Question

Prove the presence of five –OH groups and a –CHO group in glucose molecule, giving chemical equations. How is silver mirror formed from glucose?

Hint:

Remember: Glucose → Penta-acetate (5 –OH groups) and reacts with Tollen’s reagent (–CHO group). This is the classical proof of its structure.

Answer 1

Step 1: Presence of five –OH groups.
When glucose is acetylated with acetic anhydride, it forms penta-acetate, showing the presence of five –OH groups. \[ C_6H_{12}O_6 + 5 (CH_3CO)_2O \;\longrightarrow\; C_6H_7O(OCOCH_3)_5 + 5CH_3COOH \] Thus, glucose contains five hydroxyl groups. Step 2: Presence of –CHO group.
Glucose reacts with hydroxylamine (NH$_2$OH) to form oxime, and with hydrogen cyanide (HCN) to form cyanohydrin. These reactions prove the presence of the –CHO group in glucose. \[ C_6H_{12}O_6 + NH_2OH \;\longrightarrow\; C_6H_{11}O_6N (Oxime) \] \[ C_6H_{12}O_6 + HCN \;\longrightarrow\; C_6H_{11}O_6CN (Cyanohydrin) \] Step 3: Silver mirror test.
When glucose is warmed with Tollen’s reagent (ammoniacal AgNO$_3$), the –CHO group is oxidised to –COOH, and metallic silver is deposited on the walls of the test tube as a shining silver mirror. \[ C_6H_{12}O_6 + 2[Ag(NH_3)_2]^+ + 3OH^- \;\longrightarrow\; C_6H_{12}O_7 + 2Ag \downarrow + 4NH_3 + H_2O \] Conclusion:
Glucose contains five alcoholic –OH groups and one aldehydic –CHO group. The aldehyde group is confirmed by Tollen’s reagent test, giving a silver mirror.