Para-iodobenzoic acid (pKa ≈ 4.9) and meta-iodobenzoic acid (pKa ≈ 4.8) have nearly…
CHEMISTRY · Single correct · CBSE · Chemistry · Organic Chemistry - Some Basic Principles and Techniques · Organic Chemistry - Some Basic Principles and Techniques
Problem
Para-iodobenzoic acid (pKa ≈ 4.9) and meta-iodobenzoic acid (pKa ≈ 4.8) have nearly identical acid strengths, whereas para-fluorobenzoic acid (pKa ≈ 4.4) is appreciably less acidic than meta-fluorobenzoic acid (pKa ≈ 4.1). Which statement best explains this pattern?
(a) Fluorine withdraws electron density through induction more strongly when para to the carboxyl group; iodine is only weakly inductive, so its meta and para acids have similar strengths.
(b) Resonance donation by halogens stabilises the conjugate base and therefore strengthens the acid; because iodine cannot donate effectively, only the para-fluoro isomer shows the additional stabilization.
(c) Fluorine donates electron density through resonance more effectively than iodine, offsetting its inductive withdrawal when placed para to the carboxyl group.
(d) Only inductive effects influence benzoic-acid acidity, and since inductive effects diminish with distance, the meta isomer is always the stronger acid regardless of the halogen involved.