Amines — NEET Notes

Structure & classification of amines

An amine is what you get when you replace one, two, or all three hydrogens of ammonia with an alkyl or aryl group. The nitrogen is sp³ hybridised, the molecule is pyramidal, and the fourth sp³ orbital carries an unshared lone pair. That lone pair is the entire personality of the functional group — it makes amines basic, nucleophilic, and reactive towards electrophiles. Because of lone-pair repulsion, the C–N–C angle is slightly compressed from the ideal tetrahedral 109.5° — trimethylamine has a C–N–C angle of about 108°.

The classification is mechanical: count how many carbons sit on nitrogen.

A fourth class — quaternary ammonium salts R₄N⁺X⁻ — has four R groups on a positively charged nitrogen with no lone pair. These are not basic, are permanently charged, and are widely used as surfactants and detergents.

IUPAC nomenclature

In the IUPAC system, primary amines are named as alkanamines: drop the final 'e' of the parent alkane and add 'amine'. CH₃NH₂ becomes methanamine, CH₃CH₂NH₂ becomes ethanamine. When the amino group is not on C-1 of a longer chain, the position is indicated by a locant — CH₃CH(NH₂)CH₃ is propan-2-amine. When two amino groups are present, the 'e' is retained and 'di' is added: H₂N–CH₂–CH₂–NH₂ is ethane-1,2-diamine.

For secondary and tertiary amines, the largest carbon chain bonded to nitrogen is the parent. Substituents on nitrogen are prefixed with the locant 'N'. So CH₃NHCH₂CH₃ becomes N-methylethanamine, and (CH₃CH₂)₃N becomes N,N-diethylethanamine. For arylamines, the IUPAC rule replaces the 'e' of the parent arene by 'amine' — C₆H₅NH₂ is benzenamine, but the common name aniline is also accepted by IUPAC.

CH₃CH₂NHCH₃  →  N-methylethanamine

Locant 'N' marks substituents on nitrogen in 2° and 3° amines

Preparation of amines — six routes

NEET asks two questions from this section almost every year, and they usually test which method gives which class of amine. Six routes appear in NCERT — three are general, three are specific to primary amines, and one strips a carbon off the chain. Memorise the table below cold.

1. Reduction of nitro compounds

Nitro compounds reduce smoothly to primary amines with the same carbon count. Reagents: H₂/Ni (or Pd, Pt), or a metal in acidic medium — Sn/HCl, Fe/HCl, Zn/HCl. Iron + HCl is preferred industrially because the FeCl₂ formed hydrolyses to regenerate HCl, so only a catalytic amount of HCl is needed. The classic example: nitrobenzene → aniline.

2. Ammonolysis of haloalkanes (Hofmann's exhaustive methylation)

The C–X bond of an alkyl halide is cleaved by ammonia in a sealed tube at 373 K. The first product is a primary amine, but the primary amine is also a nucleophile, so it attacks another molecule of alkyl halide to give a secondary amine. The secondary amine then keeps going — tertiary, then quaternary ammonium salt. Ammonolysis thus delivers a mixture of 1°, 2°, 3°, and 4° amines. Using a large excess of ammonia shifts the product distribution toward the primary amine. Order of reactivity: RI > RBr > RCl. This is exactly what NCERT calls Hofmann's exhaustive methylation when continued to the quaternary salt.

3. Reduction of nitriles

Nitriles (R–C≡N) reduce to primary amines with one extra carbon. Reagents: LiAlH₄ followed by aqueous workup, or H₂/Ni (catalytic hydrogenation). This is called ascent of the amine series because the amine contains one carbon more than the starting alkyl halide that was used to make the nitrile. NEET 2023 tested this: CH₃CN + LiAlH₄ → CH₃CH₂NH₂.

4. Reduction of amides

Amides (R–CONH₂) reduce with LiAlH₄ to give primary amines with the same carbon count. CH₃CONH₂ → CH₃CH₂NH₂. NEET 2023 again tested this: among the four reactions in Q.60, this was one of the three that gave a primary amine. The contrast with Hofmann bromamide is critical: amide + LiAlH₄ keeps carbons; amide + Br₂/KOH loses one.

5. Gabriel phthalimide synthesis

Phthalimide is treated with ethanolic KOH to give the potassium salt (the phthalimide anion). This anion is alkylated by an alkyl halide, and the resulting N-alkyl phthalimide is hydrolysed by alkali to release the primary amine. Gabriel synthesis gives only primary amines — never secondary, never tertiary, never aromatic.

6. Hofmann bromamide degradation

An amide (RCONH₂) reacts with Br₂ in aqueous or ethanolic NaOH (or KOH) to give a primary amine with one carbon less than the starting amide. Mechanistically, the alkyl/aryl group migrates from the carbonyl carbon to nitrogen as CO₂ is lost. Acetamide CH₃CONH₂ → methanamine CH₃NH₂. NEET 2017 Q.45 tested this exact conversion: "Which reaction is appropriate for converting acetamide to methanamine?" The answer is Hofmann bromamide.

