Chapter 27 — Quiz

Twenty-five questions on enols, enolates, and α-carbon chemistry. ∗ marks questions answered in the answer key.

Multiple choice

1.∗ The α-H of a simple ketone has pKa approximately: (a) 5 (b) 13 (c) 20 (d) 50

2.∗ The enolate of a ketone has its negative charge: (a) on the α-carbon only (b) on the carbonyl O only (c) delocalized between α-C and carbonyl O via resonance (d) on a remote carbon

3.∗ Keto-enol tautomerism in simple ketones favors: (a) keto form (>99% at equilibrium) (b) enol form (>99% at equilibrium) (c) ~50:50 (d) varies wildly

4.∗ 2,4-Pentanedione (acetylacetone) has ~80% enol in pure liquid because: (a) the enol has aromatic stabilization (b) two carbonyls plus chelating intramolecular H-bond stabilize the enol (c) the keto form is sterically strained (d) it's actually 100% keto

5.∗ LDA (lithium diisopropylamide) is: (a) a small, weak base (b) a bulky, strong base that does not add to C=O (c) a hydride donor (d) a Lewis acid

6.∗ The kinetic enolate of an unsymmetrical ketone forms: (a) at the more-substituted α-C (b) at the less-hindered α-C (c) only with strong base (d) only at high temperature

7.∗ α-Halogenation of a ketone: (a) the α-C attacks X₂ via the enolate or enol (b) X attacks the carbonyl C directly (c) requires a metal catalyst (d) only works in protic solvents

8.∗ The haloform reaction of a methyl ketone gives: (a) trihalogenated ketone (3 X on the methyl) (b) haloform (CHX₃) + carboxylate (c) only mono-halogenated product (d) only the alcohol

9.∗ α-Alkylation works best with: (a) tertiary alkyl halides (SN1) (b) primary alkyl halides (SN2) (c) any alkyl halide (d) only acid chlorides

10.∗ Thalidomide racemizes in vivo at: (a) the C adjacent to one carbonyl (pKa ~20) (b) the C between two carbonyls (pKa ~10) (c) does not racemize (d) only in liver

11.∗ Phenol is a special case because: (a) it's a regular ester (b) it's an enol that cannot tautomerize back to a stable keto form (the keto would lose aromaticity) (c) it's a regular ketone (d) phenol does not exist

12.∗ The Stork enamine synthesis uses: (a) sodium hydride and an alkyl halide (b) a secondary amine + ketone (form enamine), then alkylate at C, then hydrolyze (c) only a Lewis acid (d) is the same as LDA enolate alkylation

13.∗ Diethyl malonate has α-H pKa of ~13 because: (a) it's a 1,3-dicarbonyl; conjugate base stabilized by two C=O (b) it's exceptionally acidic (c) it's an alcohol (d) of an inductive effect alone

14.∗ A silyl enol ether (TMS or TBS) is: (a) a stable, isolable form of the enolate (b) only reactive with Lewis acid catalysis (Mukaiyama) (c) both (a) and (b) (d) a starting material for SN2

15.∗ The acetoacetic ester synthesis makes: (a) α-alkyl methyl ketones (after hydrolysis + decarboxylation) (b) α-alkyl carboxylic acids (c) α-alkyl amines (d) α-alkyl alcohols

16.∗ The malonic ester synthesis makes: (a) α-alkyl methyl ketones (b) α-alkyl carboxylic acids (after hydrolysis + decarboxylation) (c) α-alkyl alcohols (d) α-alkyl amines

17.∗ A 1,3-dicarbonyl is more reactive at α-C than a simple ketone because: (a) extra steric bulk (b) double resonance stabilization of the enolate (c) lower temperature reactivity (d) different solvent

18.∗ PLP-dependent enzymes catalyze α-C reactions on amino acids by: (a) forming a Schiff base, which makes the α-H more acidic (enol-like in the imine-cation system) (b) breaking the amino acid backbone (c) catalyzing only addition (d) using metal cofactors only

19.∗ A student needs to alkylate the α-C of a complex molecule with an OH and an NH already present. Which approach is best? (a) LDA in THF (would deprotonate everything) (b) Stork enamine method (mild, selective for the α-C) (c) NaOH (too weak to deprotonate) (d) HBr (no selectivity)

20.∗ The α-H of an amide (e.g., DMF, CH₃NHCHO) has pKa: (a) ~30 (much less acidic than a ketone) (b) ~20 (same as ketone) (c) ~10 (same as 1,3-diketone) (d) ~5 (very acidic)

Short answer

21. Draw the resonance structures of the enolate of acetone. Indicate which atom carries the negative charge in each.

22. Explain why the α-H of a 1,3-dicarbonyl (like ethyl acetoacetate) has pKa ~11 while a simple ester (like ethyl acetate) has pKa ~25.

