Amines: common doubts, answered
The questions students ask most often about Amines, each with a short answer. For the full chapter, read the Amines notes.
About the chapter
How is the Amines chapter usually tested in NEET?
Expect basicity order questions in water and the gas phase, and comparisons of aniline with aliphatic amines. Distinguishing tests, carbylamine, Hinsberg and nitrous acid, appear often, as do Gabriel and Hoffmann reactions with their carbon counts. Conversion chains through diazonium salts are common, so learn which reagent introduces each group.
Structure, classes and names
Read this section in the notes →How is an amine classed as primary, secondary or tertiary?
By the number of carbon groups bonded to the nitrogen, not by the carbon the nitrogen sits on. One group makes a primary amine, RNH₂; two make a secondary amine, R₂NH; three make a tertiary amine, R₃N. So tert-butylamine, (CH₃)₃CNH₂, is a primary amine even though its nitrogen is on a tertiary carbon, unlike the rule used for alcohols.
What is the shape of an amine molecule?
The nitrogen in an amine is sp³ hybridised, with three bonds and one lone pair, so the molecule is pyramidal, like ammonia. The lone pair repels the bonding pairs more strongly, so the C–N–C angle in trimethylamine is about 108°, slightly below the tetrahedral value. That lone pair is responsible for the basic and nucleophilic behaviour of amines.
Preparing amines by reduction and ammonolysis
Read this section in the notes →Why is iron with HCl preferred over tin for reducing nitrobenzene to aniline?
With iron, the FeCl₂ formed is hydrolysed and regenerates HCl, so only a small amount of acid is needed to start the reaction. That makes it cheaper and more convenient for industry. Tin with HCl also reduces nitro compounds to amines, as does catalytic hydrogenation with nickel, palladium or platinum.
Why does ammonolysis of an alkyl halide give a mixture of amines?
The primary amine formed first is itself a nucleophile, often a stronger one than ammonia, so it attacks more alkyl halide and gives a secondary amine. That reacts again to give a tertiary amine and finally a quaternary ammonium salt. Using a large excess of ammonia favours the primary amine, but separation is still needed.
Gabriel and Hoffmann methods
Read this section in the notes →Why can't the Gabriel phthalimide synthesis make aniline?
The key step is an SN2 attack by the phthalimide anion on an alkyl halide. Aryl halides do not undergo ordinary nucleophilic substitution, because their C–X bond is strengthened by resonance and the ring repels nucleophiles. So the method gives only primary aliphatic amines, which come out pure with no secondary or tertiary amine mixed in.
Why does Hoffmann bromamide degradation give an amine with one carbon fewer?
An amide treated with bromine and sodium hydroxide rearranges so that the alkyl or aryl group moves from the carbonyl carbon onto the nitrogen. The carbonyl carbon is then lost as carbonate. So ethanamide, with two carbons, gives methanamine, with one, and benzamide gives aniline. This makes it a useful way to shorten a carbon chain.
Physical properties of amines
Read this section in the notes →Why do primary amines boil higher than tertiary amines of the same mass?
Primary amines have two hydrogens on nitrogen and secondary amines one, so their molecules form hydrogen bonds with each other. Tertiary amines have no N–H hydrogen and cannot. So boiling points run primary > secondary > tertiary for isomers. Amines boil lower than alcohols of similar mass, because nitrogen is less electronegative than oxygen and forms weaker hydrogen bonds.
Basic character of amines
Read this section in the notes →Why are amines basic?
The nitrogen in an amine carries an unshared electron pair, which it can donate to a proton. So amines accept H⁺ from acids to form ammonium salts, and they turn red litmus blue in water. Basic strength is measured by Kb or pKb, and a smaller pKb means a stronger base. Ammonium salts release the free amine when treated with NaOH.
Why isn't trimethylamine the strongest base among methylamines in water?
In the gas phase, more alkyl groups mean more electron donation, so the order is 3° > 2° > 1°. In water, the protonated amine is also stabilised by hydrogen bonding with water, which needs N–H hydrogens, and bulky groups hinder it. Combining these effects, the aqueous order is (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃.
Why is the order of basicity of ethylamines different from that of methylamines in water?
Larger ethyl groups increase steric hindrance around nitrogen and change the balance between electron donation, hydration and crowding. For the ethylamines the aqueous order becomes (C₂H₅)₂NH > (C₂H₅)₃N > C₂H₅NH₂ > NH₃, so the tertiary amine moves ahead of the primary one. Remember both orders separately, as they are often compared in exam questions.
Why is aniline a much weaker base than methylamine and ammonia?
In aniline the nitrogen's lone pair is delocalised into the benzene ring by resonance, so it is less available to accept a proton. The anilinium ion has no such resonance stabilisation, so protonation costs aniline its extra stability. That makes aniline's pKb 9.38, weaker than ammonia. Benzylamine, whose nitrogen is not on the ring, has pKb 4.70 and is a stronger base.
