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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)

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Answer:

After deprotonation of tannin, tannin (Ar-OH) is converted to [tex]Ar-O^{-}Na^{+}[/tex] salt which is soluble in water and [tex]CO_{3}^{2-}[/tex] is converted to [tex]HCO_{3}^{-}Na^{+}[/tex] 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 [tex]HCO_{3}^{-}[/tex]
  • After deprotonation of tannin, tannin (Ar-OH) is converted to [tex]Ar-O^{-}Na^{+}[/tex] salt which is soluble in water and [tex]CO_{3}^{2-}[/tex] is converted to [tex]HCO_{3}^{-}Na^{+}[/tex] which is also soluble in water
  • Balanced equation: [tex]Ar-OH+Na_{2}CO_{3}\rightarrow Ar-O^{-}Na^{+}+Na^{+}HCO_{3}^{-}[/tex]
  • Mechanism has been shown below.
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