Question

Write a mechanism (using curved-arrow notation) for the deprotonation of tannins in base. Use Ar-OH as a generic form of a tannin and use sodium carbonate (Na2CO3) as the base. Balance the chemical equation. Comment on the aqueous solubilities of each species (assume that the tannin is insoluble in water initially)

Answer 1

Answer:

After deprotonation of tannin, tannin (Ar-OH) is converted to $Ar-O^{-}Na^{+}$ salt which is soluble in water and $CO_{3}^{2-}$ is converted to $HCO_{3}^{-}Na^{+}$ which is also soluble in water

Explanation:

• Sodium carbonate is a strong electrolyte which produces carbonate anion upon dissociation.
• Carbonate ion is a strong conjugate base of weak acid $HCO_{3}^{-}$
• After deprotonation of tannin, tannin (Ar-OH) is converted to $Ar-O^{-}Na^{+}$ salt which is soluble in water and $CO_{3}^{2-}$ is converted to $HCO_{3}^{-}Na^{+}$ which is also soluble in water
• Balanced equation: $Ar-OH+Na_{2}CO_{3}\rightarrow Ar-O^{-}Na^{+}+Na^{+}HCO_{3}^{-}$
• Mechanism has been shown below.