Acylation, alkylation and the carbylamine test
Read this section in the notes →Which amines give the carbylamine test?
Only primary amines, aliphatic or aromatic. Warming a primary amine with chloroform and alcoholic KOH produces an isocyanide, which has a very unpleasant smell. Secondary and tertiary amines do not react, so the test identifies a primary amine. It is often used to tell apart isomers such as ethanamine and N-methylmethanamine.
Why is acetylation done before nitrating or brominating aniline?
The –NH₂ group activates the ring so strongly that substitution is hard to control, giving polysubstitution with bromine and oxidation with nitric acid. Converting it to acetanilide, –NHCOCH₃, reduces the activation because the nitrogen's lone pair is now shared with the carbonyl. Substitution then stops at one group, mainly para, and the acetyl group is removed by hydrolysis afterwards.
Nitrous acid and Hinsberg's reagent
Read this section in the notes →How does nitrous acid distinguish primary aliphatic and aromatic amines?
Nitrous acid, made from NaNO₂ and HCl, converts primary aliphatic amines into unstable diazonium salts that break down at once, releasing nitrogen gas and giving alcohols; measuring the nitrogen is used to estimate amino acids and proteins. Primary aromatic amines at 273–278 K give stable arenediazonium salts, such as benzenediazonium chloride from aniline, which are used for further synthesis.
How does Hinsberg's reagent distinguish primary, secondary and tertiary amines?
Hinsberg's reagent is benzenesulphonyl chloride. A primary amine forms a sulphonamide that still has an acidic N–H hydrogen, so it dissolves in alkali. A secondary amine forms a sulphonamide with no hydrogen on nitrogen, so it is insoluble in alkali. A tertiary amine does not react at all. Today p-toluenesulphonyl chloride is often used in place of benzenesulphonyl chloride.
Electrophilic substitution in aniline
Read this section in the notes →Why does direct nitration of aniline give a large amount of the meta product?
Nitration needs strongly acidic conditions, which protonate much of the aniline to the anilinium ion, –NH₃⁺. That ion withdraws electrons and directs substitution to the meta position. So the product is about 51% para, 47% meta and only 2% ortho, plus oxidation products. Protecting the amine as acetanilide first gives mainly the para product instead.
Why doesn't aniline undergo Friedel-Crafts reactions?
Aniline is a Lewis base and reacts with the catalyst, aluminium chloride, which is a Lewis acid, forming a salt. The nitrogen then carries a positive charge, which strongly deactivates the ring, so the Friedel-Crafts reaction cannot take place. This is a standard reason-type question in the chapter.
Diazonium salts: making and handling
Read this section in the notes →Why are arenediazonium salts more stable than alkyldiazonium salts?
In benzenediazonium ions the positive charge of the –N₂⁺ group is spread into the benzene ring by resonance, which stabilises them. Alkyldiazonium ions have no such delocalisation, so they lose nitrogen immediately. Even arenediazonium salts are stable only in cold solution, around 273–278 K, and are used straight away instead of being stored.
Replacing the diazonium group
Read this section in the notes →What is the difference between the Sandmeyer and Gatterman reactions?
Both swap the diazonium group for a halogen, but with different copper reagents. The Sandmeyer reaction uses copper(I) salts, CuCl with HCl or CuBr with HBr, and CuCN with KCN also brings in a cyano group. The Gatterman reaction uses copper powder together with HCl or HBr instead of a copper salt. The Sandmeyer method usually gives better yields.
How can the diazonium group be replaced by iodine, fluorine, hydrogen or –OH?
Iodobenzene forms when the diazonium salt solution is simply warmed with KI, with no copper needed. Fluorobenzene comes from heating the fluoroborate salt made with HBF₄. Hypophosphorous acid or ethanol replaces the group with hydrogen, removing it completely. Warming the solution with water gives phenol. These replacements make diazonium salts valuable for placing groups on a ring.
Coupling and uses in synthesis
Read this section in the notes →What happens in diazonium coupling reactions?
The diazonium ion acts as a weak electrophile and attacks the para position of a strongly activated ring, such as phenol or aniline, keeping both nitrogen atoms. The result is an azo compound, with an extended conjugated system that absorbs visible light. With phenol it gives orange p-hydroxyazobenzene, and with aniline yellow p-aminoazobenzene, which is why these reactions are used to make dyes.
Lumi is not affiliated with or endorsed by NCERT. The official NCERT textbooks are free to read and download from NCERT's own website, ncert.nic.in. These notes and simulations are original work by Lumi (Aikolumi Software Pvt Ltd), © 2026, shared under CC BY-NC 4.0: copy, print, share and adapt them for any non-commercial use, with credit to Lumi and a link to lumineet.com.