See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1 - Starting material: 1-indanone is a five-membered ketone fused to a benzene ring (cyclopentanone ring fused to benzene). Step 2 - Reaction 1 (HCN): HCN adds to the carbonyl of 1-indanone via nucleophilic addition to give a cyanohydrin. The CN group adds to the carbonyl carbon (C1 of the cyclopentanone), and OH is also on C1. This converts the five-membered ring ketone into a five-membered ring bearing both OH and CN at C1, effectively inserting one carbon (as CN) at the carbonyl position. The cyanohydrin has the structure: benzene fused to cyclopentane ring with C1 bearing OH and CN. Step 3 - Reaction 2 (LiAlH4): LiAlH4 is a strong reducing agent that reduces both the nitrile (CN) to a primary amine (CH2NH2) and also reduces the OH-bearing carbon... Actually, LiAlH4 reduces the nitrile group (C≡N) to a primary amine (-CH2NH2). The OH from the cyanohydrin is not reduced by LiAlH4 under normal conditions (secondary alcohols are stable to LiAlH4). So after LiAlH4, we have: benzene fused to cyclopentane ring with C1 bearing OH and CH2NH2. This gives a compound with a primary amine tethered one carbon away from C1 of the indane skeleton, and a hydroxyl at C1. The molecule now has an aminomethyl group (-CH2NH2) and hydroxyl (-OH) on the same carbon of the cyclopentane ring fused to benzene, effectively making the chain: Ph-fused-cyclopentane-C1(OH)(CH2NH2). Step 4 - Reaction 3 (NaNO2/H+): This is a diazotization condition. NaNO2/H+ converts a primary amine to a diazonium ion. The primary amine -CH2NH2 becomes -CH2N2+. For a primary alkyl diazonium salt, the diazonium is extremely unstable and immediately loses N2 to form a primary carbocation (or proceeds via SN1/rearrangement). The carbocation -CH2+ is a primary carbocation adjacent to a carbon bearing an OH group. This undergoes a 1,2-hydride or 1,2-alkyl shift (ring expansion). The carbon bearing OH (C1 of cyclopentane) migrates to the adjacent carbocation center, resulting in ring expansion of the five-membered ring to a six-membered ring. The ring expansion converts the indanone-derived skeleton into a tetralin (tetrahydronaphthalene) skeleton. The migration of the C1-OH carbon to the carbocation gives a six-membered ring with the oxygen (as a ketone after loss of H+ from the oxocarbenium ion or rearrangement of the alcohol to ketone). The product of this ring expansion is a six-membered ring ketone fused to benzene, i.e., a tetralone. The hydroxyl at C1 becomes the carbonyl after the 1,2-shift and ring expansion, and the new carbonyl ends up at the position alpha to the benzene ring, giving 1-tetralone (option a). This is because the carbon bearing OH (originally C1, alpha to benzene) migrates, placing the carbonyl alpha to benzene in the new six-membered ring. Why other options fail: - (b) 2-tetralone would require the carbonyl to be beta to benzene, which is not consistent with the regiochemistry of this ring expansion. - (c) 1,2,3,4-tetrahydronaphthalene has no carbonyl; we would need the OH to be lost without forming a ketone, which is not the outcome here. - (d) naphthalene requires full aromatization, which does not occur under these conditions. Therefore, the correct answer is A.