Question

Which of the following statements is not correct for glucose?

• A. Glucose does not give Schiff's test.
• B. Glucose exists in two crystalline forms \(\alpha\)- and \(\beta\)-.
• C. The pentaacetate of glucose does not react with \( \mathrm{NH_2OH} \).
• D. Glucose forms addition product with \( \mathrm{NaHSO_3} \).

Hint:

Pentaacetate of glucose can still react with reagents like \( \mathrm{NH_2OH} \), indicating the presence of a carbonyl group.

Answer 1

The Correct Option is C

Step 1: Glucose contains an aldehyde group in its open-chain form and can react with hydroxylamine (\( \mathrm{NH_2OH} \)) to form an oxime. Step 2: The pentaacetate of glucose does not have a free –CHO group, but it still retains the carbonyl nature and can react with hydroxylamine. Step 3: Hence, the statement that pentaacetate of glucose does not react with \( \mathrm{NH_2OH} \) is incorrect.