Chapter 29 — Quiz

Twenty-five questions on conjugate (Michael) addition and Robinson annulation. ∗ marks questions answered in the answer key.

Multiple choice

1.∗ An α,β-unsaturated carbonyl has electrophilic positions at: (a) only the C=O carbon (b) only the β-C of the C=C (c) both the C=O carbon and the β-C of the C=C (d) only at the α-C

2.∗ Hard nucleophiles (e.g., Grignard, hydride) prefer: (a) 1,2-addition (b) 1,4-addition (c) random (d) only the α-C

3.∗ Soft nucleophiles (e.g., enolates, organocuprates) prefer: (a) 1,2-addition (b) 1,4-addition (conjugate addition) (c) random (d) only attack the carbonyl O

4.∗ Michael addition is: (a) 1,2-addition with a hard nucleophile (b) 1,4-addition with an enolate as the nucleophile (c) ester hydrolysis (d) α-halogenation

5.∗ Robinson annulation makes: (a) a 5-membered ring (b) a 6-membered enone (c) an open-chain product (d) an aromatic ring

6.∗ The Stork enamine method uses: (a) a strong-base enolate (b) an enamine (from ketone + 2° amine) as a milder, softer nucleophile (c) a Lewis acid only (d) a radical

7.∗ A Grignard reagent adding to cyclohex-2-enone at -78 °C gives: (a) 1,2-addition (allylic alcohol) (b) 1,4-addition (saturated ketone) (c) reduction (d) racemization

8.∗ A Gilman reagent (R₂CuLi) adding to cyclohex-2-enone gives: (a) 1,2-addition (b) 1,4-addition (saturated ketone with R at β-C) (c) reduction (d) inversion

9.∗ A Michael donor must have: (a) a low α-H pKa (active methylene) (b) a strong base (c) a Lewis acid (d) only carbonyls

10.∗ A Michael acceptor is: (a) electrophilic at the β-C (b) nucleophilic at the β-C (c) only an aldehyde (d) only a ketone

11.∗ The 1,5-dicarbonyl pattern in a product indicates: (a) it is the result of a Michael addition (b) random arrangement (c) it is an alcohol (d) it is an amine

12.∗ Why does the C=C of an α,β-unsaturated carbonyl have its largest LUMO coefficient at the β-C? (a) the β-C is the position farthest from the C=O, where the conjugated π* peaks (b) electronic charge accumulates at the β-C (c) it's just the way it works empirically (d) all of the above

13.∗ The HSAB principle tells you to pair: (a) hard with hard, soft with soft (b) hard with soft (c) random (d) only soft is reactive

14.∗ Steroid biosynthesis uses Robinson annulation for: (a) constructing 6-membered rings of the steroid skeleton (b) the carbon C28 carbon (c) only side-chain reactions (d) photochemistry

15.∗ A drug that covalently modifies a cysteine in a target protein typically uses: (a) a thia-Michael reaction (cysteine thiolate attacks an α,β-unsaturated electrophile) (b) only ester chemistry (c) only amide chemistry (d) only photochemistry

16.∗ Ibrutinib's mechanism involves: (a) reversible binding to BTK kinase (b) covalent modification of Cys481 of BTK via Michael addition (c) only ATP-competitive inhibition (d) photochemistry

17.∗ Asymmetric Michael with a proline catalyst gives: (a) racemic product (b) enantioenriched product (proline forms an iminium ion that delivers face-selectively) (c) opposite stereochemistry to the substrate (d) no product

18.∗ A typical Michael acceptor is: (a) acrylonitrile (CH₂=CH-CN) (b) cyclohex-2-enone (c) methyl acrylate (CH₂=CH-CO₂CH₃) (d) all of the above

19.∗ A Michael donor with very low α-H pKa (e.g., diethyl malonate, pKa 13) reacts: (a) faster than a regular ketone donor (b) slower (c) the same (d) does not react

20.∗ When does the aldol step in Robinson annulation occur? (a) before the Michael step (b) after the Michael step (intramolecularly, forming the 6-membered ring) (c) it doesn't, the Michael alone is the Robinson (d) only in the absence of base

Short answer

21. Predict the products of: (a) cyclohex-2-enone + CH₃MgBr → ?; (b) cyclohex-2-enone + Me₂CuLi → ?. Justify the difference.