23. Predict the product of: 2-methylcyclohexanone + LDA in THF at -78 °C, then methyl iodide. Specify the regiochemistry and identify the kinetic enolate.

24. Outline the steps of the malonic ester synthesis to make 2-methylpentanoic acid. Show all intermediates.

25. Explain the chemical reason that thalidomide racemizes in the body. Why does this make single-enantiomer drug administration ineffective?

Answer key

  1. c — Simple ketone α-H pKa is ~20.
  2. c — Resonance delocalizes the charge.
  3. a — Keto form dominates for simple ketones.
  4. b — Two C=O plus an intramolecular H-bond.
  5. b — LDA is bulky and strong.
  6. b — Kinetic enolate is the less-hindered (less-substituted) one.
  7. a — α-C attacks X₂ via the enol or enolate.
  8. b — Haloform reaction yields haloform + carboxylate.
  9. b — Primary halides via SN2 work best.
  10. b — Thalidomide's α-C is between two C=O of an imide; pKa ~10.
  11. b — Phenol is a trapped enol because the keto form loses aromaticity.
  12. b — Stork enamine procedure.
  13. a — 1,3-Dicarbonyl: doubly stabilized enolate.
  14. c — Both (a) and (b) are correct.
  15. a — Acetoacetic ester → α-alkyl methyl ketones.
  16. b — Malonic ester → α-alkyl carboxylic acids.
  17. b — Double resonance stabilization.
  18. a — Schiff base + ring system makes the α-H more acidic.
  19. b — Stork enamine is mild and selective.
  20. a — Amide α-H is ~30 (less acidic because the lone pair on N is already donated into C=O, leaving less stabilization for the enolate).

21. Form 1: $CH_3-CO-CH_2^-$ — negative charge on α-C, C=O bond intact. Form 2: $CH_3-C(O^-)=CH_2$ — negative charge on O, double bond between former α-C and carbonyl C. The hybrid is closer to form 2 (negative on more electronegative O).

22. A simple ester (ethyl acetate) has a pKa of ~25 because the conjugate base is stabilized by only one C=O (one resonance structure with negative on O). A 1,3-dicarbonyl like ethyl acetoacetate has its α-C between two carbonyls, so the conjugate base has TWO resonance structures (negative on each O), each contributing to stabilization. The double stabilization shifts the pKa down by ~14 units (from ~25 to ~11).

23. Under LDA at -78 °C, the kinetic enolate forms — deprotonation at the less-hindered C6 (the methylene side, not the side with the methyl substituent). Subsequent methyl iodide alkylation gives 2,6-dimethylcyclohexanone (the kinetic alkylation product). Under thermodynamic conditions (NaOEt, room temperature), the enolate equilibrates to the more-substituted side (C2 with the methyl); alkylation gives 2,2-dimethylcyclohexanone.

24. (1) Diethyl malonate + NaOEt → enolate. (2) Add R-X (here, R = CH₂-CH(CH₃)CH₂Br for 2-methylpentanoic synthesis... actually let me reconsider. For 2-methylpentanoic acid, the structure is CH₃CH₂CH₂CH(CH₃)COOH. So we want R = CH₃ and R' = CH₃CH₂CH₂. (1) malonate + NaOEt → enolate; (2) + CH₃I → α-methyl malonate; (3) + NaOEt + propyl bromide → α-methyl-α-propyl malonate; (4) NaOH/H₂O hydrolysis → 2-methyl-2-propylmalonic acid; (5) heat (decarboxylation of one COOH) → 2-methylpentanoic acid + CO₂.

25. Thalidomide has a chiral center at the α-position of two carbonyls (an imide). The α-H pKa is ~10 — comparable to a 1,3-dicarbonyl. At physiological pH 7.4, the α-H is removed by water (or OH⁻) at a rate fast enough to racemize the molecule with a half-life of ~8 hours. Even if pure (R)-thalidomide is administered, by 24 hours the (S)-enantiomer (the teratogenic form) is also present in significant amount. This is why thalidomide cannot be safely administered as a single enantiomer.