Previous year questions: Amines

The questions below come from official CBSE board papers and official Sample Question Papers. The year label in each heading is printed on that source, and the marks are the marks printed for that question or part.

Q1 (CBSE Board 2024, 1 mark)

Assertion (A): Aniline is a stronger base than ammonia. Reason (R): The unshared electron pair on the nitrogen atom in aniline becomes less available for protonation due to resonance.

Answer: A is false but R is true. Aniline is a weaker base than ammonia, not stronger, because resonance spreads the nitrogen lone pair into the ring and makes it less available to a proton. The reason given is the correct explanation of why aniline is weak, which is exactly why the assertion fails. Assertion and reason marks come from judging A and R separately, so state that A is false and R is true before explaining the resonance point.

Q2 (CBSE Board 2024, 5 marks)

Answer any five of the following. (a) N,N-diethylbenzenesulphonamide is insoluble in alkali. Give reason. (b) Aniline does not undergo the Friedel Crafts reaction. Why? (c) Write a simple chemical test to distinguish methylamine from aniline. (d) Write the chemical reaction involved in the Gabriel phthalimide synthesis. (e) How will you convert aniline to p-bromoaniline?

Answer: (a) In N,N-diethylbenzenesulphonamide the nitrogen carries no hydrogen, because both hydrogens of the sulphonamide were replaced by ethyl groups, so there is no acidic hydrogen to lose and the compound cannot form a soluble salt with alkali. (b) The amino nitrogen donates its lone pair to the Lewis acid aluminium chloride and forms a complex. That converts the activating NH2 group into a deactivated complexed group, so the ring does not react. (c) Use the carbylamine test: methylamine with chloroform and alcoholic KOH gives the foul smelling methyl isocyanide, while aniline does not. The Hinsberg test works too. (d) Phthalimide with KOH gives potassium phthalimide; with an alkyl halide and then hydrolysis it gives a primary amine. (e) Acetylate aniline to acetanilide, brominate to p-bromoacetanilide, then hydrolyse to p-bromoaniline. Each of the five parts carries one mark, and every answer needs the observation or the reagent along with the reason, so write both.

Q3 (CBSE Board 2025, 1 mark)

The IUPAC name of CH3CH2N(CH3)CH2CH2CH3 is (A) N-methylpentan-2-amine (B) N-ethyl-N-methylpropan-1-amine (C) N,N-diethylpropan-1-amine (D) N,N-dimethylpropan-1-amine

Answer: (B) N-ethyl-N-methylpropan-1-amine. The longest chain attached to nitrogen gives the parent amine, which is propan-1-amine through the three carbon chain. The other two groups on nitrogen are ethyl and methyl, and they take the N locants in alphabetical order. The longest chain gives the parent amine and the N substituents are arranged alphabetically, so show both steps in the answer.

Q4 (CBSE Board 2025, 1 mark)

An amine X with molecular formula C2H7N reacts with C6H5SO2Cl to give a compound that dissolves in NaOH solution. Identify X.

Answer: X is ethanamine, CH3CH2NH2. The reagent is benzenesulphonyl chloride, the Hinsberg reagent. A primary amine gives a sulphonamide that still has a hydrogen on nitrogen, so the product is acidic enough to dissolve in sodium hydroxide. A secondary amine would give an insoluble product and a tertiary amine would not react. The test name and the solubility observation are the marks, so mention the Hinsberg reagent and the alkali soluble sulphonamide.

Q5 (CBSE Board 2023, 2 marks)

(a)(i) Draw the zwitter ion structure for sulphanilic acid. (ii) How can the activating effect of the NH2 group in aniline be controlled?

Answer: (i) Sulphanilic acid is p-aminobenzenesulphonic acid. The molecule exists as the inner salt H3N+-C6H4-SO3-, with a positive charge on the ammonium nitrogen and a negative charge on the sulphonate group, so it is neutral overall. (ii) The NH2 group is strongly activating and can be tamed by acetylation. Acetyl chloride converts aniline into acetanilide, whose amide nitrogen is much less activating, and after the reaction the acetyl group is removed by hydrolysis. Part (i) needs the charges marked on the zwitterion, and part (ii) needs acetylation with the reason it lowers the activation of the ring.

