Chapter 16 — Quiz

Twenty-five questions on the alkene addition toolbox. ∗ marks questions answered in the answer key.

Multiple choice

1.∗ Hydroboration-oxidation gives: (a) Markovnikov alcohol (b) anti-Markovnikov alcohol with syn stereochemistry (c) racemic alcohol (d) ester

2.∗ mCPBA + alkene gives: (a) diol (b) epoxide (3-membered O-ring) (c) alcohol (d) carbonyl

3.∗ OsO₄ + NMO + alkene gives: (a) syn-1,2-diol (cis) (b) anti-1,2-diol (c) two carbonyls (d) epoxide

4.∗ Ozonolysis (O₃ + alkene + reductive workup) gives: (a) diol (b) two carbonyls (cleavage of C=C) (c) epoxide (d) alkane

5.∗ Hydrogenation (H₂/Pd/C + alkene) gives: (a) alkane with syn addition (b) alkane with anti addition (c) alkene unchanged (d) two carbonyls

6.∗ Oxymercuration-demercuration gives: (a) anti-Markovnikov alcohol (b) Markovnikov alcohol with no rearrangement (c) syn-1,2-diol (d) epoxide

7.∗ Br₂ + alkene + H₂O gives a bromohydrin with: (a) Br on less-substituted C, OH on more-substituted C (Markovnikov-like, anti) (b) Br on more-substituted C (c) random regiochemistry (d) only one stereoisomer

8.∗ Peroxyacid (e.g., mCPBA) adds: (a) one O atom across the C=C, giving a 3-membered epoxide (b) two O atoms (giving a diol) (c) the OH group only (d) two H atoms

9.∗ NaBH₄ in oxymercuration removes the Hg group with: (a) retention of stereochemistry (b) inversion (c) random (d) the Hg stays attached

10.∗ Catalytic hydrogenation typically uses: (a) Pd/C, PtO₂, or Raney Ni (b) LiAlH₄ (c) NaBH₄ (d) H₂SO₄

11.∗ Why does hydroboration give anti-Markovnikov alcohol? (a) B has empty p orbital; B is electrophilic; alkene's more-substituted C bonds to H (less-electrophilic position relative to B); so B ends up at less-substituted C; later oxidation gives OH at less-substituted C (b) random (c) only thermodynamic (d) only with peroxides

12.∗ Why does Br₂ + H₂O give a halohydrin instead of dibromide? (a) water is more abundant than Br⁻; water attacks the bromonium ion (b) bromonium isn't formed (c) it's a different reagent (d) acid catalysis is needed

13.∗ Sharpless asymmetric dihydroxylation: (a) enantioselective version of OsO₄ oxidation; uses chiral cinchona ligand (b) achiral (c) only on alkynes (d) only with H₂O₂

14.∗ mCPBA + cis-2-butene gives: (a) cis-2,3-epoxybutane (the cis configuration is preserved) (b) trans-2,3-epoxybutane (c) diol (d) ketone

15.∗ Carbocation rearrangement is a concern in: (a) acid-catalyzed hydration (cation intermediate) (b) hydroboration-oxidation (no cation; concerted) (c) OsO₄ dihydroxylation (concerted; no cation) (d) only oxymercuration

16.∗ Why does oxymercuration give Markovnikov without rearrangement? (a) the mercurinium ion (3-membered ring) prevents the open-chain cation from forming; no 1,2-shifts possible (b) it's not Markovnikov (c) it's stereospecific (d) random

17.∗ Ozonolysis with H₂O₂ workup gives: (a) carboxylic acids (oxidative; aldehydes are oxidized to COOHs) (b) only aldehydes (c) only ketones (d) diols

18.∗ Lindlar Pd is: (a) a poisoned palladium catalyst that selectively reduces alkynes to cis-alkenes (and stops there) (b) the same as Pd/C (c) for alcohols (d) only for aromatic rings

19.∗ Asymmetric hydrogenation (Knowles, Noyori, 2001 Nobel) uses: (a) chiral Rh or Ru catalyst with chiral phosphine ligand (BINAP, DiPAMP) (b) only Pd (c) only enzymes (d) only photocatalysis

20.∗ Modern margarine production uses: (a) catalytic hydrogenation of vegetable oils (b) only saturation by chemical reduction (c) only enzymatic reduction (d) only physical fractionation

Short answer

21. Match each reagent to its product (regio- and stereochemistry): (a) BH₃, then H₂O₂/NaOH → ? (b) Hg(OAc)₂/H₂O, then NaBH₄ → ? (c) Br₂ in CCl₄ → ? (d) mCPBA → ? (e) OsO₄/NMO → ?

