See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the starting material: The starting material is methylenecyclohexane, which is a cyclohexane ring bearing an exocyclic methylene group (=CH2) at C1. It is a terminal alkene (exocyclic double bond). Step 2 - Reaction with HNO3 (oxidative cleavage): Concentrated or dilute HNO3 can act as an oxidizing agent toward alkenes. For a cyclic alkene with an exocyclic double bond, oxidative cleavage of the C=C bond occurs. The exocyclic =CH2 is cleaved off as CO2 (or formaldehyde/formic acid level oxidation product), and the ring carbon that bore the double bond is oxidized to a carbonyl group. However, since the double bond is exocyclic, cleavage opens the exocyclic bond: the =CH2 end gives CO2 (fully oxidized) and the ring carbon C1 becomes a ketone carbon. But C1 is part of the ring, so it becomes a ring carbonyl. Yet, the ring itself is intact with 6 carbons. Wait - let me reconsider. Step 3 - Reconsider the mechanism: Methylenecyclohexane has the structure where C1 of cyclohexane has an exo =CH2. Oxidative cleavage of this exocyclic double bond by HNO3 gives: the =CH2 fragment oxidized to CO2 (or HCOOH), and C1 of the ring becomes a carbonyl. Since C1 is within the six-membered ring and now bears a C=O with two ring carbons adjacent, this would give cyclohexanone... but that is a 6-membered ring ketone (not listed as an option matching answer C). Step 4 - Alternative interpretation (ring-opening oxidation): HNO3 can oxidize cyclic systems more aggressively. Consider that HNO3 oxidizes methylenecyclohexane via oxidation of the exocyclic double bond leading to ring contraction or ring cleavage. The nitric acid oxidation of methylenecyclohexane gives a dicarboxylic acid (A) by oxidative cleavage of the ring through the exocyclic alkene: the double bond cleaves to give a keto-acid or diacid. Specifically, oxidative cleavage of the exocyclic C=C of methylenecyclohexane gives OHC-(CH2)5-COOH (6-oxoheptanedioic acid type) or more precisely opens to give a 7-carbon chain with terminal acid/aldehyde groups: HOOC-(CH2)4-CO-COOH or adipaldehyde-acid. Actually the cleavage of the exocyclic double bond in methylenecyclohexane gives: one fragment is HCOOH (from =CH2) and the other is cyclohexanone. But if HNO3 further oxidizes and opens the ring: the ring double bond being attacked leads to a dicarboxylic acid. Step 5 - Correct pathway: Methylenecyclohexane treated with HNO3 undergoes oxidation. The exocyclic alkene is oxidized; the =CH2 becomes CO2/HCOOH, and the ring C1 carbonyl with the ring intact gives cyclohexanone initially, but further oxidation by excess HNO3 opens the ring at C1 (the most oxidized site) to give a linear dicarboxylic acid. The ring opening of cyclohexanone by HNO3 gives adipic acid (hexanedioic acid, HOOC-(CH2)4-COOH) - this is product A. Step 6 - Reaction with Ca(OH)2: Adipic acid (a dicarboxylic acid) reacts with Ca(OH)2 to form calcium adipate (the calcium salt). Upon dry distillation (pyrolysis) of calcium salts of dicarboxylic acids, cyclic ketones are formed. Calcium adipate (calcium salt of hexanedioic acid, 6 carbons) upon pyrolysis loses CaCO3 and forms cyclopentanone (5-membered ring ketone). This is the classic industrial synthesis of cyclopentanone: calcium adipate --heat--> cyclopentanone + CaCO3. Step 7 - Why other options fail: (a) cyclobutanone would require a 5-carbon diacid (glutaric acid); (b) cyclopropenone is not formed this way; (d) cycloheptanone would require a 8-carbon diacid (suberic acid). Adipic acid from cyclohexane ring-opening gives cyclopentanone specifically. Therefore, the correct answer is C.