Chapter 17 — Quiz

Twenty-five questions on alkyne chemistry. ∗ marks questions answered in the answer key.

Multiple choice

1.∗ Terminal alkyne pKa is approximately: (a) 5 (b) 10 (c) 25 (uniquely acidic; sp orbital character) (d) 50

2.∗ NaNH₂ can deprotonate: (a) alkane (b) alkene (c) terminal alkyne (pKa 25 < pKaH of NH₃ which is 38) (d) any of the above

3.∗ Lindlar catalyst reduces alkyne to: (a) trans-alkene (b) cis-alkene (poisoned Pd; syn addition; doesn't continue to alkane) (c) alkane (d) alkene of mixed geometry

4.∗ Na in liquid NH₃ reduces alkyne to: (a) trans-alkene (anti addition via radical anion mechanism) (b) cis-alkene (c) alkane (d) alkynide

5.∗ Hydration of a terminal alkyne (with HgSO₄/H₂SO₄/H₂O) gives: (a) aldehyde (b) methyl ketone (Markovnikov; vinyl alcohol → ketone via tautomerism) (c) carboxylic acid (d) primary alcohol

6.∗ Hydroboration-oxidation of a terminal alkyne gives: (a) aldehyde (anti-Markovnikov; B at less-subs C; vinyl alcohol → aldehyde) (b) methyl ketone (c) carboxylic acid (d) primary alcohol

7.∗ An alkynide anion (RC≡C⁻) is: (a) a proton donor (b) a strong carbon nucleophile (sp carbanion; reacts with alkyl halides via SN2) (c) only a base, not a nucleophile (d) an oxidizing agent

8.∗ A C≡C triple bond consists of: (a) 3 σ bonds (b) 1 σ + 2 π bonds (sp hybridized; perpendicular π bonds) (c) 3 π bonds (d) 2 σ + 1 π

9.∗ Alkyne + HX twice gives: (a) geminal dihalide (both X on the same C; Markovnikov) (b) vicinal dihalide (c) tetrahalide (d) random

10.∗ Alkyne + Br₂ once gives: (a) cis-dihaloalkene (b) trans-dihaloalkene (anti addition; bromonium-ion-like) (c) random (d) tetrabromide

11.∗ Why is the terminal alkyne C-H more acidic than alkene C-H? (a) sp orbital has more s-character (50%) than sp² (33%); the conjugate base (carbanion) is more stable in the sp orbital (b) the C-H bond is weaker (c) it's just empirical (d) hyperconjugation

12.∗ Alkynide anion + primary alkyl halide gives: (a) substituted alkyne (SN2; new C-C bond) (b) elimination product (c) only deprotonation (d) no reaction

13.∗ Alkynide anion + tertiary alkyl halide gives: (a) substituted alkyne (SN2) (b) elimination product (E2; tertiary halide cannot do SN2) (c) clean SN1 (d) random

14.∗ Sonogashira coupling: (a) aryl halide + terminal alkyne + Pd + Cu → aryl alkyne (b) only with primary alkyl halides (c) only with alkenes (d) photochemical

15.∗ Acetylene + 2 HCl (excess) gives: (a) 1,1-dichloroethane (gem-dichloride; Markovnikov on each addition) (b) 1,2-dichloroethane (c) chloroethane (d) chloroform

16.∗ Alkyne hydroboration with disiamylborane stops after one addition because: (a) the bulky borane prevents a second addition (steric) (b) the alkyne is unreactive after one addition (c) it's photochemical (d) it's inhibited by acid

17.∗ Why does acetylene + Hg(II) catalyst give vinyl alcohol that tautomerizes to acetaldehyde? (a) Markovnikov hydration of the terminal alkyne; the resulting enol tautomerizes to the keto form (b) it's an oxidation (c) it's a reduction (d) random

18.∗ Linear geometry of alkynes (~180°): (a) results from sp hybridization at each C (b) is sp² (c) is sp³ (d) varies with substituents

19.∗ Cyclic alkynes (e.g., cyclooctyne) are: (a) commonly stable for ring sizes 8 and larger (b) easy to make for any size (c) only stable for ring sizes 14 and larger (d) impossible

20.∗ Click chemistry (CuAAC, Sharpless 2022 Nobel) is: (a) Cu-catalyzed alkyne + azide cycloaddition giving a 1,2,3-triazole (b) only with aryl halides (c) only photochemical (d) only on alkenes

Short answer

21. Why is the terminal alkyne C-H so acidic (pKa 25) compared to alkene (pKa 44) and alkane (pKa 50)? Use hybridization arguments.

