Chapter 26 — Quiz
Twenty-five questions on nucleophilic acyl substitution, including mechanism, products, equilibria, leaving groups, and biological context. ∗ marks questions answered in the answer key at the end.
Multiple choice
1.∗ The most reactive substrate in nucleophilic acyl substitution is: (a) amide (b) ester (c) acid halide (d) anhydride
2.∗ Saponification is: (a) reversible (b) irreversible (c) catalyst-dependent (d) only at low pH
3.∗ Fischer esterification is: (a) forward only (irreversible) (b) reversible (controlled by Le Chatelier's principle) (c) base catalyzed (d) thermodynamically forbidden
4.∗ Amide hydrolysis requires: (a) mild aqueous conditions at room temperature (b) strong acid or base + heat for extended time (c) catalysis by Lewis acid only (d) light
5.∗ Acid chloride + alcohol gives: (a) ester (b) amide (c) anhydride (d) ketone
6.∗ Acid chloride + primary amine gives: (a) ester (b) amide (c) imine (d) carboxylic acid
7.∗ To convert COOH to ester, the easiest practical approach is: (a) direct reaction with alcohol (no catalyst) (b) Fischer (H⁺ + alcohol + heat) or via acid chloride (c) reduction with NaBH₄ (d) reaction with HCl
8.∗ The aspirin synthesis uses salicylic acid plus: (a) methanol (Fischer) (b) acetic anhydride (c) chloroacetic acid (d) acetyl-CoA
9.∗ The tetrahedral intermediate in acyl substitution: (a) does not exist (concerted mechanism) (b) is a real, sometimes detectable intermediate (c) only exists in solution (d) is identical to the SN2 transition state
10.∗ Transesterification is: (a) one ester + another alcohol → different ester (acid catalyst) (b) only happens with acid chlorides (c) requires strong base (d) impossible
11.∗ Why are amides such poor electrophiles? (a) The C is sterically blocked (b) The N donates a lone pair into C=O via resonance (c) Amides are aromatic (d) Amides have no C=O bond
12.∗ The order of leaving group ability for acyl substitution is: (a) Cl⁻ > RCOO⁻ > RO⁻ > NR₂⁻ (b) NR₂⁻ > RO⁻ > RCOO⁻ > Cl⁻ (c) RO⁻ = NR₂⁻ > Cl⁻ > RCOO⁻ (d) all are about equal
13.∗ A thioester (acetyl-CoA) is more reactive than an oxoester (methyl acetate) by: (a) ~10⁵ × faster (b) ~100 × slower (c) about the same (d) only reactive in water
14.∗ Aspirin's mechanism on COX is: (a) reversible binding (b) acyl transfer of acetyl group from aspirin's ester to COX serine (c) C-H activation (d) covalent binding to a cysteine
15.∗ DCC coupling is preferred over acid chloride for peptide synthesis because: (a) it is faster (b) it works at low temperature without racemization at the α-carbon (c) it is cheaper (d) it is reversible
16.∗ A 4-nitrobenzoate ester is more reactive than a regular benzoate because: (a) the nitro group donates electrons to the C=O (b) the nitro group withdraws electrons from the C=O, making C more electrophilic (c) the nitro group adds steric bulk (d) the nitro group is a leaving group
17.∗ The half-life of an amide in water at 25 °C without catalyst is approximately: (a) 1 second (b) 1 hour (c) 1 day (d) ~600 years
18.∗ Penicillin's β-lactam is unusually reactive because: (a) the 4-membered ring strains the C=O bond, raising the C=O wavenumber and increasing electrophilicity (b) penicillin contains free SH groups (c) penicillin's amide is aromatic (d) the amide is open-chain
19.∗ A Gilman reagent (R₂CuLi) reacts with an acyl chloride to give: (a) tertiary alcohol (b) ketone (stops there) (c) amide (d) ester
20.∗ Fatty acid biosynthesis uses thioesters because: (a) they are unreactive enough to circulate in cells but become reactive when activated (b) they are 10⁵× more reactive than oxoesters (c) the body cannot make oxoesters (d) all of the above except (c)
Short answer
21. Draw the full mechanism for: ethyl acetate + NaOH + water → sodium acetate + ethanol. Show the tetrahedral intermediate. Why is this irreversible?
22. Write the mechanism for the aspirin synthesis: salicylic acid + acetic anhydride + H₂SO₄ → aspirin + acetic acid. Identify each step.