22. Draw the full mechanism for: ethyl acetoacetate + methyl vinyl ketone + NaOEt → 1,5-dicarbonyl Michael adduct.

23. Design a Robinson annulation: 2-methylcyclohexanone + methyl vinyl ketone → ? (a fused 6-6 ring system).

24. Sketch the Michael chemistry of ibrutinib + Cys481-SH of BTK kinase. Identify the donor, the acceptor, and the leaving group (none — covalent bond formed).

25. Why does an organocuprate ($R_2CuLi$) preferentially do 1,4-addition while a Grignard does 1,2-addition on the same substrate?

Answer key

  1. c — Both the C=O carbon and the β-C of C=C are electrophilic.
  2. a — Hard Nu → 1,2.
  3. b — Soft Nu → 1,4.
  4. b — Michael = 1,4 with enolate.
  5. b — Robinson product is a 6-membered enone.
  6. b — Stork enamine.
  7. a — Grignard at low T usually 1,2.
  8. b — Gilman gives 1,4.
  9. a — Active methylene with low pKa.
  10. a — β-C of acceptor is the electrophile.
  11. a — 1,5-Dicarbonyl pattern = Michael product.
  12. a — β-C has the largest LUMO coefficient (and the (a) reasoning is correct, the others are also correct).
  13. a — HSAB: hard-hard, soft-soft.
  14. a — Robinson builds 6-membered rings of the steroid skeleton.
  15. a — Cysteine-targeting drugs use thia-Michael.
  16. b — Ibrutinib + BTK Cys481 = covalent thia-Michael.
  17. b — Proline organocatalyst gives enantioenrichment.
  18. d — All three are Michael acceptors.
  19. a — Faster (more acidic α-H = more enolate available).
  20. b — The aldol is the second step of Robinson, after the Michael.

21. (a) Cyclohex-2-enone + CH₃MgBr at -78 °C → 1-methylcyclohex-2-enol (1,2-product, allylic alcohol). The Grignard is a hard nucleophile attacking the carbonyl C. (b) Cyclohex-2-enone + Me₂CuLi → 3-methylcyclohexanone (1,4-product, saturated ketone). The Gilman is a soft nucleophile attacking the β-C; after enol-keto tautomerism, the saturated ketone is restored.

22. Step 1: NaOEt removes the α-H of ethyl acetoacetate (pKa ~11) to give the stabilized enolate. Step 2: the enolate attacks the β-C of methyl vinyl ketone. Step 3: the C=C π electrons collapse onto the α-C, generating an enolate of the methyl vinyl ketone half. Step 4: protonation of the new enolate gives the keto form. Final product: a 1,5-dicarbonyl (with one ester carbonyl and one ketone carbonyl, separated by a -CH(CH₃)-CH₂-CH₂- chain).

23. Step 1: 2-Methylcyclohexanone + NaOH → enolate (probably at the more-substituted α-C as the thermodynamic; or the less-substituted as the kinetic — depends on conditions). Step 2: Enolate attacks methyl vinyl ketone's β-C → 1,5-dicarbonyl. Step 3: The new α-C (next to the former MVK's C=O) is deprotonated. Step 4: Intramolecular aldol: the new enolate attacks the cyclohexanone's C=O. Step 5: Tetrahedral alkoxide → β-hydroxy ketone. Step 6: Dehydration → 6-membered enone fused to the original cyclohexane ring. Final product: a 4a-methyloctahydronaphthalen-1(2H)-one (an octalin enone, the kind used in steroid synthesis).

24. Donor: Cys481-SH of BTK (the cysteine thiolate). Acceptor: ibrutinib's α,β-unsaturated amide (acrylamide). Mechanism: the cysteine thiolate attacks the β-C of the acrylamide; the C=C π electrons collapse onto the α-C; the new enolate is protonated to the keto form. The result: a covalent thioether bond between BTK's Cys481-S and the β-C of ibrutinib's acrylamide. No leaving group is required; this is a pure Michael addition with covalent product.

25. A Grignard (R-MgX) is a hard nucleophile: the R-Mg bond is polar, with a concentrated negative charge on R. Hard nucleophiles attack hard electrophiles — the C=O carbon of the enone (where the partial positive charge is concentrated). Result: 1,2-addition.

A Gilman reagent ($R_2CuLi$) has the R-Cu bond as the reactive site. The Cu(I) is a soft Lewis acid that interacts with the soft C=C π bond of the enone. The Cu coordinates to the C=C; R⁻ is delivered to the β-C. Soft nucleophile + soft electrophile = 1,4-product. Result: saturated ketone with R at β-C.

The HSAB principle predicts both outcomes correctly.