Question:
Glucose does not react with which of the following?
Glucose does not react with which of the following?
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Which classic reagent fails because glucose has no free aldehyde group (cyclic hemiacetal)?
Updated On: Jul 10, 2026
- Bromine water
- \( NaHSO_3 \)
- \( C_6H_5NHNH_2 \)
- \( (CH_3CO)_2O \)
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The Correct Option is B
Solution and Explanation
Step 1: Test each reagent against the known reactions of glucose.
Step 2: Bromine water oxidises the \( -CHO \) group of glucose to gluconic acid, so a reaction occurs.
Step 3: Phenylhydrazine \( (C_6H_5NHNH_2) \) reacts with glucose to give the yellow crystalline osazone, so a reaction occurs.
Step 4: Acetic anhydride \( (CH_3CO)_2O \) acetylates the five \( -OH \) groups to give glucose pentaacetate, so a reaction occurs.
Step 5: Glucose does NOT form the aldehyde-bisulphite addition product with \( NaHSO_3 \). This failure shows that the free \( -CHO \) group is largely absent because glucose exists mainly as a cyclic hemiacetal.
Answer: \( NaHSO_3 \).
Step 2: Bromine water oxidises the \( -CHO \) group of glucose to gluconic acid, so a reaction occurs.
Step 3: Phenylhydrazine \( (C_6H_5NHNH_2) \) reacts with glucose to give the yellow crystalline osazone, so a reaction occurs.
Step 4: Acetic anhydride \( (CH_3CO)_2O \) acetylates the five \( -OH \) groups to give glucose pentaacetate, so a reaction occurs.
Step 5: Glucose does NOT form the aldehyde-bisulphite addition product with \( NaHSO_3 \). This failure shows that the free \( -CHO \) group is largely absent because glucose exists mainly as a cyclic hemiacetal.
Answer: \( NaHSO_3 \).
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