Question

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

• A. Both statements I and II are incorrect
• B. Both statements I and II are correct
• C. Statement I is correct but statement II is incorrect
• D. Statement I is incorrect but statement II is correct

Hint:

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.

Answer 1

The Correct Option is B

Step 1: Analyze Statement-I about glucose and Schiff's test.
Schiff's test detects free aldehyde groups. Glucose has an aldehyde group ($-CHO$) in its open-chain form. However, in aqueous solution, glucose exists overwhelmingly in cyclic hemiacetal (pyranose) forms, with only a tiny fraction as the free open-chain aldehyde.
Step 2: Explain why glucose gives a negative Schiff's test.
Schiff's reagent needs a sufficient concentration of free aldehyde to give a visible color change. Since the free aldehyde form of glucose is present in extremely low concentration, the test is negative. Glucose still reduces Fehling's solution because that reaction is slow enough to shift the equilibrium gradually. Statement-I is correct.
Step 3: Analyze Statement-II about glucose existing in alpha and beta forms.
When glucose forms a pyranose ring, the anomeric carbon ($C_1$) becomes a new chiral center, giving two distinct cyclic forms: alpha-D-glucose and beta-D-glucose, differing in the orientation of $-OH$ at $C_1$.
Step 4: Confirm the existence of both forms.
Both alpha-D-glucose and beta-D-glucose are well-characterized crystalline compounds. In solution, they interconvert through the open-chain form (mutarotation), reaching an equilibrium mixture. Statement-II is correct.
Step 5: Evaluate the answer options.
Statement-I: Correct. Statement-II: Correct. Both statements are independently true.
Step 6: State the final answer.
\[ \boxed{\text{Both statements I and II are correct.}} \]