Board Previous Year Questions — Class 12 Chemistry Notes

How to Score Full Marks in the Board Exam

A complete bank of board-style questions with model answers for Amines. In the exam: write the IUPAC name carefully, draw structures with reagents over the arrow, name the mechanism/reaction, and give the NCERT-canonical reason (lone-pair delocalisation for aniline's weak basicity; solvation + steric effects for the aqueous basicity order; carbon-count changes for Gabriel/Hofmann). For distinguishing tests, state the reagent and the observation for each compound, and remember to protect -NH $_2$ by acetylation before nitrating aniline.

1-Mark Questions (Definitions & Direct)

Q1. Write the IUPAC name of (CH $_3$ ) $_2$ NH. Answer: N-methylmethanamine (common name dimethylamine).

Q2. What is the Hinsberg reagent? Answer: Benzenesulphonyl chloride (C $_6$ H $_5$ SO $_2$ Cl), used to distinguish 1°, 2° and 3° amines.

Q3. Name the reaction: C $_6$ H $_5$ N $_2$ +Cl- + CuCN → C $_6$ H $_5$ CN + N $_2$ . Answer: Sandmeyer reaction.

Q4. Why is aniline diazotised at 273-278 K? Answer: Because the diazonium salt is unstable at higher temperatures and decomposes (to phenol + N $_2$ ), so it is kept cold.

Q5. Which amine gives the carbylamine reaction? Answer: A primary amine (aliphatic or aromatic).

Q6. Name the product of Hofmann bromamide degradation of acetamide. Answer: Methanamine (methylamine), with one fewer carbon.

2-Mark Questions

Q7. Why are primary amines higher boiling than tertiary amines of comparable molecular mass? Answer: Primary amines have two N-H bonds and form intermolecular N-H…N hydrogen bonds, which require extra energy to break. Tertiary amines have no N-H and cannot form such hydrogen bonds, so they boil lower.

Q8. Why is aniline a weaker base than ethylamine? Answer: In aniline the nitrogen lone pair is delocalised into the benzene ring by resonance, so it is much less available to accept a proton. In ethylamine the lone pair is fully available (and the alkyl group donates electrons), so it is a much stronger base.

Q9. Arrange in increasing basic strength in aqueous solution: NH $_3$ , CH $_3$ NH $_2$ , (CH $_3$ ) $_2$ NH, (CH $_3$ ) $_3$ N. Answer: NH $_3$ < (CH $_3$ ) $_3$ N < CH $_3$ NH $_2$ < (CH $_3$ ) $_2$ NH. In water, the balance of electron donation, cation solvation and steric hindrance makes the secondary amine the strongest and drops the tertiary amine below the primary.

Q10. How will you distinguish a primary amine from a tertiary amine? Answer: By the carbylamine test: a primary amine heated with chloroform and alcoholic KOH gives a foul-smelling isocyanide; a tertiary amine does not react.

2-Mark Questions (continued)

Q11. Why does aniline not undergo the Friedel-Crafts reaction? Answer: The Lewis acid AlCl $_3$ reacts with the lone pair on the nitrogen of aniline to form a salt; the nitrogen acquires a positive charge, which makes the group strongly deactivating (meta-directing), so the Friedel-Crafts reaction does not take place as intended.

Q12. Give the products of the reaction of (i) a primary aliphatic amine and (ii) a primary aromatic amine with nitrous acid. Answer: (i) A primary aliphatic amine gives an alcohol and nitrogen gas (the aliphatic diazonium salt is unstable and decomposes). (ii) A primary aromatic amine gives a stable diazonium salt (at 273-278 K).

Q13. Why must aniline be acetylated before nitration? Answer: The strongly acidic nitrating mixture would protonate the -NH $_2$ group to -NH $_3$ +, which is meta-directing. Acetylation converts -NH $_2$ to -NHCOCH $_3$ , which is less basic (not protonated) and still ortho/para-directing, giving mainly the para nitro product.

Q14. Identify A and B: aniline --(NaNO $_2$ /HCl, 278 K)--> A --(CuCl)--> B. Answer: A is benzene diazonium chloride (C $_6$ H $_5$ N $_2$ +Cl-); B is chlorobenzene (C $_6$ H $_5$ Cl), formed by the Sandmeyer reaction.

