JEE Main · 2011 · 10 Apr 2011 · Offline · Aldehydes Ketones and Carboxylic Acids

Acid strength here is set by how well the conjugate base is stabilised by electron withdrawal, and the $-\mathrm{I}$ effect is strongest when the withdrawing group sits closest to the carboxyl. Formic acid is moderately strong, acetic acid is weak because the methyl group pushes electrons in, and in $\mathrm{ClC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{COOH}$ the chlorine is three carbons away so its effect is faint. In $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{CH}(\mathrm{Cl})\mathrm{C}{\mathrm{O}}_2\mathrm{H}$ the chlorine sits on the $\alpha$-carbon, right next to the $-\mathrm{COOH}$, so its $-\mathrm{I}$ pull stabilises the carboxylate most and makes this the strongest acid of the set.