See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the starting material: 1-indanone is a bicyclic compound consisting of a benzene ring fused to a cyclopentanone (5-membered ring with a ketone at C1). Step 2 - Reaction with NH2OH (hydroxylamine): The ketone reacts with hydroxylamine to form an oxime (C=N-OH). This is a standard ketone-to-oxime conversion. Step 3 - Treatment with H+ and heat (Beckmann rearrangement conditions): Under acidic conditions with heat, the oxime undergoes the Beckmann rearrangement. In this rearrangement, the group anti to the OH migrates to nitrogen, the N-O bond breaks, and a nitrilium ion is formed which is trapped by water to give a lactam (cyclic amide). For 1-indanone oxime, the Beckmann rearrangement expands the 5-membered ring to a 6-membered lactam. Since the oxime can exist as two geometric isomers (syn and anti), two different lactams are possible: one where the benzene-C bond migrates giving a 6-membered lactam fused to benzene (i.e., a benzazepinone or isoquinolinone framework) - specifically, rearrangement gives either 2,3-dihydro-1H-isoindol-1-one type or more correctly the ring expansion gives a 7-membered or 6-membered lactam. For indanone, the two oxime isomers give two lactams upon Beckmann rearrangement: one is a 6-membered lactam (3,4-dihydroisocarbostyril / 3,4-dihydroisoquinolin-1(2H)-one) and the other is also a lactam isomer depending on which group migrates. Step 4 - Reduction with LiAlH4: LiAlH4 reduces lactams (cyclic amides) to cyclic amines. Reduction of a 6-membered lactam fused to benzene gives a tetrahydroisoquinoline. The two lactam isomers from the two Beckmann rearrangement products give two amine products upon LiAlH4 reduction. Step 5 - Identifying products A and B: The two oxime isomers of 1-indanone oxime undergo Beckmann rearrangement to give two ring-expanded lactams. LiAlH4 reduction of these lactams yields 1,2,3,4-tetrahydroisoquinoline (from one lactam) and its isomer where nitrogen is positioned differently - i.e., 2,3-dihydro-1H-isoindole type or the other tetrahydroisoquinoline regioisomer. The products shown in option (a) are 1,2,3,4-tetrahydroisoquinoline and its isomer (2,3-dihydro-1H-isoindole / indoline framework with an extra CH2), which correspond to the two possible Beckmann rearrangement/reduction pathways. Step 6 - Elimination of other options: Option (b) gives carbocyclic products with no nitrogen, inconsistent with NH2OH reaction. Option (c) gives amino alcohols, but LiAlH4 reduces amides fully to amines without retaining OH. Option (d) gives indane and a naphthalene derivative with no nitrogen. Therefore, the correct answer is A.