Level 2 — RecallNitrogen-Containing Compounds

Nitrogen-Containing Compounds

⏱ 30 minutes40 marksprintable — key stays hidden on paper

Level 2 — Recall (definitions, standard textbook problems, short derivations) Time Limit: 30 minutes Total Marks: 40

Instructions: Answer all questions. Show reagents/conditions clearly on all reaction arrows.

Q1. Arrange the following in decreasing order of basicity in aqueous solution and give a one-line reason: $\text{NH}_3$ , $\text{CH}_3\text{NH}_2$ , $\text{C}_6\text{H}_5\text{NH}_2$ . (3 marks)

Q2. State how the basicity order of methylamine and ammonia reverses in the gas phase compared to aqueous solution. Explain briefly. (3 marks)

Q3. Write the reaction and name of the test used to detect a primary amine using chloroform and alcoholic KOH. Give the product for $\text{CH}_3\text{NH}_2$ . (4 marks)

Q4. Describe the Hinsberg test and explain how it distinguishes primary, secondary and tertiary amines. (5 marks)

Q5. Give the product and conditions for the Hofmann elimination (exhaustive methylation) of the quaternary ammonium hydroxide derived from propan-1-amine. State the Hofmann (regioselectivity) rule. (4 marks)

Q6. Write the equation for the preparation of benzene diazonium chloride from aniline. State the temperature range required and why. (4 marks)

Q7. Complete the following conversions from benzene diazonium chloride, naming the reaction type: (6 marks) (a) $\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- \xrightarrow{\text{CuCl / HCl}} ?$ (b) $\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- \xrightarrow{\text{CuCN / KCN}} ?$ (c) $\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- \xrightarrow{\text{Cu powder / HCl}} ?$ (Gattermann)

Q8. Write the azo coupling reaction of benzene diazonium chloride with phenol (in mildly alkaline medium). Name the class of product. (4 marks)

Q9. Give reagents for reducing nitrobenzene to aniline by three different methods. Write the balanced skeletal reaction for one. (4 marks)

Q10. Distinguish between cyanides (nitriles) and isocyanides in terms of (a) structure of the functional group and (b) product on acid hydrolysis. (3 marks)

End of Paper

Answer keyMark scheme & solutions

Q1. (3 marks) Order: $\text{CH}_3\text{NH}_2 > \text{NH}_3 > \text{C}_6\text{H}_5\text{NH}_2$ (2) Reason: In alkylamine, +I effect of methyl increases electron density on N (more basic); in aniline the lone pair is delocalised into the ring, reducing availability, making it least basic. (1)

Q2. (3 marks) In gas phase: $\text{CH}_3\text{NH}_2 > \text{NH}_3$ becomes dominated only by the intrinsic electron-donating (+I) effect — (1); without water there is no solvation/H-bond stabilisation of the cation which in solution favours smaller/less substituted cations. (1) In gas phase basicity is governed purely by inductive/polarisability effects, so more alkyl substitution = more basic (order: 3° > 2° > 1° > NH₃). (1)

Q3. (4 marks) Carbylamine (isocyanide) test. (1) $\text{CH}_3\text{NH}_2 + \text{CHCl}_3 + 3\text{KOH} \xrightarrow{\Delta} \text{CH}_3\text{NC} + 3\text{KCl} + 3\text{H}_2\text{O}$ Balanced equation (2); product methyl isocyanide (foul smelling) confirms 1° amine (1).

Q4. (5 marks) Hinsberg reagent = benzenesulphonyl chloride ( $\text{C}_6\text{H}_5\text{SO}_2\text{Cl}$ ). (1)

1° amine → N-substituted sulphonamide with acidic N–H; soluble in KOH (clear solution). (1.5)
2° amine → sulphonamide with no N–H; insoluble in KOH (precipitate). (1.5)
3° amine → does not react (no replaceable H on N). (1)

