Question:
Glucose reacts with acetic anhydride to give pentaacetyl derivative. Which of the following is true about that?
Glucose reacts with acetic anhydride to give pentaacetyl derivative. Which of the following is true about that?
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The solubility of acetylated glucose in NaOH is due to the ester groups, which are polar and interact with the aqueous solution.
Updated On: Apr 22, 2026
- That can reduce Fehling or Tollen’s reagent
- That’s soluble in dilute NaOH solution
- That consumes one mole of \( \mathrm{HIO_4} \)
- \( \mathrm{Br_2/H_2O} \) can oxidize
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The Correct Option is B
Solution and Explanation
Step 1: Understand the reaction with acetic anhydride.
When glucose reacts with acetic anhydride, it forms pentaacetyl glucose, which is an ester derivative. This reaction involves the acylation of the hydroxyl groups in glucose with acetic acid.
Step 2: Analyze the options.
- (1) That can reduce Fehling or Tollenâs reagent: This is incorrect. The acetylated form of glucose no longer has the free aldehyde group that can reduce these reagents.
- (2) Thatâs soluble in dilute NaOH solution: This is correct. The esterified glucose is soluble in dilute NaOH due to the polarity of the ester group.
- (3) That consumes one mole of \( \text{HIO}_4 \): This is incorrect. The oxidation of glucose with \( \text{HIO}_4 \) consumes two moles, not one.
- (4) \( \text{Br}_2/\text{H}_2\text{O} \) can oxidize: This is incorrect. The acetylated glucose will not undergo oxidation by bromine water.
Step 3: Conclusion.
The correct answer is (2), as the pentaacetyl derivative of glucose is soluble in dilute NaOH.
When glucose reacts with acetic anhydride, it forms pentaacetyl glucose, which is an ester derivative. This reaction involves the acylation of the hydroxyl groups in glucose with acetic acid.
Step 2: Analyze the options.
- (1) That can reduce Fehling or Tollenâs reagent: This is incorrect. The acetylated form of glucose no longer has the free aldehyde group that can reduce these reagents.
- (2) Thatâs soluble in dilute NaOH solution: This is correct. The esterified glucose is soluble in dilute NaOH due to the polarity of the ester group.
- (3) That consumes one mole of \( \text{HIO}_4 \): This is incorrect. The oxidation of glucose with \( \text{HIO}_4 \) consumes two moles, not one.
- (4) \( \text{Br}_2/\text{H}_2\text{O} \) can oxidize: This is incorrect. The acetylated glucose will not undergo oxidation by bromine water.
Step 3: Conclusion.
The correct answer is (2), as the pentaacetyl derivative of glucose is soluble in dilute NaOH.
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