Question:
D-Glucose does not react with which of the following reagents?
I- $\text{NaHSO}_3$
II- $\text{NH}_2\text{OH}$
III- $(\text{CH}_3\text{CO})_2\text{O}$
IV- Schiff's reagent
D-Glucose does not react with which of the following reagents?
I- $\text{NaHSO}_3$
II- $\text{NH}_2\text{OH}$
III- $(\text{CH}_3\text{CO})_2\text{O}$
IV- Schiff's reagent
I- $\text{NaHSO}_3$
II- $\text{NH}_2\text{OH}$
III- $(\text{CH}_3\text{CO})_2\text{O}$
IV- Schiff's reagent
Show Hint
The absence of reaction with $\text{NaHSO}_3$ and Schiff's reagent was the main historical reason for proposing the cyclic structure of glucose over the open-chain structure.
Updated On: Jul 22, 2026
- II, III only
- I, II, III only
- I, II only
- I, IV only
Show Solution
Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Step 1: Understanding the Question:
The question is from biomolecules.
We need to identify which of the given reagents do not react with D-glucose, which is a key piece of evidence for its cyclic structure.
Step 2: Key Formula or Approach:
D-Glucose exists primarily in cyclic hemiacetal forms ($\alpha$- and $\beta$-anomers) in equilibrium with a very small concentration of the open-chain form ($\lt 1\%$).
Reagents that require a free, high-concentration carbonyl group will fail to react with glucose, whereas stronger nucleophiles can shift the equilibrium to open the ring.
Step 3: Detailed Explanation:
• Let us analyze the reaction of D-glucose with each reagent:
• I. Sodium bisulfite ($\text{NaHSO}_3$):
D-Glucose does not form a bisulfite addition product. The nucleophile is not strong enough to force the cyclic hemiacetal ring to open, so it does not react.
• II. Hydroxylamine ($\text{NH}_2\text{OH}$):
D-Glucose reacts with hydroxylamine to form an oxime. This is because hydroxylamine is a strong enough nucleophile to open the hemiacetal ring.
• III. Acetic anhydride ($(\text{CH}_3\text{CO})_2\text{O}$):
D-Glucose reacts with acetic anhydride to form a pentaacetate, confirming the presence of five hydroxyl groups. This reaction proceeds readily.
• IV. Schiff's reagent:
D-Glucose does not restore the pink color of Schiff's reagent, indicating the absence of a free, readily available aldehyde group in its stable cyclic form.
• Therefore, D-glucose does not react with $\text{NaHSO}_3$ (I) and Schiff's reagent (IV).
Step 4: Final Answer:
D-Glucose does not react with reagents I and IV.
The question is from biomolecules.
We need to identify which of the given reagents do not react with D-glucose, which is a key piece of evidence for its cyclic structure.
Step 2: Key Formula or Approach:
D-Glucose exists primarily in cyclic hemiacetal forms ($\alpha$- and $\beta$-anomers) in equilibrium with a very small concentration of the open-chain form ($\lt 1\%$).
Reagents that require a free, high-concentration carbonyl group will fail to react with glucose, whereas stronger nucleophiles can shift the equilibrium to open the ring.
Step 3: Detailed Explanation:
• Let us analyze the reaction of D-glucose with each reagent:
• I. Sodium bisulfite ($\text{NaHSO}_3$):
D-Glucose does not form a bisulfite addition product. The nucleophile is not strong enough to force the cyclic hemiacetal ring to open, so it does not react.
• II. Hydroxylamine ($\text{NH}_2\text{OH}$):
D-Glucose reacts with hydroxylamine to form an oxime. This is because hydroxylamine is a strong enough nucleophile to open the hemiacetal ring.
• III. Acetic anhydride ($(\text{CH}_3\text{CO})_2\text{O}$):
D-Glucose reacts with acetic anhydride to form a pentaacetate, confirming the presence of five hydroxyl groups. This reaction proceeds readily.
• IV. Schiff's reagent:
D-Glucose does not restore the pink color of Schiff's reagent, indicating the absence of a free, readily available aldehyde group in its stable cyclic form.
• Therefore, D-glucose does not react with $\text{NaHSO}_3$ (I) and Schiff's reagent (IV).
Step 4: Final Answer:
D-Glucose does not react with reagents I and IV.
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