Which will undergo deprotonation most readily in the basic medium?

Solution

Chemistry ma’am · 3 steps

Look at the CH₂ between two C=O groups: ketones stabilise its carbanion better than esters.

1. Active methylene. In each compound the CH₂ sits between two C=O groups. Removing an H⁺ there gives a carbanion whose negative charge spreads onto both carbonyl oxygens. The more stable this enolate, the faster the deprotonation.
2. Ester carbonyl is weaker. In an ester, the lone pair on OCH₃ already pushes into its own C=O by resonance (+R). That carbonyl is partly satisfied, so it pulls the carbanion's charge less strongly than a ketone carbonyl does.

   Compound	Groups flanking CH₂	Approx. pKₐ
   a, pentane-2,4-dione	ketone + ketone	about 9, most acidic
   c, methyl acetoacetate	ketone + ester	about 11
   b, dimethyl malonate	ester + ester	about 13

3. So. The acidity order is a > c > b. Compound a, with two ketone groups and no +R donor competing, is deprotonated most readily, which is option (a).

The idea to keep. For active methylene compounds, every ester OR in place of an alkyl lowers acidity: diketone > keto ester > diester. A +R group on the carbonyl weakens its pull on the carbanion.

Answer from the NTA final answer key. Solution by PrepWiser.