Which answer choice explains why furanoses and pyranoses can be reduced by sodium borohydride despite having no carbonyl functional groups.

Question

BlackTom AI · Accepted Answer

Mutarotation converts cyclic hemiacetals/hemiketals into their open-chain aldehyde or ketone forms, which is the key to understanding why NaBH4 can reduce furanoses and pyranoses.

1) Relevant concepts
- Many monosaccharides in solution exist predominantly as cyclic hemiacetals (furanoses/pyranoses) but are in rapid equilibrium with a small fraction of the open-chain aldehyde (or ketone) form; this interconversion is called mutarotation.
- Sodium borohydride (NaBH4) is a hydride donor that selectively reduces carbonyl groups (aldehydes and ketones) to alcohols under mild conditions; it does not directly reduce stable ethers or typical hemiacetals under normal conditions.
- Le Chatelier’s principle: if an open-chain carbonyl is consumed by reduction, the equilibrium shifts to produce more open-chain form from cyclic hemiacetals.

2) Step-by-step reasoning
- Step 1: A cyclic furanose/pyranose exists in equilibrium with its open-chain aldehyde/ketone form via ring opening and closing (mutarotation).
- Step 2: NaBH4 reduces the small population of open-chain carbonyl species to the corresponding alcohol (e.g., an aldose → alditol).
- Step 3: Because the carbonyl is consumed, the equilibrium shifts, causing additional ring-opening of cyclic hemiacetals to replenish the open-chain form.
- Step 4: Repetition of this cycle leads to net reduction of the sugar, even though the predominant isolated species at any moment may be the cyclic form.

3) Calculations
- No quantitative calculation is required; the argument depends on equilibrium and stoichiometry: consumption of trace carbonyl by NaBH4 drives conversion of more cyclic sugar into the open form until reduction is complete.

4) Why the correct answer is correct
- The stated choice identifies mutarotation producing aldehyde/ketone intermediates; these are directly reducible by NaBH4, explaining the observed reduction of cyclic sugars.

5) Why other options are incorrect
- “Furanoses and pyranoses cannot be reduced by NaBH4” is false (they are reduced via the open-chain form).
- “NaBH4 is a strong reducing agent…reduce hemiacetals” is misleading: NaBH4 is mild and reduces carbonyls, not stable hemiacetals directly.
- “NaBH4 is also a strong base…” is incorrect: NaBH4 is not a base that converts hemiacetals to carbonyls; ring-opening is typically acid/base-catalyzed but not driven by NaBH4 acting as base.
- “NaBH4 is a weak reducing agent…reduce hemiacetals” is contradictory and incorrect for the same reason.

6) Conclusion
Mutarotation producing an aldehyde/ketone intermediate (the provided answer) correctly explains why furanoses and pyranoses can be reduced by NaBH4. This matches the given correct answer.

Which answer choice explains why furanoses and pyranoses can be reduced by sodium borohydride despite having no carbonyl functional groups.Single choice

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