Chapter 28 — Quiz

Twenty-five questions on aldol and Claisen reactions. ∗ marks questions answered in the answer key.

Multiple choice

1.∗ The aldol reaction: (a) enolate + alcohol → ester (b) enolate + carbonyl → β-hydroxy carbonyl (c) two alcohols + carbonyl → ester (d) acid + alcohol → ester

2.∗ The aldol condensation: (a) loss of water gives an α,β-unsaturated carbonyl (enone) (b) no dehydration occurs (c) gives a tertiary alcohol (d) gives an amine

3.∗ A crossed aldol with an unenolizable aldehyde (e.g., benzaldehyde) as electrophile: (a) is best because there's only one possible enolate (b) always gives mixtures (c) requires LDA (d) does not work

4.∗ The Claisen condensation product is: (a) β-keto ester (b) β-hydroxy ester (c) α,β-unsaturated ester (d) amide

5.∗ The Claisen mechanism is: (a) substitution + addition (nucleophilic acyl substitution + enolate) (b) elimination + addition (c) only nucleophilic acyl substitution (d) only nucleophilic addition

6.∗ LDA is used at low T (-78 °C) for aldol because: (a) it forms the kinetic enolate selectively (b) it heats the reaction (c) it makes the aldehyde more reactive (d) it doesn't matter

7.∗ Fatty acid biosynthesis uses: (a) Claisen-like C-C bond formation (decarboxylative Claisen) (b) radical reactions (c) only enolate alkylations (d) photochemistry

8.∗ The α-H of a β-keto ester (Claisen product) has pKa: (a) ~11 (very acidic; doubly stabilized by two C=O) (b) ~20 (regular ketone) (c) ~25 (regular ester) (d) ~50 (nonacidic)

9.∗ Dieckmann cyclization is: (a) intramolecular Claisen → cyclic β-keto ester (b) intermolecular Claisen (c) aldol condensation (d) Michael addition

10.∗ A pre-formed enolate (LDA at low T) is used in crossed aldol to: (a) avoid mixtures by selecting which carbonyl forms the enolate (b) speed the reaction up (c) require less base (d) it doesn't work

11.∗ The Zimmerman-Traxler transition state for aldol is: (a) a 4-membered ring (b) a chair-like 6-membered ring (c) acyclic (d) only relevant for Claisen

12.∗ A Z-enolate gives a syn-aldol product because: (a) the substituent and aldehyde substituent are gauche in the chair TS (b) it's electronic (c) it's never preferred (d) actually it's the E-enolate that gives syn

13.∗ The Mukaiyama aldol: (a) uses a silyl enol ether + Lewis acid + carbonyl (b) uses LDA only (c) uses NaOH + heat (d) uses radical conditions

14.∗ Citrate synthase catalyzes: (a) aldol condensation between acetyl-CoA and oxaloacetate (b) Claisen condensation (c) ester hydrolysis (d) Michael addition

15.∗ Knoevenagel condensation: (a) 1,3-dicarbonyl + aldehyde + base → α,β-unsaturated dicarbonyl + water (b) acid + alcohol (c) only with esters (d) only with amides

16.∗ Mannich reaction: (a) aldehyde + amine + 1,3-dicarbonyl → β-amino dicarbonyl (b) only with primary amines (c) requires Lewis acid (d) photochemical

17.∗ Why does the Claisen condensation need a base specific to the ester's leaving group (e.g., NaOEt for ethyl ester)? (a) to avoid transesterification competing with the desired Claisen (b) for purity (c) to slow the reaction (d) it doesn't matter which base

18.∗ What is the driving force for the Claisen condensation forward reaction? (a) the product β-keto ester is dramatically more acidic; deprotonation by NaOR pulls the equilibrium (b) high temperature (c) Lewis acid catalyst (d) entropy

19.∗ A retro-aldol means: (a) the reverse of an aldol (β-hydroxy carbonyl → enolate + carbonyl) (b) failed aldol (c) photo-aldol (d) only at very low T

20.∗ Glycolytic aldolase catalyzes: (a) cleavage of fructose-1,6-bisphosphate into DHAP + G3P (a retro-aldol) (b) production of fructose (c) glycolysis end step (d) oxidative phosphorylation

Short answer

21. Draw the full mechanism for the self-aldol of acetone under NaOH catalysis. Show every step and identify the rate-determining step.

22. Predict the major product of the crossed aldol between acetone and benzaldehyde under NaOH/EtOH at room temperature, then heat. Explain the selectivity.

