Which one of the following compounds possesses the most acidic hydrogen?

The acidity of an organic compound is governed by the stability of the conjugate base that remains after removal of a proton. Greater the stabilisation (-I effect, ‑R effect, hybridisation, resonance delocalisation, etc.), lower the $$pK_a$$ and hence stronger the acid.

Step 1 — Locate the hydrogen that can be lost as $$H^+$$ in each option.
A: alcoholic $$O\!-\!H$$ hydrogen of an alkanol.
B: terminal $$C\!\equiv\!C\!-\!H$$ hydrogen of a terminal alkyne.
C: no hetero-atom; the only hydrogens are on an sp² carbon of an alkene.
D: the methylene $$-CH_2-$$ sandwiched between two carbonyl groups in $$CH_3CO\!-\!CH_2\!-\!COCH_3$$ (2,4-pentanedione).

Step 2 — Write the conjugate bases and examine their stabilisation.

• Compound A (ethanol type): $$CH_3CH_2O^-$$ is stabilised only by the electronegative oxygen (no extra resonance with another group). $$pK_a \approx 16$$.

• Compound B (propyne): $$CH_3C\!\equiv\!C^-$$ is an sp hybrid carbanion (50 % s-character), so the negative charge is held closer to the nucleus than in sp²/sp³ cases. $$pK_a \approx 25$$ (less acidic than alcohols).

• Compound C (propene/alkene type): removal of a vinylic hydrogen would give an sp² carbanion with no resonance; $$pK_a \gt 40$$, so it is the least acidic among the list.

• Compound D (2,4-pentanedione): deprotonation at the central $$CH_2$$ gives an enolate ion

$$CH_3CO-CH^{-}-COCH_3 \;\;\longrightarrow\;\; \underset{\text{(I)}}{CH_3CO\!-\!CH^{-}\!-\!COCH_3} \;\;\rightleftharpoons\;\; \underset{\text{(II)}}{CH_3C O^{-}=CH-COCH_3} \;\;\rightleftharpoons\;\; \underset{\text{(III)}}{CH_3CO-CH=CO^{-}CH_3}$$

The negative charge is delocalised over TWO carbonyl oxygens through resonance structures (I)-(III). In addition, each carbonyl group withdraws electron density by the inductive (-I) effect. The combined resonance and inductive stabilisation makes this conjugate base far more stable than the others. Consequently, the hydrogen on that $$CH_2$$ has a $$pK_a \approx 9$$, which is markedly lower than for the other compounds.

Step 3 — Compare the $$pK_a$$ (acid strength) values.
$$CH_3COCH_2COCH_3 \;(pK_a \approx 9) \lt CH_3CH_2OH \;(16) \lt CH_3C\!\equiv\!CH \;(25) \lt \text{alkene} \;( \gt 40)$$

Lower $$pK_a$$ means stronger acid; hence compound D possesses the most acidic hydrogen.

Option D which is: $$CH_3CO-CH_2-COCH_3$$