To compare the acidity of the $\alpha$-hydrogen, we use the stability of the conjugate base formed after removal of an $\alpha$-H.

1. Key idea

The more stable the enolate/carbanion formed, the more acidic the $\alpha$-hydrogen.

Stability of the conjugate base increases when:

1. The negative charge is better delocalized by resonance.
2. Electron-withdrawing groups are present.
3. More carbonyl groups are available to stabilize the anion.

2. General acidity trend of compounds with $\alpha$-hydrogen

Typical order is:

• Simple ketone/aldehyde: less acidic
• Ester: somewhat acidic
• 1,3-dicarbonyl compounds: much more acidic
• Compounds where the anion is stabilized by two strong electron-withdrawing carbonyl groups are the most acidic

Thus, among the given compounds, the one with least stabilization of the conjugate base will have the least acidic $\alpha$-H, and the one with maximum stabilization will have the most acidic $\alpha$-H.

From the structures implied in the question, the increasing order comes out as:

3. Checking options

• Option A: $(C) < (A) < (B) < (D)$ — incorrect
• Option B: $(B) < (C) < (A) < (D)$ — incorrect
• Option C: $(A) < (C) < (D) < (B)$ — incorrect
• Option D: $(D) < (C) < (A) < (B)$ — correct

4. Final answer

The increasing order of acidity of the $\alpha$-hydrogen is

So, the correct option is D.