22. Why does hydroboration give syn addition? Sketch the four-centered transition state.

23. Predict the product (with stereochemistry) of cyclohexene + Br₂ + H₂O. Sketch the bromonium ion intermediate.

24. Compare OsO₄ + NMO and KMnO₄ (cold, dilute) for syn-dihydroxylation. Which is preferred for selective synthesis?

25. Design a synthesis of trans-1,2-dimethoxycyclohexane from cyclohexene. Use a multistep route involving alkene reagents.

Answer key

  1. b — Hydroboration-oxidation = anti-Markovnikov, syn.
  2. b — mCPBA → epoxide.
  3. a — OsO₄/NMO → syn-diol.
  4. b — Ozonolysis cleaves C=C to two carbonyls.
  5. a — Hydrogenation gives syn alkane.
  6. b — Oxymercuration = Markovnikov, no rearrangement.
  7. a — Bromohydrin: Br at less-subst, OH at more-subst.
  8. a — Peroxyacid = epoxide.
  9. a — NaBH₄ removal of Hg = retention.
  10. a — Various Pd/Pt/Ni catalysts.
  11. a — B is electrophilic with empty p; ends up at less-subst C.
  12. a — Water more abundant; opens bromonium.
  13. a — Sharpless AD = enantioselective.
  14. a — mCPBA preserves cis.
  15. a — Cation rearrangement in acid-catalyzed hydration.
  16. a — Mercurinium prevents 1,2-shifts.
  17. a — H₂O₂ ozonolysis → COOHs.
  18. a — Lindlar Pd = alkyne to cis-alkene.
  19. a — Asymmetric hydrogenation with chiral Rh/Ru.
  20. a — Margarine via hydrogenation.

21. (a) BH₃, then H₂O₂/NaOH → anti-Markovnikov alcohol; syn stereochem (the new -OH and the new -H are on the same face of the original alkene). (b) Hg(OAc)₂/H₂O, then NaBH₄ → Markovnikov alcohol; effectively anti stereochem from the mercurinium opening; no rearrangement. (c) Br₂ in CCl₄ → vicinal dibromide; anti. (d) mCPBA → epoxide; cis-alkene gives cis-epoxide; trans gives trans (stereospecific). (e) OsO₄/NMO → syn-1,2-diol (cis).

22. Hydroboration TS: - A 4-membered cyclic TS with: B, H, alkene-C₁, alkene-C₂. - The B has empty p orbital; H has its 1s; alkene has its π MO. - All four atoms are arranged in a square planar geometry briefly. - The two new bonds (B-C₁ and H-C₂) form simultaneously, both on the same face of the alkene. - This concerted, syn addition gives the syn stereochemistry. No carbocation intermediate; no rearrangement.

23. Cyclohexene + Br₂ + H₂O: - Bromonium ion forms first: a 3-membered ring with Br at the apex on the cyclohexene face. - Water attacks the more-substituted (or both — they are equivalent for cyclohexene) C from the opposite face of the Br. - Net result: trans-2-bromo-1-hydroxycyclohexane (anti orientation; the Br and OH are on opposite faces of the ring).

24. OsO₄ + NMO vs KMnO₄ (cold, dilute): OsO₄ + NMO: - Catalytic OsO₄ with NMO recycling the Os(VI) → Os(VIII). - Mild conditions; doesn't over-oxidize. - Highly selective for the syn-diol. - Compatible with most other functional groups. - Industrial scale possible (Sharpless AD with chiral ligand). KMnO₄ (cold, dilute): - Stoichiometric. - More aggressive: can over-oxidize (especially at higher T or higher concentration). - Cleaves C=C in some conditions. - Cheaper than Os. - Not stereospecific in the same way.

OsO₄/NMO is preferred for selective synthesis where chemoselectivity matters. KMnO₄ is acceptable for simple substrates where cost matters.

25. Design synthesis of trans-1,2-dimethoxycyclohexane from cyclohexene: - Step 1: cyclohexene + Br₂ → trans-1,2-dibromocyclohexane (anti addition via bromonium). - Step 2: trans-1,2-dibromocyclohexane + 2 NaOMe → trans-1,2-dimethoxycyclohexane (SN2 on each C-Br; inversion of stereochemistry at each carbon; anti relationship is preserved as trans-diether). - Alternative: cyclohexene + mCPBA → epoxide; then MeOH + acid → trans-1-methoxy-2-hydroxy + repeat for the other side. Or use 2 MeOH on the epoxide for trans-diether. The first approach (Br₂ then SN2) gives the desired trans-diether efficiently.