23. Why does amide hydrolysis require strong acid + heat? Identify the two factors that contribute to amide kinetic stability.
24. Sketch the COX enzyme mechanism with aspirin: aspirin + COX-Ser-OH → salicylate + acetyl-COX-Ser. Identify the carbonyl class involved (ester) and the nucleophile (Ser-OH).
25. Compare the reactivity of acetyl-CoA vs. methyl acetate. Why is acetyl-CoA preferred in biology for acyl transfer reactions?
Answer key
- c — Acid halide is the most reactive (Cl is the best leaving group; minimal π donation).
- b — Saponification is irreversible because the carboxylate product is unreactive.
- b — Fischer is reversible; equilibrium constant is ~4 with stoichiometric reactants.
- b — Amide hydrolysis requires harsh conditions (strong acid/base + heat).
- a — Acid chloride + ROH → ester + HCl.
- b — Acid chloride + RNH₂ → amide + HCl.
- b — Direct reaction (no catalyst) is too slow; Fischer or via acid chloride is the practical choice.
- b — Aspirin synthesis uses acetic anhydride.
- b — Tetrahedral intermediates are real and have been detected experimentally for slow substrates (e.g., amides, esters).
- a — Transesterification is the standard ester-to-ester swap, acid catalyzed.
- b — N's lone pair donates into the C=O, lowering the C electrophilicity.
- a — Cl⁻ is the best leaving group; amide N⁻ is the worst.
- a — Thioesters are 10⁵× more reactive than oxoesters.
- b — Acyl transfer; aspirin's ester C=O is attacked by COX serine OH.
- b — DCC coupling avoids the harsh conditions of acid chloride that would epimerize the α-carbon (racemization).
- b — Nitro is electron-withdrawing; pulls electron density off the C=O, making C more electrophilic.
- d — ~600 years (Wolfenden 2011 measurement).
- a — Ring strain forces the C=O to bond differently; the bond is "more sp²-like" and more reactive.
- b — Gilman selectively gives ketone from acyl chloride; doesn't continue to attack the ketone.
- d — Both (a) and (b) are correct: thioesters are 10⁵× more reactive AND they don't randomly hydrolyze in cells.
21. Step 1: OH⁻ attacks the C=O of ethyl acetate, forming a tetrahedral alkoxide intermediate (with -O⁻, -OEt, -CH₃, -OH on the C). Step 2: ethoxide (EtO⁻) leaves, regenerating C=O of acetic acid. Step 3: ethoxide deprotonates the COOH → acetate (RCOO⁻) + ethanol. Why irreversible: acetate is a much weaker electrophile than the ester; the reverse reaction (acetate + ethanol → ester) has zero rate at room temperature. The system is thermodynamically locked.
22. Step 1: H⁺ protonates the carbonyl O of acetic anhydride. Step 2: phenol OH of salicylic acid attacks the protonated C=O. Tetrahedral intermediate. Step 3: acetate leaves (with the H now on it, as acetic acid). Step 4: deprotonation of the new ester gives aspirin.
23. (1) The amide C is barely electrophilic — N's lone pair donation into C=O makes the carbonyl carbon a poor electrophile. (2) The amide N is a terrible leaving group — would have to leave as an amide anion (very high energy). Together, the rate of hydrolysis is ~10¹⁰× slower than ester hydrolysis. Strong acid protonates C=O (making C more electrophilic) AND protonates the amide N (making it a better leaving group). Heat overcomes the activation barrier.
24. Aspirin has two carbonyls: the ester (acetyl-O-Ar) and the COOH. The ester is the more reactive one (less resonance dampening). COX's Ser530-OH attacks the acetyl C=O; the salicylate is the leaving group. The acetyl group transfers to the serine, blocking the substrate (arachidonic acid) from binding. The COX is permanently modified.
25. Acetyl-CoA is a thioester (R-CO-S-CoA); methyl acetate is an oxoester (R-CO-O-Me). Sulfur is a worse π donor than oxygen (the S lone pair is at higher energy), so acetyl-CoA's C=O is less dampened by resonance — more electrophilic. Sulfur is also a better leaving group than oxygen (weaker C-S bond, more stable thiolate anion). Combined effect: acetyl-CoA is ~10⁵× faster than methyl acetate at acyl transfer. Biology uses thioesters specifically for this rate advantage; the high-energy thioester bond is the "activated form" of an acyl group.