Question:

Statement-I : Despite having aldehyde group, glucose does not give Schiff test.
Statement-II : Glucose exists in \(\alpha\) and \(\beta\) crystalline forms.

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Glucose mainly exists in cyclic hemiacetal forms (\(\alpha\)- and \(\beta\)-glucose). Because the free aldehyde form is present only in a very small amount, glucose does not give Schiff's test even though it is a reducing sugar.
Updated On: Jun 26, 2026
  • Both statements I and II are incorrect
  • Both statements I and II are correct
  • Statement I is correct but statement II is incorrect
  • Statement I is incorrect but statement II is correct
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The Correct Option is B

Solution and Explanation

Step 1: Examine Statement-I.
Glucose contains an aldehyde group in its open-chain structure.
However, in aqueous solution and in crystalline form, glucose predominantly exists in cyclic hemiacetal forms rather than in the open-chain aldehyde form.
The concentration of the free aldehyde form is extremely small.
Therefore, glucose does not give Schiff's test, although it can reduce Fehling's solution and Tollens' reagent due to equilibrium with the open-chain form.
Hence, Statement-I is correct.

Step 2: Examine Statement-II.
Glucose exists in two cyclic crystalline forms known as: \[ \alpha\text{-D-glucose} \] and \[ \beta\text{-D-glucose} \] These two forms differ in the configuration of the hydroxyl group attached to the anomeric carbon (\(C_1\)).
They are called anomers and interconvert in solution through mutarotation.
Hence, Statement-II is also correct.

Step 3: Final conclusion.
Both statements are correct: \[ \boxed{\text{Statement-I is correct}} \] and \[ \boxed{\text{Statement-II is correct}} \] Therefore, \[ \boxed{\text{Both statements I and II are correct}} \] Hence, the correct option is \[ \boxed{(2)} \]
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