04-BS-12 · December 2019
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
National Exam 04-BS-12, Organic Chemistry — December 2019. 3 hours, closed-book examination (no calculator; 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. (functional groups, stereochemistry, SN1/SN2 mechanisms and stereochemistry, steroid/bile-acid amphiphilicity, named-drug synthesis design, reaction-energy diagrams, polymer chemistry, arene-oxide metabolism, and mass-spectral/IR/NMR structure elucidation). Every molecular formula, exact mass, and stereochemical (R/S, cis/trans, meso/chiral) 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.
The starting alcohol is (R)-hexan-2-ol (priority order at C2 is OH > butyl > methyl > H). Three reagent sequences are applied, and the key question for each step is whether the C–O(or C– leaving-group) bond at the stereocentre is ever broken:
Path 1 (NaH, then CH3I): NaH simply deprotonates the O–H to give the sodium alkoxide A — the C–O bond at the stereocentre is untouched, so configuration is retained. CH3I then alkylates the alkoxide oxygen (Williamson ether synthesis, SN2 at the methyl carbon, not at the stereocentre) to give B = (R)-2-methoxyhexane — again the stereocentre's bonds are never broken, so the configuration is retained.
Path 2 (TsCl/pyridine, then CH3O-): TsCl/pyridine converts the O–H into a tosylate ester C (again only the O–H bond is touched, retention). Methoxide then performs a genuine back-side SN2 attack directly on the stereocentre, displacing OTs- — this inverts the configuration, giving D = (S)-2-methoxyhexane.
Path 3 (PBr3, then CH3O-): PBr3 converts the alcohol directly to the bromide via an SN2-type mechanism at the stereocentre (through a protonated dibromophosphite intermediate that is displaced by bromide) — this is a bond-breaking event at the stereocentre and inverts configuration, giving E = (S)-2-bromohexane. Methoxide then displaces bromide by a second back-side SN2 attack at the same carbon — a second inversion, giving F = (R)-2-methoxyhexane.
B and D are the same constitution (both 2-methoxyhexane) but opposite configuration at C2 (R vs. S) — B and D are enantiomers. B and F are both (R)-2-methoxyhexane, produced by pathways that undergo zero and two inversions respectively — net retention either way — so B and F are the identical compound.
