Chapter 30 — Quiz

Twenty-five questions on amine chemistry, basicity, nucleophilicity, and synthesis. ∗ marks questions answered in the answer key.

Multiple choice

1.∗ An amine + HCl gives: (a) an amide (b) an ammonium salt (ionic, water-soluble) (c) an alcohol (d) no reaction

2.∗ A simple aliphatic primary amine has pKaH approximately: (a) ~5 (b) ~10 (c) ~15 (d) ~20

3.∗ Aniline has pKaH: (a) ~10 (same as aliphatic) (b) ~4.6 (much lower than aliphatic) (c) ~15 (d) > 14

4.∗ Pyridine has pKaH: (a) ~10 (b) ~5.2 (c) ~15 (d) ~0

5.∗ A primary amine + aldehyde + cat. acid gives: (a) an imine (Schiff base) (b) an amide (c) an enamine (d) a tertiary amine

6.∗ A secondary amine + ketone + cat. acid gives: (a) an imine (b) an amide (c) an enamine (d) a quaternary salt

7.∗ Direct alkylation of NH₃ by R-X gives: (a) clean primary amine (b) a mixture of 1°, 2°, 3°, and 4° amines (over-alkylation problem) (c) only secondary amine (d) an alcohol

8.∗ The Gabriel synthesis gives: (a) primary amine (clean, no over-alkylation) (b) secondary amine (c) tertiary amine (d) quaternary ammonium

9.∗ The standard reagent for reductive amination is: (a) LiAlH₄ (b) NaBH₃CN or NaBH(OAc)₃ (selective for iminium over aldehyde) (c) H₂/Pd (d) Zn dust

10.∗ Diazonium salt + KI gives: (a) ArI (aromatic iodide) (b) ArOH (c) ArH (no halide) (d) AraOH

11.∗ Why is aniline less basic than methylamine? (a) the lone pair is delocalized into the benzene ring (b) aniline has more steric strain (c) aniline is heavier (d) aniline has different hybridization

12.∗ Why is an amide essentially non-basic? (a) the N lone pair is donated into the C=O π system (b) the N is too sp² (c) the N is too crowded (d) the amide is acidic

13.∗ The Hofmann elimination of a quaternary ammonium gives: (a) Zaitsev (more-substituted) alkene (b) Hofmann (less-substituted) alkene (c) gives no alkene (d) gives a new amine

14.∗ The Sandmeyer reaction is: (a) ArN₂⁺ + CuX → ArX (where X = Cl, Br, CN) (b) ArN₂⁺ + H₂O → ArOH (c) only with iodine (d) photochemistry

15.∗ Why is imidazole's "pyridine-like" N basic and its "pyrrole-like" N acidic? (a) different electronegativity (b) the basic N has its lone pair available; the acidic N has its lone pair locked into the aromatic π system (c) different hybridization (d) different N-H bond strengths

16.∗ Histamine has the structure: (a) imidazoleethylamine (b) phenethylamine (c) tryptamine (d) GABA

17.∗ Most drugs contain at least one amine because: (a) amine basicity allows partial protonation at pH 7.4, balancing solubility and membrane crossing (b) amines are easy to handle (c) amines are cheap (d) all of the above except (c)

18.∗ A tertiary amine adding to an alkyl halide gives: (a) tertiary amine (b) quaternary ammonium salt (the only step possible) (c) primary amine (d) an alcohol

19.∗ Why does NaBH₃CN reduce iminium ions but not the parent aldehydes? (a) iminium is more electrophilic than aldehyde (b) NaBH₃CN is bulky (c) iminium has more H atoms (d) aldehydes are too reactive

20.∗ A diazonium salt is stable at: (a) room temperature for hours (b) 0–5 °C; decomposes above ~5 °C (c) only in vacuum (d) only in light

Short answer

21. Predict the pKaH of: methylamine, dimethylamine, trimethylamine, aniline, pyridine. Rank from most basic to least basic.