Physical properties

Lower aliphatic amines are gases with a fishy odour (methylamine, ethylamine). Primary amines with three or more carbons are liquids; higher ones are solids. Aniline and other arylamines are usually colourless when pure but darken on storage because of atmospheric oxidation.

Lower amines are soluble in water because they form hydrogen bonds with water through the N–H bonds and the lone pair on nitrogen. Solubility falls as the hydrophobic alkyl chain grows. Higher amines are essentially water-insoluble. All amines dissolve in organic solvents like alcohol, ether, and benzene.

Among amines themselves, intermolecular hydrogen bonding sets the boiling-point trend. Primary amines have two N–H bonds and can form the most extensive H-bond network. Secondary amines have one N–H. Tertiary amines have no N–H bond and cannot self-associate. The boiling-point order of isomeric amines is therefore:

Primary > Secondary > Tertiary

Boiling-point order of isomeric amines — driven by N–H count

Compared with alcohols of similar molar mass, amines have lower boiling points. Oxygen (electronegativity 3.5) forms stronger hydrogen bonds than nitrogen (3.0), so alcohols associate more tightly than amines. NCERT Table 9.2 makes this concrete: n-C₄H₉NH₂ (b.p. 350.8 K, M = 73) vs n-C₄H₉OH (b.p. 390.3 K, M = 74).

Basicity of amines

Amines are Lewis bases because the lone pair on nitrogen accepts a proton. The basicity is measured by K_b (or its negative logarithm pK_b) — larger K_b (or smaller pK_b) means a stronger base. Ammonia has pK_b = 4.75. The picture splits into three comparisons that NEET tests every year.

(a) Aliphatic amines vs ammonia

Alkyl groups are electron-releasing (+I effect). They push electron density toward nitrogen, making the lone pair more available for protonation, and they also stabilise the resulting alkylammonium cation by dispersing the positive charge. Conclusion: aliphatic amines are stronger bases than ammonia. NCERT lists pK_b values in the range 3.00–4.22 — well below ammonia's 4.75.

(b) Gas phase vs aqueous phase — the inversion

In the gas phase only the inductive effect operates. More alkyl groups means more electron donation, so the order is clean and expected: 3° > 2° > 1° > NH₃.

In the aqueous phase three forces compete:

• Inductive (+I) effect: favours more alkyl groups (3° > 2° > 1°).
• Solvation by water: the protonated ammonium cation is stabilised by H-bonding to water. Cations with more N–H bonds form more H-bonds and are more stable. This favours fewer alkyl groups (1° > 2° > 3°).
• Steric hindrance: bulky alkyl groups crowd the lone pair, making it harder for a proton to reach nitrogen. This penalises 3° more than 2°.

The interplay shifts the order depending on alkyl group size. NCERT gives the experimental orders explicitly:

(c) Aromatic amines vs aliphatic amines

Aniline has pK_b = 9.38 — far weaker than methylamine (3.38). Why? In aniline, the lone pair on nitrogen is in conjugation with the benzene ring. Aniline is a resonance hybrid of five contributing structures — the lone pair is partly delocalised into the ortho and para positions of the ring. The anilinium cation (formed when N is protonated) has only two resonance structures (the two Kekulé forms). Aniline is more stable than its conjugate acid; protonation is unfavourable.

Chemical reactions of amines

The lone pair on nitrogen drives every reaction in this section. Amines act as nucleophiles in alkylation, acylation, the Hinsberg reaction, and the carbylamine reaction — all of which test the number of N–H bonds and tell us whether the amine is 1°, 2°, or 3°.

Alkylation

Amines act as nucleophiles toward alkyl halides. A primary amine adds an alkyl group to become secondary, then tertiary, then quaternary. With an excess of methyl iodide and a base like Na₂CO₃, aniline is exhaustively methylated all the way to the quaternary ammonium iodide.

Acylation

Primary and secondary amines react with acyl chlorides, acid anhydrides, or esters to form amides. An H on nitrogen is replaced by an acyl group (RCO–). The reaction is run in pyridine, which removes HCl and shifts the equilibrium. With benzoyl chloride C₆H₅COCl, the reaction is called benzoylation — methanamine + benzoyl chloride → N-methylbenzamide. Tertiary amines, having no N–H, cannot be acylated.

Carbylamine reaction (isocyanide test)

Primary amines (aliphatic or aromatic) heated with chloroform (CHCl₃) and ethanolic KOH form isocyanides (RNC), also called carbylamines, which have an extraordinarily foul odour. Secondary and tertiary amines do not give this reaction. The carbylamine test is therefore a specific test for primary amines. NEET 2020 Q.180 tested this directly: of four anilines (1°, 2°, 3°), only the primary one gave the carbylamine test.

Reaction with nitrous acid (HNO₂)

Nitrous acid is prepared in situ from NaNO₂ and a mineral acid (usually HCl). All three amine classes react differently — and this is one of the most heavily tested patterns in the chapter.

