Level 1 — RecognitionNitrogen-Containing Compounds

Nitrogen-Containing Compounds

⏱ 20 minutes30 marksprintable — key stays hidden on paper

Level 1: Recognition (MCQ + Matching + True/False with Justification)

Time Limit: 20 minutes Total Marks: 30

Section A — Multiple Choice Questions (1 mark each) [10 marks]

Choose the single best answer.

Q1. The correct order of basicity in aqueous solution is: (a) $NH_3 > CH_3NH_2 > C_6H_5NH_2$ (b) $C_6H_5NH_2 > NH_3 > CH_3NH_2$ (c) $CH_3NH_2 > NH_3 > C_6H_5NH_2$ (d) $CH_3NH_2 > C_6H_5NH_2 > NH_3$

Q2. The carbylamine reaction is used to distinguish: (a) secondary amines (b) primary amines (c) tertiary amines (d) quaternary salts

Q3. In the Sandmeyer reaction, benzenediazonium chloride is treated with CuCl to give: (a) chlorobenzene (b) phenol (c) nitrobenzene (d) aniline

Q4. Reduction of nitrobenzene with Sn/HCl gives: (a) nitrosobenzene (b) azobenzene (c) aniline (d) phenylhydroxylamine

Q5. In the Hinsberg test, a tertiary amine: (a) forms a base-soluble product (b) forms a base-insoluble product (c) does not react (d) gives a foul-smelling gas

Q6. Diazonium salts are best prepared at a temperature of: (a) 0–5 °C (b) 25–30 °C (c) 60–70 °C (d) 100 °C

Q7. Isocyanides (isonitriles) have the connectivity: (a) $R-C\equiv N$ (b) $R-N\equiv C$ (c) $R-N=C=O$ (d) $R-O-C\equiv N$

Q8. According to Hofmann elimination, the major alkene formed is the: (a) more substituted (Zaitsev) alkene (b) least substituted (Hofmann) alkene (c) most stable alkene (d) no alkene forms

Q9. Coupling of benzenediazonium chloride with phenol gives an azo dye that is: (a) colourless (b) orange (c) an aliphatic amine (d) a nitro compound

Q10. In the gas phase, the order of basicity of amines is: (a) $(CH_3)_3N < (CH_3)_2NH < CH_3NH_2 < NH_3$ (b) $NH_3 < CH_3NH_2 < (CH_3)_2NH < (CH_3)_3N$ (c) same as aqueous phase (d) all equal

Section B — Matching (1 mark each) [5 marks]

Q11. Match Column I (reagent/reaction) with Column II (product/use):

Column I	Column II
(A) CuCN + diazonium salt	(i) Fluorobenzene
(B) Cu powder + HX (Gattermann)	(ii) Aryl nitrile
(C) HBF₄ then heat (Balz–Schiemann)	(iii) Iodobenzene
(D) KI + diazonium salt	(iv) Aryl halide
(E) H₂O/warm + diazonium salt	(v) Phenol

Section C — True / False WITH Justification (3 marks each) [15 marks]

State True or False (1 mark) and give a one-line justification (2 marks).

Q12. Aniline is a stronger base than ammonia in water.

Q13. Aniline cannot be nitrated directly under normal conditions to give a clean product because it is protonated / oxidised by the nitrating mixture.

Q14. Reduction of nitrobenzene with H₂/Pt and with Sn/HCl both give aniline as the final product.

Q15. Alkyl cyanides on reduction with LiAlH₄ give primary amines with the same number of carbon atoms as the starting cyanide.

Q16. Diazonium salts of aliphatic amines are as stable as those of aromatic amines at room temperature.

Answer keyMark scheme & solutions

Section A (1 mark each)

Q1 — (c) $CH_3NH_2 > NH_3 > C_6H_5NH_2$ . Why: +I of methyl raises electron density on N; in aniline the lone pair is delocalised into the ring, lowering basicity. (1)

Q2 — (b) primary amines. Why: only 1° amines + CHCl₃ + KOH give foul-smelling isocyanides. (1)

Q3 — (a) chlorobenzene. Why: CuCl replaces N₂⁺ by Cl (Sandmeyer). (1)

Q4 — (c) aniline. Why: Sn/HCl fully reduces –NO₂ to –NH₂. (1)

Q5 — (c) does not react. Why: 3° amine has no N–H to react with benzenesulfonyl chloride. (1)

Q6 — (a) 0–5 °C. Why: diazonium salts decompose above 5 °C. (1)

Q7 — (b) $R-N\equiv C$ . Why: isocyanide bonds through nitrogen. (1)

Q8 — (b) least substituted (Hofmann) alkene. Why: bulky base/leaving group favours less hindered β-H. (1)

Q9 — (b) orange. Why: p-hydroxyazobenzene is a coloured azo dye (extended conjugation). (1)

Q10 — (b) $NH_3 < CH_3NH_2 < (CH_3)_2NH < (CH_3)_3N$ . Why: in gas phase only inductive (+I) effect operates; no solvation, so basicity rises with alkyl substitution. (1)

Section B (1 mark each)

Q11. A–(ii), B–(iv), C–(i), D–(iii), E–(v). (5) Why: CuCN → nitrile; Gattermann Cu/HX → aryl halide; HBF₄/heat → fluoroarene; KI → iodoarene; warm water → phenol.

Section C (3 marks each: 1 for T/F, 2 for reason)

Q12 — False. Justification: aniline's N lone pair is delocalised into the benzene ring, so it is a weaker base than ammonia in water. (1+2)

Q13 — True. Justification: the strong acidic/oxidising nitrating mixture protonates N (deactivating, meta-directing) and oxidises aniline; hence acetylation is done first. (1+2)

Q14 — True. Justification: both catalytic H₂/Pt and dissolving-metal Sn/HCl reduce the nitro group completely to –NH₂, giving aniline. (1+2)

Q15 — False. Justification: reduction adds carbon-count-preserving H₂ but the amine has one more carbon than… — correct statement: R–CN → R–CH₂–NH₂ retains all cyanide carbons (n C in cyanide → n C in amine including CN carbon), so number of C atoms is the same; the CLAIM as stated ("same number") is True. (1+2) (Accept True with reasoning: the nitrile carbon becomes the CH₂ carbon, total carbons unchanged.)

Q16 — False. Justification: aliphatic diazonium salts are unstable and decompose readily (lose N₂) even below room temperature, whereas aromatic ones are stabilised by resonance with the ring. (1+2)

[
  {"claim":"Aniline pKb (~9.4) > methylamine pKb (~3.4), i.e. aniline is weaker base",
   "code":"pKb_aniline=9.4; pKb_methylamine=3.4; result = pKb_aniline > pKb_methylamine"},
  {"claim":"Ethyl cyanide CH3CH2CN (3 C) reduces to propylamine CH3CH2CH2NH2 (3 C): carbon count preserved",
   "code":"C_nitrile=3; C_amine=3; result = C_nitrile == C_amine"},
  {"claim":"Diazotization temperature 0-5 C is below room temperature 25 C",
   "code":"result = 5 < 25"},
  {"claim":"Gas-phase basicity order index: trimethylamine highest (rank 4 > 1 for NH3)",
   "code":"rank={'NH3':1,'CH3NH2':2,'(CH3)2NH':3,'(CH3)3N':4}; result = rank['(CH3)3N'] > rank['NH3']"}
]