See image — Alcohols Phenols and Ethers Chemistry Question

💡 Solution & Explanation

Step 1: Identify the product. The product shown is a cyclohexane-1,4-dione bearing two aldehyde (CHO) groups at C2 and C5 (or equivalently positions 2 and 5 on the ring). This is 2,5-diformylcyclohexane-1,4-dione. Step 2: Work backward from the product using PCC. PCC oxidizes primary alcohols to aldehydes and secondary alcohols to ketones (it does not over-oxidize). Since 4 moles of PCC are used and the product has 2 ketone carbonyls and 2 aldehyde carbonyls, reactant (A) must have had 4 hydroxyl groups that were oxidized: 2 secondary alcohols (giving the two ring ketones) and 2 primary alcohols (giving the two aldehydes). Therefore, (A) is 2,5-bis(hydroxymethyl)cyclohexane-1,4-diol — i.e., a cyclohexane ring with OH at C1 and C4 (secondary), and CH2OH groups at C2 and C5 (primary). This accounts for exactly 4 moles of PCC. Step 3: Count the OH groups in (A) available to react with Ac2O. Ac2O acetylates hydroxyl groups (and any other nucleophilic -OH). Reactant (A) has: - 2 secondary alcohols (at C1 and C4 of the ring) - 2 primary alcohols (the CH2OH groups at C2 and C5) Total OH groups = 4. Step 4: Each mole of Ac2O acetylates one OH group (one equivalent of Ac2O per OH), consuming 1 mole of Ac2O per OH. Therefore, 4 moles of OH groups require 4 moles of Ac2O. Step 5: Check for any other nucleophilic sites. There are no amine groups or other readily acetylatable functionalities in (A), so the maximum number of moles of Ac2O consumed is 4. Therefore, the correct answer is 4.