Class 12 Chemistry Chapter 9 Amines – Extra Questions with Answers

These extra practice questions for Class 12 Chemistry Chapter 9 – Amines go beyond the NCERT textbook exercises to reinforce basicity trends, diazonium salt chemistry, and the tests used to identify and distinguish amines. Useful for board exam revision and quick concept checks.

Last Updated: September 23, 2026

Very Short Answer Type Questions (1 Mark)

Q1. Name the test used to distinguish a primary amine from secondary and tertiary amines.
Ans: The carbylamine test (CHCl₃+alc. KOH, heat) — only primary amines (aliphatic or aromatic) give the foul-smelling isocyanide; secondary and tertiary amines give no reaction.

Q2. At what temperature is a diazonium salt prepared, and why?
Ans: 273–278K (0–5°C). Diazonium salts are thermally unstable and decompose readily above this range, releasing N₂ gas; low temperature keeps decomposition slow enough for the salt to be isolated and used.

Q3. Which reagent converts a benzenediazonium salt directly into fluorobenzene?
Ans: HBF₄ (fluoroboric acid), followed by heating the dry diazonium fluoroborate salt — the Balz–Schiemann reaction.

Q4. Arrange in increasing order of basic strength (aqueous): (C₂H₅)₃N, C₂H₅NH₂, (C₂H₅)₂NH.
Ans: (C₂H₅)₃N<C₂H₅NH₂<(C₂H₅)₂NH. In water, the tertiary amine’s bulky, poorly-solvated ammonium ion makes it the weakest base of the three despite having the most alkyl groups.

Short Answer Type Questions (2–3 Marks)

Q5. Why does aniline not undergo Friedel-Crafts alkylation or acylation, even though –NH₂ is an activating group?
Ans: The Lewis acid catalyst (AlCl₃) used in Friedel-Crafts reactions is itself an electrophile that coordinates strongly with aniline’s basic nitrogen lone pair, forming a salt-like –NH₂→AlCl₃ complex. This converts the ring-activating –NH₂ group into a strongly deactivating, bulky substituent, so the reaction fails (or gives very poor yields) instead of proceeding as it would for a typical activated arene.

Q6. Explain why the coupling reaction of a diazonium salt with phenol is carried out in mildly alkaline medium, while with aniline it is carried out in mildly acidic medium.
Ans: With phenol, a mildly alkaline medium converts some phenol into the more strongly activating phenoxide ion (C₆H₅O⁻), which couples efficiently with the diazonium electrophile; too strongly alkaline conditions instead destroy the diazonium salt itself (converting it to a diazotate). With aniline, a mildly acidic medium keeps most of the aniline as the free (unprotonated, still nucleophilic-at-the-ring) amine rather than the fully protonated, deactivated anilinium ion, while still being acidic enough to prevent the diazonium salt from decomposing.

Q7. What happens when benzenediazonium chloride is warmed with water? Write the equation.
Ans: The diazonium group is hydrolysed and replaced by –OH, releasing nitrogen gas: C₆H₅N₂Cl+H₂O→(Δ)→C₆H₅OH+N₂+HCl. This is a standard route to phenol from aniline via diazotisation.

Higher Order Thinking Skills (HOTS)

Q8. A student says: “Since aniline’s nitrogen lone pair is delocalised into the ring, the ring should be deactivated towards electrophilic substitution, just as it is deactivated towards forming a stronger base.” Is this reasoning correct? Explain with reference to aniline’s actual behaviour in bromination.
Ans: The reasoning is incorrect. Delocalisation of the nitrogen lone pair into the ring is exactly what makes aniline strongly activating (electron-rich) towards electrophilic aromatic substitution — the same delocalisation that reduces the lone pair’s availability for protonation (lowering basicity) simultaneously increases electron density at the ortho/para ring positions (raising reactivity towards electrophiles). This is why aniline reacts instantly with bromine water at room temperature, with no catalyst, to give a white precipitate of 2,4,6-tribromoaniline — a textbook demonstration of the ring being strongly activated even as the nitrogen itself is a weaker base than an aliphatic amine.

Q9. Two isomeric amines, C₃H₉N, are treated separately with HNO₂ at 273K. One sample effervesces briskly with N₂ gas; the other shows no visible gas evolution but forms a yellow oily layer. Identify the class of each amine and explain the difference in behaviour.
Ans: The sample effervescing briskly with N₂ is a primary aliphatic amine (e.g. propan-1-amine or propan-2-amine) — its diazonium salt is too unstable to exist even briefly and decomposes instantly, releasing nitrogen gas and forming a carbocation that goes on to alcohol/alkene products. The sample forming a yellow oily layer with no gas is a secondary amine (e.g. N-methylethanamine) — it reacts with HNO₂ to form a stable N-nitrosamine (a yellow oil) rather than releasing nitrogen, since it has no N–H available for the diazotisation pathway that primary amines follow.

Written by Satish

NCERTBooks.org is an independent educational resource run by a small team focused on making official NCERT textbooks easy to find, read, and download for students, parents, and teachers across India. We are not affiliated with NCERT or the Ministry of Education -- we organise publicly available NCERT content by class and subject, verify links against official sources, and build tools (like our in-browser reader) that make studying more convenient. Every guide we publish is written and reviewed by our team based on the actual NCERT curriculum and syllabus.