Q6 (CBSE Board 2023, 1 mark)

Convert bromoethane to propanamine.

Answer: Bromoethane is treated with potassium cyanide to give propanenitrile, CH3CH2CN, which has one carbon more. The nitrile is then reduced with lithium aluminium hydride, or with hydrogen over nickel, to give propan-1-amine, CH3CH2CH2NH2. The two steps lengthen the chain by one carbon and then reduce the nitrile to the amine. Two marks need the two steps, the nitrile formation with potassium cyanide and then the reduction with named reagents.

Q7 (CBSE SQP 2024-25, 1 mark)

Ammonolysis of ethyl chloride followed by reaction of the amine so formed with one mole of methyl chloride gives an amine that (a) reacts with Hinsberg reagent to form a product soluble in alkali (b) on reaction with nitrous acid produces nitrogen gas (c) reacts with benzenesulphonyl chloride to form a product insoluble in alkali (d) does not react with Hinsberg reagent

Answer: Option (c). Ethyl chloride with ammonia gives ethanamine, and that with methyl chloride gives N-methylethanamine, a secondary amine. A secondary amine with benzenesulphonyl chloride gives a sulphonamide with no hydrogen on nitrogen, so it does not dissolve in alkali. Identifying the amine as secondary is the mark, so explain why its sulphonamide has no hydrogen and therefore does not dissolve in alkali.

Q8 (CBSE SQP 2025-26, 2 marks)

Carry out the following conversions. I. Nitrobenzene to 4-bromobenzenamine. II. Chlorophenylmethane to 2-phenylethanamine.

Answer: I. Reduce nitrobenzene to aniline with tin and hydrochloric acid or hydrogen over nickel, then brominate with bromine water to get 2,4,6-tribromoaniline, and remove the extra bromines by reduction with zinc and hydrochloric acid. A cleaner route protects the amine by acetylation before brominating. II. Treat benzyl chloride with potassium cyanide to get phenylacetonitrile, C6H5CH2CN, then reduce it with lithium aluminium hydride to 2-phenylethanamine, C6H5CH2CH2NH2. Each conversion carries one mark, so write every step with its reagent and condition rather than the final product alone.

Q9 (CBSE SQP 2023-24, 1 mark)

In which of the following solvents is the salt C4H8NH3+X- soluble? (a) ether (b) acetone (c) water (d) bromine water

Answer: (c) water. The compound is an ammonium halide, an ionic salt, and ionic salts dissolve in a polar solvent that can solvate the ions. Water is the most polar of the choices and it hydrogen bonds to the ammonium ion; ether, acetone and bromine water cannot dissolve the salt well. The ionic nature of the salt is the reason, so link it to the polarity of water and to hydrogen bonding with the ammonium ion.

Q10 (CBSE SQP 2025-26, 1 mark)

Arun heated a mixture of ethylamine and chloroform with ethanolic KOH and obtained a foul smelling gas. Write the chemical equation involved.

Answer: This is the carbylamine reaction. Ethylamine with chloroform and alcoholic potassium hydroxide gives ethyl isocyanide, which is the foul smelling gas: C2H5NH2 + CHCl3 + 3KOH gives C2H5NC + 3KCl + 3H2O. Only a primary amine gives this test, so it is used to distinguish primary amines from secondary and tertiary amines. The equation must be balanced and the isocyanide named, and the foul smell observation is the identification point that carries the second half of the mark.

How marks are usually split

A one mark assertion and reason question is answered by two separate judgements, so write whether A is true and whether R is true before touching the explanation. Two mark questions in this chapter usually expect a test and an observation, or a conversion in two steps with reagents, so keep both halves visible.

A five mark question on amines is often a set of short parts, and each part carries one mark. Answer every part briefly rather than one part at length, because an empty part cannot be compensated by a long answer elsewhere. For the diazonium and Hinsberg questions, name the reaction, give the equation and add the observation in one line. Keep a small list of the reagents that appear again and again, lithium aluminium hydride, bromine with sodium hydroxide, benzenesulphonyl chloride and sodium nitrite with hydrochloric acid, and attach each one to its reaction name so that a one line question can be answered in one line.