See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: Reaction of ketones with hydroxylamine (NH2OH) under acidic conditions is a classic condensation reaction that forms an oxime. Step 1 - Identify the reagents and substrate: Cyclopentanone (a cyclic ketone) is treated with hydroxylamine (NH2OH) in the presence of an acid catalyst (H+). Step 2 - Mechanism: Under acidic conditions, the nitrogen of NH2OH acts as a nucleophile and attacks the electrophilic carbonyl carbon of cyclopentanone. This forms a tetrahedral hemiorthoamide intermediate (carbinolamine). Under acidic conditions, the OH group of this intermediate is protonated and leaves as water. This elimination step produces a C=N double bond, yielding an oxime. Step 3 - Product identification: The product is cyclopentanone oxime, where the C=O of cyclopentanone is converted to C=N-OH. This corresponds to option (d), which shows a cyclopentane ring with a C=N-OH (oxime) functional group. Step 4 - Why other options fail: - Option (a) shows a product where both OH and NHOH are added to the carbonyl carbon without loss of water - this would be the carbinolamine intermediate, not the final product. - Option (b) shows 1-hydroxy-1-(aminooxy)cyclopentane, which is not a product of this reaction. - Option (c) shows an ethoxy group, which cannot come from NH2OH; there is no ethanol in the reaction. - Option (d) correctly shows the oxime formed after condensation (addition + dehydration). Therefore, the correct answer is D.