Question 12 of 13: Explaining Relative Acidities (pK a )

Question 12: Explaining Relative Acidities (pK_a) (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

1. a) p-Nitrophenol (7.2) < phenol (10) — resonance withdrawal by a para substituent. Deprotonating phenol gives a phenolate anion whose negative charge is already resonance-delocalised onto the ring oxygen and the ortho/para ring carbons. When a nitro group sits at the para carbon, one of those very resonance structures places the negative charge directly on a ring carbon adjacent to the nitro group's own electron-withdrawing π-system, allowing a further resonance structure that pushes the charge all the way onto a nitro oxygen: $$\mathrm{ArO^- \; (\text{charge on ring C para to NO}_2) \; \longleftrightarrow \; \text{charge delocalised onto a nitro O}}$$ This extra resonance stabilisation of the conjugate base (not available to plain phenol, which has no nitro group to accept the charge) lowers the free energy of the anion substantially, making p-nitrophenol a much stronger acid (lower pK_a).
2. b) p-Nitrophenol (7.2) < m-nitrophenol (8.3) — only ortho/para substituents can resonance-stabilise the phenolate. The key resonance structures of the phenolate place negative charge only on the ring carbons ortho and para to the oxygen — never on a meta carbon. A nitro group sitting at the meta position therefore cannot participate in that direct resonance delocalisation at all; it can only stabilise the anion by a weaker, through-bond inductive effect. Since resonance withdrawal is much more powerful than induction alone, p-nitrophenol (nitro at a resonance-accessible position) is more acidic than m-nitrophenol (nitro only inductively active), even though both isomers have exactly the same electron-withdrawing group and the same molecular formula.
3. c) Carboxylic acids A (2-chloropropanoic acid) and B (3-chloropropanoic acid) vs. propanoic acid — inductive withdrawal, falling off with distance. A chlorine atom is strongly electronegative and withdraws electron density inductively through the σ-bond framework, stabilising the carboxylate conjugate base (spreading out its negative charge) regardless of whether the resulting position is capable of resonance with the carboxyl (induction works through any number of σ-bonds, just progressively more weakly). Both A and B therefore have a lower pK_a than unsubstituted propanoic acid. Because inductive effects fall off rapidly with distance (roughly with the number of intervening bonds), A (Cl on C2, directly adjacent to the carboxyl carbon) is stabilised more than B (Cl on C3, one carbon further away), so A is more acidic than B, and both are more acidic than plain propanoic acid (no chlorine at all). $$\mathrm{pK_a:\quad \underbrace{2\text{-chloropropanoic acid (A)}}_{\approx 2.8} \;<\; \underbrace{3\text{-chloropropanoic acid (B)}}_{\approx 4.0} \;<\; \underbrace{\text{propanoic acid}}_{\approx 4.9}}$$