04-BS-12 · May 2018
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
National Exam 04-BS-12, Organic Chemistry — May 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. (drug acid–base/salt pharmacokinetics, steroid/bile-acid amphiphilicity, arene-oxide metabolism, cyclopropane stereochemistry and CIP assignment, reaction-energy diagrams, ester equilibria and intramolecular effective molarity, SN2 stereochemistry at a common stereocentre, named-drug synthesis design, epoxide ring-opening stereochemistry, mass-spectral formula discrimination, opioid IR/NMR structure elucidation, keto–enol tautomerism and conjugation/acidity, and condensation-polymer monomer identification). Every molecular formula, mass-balance, exact-mass, and stereochemical (R/S) 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.
| Branch | Step 1 | Intermediate | Step 2 | Product |
|---|---|---|---|---|
| Top | NaH | A = sodium (R)-hexan-2-olate | CH3I | B = (R)-2-methoxyhexane |
| Middle | TsCl, pyridine | C = (R)-hexan-2-yl tosylate | CH3O− | D = (S)-2-methoxyhexane |
| Bottom | PBr3 | E = (S)-2-bromohexane | CH3O− | F = (R)-2-methoxyhexane |
Why each step retains or inverts. The stereocentre is C2. In the top branch, NaH simply deprotonates the O–H (no bond to C2 is touched, so configuration is retained in the alkoxide A); the alkoxide oxygen then performs the SN2 substitution as the nucleophile attacking CH3I — the new bond forms at the methyl carbon of CH3I, not at C2, so C2's configuration is again completely undisturbed. B is therefore (R), the same configuration as the starting alcohol (net retention, because the stereocentre was never the site of bond-breaking).
In the middle branch, TsCl/pyridine only replaces the O–H hydrogen with a tosyl group (again, no bond to C2 is broken); C2 is still (R) in tosylate C. But the second step is different in kind: methoxide now attacks directly at C2, displacing the tosylate leaving group in a single-step, backside-attack SN2 — this is a classic Walden inversion, so D is (S), the enantiomer of B.
In the bottom branch, PBr3 converts the alcohol to the bromide via attack of bromide at C2 (with the oxygen, now activated as O–PBr2, as the leaving group) — this bond-breaking-at-C2 event inverts the centre, giving E = (S)-2-bromohexane. The second step, methoxide displacing bromide, is again a direct SN2 at C2 and inverts a second time, giving F = (R)-2-methoxyhexane — back to the original spatial arrangement, because two inversions at the same centre net out to overall retention.
B and D: both are 2-methoxyhexane (identical connectivity) but opposite absolute configuration — B and D are enantiomers. The mechanistic reason is that B was made by a route that never touches C2 (retention), while D was made by a route with a single inversion at C2.
B and F: both routes leave C2 as (R) — B by never disturbing the stereocentre at all, F by inverting it twice (once via PBr3, once via the second SN2) — two inversions restore the original handedness. B and F are therefore the same compound (identical constitution and identical absolute configuration), even though they were synthesised by mechanistically very different paths.