23. Outline the mechanism of citrate synthase: acetyl-CoA's enolate attacks oxaloacetate's central ketone. Identify the Family I/II/III chemistry involved.

24. Explain why the Claisen condensation is essentially irreversible while a typical aldol is reversible.

25. Design a synthesis of cyclohexanone using a Dieckmann cyclization on a 6-carbon diester.

Answer key

  1. b — Aldol = enolate + carbonyl → β-hydroxy carbonyl.
  2. a — Aldol condensation includes the dehydration step.
  3. a — Unenolizable aldehyde restricts to one possible enolate (from the other carbonyl).
  4. a — Claisen gives β-keto ester.
  5. a — Mechanism is substitution at the second ester's C=O (Family II) using an enolate (Family III).
  6. a — LDA at low T gives kinetic enolate.
  7. a — Fatty acid biosynthesis uses decarboxylative Claisen.
  8. a — β-Keto ester is dramatically more acidic (pKa 11) than a simple ester (pKa 25).
  9. a — Dieckmann is intramolecular Claisen.
  10. a — Pre-formed enolate restricts to one nucleophile.
  11. b — Zimmerman-Traxler is a chair-like TS.
  12. a — Z-enolate puts substituent in chair-like TS so as to be gauche → syn product.
  13. a — Mukaiyama uses silyl enol ether + Lewis acid + carbonyl.
  14. a — Citrate synthase is an aldol of acetyl-CoA's enolate onto oxaloacetate.
  15. a — Knoevenagel description.
  16. a — Mannich description.
  17. a — Avoiding transesterification.
  18. a — The product β-keto ester is much more acidic; deprotonation by NaOR drives the equilibrium forward.
  19. a — Retro-aldol = reverse aldol.
  20. a — Glycolytic aldolase = retro-aldol.

21. Step 1: NaOH removes one of acetone's α-H → acetone enolate (negative on α-C, partially on O). Step 2: enolate attacks the C=O carbon of another acetone molecule. Step 3: tetrahedral alkoxide forms (with -O⁻, -CH₃, and -CH₂-COCH₃ on the central C). Step 4: water protonates the alkoxide → 4-hydroxy-4-methyl-2-pentanone (the aldol product). Rate-determining step: step 2 (the enolate attack on C=O). Equilibrium favors back-aldol slightly without dehydration; warming favors dehydration to mesityl oxide ((E)-4-methyl-3-penten-2-one).

22. Acetone has α-H; benzaldehyde does not. So benzaldehyde cannot form an enolate. The crossed aldol gives 4-phenyl-3-buten-2-one (E-isomer). At higher temperature, water is eliminated, giving the α,β-unsaturated ketone (also called benzylideneacetone). The (E)-isomer is preferred due to steric reasons in the dehydration TS.

23. Step 1: Acetyl-CoA's α-H is removed by an enzyme base (Asp375 in citrate synthase) → acetyl-CoA enolate. Step 2: the enolate attacks oxaloacetate's central ketone (the C2 carbonyl) — Family I/III combined: enolate (Family III) attacking carbonyl C (Family I substrate). Step 3: tetrahedral intermediate; protonation gives citryl-CoA. Step 4: a water molecule attacks the thioester C=O → tetrahedral intermediate → CoA-S⁻ leaves → citrate. Step 4 is Family II (acyl substitution).

24. A typical aldol is reversible: $\Delta G \approx 0$ for ketone + enolate ↔ β-hydroxy ketone. Both starting materials and product are present at significant amounts at equilibrium. The Claisen condensation, in contrast, gives a β-keto ester product whose α-H (between two C=Os) has pKa ~11 — about 14 pKa units more acidic than a regular ester. Under NaOEt (pKaH 16), the β-keto ester is essentially fully deprotonated (>99%), removing it from the equilibrium and pulling the reaction forward. The deprotonation cycle drives the Claisen to completion.

25. Use diethyl pimelate (a 1,7-diester: $\text{(EtOOC)-(CH}_2\text{)}_5\text{-(COOEt)}$, with 5 CH₂ between the two esters). Treat with NaOEt → the α-H of one ester is removed; the other ester is attacked intramolecularly. This forms a 6-membered ring with a β-keto ester functionality. Hydrolyze the ester (NaOH/H₂O) → 1,3-cyclohexanedione mono-ester acid. Heat → decarboxylation gives cyclohexanone. Total: 3 steps from diethyl pimelate.