See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the two compounds: The first compound is 2-methoxytetrahydrofuran (a cyclic acetal/hemiacetal methyl ether), and the second compound is 2-hydroxytetrahydrofuran (a cyclic hemiacetal). Step 2 - Understand the key chemical difference: 2-Hydroxytetrahydrofuran is a cyclic hemiacetal, meaning it exists in equilibrium with its open-chain form, which is 4-hydroxybutanal (an aldehyde). This open-chain aldehyde form is available in solution. In contrast, 2-methoxytetrahydrofuran is a full acetal (both oxygens are OR groups), which does not readily open to give a free aldehyde under mild conditions. Step 3 - Apply Tollen's reagent: Tollen's reagent (ammoniacal silver nitrate, [Ag(NH3)2]+) is a mild oxidizing agent that detects aldehydes by oxidizing them to carboxylic acids while reducing Ag+ to Ag0, producing a silver mirror. Since 2-hydroxytetrahydrofuran (cyclic hemiacetal) is in equilibrium with the free aldehyde (4-hydroxybutanal), it will give a positive silver mirror test with Tollen's reagent. The acetal (2-methoxytetrahydrofuran) has no free aldehyde in equilibrium and will NOT react with Tollen's reagent under mild conditions. Step 4 - Why other options fail: (a) 2,4-DNP (Brady reagent) reacts with aldehydes and ketones (carbonyl compounds), but because the hemiacetal equilibrium provides only a small amount of free aldehyde, DNP is less reliable and specific for distinguishing these two; moreover, both might appear similar. More importantly, Tollen's reagent is the classic, definitive test here. (c) Lucas reagent differentiates primary, secondary, and tertiary alcohols - it does not specifically distinguish a hemiacetal from an acetal. (d) NaHSO3 reacts with aldehydes and methyl ketones to form bisulfite addition products, but this test is less clean for cyclic hemiacetals and is not the standard test for distinguishing these two functional groups. Step 5 - Conclusion: Tollen's reagent uniquely identifies the cyclic hemiacetal (2-hydroxytetrahydrofuran) via the silver mirror reaction, while the acetal (2-methoxytetrahydrofuran) gives no reaction, thus differentiating the two compounds. Therefore, the correct answer is B.