Question

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

Hint:

A very important structural proof of glucose is: \[ \text{Glucose} \xrightarrow[\text{Pyridine}]{(CH_3CO)_2O} \text{Glucose Pentaacetate} \] Formation of a pentaacetate derivative confirms the presence of five alcoholic \((-OH)\) groups in glucose.

Answer 1

Step 1: Principle of acetylation for counting $-OH$ groups.
Each $-OH$ group reacts with one molecule of acetic anhydride $(CH_3CO)_2O$ in pyridine, replacing the $-H$ with an acetyl group $(-COCH_3)$. Counting the acetyl groups introduced equals counting the $-OH$ groups present.
Step 2: React glucose with excess acetic anhydride.
\[ \text{Glucose} + 5(CH_3CO)_2O \xrightarrow{\text{Pyridine}} \text{Glucose Pentaacetate} + 5CH_3COOH \] Exactly five acetyl groups are introduced, proving glucose contains five $-OH$ groups.
Step 3: Confirm $-OH$ groups are on different carbon atoms.
If any two $-OH$ groups shared the same carbon (geminal diol), the arrangement would be unstable. The formation of a stable pentaacetate derivative confirms all five $-OH$ groups are on five different carbon atoms.
\[ \boxed{\text{Glucose pentaacetate formation confirms 5 } -OH \text{ groups on different carbons}} \]