Case Study
Passage with linked questions
Case Set 1
Case AnalysisPassage
Ravi is a nutrition student studying the role of carbohydrates in the human diet. He observed that athletes often consume glucose drinks for instant energy during competitions. Glucose is an aldohexose with the molecular formula C6H12O6. It is found in sweet fruits, honey, and ripe grapes. In the laboratory, Ravi prepared glucose by boiling starch with dilute H2SO4 at 393 K under 2-3 atm pressure. He also noted that glucose exists in two cyclic forms, α and β, due to the hemiacetal ring formation, and that only the open-chain form reacts with certain reagents like hydroxylamine. He further confirmed the presence of five hydroxyl groups by acetylating glucose with acetic anhydride, yielding glucose pentaacetate.
Question 1: What is the molecular formula of glucose and to which class of carbohydrates does it belong?
- Molecular formula of glucose is C6H12O6.
- Glucose is a monosaccharide and belongs to the class of aldohexoses.
Question 2: What is the significance of glucose pentaacetate formation, and what does it confirm about the structure of glucose?
- Acetylation of glucose with acetic anhydride gives glucose pentaacetate.
- This confirms the presence of five free –OH groups, each attached to a different carbon atom in the glucose molecule.
Question 3: Why does glucose not give Schiff's test even though it contains an aldehyde group? Explain the cyclic structure formation that accounts for this behaviour.
- Glucose exists predominantly in cyclic hemiacetal forms (α and β) rather than the open-chain aldehyde form.
- In the cyclic form, the –OH at C-5 adds to the –CHO group at C-1, forming a six-membered pyranose ring.
- The free aldehyde group is thus absent in the pentaacetate of glucose, so it does not give Schiff's test or form hydrogen sulphite addition product with NaHSO3.
- The two cyclic forms differ in configuration at C-1 (anomeric carbon) and are called α-D-(+)-glucopyranose and β-D-(+)-glucopyranose.