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.
Reading the wedge/hash bonds directly off the structures: A has both methyl groups on the same ring carbon (1,1-disubstituted, a gem-dimethyl compound); B has one methyl wedge and one methyl wedge on the two different substituted ring carbons (cis-1,2); C and D each have one wedge and one hash on the two different ring carbons (trans-1,2), but with the wedge/hash assignment swapped between them.
a) Relationships. A has a different connectivity (1,1- vs. 1,2-substitution) from every other compound, so A–B and A–C are constitutional (structural) isomers. B and C share the same 1,2-connectivity but differ in relative (cis vs. trans) configuration, so B–C are diastereomers. C and D share the same 1,2-connectivity and are non-superimposable mirror images, so C–D are enantiomers (confirmed: inverting every stereocentre of C's canonical SMILES reproduces D's canonical SMILES exactly).
b) Chiral or achiral. A has no stereocentre at all (the two methyls on one carbon are identical substituents) — achiral. B has two stereocentres but an internal mirror plane relates them — B is a meso compound, achiral. C and D have no such internal symmetry (both stereocentres carry the same descriptor, S,S for C and R,R for D) — both are chiral.
c) Optical activity alone. Only chiral, non-meso compounds rotate plane-polarised light, so only C and D are individually optically active; A and B are not.
d) Plane of symmetry. A and B both possess a plane of symmetry (through C3 and, for B, bisecting the C1–C2 bond) — consistent with both being achiral. C and D have no symmetry plane.
f) Meso compounds. Only B has stereocentres and a symmetry plane, which is the definition of a meso compound; A is achiral simply because it never had a stereocentre to begin with, so A is not classed as meso.
e) Boiling-point trends. Two physical effects operate here: (i) branching/compactness lowers the boiling point (less surface area for van der Waals contact, exactly as isoalkanes boil lower than their straight-chain isomers), and (ii) a cis-disubstituted ring has a net molecular dipole (the two C–CH3 bond dipoles do not cancel) while the trans isomer is more symmetric and less polar, so cis isomers pack with somewhat stronger dipole–dipole attraction than their trans counterparts. Putting these together: the compact, gem-disubstituted A has the lowest boiling point; the polar cis isomer B has the highest; the trans enantiomers C and D are identical in every physical property, including boiling point (enantiomers only differ in a chiral environment), and sit between A and B. Overall: A < C = D < B.
g) Mixtures. An equal amount of C and D is a 1:1 mixture of enantiomers, i.e. a racemic mixture — the equal and opposite rotations exactly cancel, so it is not optically active. An equal amount of B and C is different: B is achiral and contributes zero rotation on its own (it is not "the opposite enantiomer" of anything), so nothing cancels C's rotation — the B/C mixture is optically active, showing the same net rotation as pure C alone.