3-Mark Questions (Conversions)

Q15. How will you convert (i) aniline to chlorobenzene, (ii) aniline to iodobenzene? Answer: (i) Diazotise aniline (NaNO $_2$ /HCl, 273-278 K) to benzene diazonium chloride, then treat with cuprous chloride (CuCl) (Sandmeyer reaction) to give chlorobenzene. (ii) Diazotise as above, then warm the diazonium salt with potassium iodide (KI) to give iodobenzene (no catalyst needed).

Q16. How will you convert (i) nitrobenzene to aniline, (ii) benzene to aniline? Answer: (i) Reduce nitrobenzene with H $_2$ /Ni (or Sn/HCl, then make alkaline) to give aniline. (ii) Nitrate benzene to nitrobenzene, then reduce it to aniline.

Q17. How will you convert (i) ethanamine to methanamine, (ii) methanamine to ethanamine? Answer: (i) Convert ethanamine to ethanamide (CH $_3$ CONH $_2$ ) and apply Hofmann bromamide degradation (Br $_2$ /NaOH) to give methanamine. (ii) To add a carbon, convert methanamine to ethanenitrile (CH $_3$ CN) via a suitable route such as CH $_3$ X + KCN, then reduce the nitrile to ethanamine; or use the cyanide/reduction route.

3-Mark Questions (Reasoning & Named Reactions)

Q18. Describe the Gabriel phthalimide synthesis and state its limitation. Answer: Phthalimide is treated with KOH to form potassium phthalimide, which is alkylated with an alkyl halide to give an N-alkylphthalimide. Hydrolysis (or reaction with hydrazine) releases the pure primary amine. Limitation: it cannot be used to prepare aromatic primary amines, because aryl halides do not undergo the required nucleophilic substitution.

Q19. Describe the Hofmann bromamide degradation reaction with an example. Answer: An amide is treated with bromine and aqueous sodium hydroxide. The carbonyl carbon is lost as carbonate, and a primary amine with one fewer carbon atom is formed. For example, ethanamide (CH $_3$ CONH $_2$ ) gives methanamine (CH $_3$ NH $_2$ ).

Q20. Explain the Hinsberg test for 1°, 2° and 3° amines. Answer: The amine is shaken with benzenesulphonyl chloride and then treated with KOH. A primary amine forms a sulphonamide with an acidic N-H that dissolves in KOH (clear solution). A secondary amine forms a sulphonamide with no N-H, which is insoluble (a precipitate). A tertiary amine has no N-H and does not react.

3-Mark Questions (Tests & Diazonium)

Q21. Give simple chemical tests to distinguish: (i) ethylamine and aniline; (ii) aniline and N,N-dimethylaniline. Answer: (i) Treat with cold nitrous acid: ethylamine releases N $_2$ gas (gives ethanol), while aniline forms a diazonium salt that couples with 2-naphthol to give a coloured dye. (ii) Carbylamine test: aniline (1°) gives a foul-smelling isocyanide; N,N-dimethylaniline (3°) does not react.

Q22. Write the reactions of benzene diazonium chloride that replace -N $_2$ + by (i) -Cl, (ii) -OH, (iii) -H. Answer: (i) With cuprous chloride (CuCl) it gives chlorobenzene (Sandmeyer). (ii) On warming with water it gives phenol. (iii) With hypophosphorous acid (H $_3$ PO $_2$ ) it gives benzene.

Q23. What are coupling reactions? Give one example. Answer: Coupling reactions are those in which the -N=N- group of a diazonium salt is retained; the diazonium ion attacks a highly activated ring (a phenol or amine) to give a coloured azo compound. For example, benzene diazonium chloride couples with phenol to give p-hydroxyazobenzene (an orange dye).

