See image — Alcohols Phenols and Ethers Chemistry Question

💡 Solution & Explanation

Concept: MnO2 is a mild, selective oxidizing agent that selectively oxidizes allylic and benzylic alcohols to their corresponding carbonyl compounds (aldehydes or ketones). It does not oxidize non-allylic or non-benzylic alcohols under normal conditions. Starting material analysis: The substrate is HO-(CH2)3-C(=CH2)-CH2-OH. This molecule contains two hydroxyl groups: 1. A primary allylic alcohol: the -CH2-OH directly attached to the exocyclic methylene carbon (C(=CH2)-CH2-OH). This is an allylic position because the CH2OH is on a carbon adjacent to the C=CH2 double bond. 2. A primary non-allylic (homoallylic/remote) alcohol: HO-(CH2)3- at the far end of the chain, which is not allylic. Selectivity of MnO2: MnO2 selectively oxidizes only the allylic alcohol (-CH2OH adjacent to C=CH2) to give the corresponding aldehyde (-CHO), while leaving the non-allylic primary alcohol (HO-(CH2)3-) untouched. Product: HO-(CH2)3-C(=CH2)-CHO, i.e., 5-hydroxy-2-methylenepentanal (the allylic -CH2OH becomes -CHO, and the remote -OH remains intact). This matches option (c), which shows HO-CH2CH2CH2-C(=CH2)-C(=O)-H, an aldehyde formed by selective oxidation of the allylic alcohol with the remote homoallylic hydroxyl group preserved. Why other options fail: - Option (a): Oxidizes the non-allylic terminal alcohol to an aldehyde while retaining the allylic alcohol — opposite selectivity, incorrect. - Option (b): Appears structurally similar to (c) but the drawing in the original shows it differently; in context option (b) may represent a different connectivity. - Option (d): Shows oxidation to a carboxylic acid, which MnO2 does not accomplish under standard conditions. Therefore, the correct answer is C.