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.
a) Exact-mass formula discrimination. Summing exact isotope masses (1H 1.007825, 12C 12.000000, 14N 14.003074, 16O 15.994915) gives C8H11NO2 = 153.0790 and C7H11N3O = 153.0902. Both round to the same nominal (integer) mass 153, which is exactly why the low-resolution spectrum cannot distinguish them — but the high-resolution measurement (153.0680) sits much closer to the C8H11NO2 value (off by 0.011) than to the C7H11N3O value (off by 0.022), so C8H11NO2 is the correct formula — consistent with dopamine's real structure, a catechol (two phenolic OH groups) with a 2-aminoethyl side chain.
b) Distinguishing morphine, heroin and oxycodone by IR.
The three share the same morphinan skeleton but differ in exactly the functional groups IR is best at telling apart. Morphine is the only one of the three with two free –OH groups (one phenolic, one secondary alcohol) — it shows a broad O–H stretch near 3200–3550 cm-1 and no ester or ketone carbonyl band. Heroin is morphine's diacetate ester: both –OH groups are capped as acetate esters, so the broad O–H stretch is essentially absent, replaced by a strong, sharp ester C=O stretch near 1740 cm-1 (and the corresponding C–O stretches). Oxycodone has no phenolic OH (it is capped as a methyl ether, so no O–H stretch there), retains one tertiary –OH (a weaker, sharper O–H band than morphine's, since it cannot donate as effectively and there is no second OH), and — distinctively — carries a ketone carbonyl (the ring is oxidised relative to morphine/heroin), giving a strong C=O stretch near 1715 cm-1 where the other two show none. In short: broad OH only/no C=O → morphine; ester C=O near 1740, weak/no OH → heroin; ketone C=O near 1715 plus a weaker OH → oxycodone.
c) Structure proposals from spectral data.
i) C4H8Br2, no OH/C=O IR band (only C–H stretches, 3000–2850), and two singlets (6H and 2H, no neighbouring H's on either carbon) fits 1-bromo-2-methyl-2-(bromomethyl)propane, i.e. (CH3)2CBr–CH2Br: the gem-dimethyl group (6H) has no adjacent H's (singlet), and the CH2Br (2H) is flanked only by the fully substituted quaternary-like bromo carbon (also a singlet); both integrations and the total formula match exactly.
ii) C3H6Br2, a central CH2 quintet (coupled to 4 neighbouring H's) and a CH2 triplet fits the symmetric 1,3-dibromopropane, BrCH2CH2CH2Br: the two equivalent terminal CH2Br groups (4H total) each couple only to the central CH2 (triplet), and the central CH2 couples to all 4 of those H's (quintet).
iii) C5H10O2 with a 1740 cm-1 ester carbonyl and two overlapping ethyl (triplet+quartet) patterns fits ethyl propanoate, CH3CH2C(=O)OCH2CH3: the propanoyl ethyl group (1.15/2.30) and the ester-oxygen ethyl group (1.24/4.72, deshielded by the adjacent ester oxygen) are two chemically distinct ethyls, each showing its own triplet/quartet pair.
iv) C6H14O, an O–H stretch (3600–3200) with no carbonyl, and the pattern 6H triplet + 4H quartet (two equivalent ethyl groups) + 3H singlet (an isolated methyl with no neighbouring H) + 1H singlet (exchangeable OH) fits the symmetric tertiary alcohol 3-methylpentan-3-ol, (CH3CH2)2C(OH)CH3: the two equivalent ethyls give one triplet + one quartet, the methyl on the quaternary alcohol carbon has no vicinal H's (singlet), and the OH proton (variable shift, singlet, no coupling to carbon-bound H's) integrates for 1H.
v) C6H14O, no OH/C=O IR band, and a clean doublet + septet pair in a 6:1 intensity ratio (the "30 units"/"5 units" are simply an arbitrary intensity scale, not literal proton counts, but their 6:1 ratio is exactly the true 12H:2H ratio) fits the fully symmetric diisopropyl ether, (CH3)2CH–O–CH(CH3)2: all four equivalent methyls (12H) appear as one doublet (coupled to their own CH), and both equivalent methine H's (2H) appear as one septet (each coupled to 6 equivalent methyl H's).