5-Mark Questions (Long Answer)

Q24. (a) Account for the basic strength order of methylamines in (i) the gas phase and (ii) aqueous solution. (b) Why is aniline a weaker base than ammonia? Answer: (a)(i) In the gas phase only the +I effect of the methyl groups operates, so the order is (CH $_3$ ) $_3$ N > (CH $_3$ ) $_2$ NH > CH $_3$ NH $_2$ > NH $_3$ . (ii) In water, the stability of the protonated cation (its solvation by hydrogen bonding, which is best when more N-H bonds are present) and steric hindrance also matter; the result is the irregular order (CH $_3$ ) $_2$ NH > CH $_3$ NH $_2$ > (CH $_3$ ) $_3$ N > NH $_3$ . (b) In aniline the nitrogen lone pair is delocalised into the benzene ring, so it is less available to accept a proton, making aniline a weaker base than ammonia.

Q25. Describe how aniline is converted to (i) p-nitroaniline and (ii) 2,4,6-tribromoaniline. Why do the conditions differ? Answer: (i) For p-nitroaniline, aniline is first acetylated to acetanilide (to protect -NH $_2$ ), then nitrated to give mainly p-nitroacetanilide, which is hydrolysed back to p-nitroaniline. Protection is needed because in the acidic nitrating mixture free -NH $_2$ becomes meta-directing -NH $_3$ +. (ii) For 2,4,6-tribromoaniline, aniline is simply treated with bromine water; the ring is so strongly activated that all three ortho/para positions are brominated at once, so no protection is needed.

5-Mark Questions (continued)

Q26. (a) How is benzene diazonium chloride prepared? (b) Give its reactions with (i) CuCN, (ii) KI, (iii) phenol. Answer: (a) Aniline is treated with sodium nitrite and dilute hydrochloric acid at 273-278 K (diazotisation) to give benzene diazonium chloride. (b)(i) With cuprous cyanide (CuCN) it gives benzonitrile (Sandmeyer). (ii) On warming with potassium iodide (KI) it gives iodobenzene. (iii) With phenol in mildly alkaline solution it couples to give p-hydroxyazobenzene (an orange azo dye).

Q27. (a) Why is the boiling point of an amine lower than that of an alcohol of comparable molecular mass? (b) Arrange ethanol, ethanamine and ethane in increasing boiling point. Answer: (a) Both amines and alcohols form intermolecular hydrogen bonds, but nitrogen is less electronegative than oxygen, so the N-H…N hydrogen bonds in amines are weaker than the O-H…O bonds in alcohols; hence amines boil lower. (b) ethane < ethanamine < ethanol (dispersion forces only < N-H hydrogen bonding < stronger O-H hydrogen bonding).

Quick-Fire Board Favourites

Q28. Why does a primary aliphatic amine give a different product from a primary aromatic amine with nitrous acid? Answer: The aliphatic diazonium salt is very unstable and decomposes at once to an alcohol and N $_2$ ; the aromatic diazonium salt is stabilised by resonance with the ring and is stable in the cold.

Q29. Name the reaction used to step down an amide to an amine by one carbon. Answer: The Hofmann bromamide degradation reaction.

Q30. Which gives a better yield: the Sandmeyer or the Gattermann reaction? Answer: The Sandmeyer reaction (using a cuprous salt) generally gives a better yield than the Gattermann reaction (copper powder + HX).

Q31. Why are lower amines soluble in water? Answer: They form hydrogen bonds with water molecules; solubility decreases as the size of the hydrocarbon part increases.

Q32. What is the product of coupling benzene diazonium chloride with aniline? Answer: p-Aminoazobenzene (aniline yellow), a yellow azo dye.

More Board Favourites

Q33. Why is the Gabriel synthesis not suitable for aniline? Answer: It needs an aryl halide to alkylate potassium phthalimide, but aryl halides do not undergo this nucleophilic substitution, so aniline cannot be prepared by this method.

Q34. Identify A: CH $_3$ CH $_2$ CN --(H $_2$ /Ni)--> A. Answer: A is propan-1-amine (CH $_3$ CH $_2$ CH $_2$ NH $_2$ ); reduction of the nitrile adds a carbon and gives a primary amine.

Q35. Why is N,N-dimethylaniline a tertiary amine? Answer: Its nitrogen carries three carbon groups (one phenyl and two methyl groups), so it is a tertiary amine.

Q36. State two important uses of amines. Answer: Amines are used in the manufacture of dyes (azo dyes, via diazonium coupling) and of drugs (e.g. the sulpha drugs); diamines are also used to make polymers such as nylon.