Among the following marked proton of which compound shows lowest pKa value?

Lower $$\mathrm{pK_a}$$ value indicates higher acidity.

Options $$\mathrm{A}$$ and $$\mathrm{B}$$ can be eliminated because the conjugate bases formed from options $$\mathrm{C}$$ and $$\mathrm{D}$$ are additionally stabilised by resonance due to their benzylic positions adjacent to carbonyl groups.

In option $$\mathrm{C}$$, the highlighted $$\mathrm{\alpha}$$-hydrogen is adjacent to a ketone group.

On removal of this proton, the resulting carbanion is strongly stabilised by resonance with the carbonyl group.

$$\mathrm{R-CO-CH_2^- \rightleftharpoons R-C(O^-)=CH_2}$$

The ketone carbonyl exerts a strong electron-withdrawing:

$$\mathrm{-M\ effect}$$

thereby stabilising the conjugate base.

In option $$\mathrm{D}$$, the highlighted hydrogen is adjacent to a carboxylic acid group:

$$\mathrm{(-COOH)}$$

The $$\mathrm{-OH}$$ group donates electron density through resonance:

$$\mathrm{+M\ effect}$$

which reduces the electron-withdrawing nature of the carbonyl group.

Hence, the carboxylic acid carbonyl stabilises the adjacent carbanion less effectively than the ketone in option $$\mathrm{C}$$.

Also, the carboxylic acid possesses a highly acidic $$\mathrm{O-H}$$ proton.

A base preferentially removes this proton first:

$$\mathrm{R-COOH + Base \longrightarrow R-COO^-}$$

Further removal of the highlighted $$\mathrm{C-H}$$ proton would generate a dianion, which is highly unstable.

Therefore, abstraction of the highlighted proton in option $$\mathrm{D}$$ is thermodynamically unfavourable.

Hence, the most acidic highlighted hydrogen is present in:

$$\boxed{\mathrm{Option\ C}}$$