Topic 34: Nitrogen Compounds

Question 1

Definitions I:
(a) Primary amine
(b) Azo compound
(c) Zwitterion

Answer 1

Definitions I:
(a) Amine with one alkyl group, e.g., CH₃NH₂.
(b) Compound with -N=N- group, e.g., C₆H₅N=NC₆H₄OH.
(c) Molecule with + and - charges, e.g., ⁺H₃NCH₂COO⁻.

Question 2

Write equations for:
(a) CH₃CH₂Cl + NH₃ → amine [1 Mark]
(b) CH₃CONH₂ + LiAlH₄ → amine [1 Mark]

Answer 2

(a) CH₃CH₂Cl + NH₃ → CH₃CH₂NH₂ + HCl (ethanol, pressure, heat) (1 mark)
(b) CH₃CONH₂ + 4[H] → CH₃CH₂NH₂ + H₂O (LiAlH₄) (1 mark)

Question 3

In a lab, nitrobenzene forms phenylamine. Write the equation and conditions. [2 Marks]

Answer 3

C₆H₅NO₂ + 6[H] → C₆H₅NH₂ + 2H₂O (Sn/HCl, heat) (1 mark)
Then + NaOH to free base (1 mark)

Question 4

Predict the product of C₆H₅NH₂ with Br₂(aq) and name it. [2 Marks]

Answer 4

C₆H₅NH₂ + 3Br₂ → C₆H₂(NH₂)Br₃ + 3HBr (aq, room temp) (1 mark)
2,4,6-Tribromophenylamine (1 mark)

Question 5

Compare the basicities of NH₃, CH₃CH₂NH₂, and C₆H₅NH₂. [3 Marks]

Answer 5

CH₃CH₂NH₂ > NH₃ > C₆H₅NH₂ (1 mark)
Ethylamine: +I effect; NH₃: no effect (1 mark)
Phenylamine: lone pair delocalised into ring (1 mark)

Question 6

Describe the coupling of C₆H₅N₂⁺Cl⁻ with phenol in NaOH and identify the azo group. [2 Marks]

Answer 6

C₆H₅N₂⁺ + C₆H₅OH → C₆H₅N=NC₆H₄OH + H⁺ (NaOH) (1 mark)
Azo group: -N=N- (1 mark)

Question 7

Why are amides weaker bases than amines? [2 Marks]

Answer 7

C=O in amides withdraws electrons from N (1 mark)
Less available lone pair than amines (1 mark)

Question 8

In a synthesis, CH₃COCl + CH₃NH₂ forms an amide. Write the equation. [2 Marks]

Answer 8

CH₃COCl + CH₃NH₂ → CH₃CONHCH₃ + HCl (room temp) (1 mark)
N-Methylacetamide formed (1 mark)

Question 9

Describe the hydrolysis of CH₃CONH₂ with NaOH(aq) and heat. [2 Marks]

Answer 9

CH₃CONH₂ + NaOH → CH₃COO⁻Na⁺ + NH₃ (heat) (1 mark)
Alkaline hydrolysis to salt, ammonia (1 mark)

Question 10

Explain how glycine (H₂NCH₂COOH) forms a zwitterion and its isoelectric point. [2 Marks]

Answer 10

H₂NCH₂COOH → ⁺H₃NCH₂COO⁻ (proton transfer) (1 mark)
Isoelectric: neutral net charge, pH ~6 (1 mark)

Question 11

Predict the electrophoresis result for glycine and H₂NCH₂CONHCH₂COOH at pH 7. [3 Marks]

Answer 11

Glycine: ⁺H₃NCH₂COO⁻, no net charge, stays (1 mark)
Dipeptide: ⁺H₃N-, -COO⁻, neutral, stays (1 mark)
pH 7, both zwitterions, no migration (1 mark)

Question 12

Show the formation of a dipeptide from alanine (CH₃CH(NH₂)COOH) and glycine. [2 Marks]

Answer 12

CH₃CH(NH₂)COOH + H₂NCH₂COOH → CH₃CH(NH₂)CONHCH₂COOH + H₂O (1 mark)
Ala-Gly dipeptide (1 mark)