Question

Which of the given statement is not true for glucose?

• A. The pentacetate glucose does not react with hydroxylamine to give oxime.
• B. Glucose reacts with hydroxylamine to form oxime.
• C. Glucose gives Schiff's test for aldehyde.
• D. Glucose exists in two crystalline forms $\alpha$ and $\beta$

Answer 1

The correct option is C Glucose gives Schiff's test for aldehyde.

Glucose exists in two crystalline forms $\alpha$ and $\beta$ which are anomers of each other.
Glucose does not react with Schiff’s reagent because after the internal cyclisation, it forms either $\alpha$-anomer or $\beta$-anomer. In these forms, free aldehydic group is not present.
Glucose forms open chain structure in aqueous solution which contains aldehyde at chain end. This aldehydic group reacts with $N{H}_{4}OH$ to form oxime. On the other hand, glucose penta acetate being a cyclic structure even in aqueous form does not have terminal carbonyl group. Therefore it will not react with $N{H}_{4}OH.$