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.
Reading the drawings first. The paper draws each substituent bond as an explicit wedge (toward viewer) or hash (away from viewer); compound A is the odd one out — both its methyl bonds (one wedge, one hash) come off the same ring carbon, i.e. A is 1,1-dimethylcyclopropane, not a 1,2-isomer. B, C and D are all 1,2-dimethylcyclopropane, differing only in which face each methyl occupies: B has both methyls wedge (same face, cis); C has the left methyl hashed and the right wedge; D has the left methyl wedged and the right hashed. Assigning CIP priorities at each ring stereocentre (O/N-free here, so the two ring-carbon neighbours and the substituent methyl are ranked by their own substituents: the ring carbon that itself bears a methyl outranks the plain –CH2– ring carbon, which outranks the exocyclic methyl) and reading the wedge/hash geometry directly gives: B = meso (1R,2S)-1,2-dimethylcyclopropane, C = (1S,2S), D = (1R,2R).
a) Pairwise relationships. A is a constitutional isomer of every other compound here (1,1- vs. 1,2-substitution is a difference in connectivity, not just spatial arrangement) — A–B: constitutional isomers; A–C: constitutional isomers. B and C share the same connectivity (1,2-dimethylcyclopropane) but are not mirror images of each other (B is cis/meso, C is one enantiomer of the trans pair) — B–C: diastereomers. C and D are the (S,S) and (R,R) forms of the same trans connectivity, non-superimposable mirror images — C–D: enantiomers.
b) Chiral or achiral. A has a plane of symmetry through the gem-dimethyl carbon and the midpoint of the opposite ring bond (its two methyls are related by that mirror, so it is achiral and has no stereocentre at all). B (the cis isomer) also has an internal mirror plane relating its two ring stereocentres to each other — it is a meso compound, achiral, despite having two stereocentres. C and D (the trans pair) have no such internal symmetry; each is chiral.
c) Optically active alone. Optical activity requires a net molecular handedness. A and B, having no net handedness (no stereocentre, and an internally-cancelling meso pair, respectively), would not rotate plane-polarised light. Only C and D would be optically active alone.
d) Plane of symmetry. A and B each possess one internal mirror plane (as described above); C and D, being chiral, have none.
e) Boiling points. A (a constitutional isomer, 1,1-substitution) and B (cis-1,2) are different compounds with different shapes and are not required to match; in practice the more compact, higher-symmetry cis-1,2 isomer (B) packs more efficiently and boils distinctly higher than the 1,1-isomer (A) or the trans-1,2 pair (literature values: cis-1,2-dimethylcyclopropane b.p. ≈ 37 °C vs. 1,1-dimethyl- and trans-1,2-dimethylcyclopropane both ≈ 20–21 °C). B and C are diastereomers with genuinely different physical properties, so B has a distinctly higher boiling point than C. C and D are enantiomers, and enantiomers always have identical physical properties (boiling point, melting point, density, refractive index) in an achiral environment — they differ only in the sign of optical rotation and in how they interact with other chiral molecules — so C and D have identical boiling points.
f) Meso compounds. Only B is a meso compound: it has two stereocentres of opposite (R,S) configuration related by an internal mirror plane, the defining feature of a meso structure. A is achiral but has no stereocentre at all (a gem-disubstituted carbon with two identical methyls is not a stereocentre), so it is simply achiral, not "meso" in the strict sense; C and D are chiral, so neither is meso.
g) Optical activity of mixtures. An equal (1:1) amount of C and D is a racemic mixture of the two trans enantiomers — their equal and opposite rotations exactly cancel, so the mixture is not optically active. An equal amount of B and C is different: B is achiral and contributes zero rotation on its own, while C is chiral and rotates light in one direction with nothing present to cancel it (D is not part of this mixture) — so a B/C mixture would be optically active, with a net rotation equal to that of pure C (diluted by the optically inactive B).
