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04-BS-12 · May 2017

Question 6 of 13: Intramolecular vs. Intermolecular Esterification — Why K eq Differs

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Notes on this paper

National Exam 04-BS-12, Organic Chemistry — May 2017. 3 hours, closed-book examination (no textbook aid beyond one double-sided aid sheet); a Casio or Sharp approved calculator is permitted. The paper prints thirteen questions using plain "Question N:" numbering; all thirteen are answered in full below.

Reference texts: McMurry, Organic Chemistry, 9th ed. (functional-group reactivity, amide/β-lactam resonance, stereochemistry and specific rotation, SN1/SN2 and epoxide-opening regiochemistry, cyclohexane conformational analysis, IR/NMR spectroscopy, electrophilic aromatic substitution and multi-step synthesis design, polymer/step-growth chemistry); Atkins, Physical Chemistry, 11th ed. (entropy of intramolecular vs. intermolecular reactions).

Question 6: Intramolecular vs. Intermolecular Esterification — Why Keq Differs

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.

[1] ethyl acetate (intermolecular product)
[2] 5-hydroxypentanoic acid
[2] δ-valerolactone (intramolecular product, 6-membered ring)

The two reactions look "apparently similar" because both break a C–OH/O–H pair and form a new C–O bond plus water, but they differ in how many independent molecules are involved, and that difference dominates the entropy of reaction.

  1. Intermolecular esterification: no net change in the number of molecules. Reaction [1] starts as two separate molecules (acetic acid + ethanol) and ends as two separate molecules (ethyl acetate + water). Because the particle count does not change (2 → 2), the reaction's standard entropy change is close to zero, and the bonds broken and made (C–OH + O–H → C–OR + H–OH) are nearly equal in energy, so ΔG° is small and Keq is only of order 1 (here 4). There is no entropic driving force either way.
  2. Intramolecular esterification gains entropy instead. In reaction [2] the carboxylic acid and the alcohol are already tethered together in one molecule (5-hydroxypentanoic acid); no intermolecular encounter is required, and the reaction converts one molecule into two (the lactone + water) — a net increase in the number of independent particles, which is entropically favourable. There is no "bring two reactants together" penalty to offset it, since the two reactive groups were never independent to begin with.
  3. The 6-membered ring product is essentially strain-free. δ-Valerolactone's ring (5 carbons + 1 oxygen) adopts a chair-like, near-strain-free geometry, so there is no enthalpic penalty counteracting the favourable entropy — both effects push Keq for reaction [2] upward. (A very small or very large ring would reintroduce ring-strain enthalpy that could partially offset this entropic advantage; this particular ring size is close to ideal.)

Put together: with essentially the same ΔH° for both esterifications, reaction [1] has ΔS° ≈ 0 (two particles in, two out), while reaction [2] has a positive ΔS° (one particle in, two out, with the reactive groups already pre-organised in one molecule). Since Keq = exp(−ΔG°/RT), the ratio 1000/4 = 250 corresponds to RT ln 250 ≈ 13.7 kJ/mol of extra driving force at 25 °C, supplied almost entirely by the TΔS° term. That is why Keq(intramolecular) = 1000 is so much larger than Keq(intermolecular) = 4, even though the bonds being made and broken are chemically the same type.