See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: When a ketone (or aldehyde) reacts with two equivalents of an alcohol in the presence of an acid catalyst (H+), it undergoes acid-catalyzed acetal formation. Step 1 - First equivalent of EtOH reacts: Cyclohexanone reacts with one molecule of ethanol under acid catalysis to form a hemiacetal (an intermediate with one -OH and one -OEt group on the former carbonyl carbon). This step is reversible. Step 2 - Second equivalent of EtOH reacts: Under acidic conditions, the hemiacetal intermediate loses water (the -OH is protonated and leaves as H2O), forming an oxocarbenium ion. The second molecule of ethanol then attacks this carbocation to give the acetal. The acetal has two -OEt groups on the carbon that was originally the carbonyl carbon of cyclohexanone, i.e., 1,1-diethoxycyclohexane. Step 3 - Why 2 equivalents are specified: The use of 2 EtOH explicitly signals that the reaction goes all the way to the acetal (full protection of the ketone), not stopping at the hemiacetal stage. Why other options fail: - (a) Hemiacetal: This is only the intermediate formed with 1 equivalent of alcohol; with 2 equivalents and acid catalyst, the reaction proceeds further. - (c) Alcohol: Reduction would be needed to form an alcohol from a ketone; EtOH is not a reducing agent here. - (d) Alkane: Complete deoxygenation requires strong reducing conditions (e.g., Clemmensen or Wolff-Kishner); not applicable here. Therefore, the correct answer is B.