The pKa of acetone, CH3COCH3, is 19.3. Which of the following bases is strong enough to de-protonate acetone?
(a) KOH (pKa of H2O = 15.7)
(b) Na+ –C ≡ CH (pKa of C2H2 = 25)
(c) NaHCO3 (pKa of H2CO3 = 6.4)
(d) NaOCH3 (pKa of CH3OH = 15.6)
The Correct Answer and Explanation is :
To determine which base is strong enough to deprotonate acetone (CH₃COCH₃, pKa = 19.3), we must compare the pKa values of acetone and the conjugate acids of the bases provided.
Concept: Acid-Base Strength and Deprotonation
A base can deprotonate an acid if the conjugate acid of the base has a higher pKa than the acid being deprotonated. This is because a larger pKa means the conjugate acid is weaker, making the base stronger.
Analyzing the Bases
- KOH (pKa of H₂O = 15.7)
- H₂O (conjugate acid) has a pKa of 15.7.
- Since 15.7 < 19.3, KOH is too weak to deprotonate acetone.
- Na⁺ –C ≡ CH (pKa of C₂H₂ = 25)
- HC≡CH (acetylene, conjugate acid) has a pKa of 25.
- Since 25 > 19.3, this base is strong enough to deprotonate acetone.
- NaHCO₃ (pKa of H₂CO₃ = 6.4)
- H₂CO₃ (carbonic acid, conjugate acid) has a pKa of 6.4.
- Since 6.4 < 19.3, NaHCO₃ is too weak to deprotonate acetone.
- NaOCH₃ (pKa of CH₃OH = 15.6)
- CH₃OH (methanol, conjugate acid) has a pKa of 15.6.
- Since 15.6 < 19.3, NaOCH₃ is too weak to deprotonate acetone.
Conclusion
Only Na⁺ –C≡CH (sodium acetylide) has a conjugate acid (HC≡CH) with a higher pKa (25) than acetone (19.3). Thus, it is strong enough to deprotonate acetone.
Final Answer:
(b) Na⁺ –C ≡ CH