04-BS-12 · May 2017
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
National Exam 04-BS-12, Organic Chemistry — May 2017. 3 hours, closed-book examination (no textbook aid beyond one double-sided aid sheet); a Casio or Sharp approved calculator is permitted. The paper prints thirteen questions using plain "Question N:" numbering; all thirteen are answered in full below.
Reference texts: McMurry, Organic Chemistry, 9th ed. (functional-group reactivity, amide/β-lactam resonance, stereochemistry and specific rotation, SN1/SN2 and epoxide-opening regiochemistry, cyclohexane conformational analysis, IR/NMR spectroscopy, electrophilic aromatic substitution and multi-step synthesis design, polymer/step-growth chemistry); Atkins, Physical Chemistry, 11th ed. (entropy of intramolecular vs. intermolecular reactions).
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) The two pKas are not describing the same kind of bond. Dimethyl ether has no O–H at all — every hydrogen on it is a C–H, so "pKa = 40" refers to deprotonating an sp3 C–H alpha to the ether oxygen (barely acidified relative to a plain alkane C–H, pKa ≈ 50, by the oxygen's modest inductive pull). Ethanol's pKa = 16 refers to its O–H. Removing a proton from an O–H puts the resulting negative charge on a small, highly electronegative atom that stabilises it well (inductively, and its lone pairs are already non-bonding); removing a proton from a C–H puts the charge on carbon, which has low electronegativity and no comparable mechanism to accommodate it — the resulting methyl/methylene carbanion is dramatically higher in energy than an alkoxide. That gap in anion stability (carbanion vs. alkoxide) is the entire 24-pKa-unit difference; it has nothing to do with the two molecules being isomers, since the isomerism only guarantees the same molecular formula, not that the same type of bond is being ionised in each.
(b) Resonance delocalisation of the resulting anion/radical. The indicated hydrogen in 1,4-pentadiene sits on C3, flanked on both sides by a vinyl group — it is doubly allylic. Removing it generates a species (radical or anion) whose non-bonding electron density is delocalised by resonance over both adjacent π systems (density appears at C1, C3, and C5), spreading the negative charge/radical character over five carbons instead of confining it to one. Pentane's C2 hydrogen has no adjacent π system on either side; removing it gives an ordinary, localised secondary carbanion/radical with zero resonance stabilisation. Delocalisation lowers the energy of the conjugate base substantially, so the doubly-allylic C–H ionises far more readily (is far more acidic) than the plain alkane C–H, even though both are formally "just C–H bonds."
