An amine's basicity comes from the lone pair of electrons on nitrogen, the same lone pair responsible for its nucleophilic behaviour: a more basic amine holds this lone pair more tightly for protonation, forming a substituted ammonium ion, and basic strength is compared quantitatively through the base dissociation constant (or ) in water, with ammonia, , as
Amines
Class 1211 previous-year questions from this chapter every option and the correct answer, free · where the marks are
1. Basicity of amines
Exam focus: In water, aliphatic amine basicity doesn't rise monotonically with substitution — the usual order is 2° > 1° > 3° > NH₃, because a 3° amine's extra alkyl groups add electron density but also add steric hindrance that weakens solvation of the resulting ammonium ion, and that loss outweighs the electronic gain. Aniline is far less basic than ammonia for a different reason: its nitrogen lone pair delocalises into the ring by resonance, and a ring nitro group weakens it further still.
Amines — Full Chapter (incl. Basicity)
Competition Wallah
Full-chapter lecture with a dedicated segment on basic strength of aliphatic/aromatic amines, bundled with preparation and other reactions.
ChemistryNEET 2025show ▾The correct order of decreasing basic strength of the given amines is:
- A.N-methylaniline > benzenamine > ethanamine > N-ethylethanamine
- B.N-ethylethanamine > ethanamine > benzenamine > N-methylaniline
- C.N-ethylethanamine > ethanamine > N-methylaniline > benzenamine✓
- D.benzenamine > ethanamine > N-methylaniline > N-ethylethanamine
Solution
Aliphatic amines are stronger bases than aromatic ones because the aryl lone pair is delocalised into the ring. Among aliphatic amines, the secondary N-ethylethanamine beats primary ethanamine (inductive effect plus solvation balance); among aromatic ones, the N-methyl group makes N-methylaniline slightly stronger than aniline. Order: N-ethylethanamine > ethanamine > N-methylaniline > benzenamine.
Open this question on its own page →
Save for spaced revision (free account)ChemistryNEET 2020show ▾Which of the following amines will give the carbylamine test?
- A.Aniline ()✓
- B.N-methylaniline ()
- C.N,N-dimethylaniline ()
- D.N-ethylaniline ()
The worked solution is part of a paid pack.
ChemistryNEET 2019show ▾The correct order of basic strength of methyl-substituted amines in aqueous solution is:
2. Preparation and reactions; diazonium salts
Exam focus: Diazonium salts are stable in solution only at 0–5°C — warm it and it hydrolyses to phenol with loss of N₂ gas, which is why every diazonium reaction is carried out cold and used immediately rather than stored. Azo-coupling with phenol needs mildly alkaline conditions and with aniline needs mildly acidic conditions, and it always occurs para to the activating group.
Amines and Diazonium Salts — Full Chapter
Lakshya NEET
Full sweep in one sitting: amine preparation methods, basicity-driven reactions, and diazonium-salt chemistry (Sandmeyer, coupling) together.
ChemistryNEET 2025show ▾Given below are two statements:
Statement I: Benzenediazonium salt is prepared by the reaction of aniline with nitrous acid at 273–278 K. It decomposes easily in the dry state.
Statement II: Insertion of iodine into the benzene ring is difficult and hence iodobenzene is prepared through the reaction of benzenediazonium salt with KI.
In the light of the above statements, choose the most appropriate answer from the options given below:
- A.Both Statement I and Statement II are correct✓
- B.Both Statement I and Statement II are incorrect
- C.Statement I is correct but Statement II is incorrect
- D.Statement I is incorrect but Statement II is correct
Solution
Diazotisation of aniline with at 273–278 K gives benzenediazonium chloride, which is unstable when dry (I correct). Direct iodination of benzene is reversible and poor, so iodobenzene is made from the diazonium salt and KI (II correct).
Open this question on its own page →
Save for spaced revision (free account)ChemistryNEET 2024show ▾Given below are two statements:
Statement I: Aniline does not undergo Friedel–Crafts alkylation reaction.
Statement II: Aniline cannot be prepared through Gabriel synthesis.
In the light of the above statements, choose the correct answer from the options given below:
- A.Both Statement I and Statement II are false
- B.Statement I is correct but Statement II is false
- C.Statement I is incorrect but Statement II is true
- D.Both Statement I and Statement II are true✓
Solution
Aniline forms a complex with the Lewis-acid catalyst through its nitrogen lone pair, so Friedel–Crafts fails. Gabriel synthesis needs an alkyl halide to undergo with phthalimide; aryl halides do not, so aniline cannot be made this way. Both true.
Open this question on its own page →
Save for spaced revision (free account)ChemistryNEET 2024show ▾Identify the major product C formed in the following reaction sequence:
- A.butylamine
- B.butanamide
- C.bromobutanoic acid
- D.propylamine✓
Solution
with NaCN gives butanenitrile (A). Partial hydrolysis converts the nitrile to the amide, butanamide (B). Hofmann bromamide degradation with removes one carbon, giving propylamine (C).
Open this question on its own page →
Save for spaced revision (free account)ChemistryNEET 2023show ▾Which of the following reactions will not give a primary amine as the product?
- A. Product
- B. Product
- C. Product✓
- D. Product
The worked solution is part of a paid pack.
ChemistryNEET 2022show ▾The product formed from the following reaction sequence is:
- A.Structure (1)✓
- B.Structure (2)
- C.Structure (3)
- D.Structure (4)
The worked solution is part of a paid pack.
ChemistryNEET 2022show ▾Given below are two statements:
Statement I: Primary aliphatic amines react with to give unstable diazonium salts.
Statement II: Primary aromatic amines react with to form diazonium salts which are stable even above 300 K.
In the light of the above statements, choose the most appropriate answer from the options given below:
- A.Statement I is incorrect but Statement II is correct
- B.Both Statement I and Statement II are correct
- C.Both Statement I and Statement II are incorrect
- D.Statement I is correct but Statement II is incorrect✓
The worked solution is part of a paid pack.
