Question:
Observe the following reactions: I. D-Glucose \(\xrightarrow{NH_2OH}\) II. D-Glucose \(\xrightarrow[(ii) NH_2OH]{(i) (CH_3CO)_2O}\). Correct statement regarding the reactions I and II is:
Observe the following reactions: I. D-Glucose \(\xrightarrow{NH_2OH}\) II. D-Glucose \(\xrightarrow[(ii) NH_2OH]{(i) (CH_3CO)_2O}\). Correct statement regarding the reactions I and II is:
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Acetylation protects the reactive carbonyl groups of monosaccharides from further nucleophilic addition.
Updated On: Jun 9, 2026
- Oxime is formed in both the reactions I, II
- Oxime is not formed in both the reactions I, II
- Oxime is formed in reaction I but oxime is not formed in reaction II
- Oxime is not formed in reaction I but oxime is formed in reaction II
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The Correct Option is C
Solution and Explanation
Concept:
The carbonyl group in glucose (\(CHO\)) reacts with hydroxylamine (\(NH_2OH\)) to form an oxime. Acetylation with acetic anhydride masks the aldehyde group.
Step 1: Analyze reaction I.
D-Glucose reacts with \(NH_2OH\) to form glucose oxime: \[ \text{Glucose} + NH_2OH \rightarrow \text{Glucose oxime} + H_2O \]
Step 2: Analyze reaction II.
D-Glucose reacts with acetic anhydride \((CH_3CO)_2O\) to form penta-acetyl glucose. The aldehyde group is now protected/acetylated (in its cyclic/acyclic form), preventing the reaction with \(NH_2OH\): \[ \text{Penta-acetyl glucose} + NH_2OH \rightarrow \text{No reaction} \] \[ \boxed{\text{Oxime is formed in reaction I but oxime is not formed in reaction II}} \]
Step 1: Analyze reaction I.
D-Glucose reacts with \(NH_2OH\) to form glucose oxime: \[ \text{Glucose} + NH_2OH \rightarrow \text{Glucose oxime} + H_2O \]
Step 2: Analyze reaction II.
D-Glucose reacts with acetic anhydride \((CH_3CO)_2O\) to form penta-acetyl glucose. The aldehyde group is now protected/acetylated (in its cyclic/acyclic form), preventing the reaction with \(NH_2OH\): \[ \text{Penta-acetyl glucose} + NH_2OH \rightarrow \text{No reaction} \] \[ \boxed{\text{Oxime is formed in reaction I but oxime is not formed in reaction II}} \]
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