22. Sketch the Gabriel synthesis of butylamine from butyl bromide. Identify the role of phthalimide and hydrazine.

23. Sketch a reductive amination: butanal + methylamine + cat. acid + NaBH₃CN → ? Predict the product.

24. Why is the Hofmann elimination preferred over Zaitsev when the leaving group is a quaternary ammonium? Identify the steric origin.

25. Sketch the diazonium chemistry: aniline → 4-iodoaniline via Sandmeyer-type chemistry.

Answer key

  1. b — Amine + HCl gives an ionic ammonium salt.
  2. b — pKaH ~10 for simple aliphatic amines.
  3. b — Aniline pKaH 4.6.
  4. b — Pyridine pKaH 5.2.
  5. a — Imine (Schiff base) from primary amine + aldehyde.
  6. c — Enamine from secondary amine + ketone.
  7. b — Direct alkylation gives over-alkylation mixture.
  8. a — Gabriel gives clean primary amine.
  9. b — NaBH₃CN or NaBH(OAc)₃ for reductive amination.
  10. a — Diazonium + KI → ArI.
  11. a — Aniline lone pair delocalized into ring.
  12. a — Amide N lone pair donated into C=O.
  13. b — Hofmann gives less-substituted alkene.
  14. a — Sandmeyer description.
  15. b — Pyridine-like N's lone pair is in the plane, available; pyrrole-like N's lone pair is in the π system, not available.
  16. a — Histamine = imidazoleethylamine.
  17. d — Both (a) and (b); (c) varies but is plausible.
  18. b — Tertiary amine + R-X = quaternary ammonium salt.
  19. a — Iminium is more electrophilic.
  20. b — Diazonium stable at 0–5 °C.

21. Most basic to least basic (highest to lowest pKaH): - Pyrrolidine 11.3 - Dimethylamine 10.7 - Methylamine 10.6 - Trimethylamine 9.8 (slightly less than dimethyl due to steric hindering of solvation of N⁺) - Imidazole 7.0 - Pyridine 5.2 - Aniline 4.6

22. Step 1: phthalimide + KOH (or NaH) → potassium phthalimide (deprotonates the cyclic imide N-H). Step 2: potassium phthalimide + butyl bromide → N-butyl phthalimide + KBr (SN2 at the alkyl halide). Step 3: N-butyl phthalimide + hydrazine + heat → butylamine + phthalhydrazide. The phthalimide acts as a "protected ammonia" — a single N-nucleophile that can be alkylated cleanly without over-alkylation.

23. Step 1: methylamine attacks butanal's C=O carbon (nucleophilic addition, Ch 25). Step 2: hemiaminal forms; loss of water gives an iminium ion. Step 3: NaBH₃CN selectively reduces the iminium → secondary amine (N-methyl-1-butylamine). Selectivity: NaBH₃CN is too mild to reduce the parent butanal but strong enough to reduce the more-electrophilic iminium ion. pH is adjusted to 5–6 to balance imine formation (acid-promoted) with amine reactivity (not over-protonated).

24. The leaving group in Hofmann elimination is the bulky $R_3N$ (a tertiary amine). For a Zaitsev (more-substituted) alkene to form, the β-H removed must be the more-substituted one, but this places more substituents close to the bulky leaving group during the TS — steric strain. The less-substituted β-H removal gives a less-strained TS and the Hofmann (less-substituted) alkene. So the Hofmann selectivity is steric: the bulky leaving group prevents a Zaitsev TS.

25. Step 1: aniline + NaNO₂ + HCl + 0 °C → arenediazonium chloride (Ar-N₂⁺Cl⁻). Step 2: add KI; the iodide attacks the diazonium ion; N₂ leaves as gas; aryl iodide forms. Net: ArNH₂ + HNO₂ + HCl + KI → ArI + N₂ + KCl + H₂O. The iodination is a one-step Sandmeyer-like reaction (no Cu catalyst needed for KI).