22. Predict the products: (a) 2-pentyne + H₂/Lindlar Pd → ? (b) 2-pentyne + Na/NH₃(l) → ?

23. Sketch the mechanism of: 1-butyne + NaNH₂, then methyl iodide. What is the product? Why does this work for primary halides?

24. Predict the product of hydration: 1-hexyne + HgSO₄/H₂SO₄/H₂O → ? (Show the enol intermediate.)

25. Compare hydration of an alkyne with hydroboration-oxidation. For a terminal alkyne, which gives an aldehyde and which gives a methyl ketone? Explain why.

Answer key

  1. c — pKa ~25.
  2. c — Terminal alkyne deprotonable by NaNH₂.
  3. b — Lindlar = cis-alkene.
  4. a — Na/NH₃ = trans-alkene.
  5. b — Methyl ketone from terminal alkyne hydration.
  6. a — Aldehyde from anti-Markovnikov hydroboration.
  7. b — Alkynide is strong C nucleophile.
  8. b — 1σ + 2π.
  9. a — Gem-dihalide.
  10. b — trans-dihaloalkene.
  11. a — sp orbital character.
  12. a — SN2 with 1° halide.
  13. b — E2 with 3° halide.
  14. a — Sonogashira.
  15. a — 1,1-dichloroethane (Markovnikov twice).
  16. a — Steric prevents second addition.
  17. a — Markovnikov hydration → enol → acetaldehyde.
  18. a — sp hybridization.
  19. a — Cyclooctyne is the smallest stable cyclic alkyne.
  20. a — Click chemistry definition.

21. Acidity follows from carbanion stability (the conjugate base). The carbanion's lone pair sits in an orbital with character determined by the parent C's hybridization: - sp³ (alkane): 25% s-character → orbital is high in energy, far from nucleus → carbanion unstable. - sp² (alkene): 33% s-character → moderate. - sp (alkyne): 50% s-character → orbital is low in energy, close to nucleus → carbanion stable. More s-character = more stable carbanion = more acidic C-H. The terminal alkyne C-H is uniquely positioned to give a stabilized sp carbanion.

22. (a) 2-pentyne + H₂/Lindlar Pd → cis-2-pentene (the (Z)-alkene; syn addition). Lindlar's poisoned Pd does syn addition and stops at the alkene (doesn't reduce further to alkane). (b) 2-pentyne + Na/NH₃(l) → trans-2-pentene (the (E)-alkene; anti addition via radical anion mechanism). The Na electrons reduce the alkyne stepwise; trans is the more-stable vinyl carbanion intermediate.

23. Mechanism of alkynide alkylation: - Step 1: NaNH₂ + 1-butyne (HC≡C-CH₂CH₃) → Na⁺C≡C-CH₂CH₃ + NH₃ (deprotonation of the terminal C-H; pKa 25 < pKaH NH₃ 38). - Step 2: methyl iodide + alkynide → CH₃-C≡C-CH₂CH₃ (2-pentyne) + NaI (SN2 at the methyl carbon; backside attack; clean inversion; alkynide is a strong nucleophile). Net: 1-butyne + NaNH₂ + CH₃I → 2-pentyne + NH₃ + NaI. This works for primary halides because SN2 is favored. Tertiary halides give E2; secondary halides may give a mix. Internal alkynes (no terminal C-H) cannot be deprotonated, so this method doesn't extend chains beyond one direction.

24. Hydration of 1-hexyne with HgSO₄/H₂SO₄/H₂O: - Hg²⁺ coordinates to the alkyne; the resulting cation is attacked by water (Markovnikov: H ends up on terminal C; OH on internal C). - The intermediate is the enol (vinyl alcohol): CH₃-CH₂-CH₂-CH₂-C(OH)=CH₂. - The enol is unstable; tautomerizes (acid-catalyzed) to the keto form: CH₃-CH₂-CH₂-CH₂-CO-CH₃ = 2-hexanone (a methyl ketone).

25. For a terminal alkyne RC≡CH: - Hydration (HgSO₄/H₂SO₄/H₂O): Markovnikov. OH adds to the more-substituted (internal) C; H adds to the terminal C. The resulting enol tautomerizes to a methyl ketone (R-CO-CH₃). - Hydroboration-oxidation ((Sia)₂BH, then H₂O₂/NaOH): anti-Markovnikov. B (and later OH) adds to the less-substituted (terminal) C; H adds to the internal C. The resulting enol tautomerizes to an aldehyde (R-CH₂-CHO). The two methods give complementary regiochemistry: hydration → ketone; hydroboration → aldehyde. Choosing between them is a key synthesis decision when the goal is a specific carbonyl from a terminal alkyne.