.e., 2,2-bis(isopropoxy)propane. Step 3: Identify the product structure. The product is an open-chain acetal: the central carbon (originally the carbonyl carbon of acetone) now bears two methyl groups and two -O-CH(CH3)2 (isopropoxy) groups. This is a quaternary carbon with two methyls and two isopropoxy substituents. Step 4: Match to options. - Option (a) shows a diethyl acetal (ethoxy groups), which would require ethanol, not isopropanol. Incorrect. - Option (b) shows a three-membered ring with two oxygens and a CH-methyl unit — this is a strained cyclic structure not consistent with simple acetal formation from a ketone and alcohol. Incorrect. - Option (c) shows a five-membered or open-chain structure where a central carbon bears two methyl groups and two isopropoxy (-O-CMe2- or -O-CHMe2) groups — this matches 2,2-bis(isopropoxy)propane, the acetal of acetone with isopropanol. Correct. - Option (d) shows a three-membered ring (epoxide-type) with isopropoxy substituents — strained and not the product of simple acetal formation. Incorrect. Step 5: The 62% yield with molecular sieves and TsOH at 0°C is consistent with ketal formation under mild conditions with water removal. Therefore, the correct answer is C.