Q5. (4 marks) Exhaustive methylation gives $\text{CH}_3\text{CH}_2\text{CH}_2\overset{+}{\text{N}}(\text{CH}_3)_3\,\text{OH}^-$ ; on heating: (1) $\text{CH}_3\text{CH}_2\text{CH}_2\overset{+}{\text{N}}(\text{CH}_3)_3\text{OH}^- \xrightarrow{\Delta} \text{CH}_3\text{CH}=\text{CH}_2 + \text{N}(\text{CH}_3)_3 + \text{H}_2\text{O}$ Product propene + trimethylamine (2). Hofmann rule: elimination gives the least substituted (Hofmann) alkene as major product. (1)

Q6. (4 marks) $\text{C}_6\text{H}_5\text{NH}_2 + \text{NaNO}_2 + 2\text{HCl} \rightarrow \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{NaCl} + 2\text{H}_2\text{O}$ Equation (2). Temperature 0–5 °C (1); because diazonium salts are unstable and decompose at higher temperature to phenol + N₂. (1)

Q7. (6 marks; 2 each) (a) $\text{C}_6\text{H}_5\text{Cl}$ (chlorobenzene) — Sandmeyer reaction. (b) $\text{C}_6\text{H}_5\text{CN}$ (benzonitrile) — Sandmeyer reaction. (c) $\text{C}_6\text{H}_5\text{Cl}$ (chlorobenzene) — Gattermann reaction.

Q8. (4 marks) $\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{C}_6\text{H}_5\text{OH} \xrightarrow{\text{OH}^-} p\text{-}\text{HO}\text{-}\text{C}_6\text{H}_4\text{-N=N-}\text{C}_6\text{H}_5 + \text{HCl}$ Product = p-hydroxyazobenzene (an azo dye), coupling at para position. Equation (3) + class (1).

Q9. (4 marks) Any three reagents: (3) — Sn/HCl; Fe/HCl; H₂/Pt (or Ni, Pd); also LiAlH₄, or catalytic hydrogenation. Reaction (1): $\text{C}_6\text{H}_5\text{NO}_2 + 6[\text{H}] \xrightarrow{\text{Sn/HCl}} \text{C}_6\text{H}_5\text{NH}_2 + 2\text{H}_2\text{O}$

Q10. (3 marks) (a) Cyanide (nitrile): $R\text{-C}\equiv\text{N}$ — carbon bonded to R; isocyanide: $R\text{-N}\equiv\overset{+}{\text{C}}{}^-$ — nitrogen bonded to R. (1.5) (b) Nitrile hydrolysis → carboxylic acid ( $R\text{COOH} + \text{NH}_3$ ); isocyanide hydrolysis → primary amine + formic acid ( $R\text{NH}_2 + \text{HCOOH}$ ). (1.5)

[
  {"claim":"Carbylamine test: CH3NH2 + CHCl3 + 3KOH balances to CH3NC + 3KCl + 3H2O (atom counts)",
   "code":"C=1+1; H=(3+1)+3*1; Cl=3; K=3; N=1; O=3; rC=1+1; rH=3+3*2; rCl=3; rK=3; rN=1; rO=3; result=(C==rC and H==rH and Cl==rCl and K==rK and N==rN and O==rO)"},
  {"claim":"Diazotisation: C6H5NH2 + NaNO2 + 2HCl -> C6H5N2Cl + NaCl + 2H2O balances for H, Cl, O, Na, N",
   "code":"H=7+2*1; Cl=2; O=2; Na=1; N=1+1; rH=5+2*2; rCl=1+1; rO=2; rNa=1; rN=2; result=(H==rH and Cl==rCl and O==rO and Na==rNa and N==rN)"},
  {"claim":"Nitrobenzene reduction: C6H5NO2 + 6[H] -> C6H5NH2 + 2H2O balances H, N, O",
   "code":"H=5+6; N=1; O=2; rH=5+2+2*2; rN=1; rO=2; result=(H==rH and N==rN and O==rO)"},
  {"claim":"Aqueous basicity order rank: methylamine(1) > ammonia(2) > aniline(3)",
   "code":"order=['CH3NH2','NH3','C6H5NH2']; result=(order[0]=='CH3NH2' and order[-1]=='C6H5NH2')"}
]