04-BS-12 · December 2018
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
National Exam 04-BS-12, Organic Chemistry — December 2018. 3 hours, closed-book examination (one Casio/Sharp-approved calculator and one hand-written aid sheet 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. (Brønsted acid–base sites in drugs, SN1/SN2 mechanism selection, Williamson ether synthesis, SN2 stereochemistry at a stereocentre, steroid/bile-acid amphiphilicity, named-drug synthesis design, fatty-acid melting-point trends, epoxide/alkene interconversion chemistry, radical stability and antioxidants, Diels–Alder stereochemistry, bicyclic-ketal pheromone synthesis, arene-oxide metabolism, and mass-spectral/IR/NMR structure elucidation). 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.
X is the ascorbate ion: ascorbic acid's more acidic enediol O–H (conjugated directly with the ring C=O through the C=C) is already removed, leaving one O– and one neutral O–H on the same C=C. Abstracting the indicated H from that remaining O–H removes the second enediol hydrogen, generating an oxygen radical that is not localised on one oxygen at all: it is delocalised by resonance across three oxygens spread over the conjugated enediol–C=O–O(ring) system (an O• ↔ O• on the far enediol oxygen ↔ a ring-oxygen-centred form), the same kind of extended, multi-atom resonance stabilisation ("captodative"-style delocalisation) that makes a semiquinone radical so unusually stable.
This is why that particular H is the easiest one in the whole molecule to remove: O–H bond homolysis anywhere else in ascorbate (the two side-chain O–H's) would give an ordinary, non-delocalised, high-energy alkoxyl radical, whereas abstracting this specific enediol H gives a radical so resonance-stabilised that it is essentially unreactive — it does not go on to abstract another H from a lipid or other biomolecule and so it terminates the radical chain reaction rather than propagating it. That "stop here, don't propagate" property is exactly what makes ascorbate an effective antioxidant.
