Question

How will you confirm the presence of five –OH groups in a glucose molecule, which are attached to different carbon atoms?

Hint:

Glucose + Acetic anhydride → Glucose pentaacetate Confirms 5 –OH groups.

Answer 1

Concept: Glucose contains multiple hydroxyl (\(-OH\)) groups. The presence and number of these groups can be confirmed through acetylation reactions.
Step 1: Acetylation of glucose. When glucose is treated with:

• Acetic anhydride (\((CH_3CO)_2O\))
• In presence of pyridine

all hydroxyl groups get acetylated.
Step 2: Formation of glucose pentaacetate. \[ \text{Glucose} + 5(CH_3CO)_2O \rightarrow \text{Glucose pentaacetate} + 5CH_3COOH \]
Step 3: Interpretation. Formation of
pentaacetate confirms:

• Presence of five hydroxyl groups
• Each attached to a different carbon atom

Conclusion: Since glucose forms a pentaacetate derivative, it confirms that glucose contains five alcoholic \(-OH\) groups. \[ \therefore \text{Glucose has five hydroxyl groups. \]