See image — Alcohols Phenols and Ethers Chemistry Question

💡 Solution & Explanation

Concept: Na2Cr2O7/H2SO4 is a strong oxidizing agent (Jones-type conditions). It oxidizes secondary alcohols to ketones and primary alcohols (including primary alcohols that would give aldehydes as intermediates) all the way to carboxylic acids under these acidic, aqueous oxidative conditions. Step 1: Identify the functional groups in the substrate. The starting material is 3-(hydroxymethyl)cyclohexan-1-ol. It contains: - A secondary alcohol at C1 of the cyclohexane ring (the ring OH) - A primary alcohol as a CH2OH group attached at C3 of the ring Step 2: Determine what Na2Cr2O7/H2SO4 does to each group. - The secondary alcohol at C1 is oxidized to a ketone (cyclohexanone). - The primary alcohol (CH2OH) is oxidized first to an aldehyde (CHO), then further oxidized to a carboxylic acid (COOH) under the strongly oxidizing aqueous acidic conditions of Na2Cr2O7/H2SO4. Step 3: Identify the product. The product is a cyclohexanone ring with a COOH group directly attached at C3, which corresponds to option (a): 3-(carboxyl)cyclohexan-1-one (cyclohexanone ring with COOH at C3). Why other options fail: - Option (b) would require oxidation of secondary alcohol to ketone but only oxidation of primary alcohol to aldehyde (not further). Na2Cr2O7/H2SO4 is a strong oxidant in aqueous acidic medium and will take primary alcohols all the way to carboxylic acids, not stop at aldehyde. - Option (c) would imply the primary alcohol is CH2OH oxidized to CH2COOH (i.e., retaining the CH2 linker). But the substrate has CH2OH directly, so oxidation of CH2OH gives COOH directly (no CH2 linker remains between ring and COOH). Option (c) has a CH2CO2H group, which would require a different substrate (one with CH2CH2OH). - Option (d) is eliminated since option (a) correctly describes the product. Therefore, the correct answer is A.