Chapter 15 — Quiz

Twenty-five questions on alkene structure, stability, and electrophilic addition. ∗ marks questions answered in the answer key.

Multiple choice

1.∗ Most stable alkene class: (a) monosubstituted (b) tetrasubstituted (most alkyl groups; most hyperconjugation) (c) both equal (d) trans-disubstituted

2.∗ Mechanism of HX addition to alkene: (a) 1 step (concerted) (b) 2 steps via carbocation intermediate (c) radical mechanism (d) photo only

3.∗ Markovnikov rule says H adds to: (a) more-substituted C (b) less-substituted C (the C with more H's; H goes where there's more H) (c) either equally (d) only in radical reactions

4.∗ Carbocation stability ranking: (a) 3° > 2° > 1° (alkyl groups stabilize via hyperconjugation) (b) 1° > 2° > 3° (c) all equal (d) only matters with halides

5.∗ Br₂ + alkene addition intermediate: (a) carbocation (b) bromonium ion (cyclic 3-membered ring with Br at apex) (c) radical (d) carbene

6.∗ Bromonium ion opening by Br⁻: (a) syn (same face) (b) anti (opposite face; backside attack) (c) random (d) gives both products equally

7.∗ Anti-Markovnikov HBr addition (with peroxides): (a) ionic mechanism (b) radical mechanism (gives the more-stable radical at the less-substituted C) (c) concerted (d) not possible

8.∗ Bond rotation around C=C is: (a) fast at room temperature (b) restricted by π bond (~60 kcal/mol cost) (c) only restricted in cyclic systems (d) never possible

9.∗ Hyperconjugation: (a) σ electrons of adjacent C-H bonds donate into π (or empty p) orbital (b) π electrons donate into σ (c) only in carbonyl chemistry (d) only in benzene

10.∗ Why does HCl give Markovnikov product? (a) the more-substituted carbocation is more stable; Hammond postulate places lower-energy TS on the lower-energy intermediate path (b) cis/trans preference (c) it's just empirical (d) random

11.∗ A 2° carbocation can rearrange to a 3° via: (a) 1,2-hydride shift or 1,2-methyl shift (b) only with light (c) Pd catalyst (d) it doesn't rearrange

12.∗ cis-2-butene is: (a) less stable than trans-2-butene (steric strain between methyls) (b) more stable than trans (c) the same as trans (d) E configuration

13.∗ E/Z nomenclature is based on: (a) CIP priority (atomic number; higher Z = higher priority) (b) cis/trans visual inspection (c) IUPAC locant rules only (d) random

14.∗ Hydroboration-oxidation gives: (a) Markovnikov alcohol (b) anti-Markovnikov alcohol (B adds to less-substituted C; OH ends up there) (c) ester (d) only ketone

15.∗ Acid-catalyzed hydration of propene gives: (a) 1-propanol (b) 2-propanol (Markovnikov) (c) propylene oxide (d) propanal

16.∗ ¹H NMR vinyl protons appear at approximately: (a) δ 0-1 ppm (b) δ 4.5-6.5 ppm (c) δ 7-9 ppm (d) δ 9-12 ppm

17.∗ Vinyl ¹H coupling constant for trans is: (a) 0-3 Hz (geminal) (b) 6-12 Hz (cis) (c) 12-18 Hz (trans, larger) (d) all the same

18.∗ Industrial alkenes: largest production volume globally is: (a) ethylene (~200 million tons/year) (b) propylene (c) butadiene (d) styrene

19.∗ Why does the alkene's π bond act as the nucleophile? (a) the two electrons in the π orbital sit above and below the plane, accessible to electrophiles (b) the alkene is positively charged (c) of the σ bond (d) of cis/trans isomerism

20.∗ Stereo-defined alkene + HX: stereochemistry of the product is: (a) preserved (anti) (b) lost (mixture; carbocation is planar; halide attacks either face) (c) inverted (d) determined by the temperature

Short answer

21. Predict and explain: 2-methyl-2-butene + HBr → ?

22. Predict and explain: cis-2-butene + Br₂ → ? (specify stereochemistry)

23. Why does HBr without peroxides give Markovnikov addition, but HBr with peroxides gives anti-Markovnikov? Explain in terms of mechanism.

24. Sketch the bromonium ion intermediate of Br₂ + ethylene. Identify the geometry.

25. Predict the product (with rearrangement!) of: 3,3-dimethyl-1-butene + HCl.

Answer key

  1. b — Tetrasubstituted = most stable.
  2. b — 2-step mechanism via cation.
  3. b — H to less-substituted C.
  4. a — 3° > 2° > 1°.
  5. b — Bromonium ion.
  6. b — Anti opening.
  7. b — Radical mechanism for anti-Markovnikov.
  8. b — Restricted rotation due to π bond.
  9. a — Hyperconjugation σ → π* (or σ → empty p for cation).
  10. a — Carbocation stability + Hammond.
  11. a — 1,2-shifts.
  12. a — cis is less stable.
  13. a — CIP priority.
  14. b — Hydroboration is anti-Markovnikov.
  15. b — 2-propanol (Markovnikov).
  16. b — Vinyl ¹H at 4.5-6.5.
  17. c — Trans coupling 12-18 Hz.
  18. a — Ethylene is the largest industrial alkene.
  19. a — π electrons are nucleophilic.
  20. b — HX gives stereochemistry mixture (cation is planar).

