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.
a) Exact-mass formula discrimination. Computing the monoisotopic exact mass of each candidate formula:
| Formula | Exact mass | |Δ| from 153.0680 |
|---|---|---|
| C8H11NO2 | 153.0790 | 11.0 mDa |
| C7H11N3O | 153.0902 | 22.2 mDa |
C8H11NO2 is roughly twice as close to the measured exact mass as C7H11N3O and is also the molecular formula that actually corresponds to dopamine's known structure (4-(2-aminoethyl)benzene-1,2-diol) — C8H11NO2 is the correct formula.
b) Distinguishing morphine, heroin, and oxycodone by IR. The three narcotics carry different combinations of O–H, phenol, and carbonyl groups, each with a diagnostic IR signature:
In short: presence/absence of O–H plus the exact carbonyl frequency and count (none / ester ≈ 1740 / ketone ≈1710) uniquely fingerprints each of the three.
c) Structure elucidation from spectral data.
i) No O–H/C=O in the IR (only C–H stretches) rules out any polar functional group — this is a saturated alkyl dibromide. A 6H singlet (no coupling partners) means two equivalent CH3 groups on a carbon bearing no H of its own; a 2H singlet further downfield (≈3.86, next to Br, also no coupling partners) means an isolated CH2Br flanked only by that same quaternary-like carbon. Structure: 1,2-dibromo-2-methylpropane, (CH3)2CBr–CH2Br.
ii) A 2H quintet (coupling to 4 equivalent neighbouring H, n+1=5) plus a 4H-equivalent triplet (two equivalent CH2Br groups, each coupling only to the 2H of the central CH2) is the classic pattern for a symmetric 1,3-disubstituted propane: 1,3-dibromopropane, Br–CH2–CH2–CH2–Br.
iii) IR 1740 cm-1 = ester carbonyl. Two triplet/quartet ethyl patterns (one pair ≈1.15/2.30, the other ≈1.24/4.72, the second shifted far downfield by the adjacent ester oxygen) is exactly the signature of an ethyl ester of a propanoyl group: ethyl propanoate, CH3CH2C(=O)OCH2CH3.
iv) IR 3600–3200 cm-1 = alcohol O–H. A 6H triplet + 4H quartet (two equivalent ethyl groups) plus a separate 3H singlet (a methyl with no neighbouring H, so attached to a carbon bearing no H) plus a 1H singlet (the variable, non-coupling O–H) fits a single symmetric tertiary alcohol: 3-methylpentan-3-ol, (CH3CH2)2C(OH)CH3.
v) No O–H in the IR (ether, not alcohol). Integration ratio 30:5 = 6:1, i.e. 12 equivalent methyl H (doublet, each coupling to one adjacent CH) for every 2 equivalent methine H (septet, each coupling to 6 equivalent neighbouring CH3 H) — the textbook signature of a symmetric diisopropyl ether: diisopropyl ether, (CH3)2CH–O–CH(CH3)2.
