04-BS-12 · Undated paper
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
National Exam 04-BS-12, Organic Chemistry — May 2019 sitting (the page-1 header and the running footer, "04-BS-12/May 2019", both give the date). 3 hours, closed-book examination; one Casio/Sharp-approved calculator permitted. NOTES on page 1 state that TEN (10) questions constitute a complete exam paper and only the first 10 as they appear in the answer book are marked, but this sitting prints 13 numbered questions — every question and sub-part below is answered in full.
Reference texts: McMurry, Organic Chemistry, 9th ed. (acid/base theory, functional-group identification, SN1/SN2 mechanisms and stereochemistry, alkyne/acetylide synthesis, electrophilic aromatic substitution, IR/NMR/mass-spectral structure elucidation, named-drug synthesis design, and step-growth polymer chemistry). Every molecular formula, exact mass, and stereochemical (R/S, cis/trans) assignment below.
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) p-Nitrophenol vs. phenol. Deprotonating either phenol gives a phenoxide ion whose negative charge is delocalised into the ring by resonance. In p-nitrophenoxide, the nitro group sits directly para to the O−, so one of the phenoxide's resonance structures places the negative charge directly on the nitro group's own oxygen (a genuine, drawable resonance structure, not merely an inductive effect) — an exceptionally effective way to delocalise and stabilise the anion, since the charge ends up on oxygen, the most electronegative atom available, rather than only on ring carbons. This large extra resonance stabilisation of the conjugate base lowers p-nitrophenol's pKa well below plain phenol's, which has no such resonance/electron-withdrawing pathway available at all.
b) p-Nitrophenol vs. m-nitrophenol. Resonance delocalisation of negative charge from a ring position onto a substituent requires that substituent to be conjugated through the ring's alternating double-bond system directly to the position bearing the charge — which only happens from the ortho and para positions relative to O−, never from meta. In m-nitrophenoxide, the nitro group can only withdraw electron density inductively (through the sigma bonds) — a real but much weaker effect with no direct resonance contribution placing charge on the nitro oxygens. m-Nitrophenol is therefore still more acidic than plain phenol (induction alone still helps), but substantially less acidic than p-nitrophenol, whose resonance pathway is unavailable to the meta isomer.
c) Carboxylic acids A and B vs. propanoic acid.
Both A (2-chloropropanoic acid, Cl on the carbon directly adjacent to the carboxyl) and B (3-chloropropanoic acid, Cl one carbon further away) are more acidic than propanoic acid because chlorine is strongly electronegative and inductively withdraws electron density through the sigma framework, which stabilises the conjugate-base carboxylate by dispersing its negative charge further. Because inductive effects fall off sharply with distance (roughly with the square of the separation), A (alpha-chloro) is substantially more acidic than B (beta-chloro), even though both are more acidic than the parent acid with no chlorine at all.