21. 2-Methyl-2-butene is a trisubstituted alkene: $(CH_3)_2C=CHCH_3$. With HBr: - Step 1: Protonation of the less-substituted C (CHCH₃ → CH₂CH₃). The 3° carbocation forms at the original $(CH_3)_2C$ position. - Step 2: Br⁻ attacks the 3° cation. - Product: 2-bromo-2-methylbutane, $(CH_3)_2CBr-CH_2CH_3$. The 3° cation is the more-stable one; Markovnikov selectivity gives the tertiary alkyl bromide.

22. cis-2-butene + Br₂: - Bromonium ion forms first (cyclic 3-membered ring with both C atoms partially bonded to Br). - Br⁻ attacks one C from the opposite face of the existing Br. - The two Br atoms end up anti (on opposite faces). - Starting from cis-2-butene (both methyls on the same face): the methyls remain on the same face (since the Br atoms are anti, and the carbon stereochemistry is set by where Br attacks). Result: meso-2,3-dibromobutane (the two Br anti, methyls anti, racemic at each C but achiral overall due to internal mirror plane). Stereospecific reaction.

23. HBr without peroxides: ionic mechanism. Step 1: π bond attacks H of HBr; carbocation forms (more-substituted = more-stable, Markovnikov). Step 2: Br⁻ attacks the cation. Markovnikov product.

HBr with peroxides: radical mechanism. Step 1: peroxide initiates radical chain by generating a Br• radical. Step 2: Br• adds to alkene to form a C-radical at the LESS-substituted C (because the more-substituted C-radical is more stable; remember: the radical is at the C where the OTHER atom didn't go). Step 3: the more-stable (more-substituted) C-radical abstracts H from another HBr molecule to give the alkyl bromide. Step 4: chain propagates. The radical mechanism gives the alkyl bromide with Br at the LESS-substituted C — anti-Markovnikov product.

The mechanism (ionic vs radical) determines the regiochemistry. Both types follow the same logic (more-stable intermediate forms preferentially) but the intermediates are different (cation vs radical).

24. Bromonium ion of Br₂ + ethylene: - Three-membered cyclic ring: 2 C atoms + 1 Br atom. - Br is at the apex; both C atoms are connected to Br via partial covalent bonds (somewhat weaker than typical C-Br bonds because of ring strain and 3-center bonding). - The CH₂ groups are still sp²-like; the geometry has flattened compared to a fully sp³ open-chain cation. - Net positive charge on the ring (mostly localized on Br but distributed). The structure is a "halonium ion" — a class that includes chloronium, bromonium, and iodonium.

25. 3,3-Dimethyl-1-butene + HCl: - Step 1: H⁺ adds to terminal CH₂ (Markovnikov-like to give the more-substituted cation). The resulting cation would be on C2: $(CH_3)_3C-CH(+)-CH_3$. But wait, this is a 2° cation; could it rearrange to a 3°? - The adjacent (CH₃)₃C group has a methyl that can migrate (1,2-methyl shift). Migration would give: $(CH_3)_2C(+)-CH(CH_3)-CH_3$. But this is also 2° in this analysis... let me re-look. - Actually: 3,3-dimethyl-1-butene = $(CH_3)_3C-CH=CH_2$. After H adds to CH₂, the cation is on $(CH_3)_3C-CH(+)-CH_3$ — wait, the H adds to CH₂ giving CH₃, and the cation is on the original CH (which becomes (CH₃)₃C-CH(+)-CH₃, a 2° cation). - Wait, let me redo: $(CH_3)_3C-CH=CH_2$. After H adds to terminal CH₂, the "formal" cation is on the CH side: $(CH_3)_3C-CH(+)-CH_3$. This 2° cation is adjacent to a quaternary carbon (the (CH₃)₃C). A 1,2-methyl shift from (CH₃)₃C-CH(+) to the cation gives a 3° cation: $(CH_3)_2C(+)-CH(CH_3)-CH_3$ — wait, that's actually still 3° but at a different carbon. Let me sketch: - Initial: (CH₃)₃C—CH(+)—CH₃ (2° cation; quaternary C adjacent). - Methyl shift: (CH₃)₂C(+)—CH(CH₃)—CH₃ (3° cation; new structure). - Cl⁻ attacks the 3° cation. - Product: 2-chloro-2,3-dimethylbutane, $(CH_3)_2CCl-CH(CH_3)-CH_3$. This shows the carbocation rearrangement giving a different product than the simple Markovnikov prediction.