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04-BS-12 · May 2018

Question 5 of 13: Reaction-Energy Diagrams

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

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 5: Reaction-Energy Diagrams (equal value)

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.

Reaction coordinateEnergySMTS‡PEa(low)ΔHo < 0
a) concerted, exothermic, low Ea — one smooth hump, product well below starting material
Reaction coordinateEnergySMTS‡PEa(high)ΔHo > 0
b) one step, endothermic, high Ea — tall hump, product above starting material
Reaction coordinateEnergyATS1‡BTS2‡C
c) two-step A→B→C, A<C<B; TS1 (A→B) is the taller barrier, so that step is rate-determining
Reaction coordinateEnergySMTS‡PEa=4 kcalΔHo=-20 kcal/mol
d) concerted, ΔHo=−20 kcal/mol (strongly exothermic), Ea=4 kcal (very low barrier)

a) A single smooth curve (one transition state, no intermediate) rises a short distance from the starting material to a low transition-state peak, then drops well below the starting-material level to the product — the short SM→TS rise is the low Ea, and the large TS→P drop reflects the negative (exothermic) ΔHo.

b) Again a single smooth curve, but now the SM→TS rise is tall (high Ea) and the product sits above the starting material (positive ΔHo, endothermic) — note that a "high Ea" and "endothermic" are independent features that both had to be drawn in, since one does not automatically imply the other.

c) A two-step profile needs two humps (TS1 for A→B, TS2 for B→C) separated by a local minimum (the intermediate, B). Placing the energies as stated, A < C < B, means: the starting material A is drawn lowest, the isolable intermediate B is drawn highest of the three stable species, and the final product C sits between them. Since B is the highest-energy species overall, the barrier up to B (TS1, over the A→B step) must be taller than the barrier down from B to C (TS2) for any physically sensible profile in which B is a real (if short-lived) intermediate sitting in its own energy well — consistent with the problem's own statement that A→B is rate-determining (the tallest overall barrier, measured from the lowest point reached beforehand, sets the rate-determining step).

d) One smooth concerted curve as in (a), but now both numeric labels are attached directly: the SM→TS rise is marked Ea = 4 kcal (a very small barrier, implying a very fast reaction at room temperature), and the overall SM→P drop is marked ΔHo = −20 kcal/mol (strongly exothermic) — drawn to scale, the product sits far below both the starting material and the low transition state.