04-BS-12 · December 2016
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
National Exam 04-BS-12, Organic Chemistry — December 2016. 3 hours, closed-book examination (no non-communicating calculator restriction beyond the standard aid sheet, 8.5×11", hand-written both sides). Ten questions constitute a complete exam paper (only the first 10 questions as they appear in the answer book are marked, each of equal value) — the source paper in fact prints thirteen questions; all thirteen are answered in full below.
Reference texts: McMurry, Organic Chemistry, 9th ed. (acid/base theory of drugs, SN1/SN2 stereochemistry, carbocation rearrangements, alkyne synthesis via acetylide alkylation, IR/NMR structure elucidation, electrophilic aromatic substitution and synthesis design, amino-acid pKa); Clayden, Organic Chemistry, 2nd ed. (amide resonance and β-lactam reactivity, radical vs. ionic HBr addition mechanisms); a standard biomaterials reference for the poly(ester amide) drug-delivery polymer chemistry of Question 13 (Katsarava-type AABB poly(ester amide)s built from diacids, diols, and protected diamino acids).
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.
(a) A para-nitro group withdraws electron density both inductively and by direct resonance delocalisation into the phenoxide oxygen, stabilising the conjugate base far more than phenol's (unsubstituted) phenoxide. Deprotonating phenol gives a phenoxide whose negative charge is delocalised only around the ring by the ring's own resonance; deprotonating p-nitrophenol gives a phenoxide where a resonance structure places the negative charge directly on a nitro oxygen, because the nitro group's position (para) allows a continuous conjugated path from the phenoxide oxygen, through the ring, into the nitro group.
(b) The para nitro group can resonance-stabilise the phenoxide directly; the meta nitro group cannot — only its (weaker) inductive effect operates. Drawing out the phenoxide resonance structures for the meta isomer shows that no resonance arrow ever places negative charge on a ring carbon adjacent to the nitro group's point of attachment, so the nitro group's powerful resonance (mesomeric) electron withdrawal is simply unavailable from the meta position — only the inductive (through-bond, distance-decaying) electron withdrawal remains, which is weaker.
(c) Both A and B are α- and β-halogen-substituted analogues of propanoic acid; the inductive electron-withdrawal from the halogen stabilises the carboxylate, with the effect falling off sharply with distance from the –COOH. a directly α-halogenated acid (A) vs. the same halogen one carbon further away, at the β-position (B) — which isolates inductive distance-dependence as the only variable and is explicitly disclosed as a reconstruction below.
Chlorine is strongly electronegative and withdraws electron density inductively through the σ-bond framework; this effect stabilises the carboxylate conjugate base (spreading its negative charge partly onto/toward the electronegative halogen) and therefore increases acidity relative to unsubstituted propanoic acid, for either position. Because inductive effects fall off rapidly with the number of intervening bonds, the α-chloro acid A (Cl directly on the carbon next to –COOH) is stabilised more, and is therefore more acidic, than the β-chloro acid B (Cl one carbon further away) — both, however, remain more acidic than propanoic acid itself, which has no halogen at all.
| Comparison | Stronger acid | Reason |
|---|---|---|
| phenol vs. p-nitrophenol | p-nitrophenol | NO2 resonance-stabilises the phenoxide directly (para) |
| m- vs. p-nitrophenol | p-nitrophenol | only para allows resonance donation of charge onto NO2; meta is inductive-only |
| propanoic acid vs. A, B | A > B > propanoic acid | inductive withdrawal by Cl, strongest closest to –COOH (A, α) > further away (B, β) |
