Give reasons as the evidence in support of cyclic structure of glucose.

Explanation

To explain the cyclic structure of glucose, we can follow these steps:

### Step 1: Understanding the Nature of Glucose
Glucose is a simple sugar (monosaccharide) that can exist in different forms, including an open-chain (linear) form and a cyclic form. The cyclic form is more stable and predominant in solution.

### Step 2: Aldehyde Characteristics
Glucose has an aldehyde group in its open-chain form. Aldehydes typically react positively in tests such as Schiff's test and the 2,4-DNP test, indicating the presence of a free aldehyde group.

### Step 3: Observing Test Results
When glucose is subjected to Schiff's test, it gives a negative result. Similarly, it also gives a negative result in the 2,4-DNP test, which usually indicates that there is no free aldehyde group present.

### Step 4: Reaction with Sodium Bisulfite
Glucose does not form a bisulfite addition product when reacted with sodium bisulfite (NaHSO3). This further supports the idea that there is no free aldehyde group available for reaction, which would be expected if glucose were in its open-chain form.

### Step 5: Reaction with Hydroxylamine
Glucose does react with hydroxylamine, which suggests the presence of a carbonyl group. However, when glucose is converted into its pentaacetate derivative, it does not react with hydroxylamine. This indicates that the carbonyl group is not free in the pentaacetate form, supporting the idea that glucose is cyclic.

### Step 6: Conclusion
The negative results from the tests for free aldehyde groups, combined with the behavior of glucose derivatives, suggest that glucose predominantly exists in a cyclic form rather than an open-chain form. The cyclic structure is more stable and is the reason for the observed negative test results.