NEET 2022 Q.69 tested this exact two-statement pattern: aliphatic 1° amines give unstable diazonium salts (correct), and aromatic 1° amines give diazonium salts stable above 300 K (incorrect — they are stable only at 273–278 K). The trap is the temperature window.

Hinsberg test — distinguishing 1°, 2°, 3° amines

The most exam-friendly test of the whole chapter. Benzenesulphonyl chloride C₆H₅SO₂Cl — known as Hinsberg's reagent — is shaken with the amine in the presence of excess aqueous KOH. The three classes give three diagnostic outcomes.

NEET 2021 Q.79 asked which amine reacts with Hinsberg's reagent to give a solid that dissolves in alkali. The answer is a primary amine — only the N-monoalkyl sulphonamide has the acidic N–H that makes it KOH-soluble.

Electrophilic substitution of aniline

The amino group is one of the most powerful activators in aromatic chemistry. Five resonance structures of aniline push electron density onto the ortho and para positions of the ring, so –NH₂ is ortho/para directing and strongly activating.

That activating power is a problem for controlled monosubstitution. Bromination of aniline with bromine water at room temperature gives 2,4,6-tribromoaniline as a white precipitate — all three ortho/para positions substitute at once. To get a monosubstituted product, the chemist must protect the –NH₂ group by acetylation: react aniline with acetic anhydride to form acetanilide. The amide nitrogen donates its lone pair to the carbonyl oxygen rather than the ring, reducing the activating power. Now nitration or bromination proceeds cleanly at the para position. Acid hydrolysis at the end removes the acetyl group and restores the –NH₂.

Nitration of aniline — the meta paradox

Direct nitration of aniline with concentrated HNO₃ / H₂SO₄ gives a surprising amount of meta-nitroaniline, even though –NH₂ is supposed to be ortho/para directing. NEET 2018 Q.47 tested this. The reason: in strongly acidic medium, aniline is protonated to the anilinium ion –NH₃⁺. The protonated nitrogen has no lone pair to donate, carries a positive charge, and acts as a strongly deactivating, meta-directing group (–I effect). The protonated and unprotonated forms exist in equilibrium, so products from both directing patterns appear.

Sulphonation

Aniline + concentrated H₂SO₄ first forms anilinium hydrogensulphate; heating at 453–473 K rearranges it to p-aminobenzene sulphonic acid (sulphanilic acid) — the building block of sulpha drugs.

Friedel–Crafts

Aniline does not undergo Friedel–Crafts alkylation or acylation. The Lewis acid catalyst (AlCl₃) coordinates to the nitrogen lone pair, putting a positive charge on N and converting the amino group into a strong deactivator — exactly like the protonation problem.

Diazonium salts — preparation

A diazonium salt has the structure Ar–N₂⁺ X⁻. The diazonium group N₂⁺ is the bridge between aromatic amine chemistry and almost every functional group you can put on a benzene ring. Diazonium salts are made by diazotisation: aniline + NaNO₂ + HCl at 273–278 K (0–5 °C). Below this range the reaction is too slow; above it the salt decomposes.

Why is the aromatic diazonium ion stable at 0–5 °C while the aliphatic one is not? The aryl ring stabilises the cation by resonance — the positive charge of N₂⁺ is shared with the ring through extended conjugation. Aliphatic diazonium salts have no such delocalisation, so they fall apart immediately, losing N₂ and giving an alcohol.

Reactions of diazonium salts — Sandmeyer and friends

The diazonium group N₂⁺ is one of the best leaving groups in all of organic chemistry — when it goes, it takes nitrogen gas with it. That makes diazonium salts a passport into every aromatic functional group: F, Cl, Br, I, CN, OH, NO₂, and even H. Two reaction classes exist:

(A) Displacement of N₂ — diazonium group is lost as N₂ gas, replaced by another group.
(B) Retention of the diazo bond — N=N–Ar' is preserved, gives azo dyes.

NEET 2023 Q.65 tested a Sandmeyer–Grignard combination: benzenediazonium chloride + Cu₂Br₂/HBr (Sandmeyer Br) → bromobenzene → Mg/dry ether → phenylmagnesium bromide → H₂O → benzene. The candidate has to identify the intermediate as PhMgBr.

Coupling reactions — azo dyes

When the diazonium ion is not allowed to lose N₂, the diazo bond is preserved and the salt couples with an electron-rich aromatic ring — phenol or aniline. The result is an azo compound Ar–N=N–Ar', with two rings joined by the N=N double bond. The extended conjugation across the azo bridge absorbs visible light, so these compounds are intensely coloured — the basis of the azo dye industry.

• With phenol (in mild alkali): coupling at the para position → p-hydroxyazobenzene (orange).
• With aniline (in mild acid): coupling at the para position → p-aminoazobenzene (yellow).

This is an electrophilic substitution where the diazonium cation is the electrophile and the phenol/aniline ring (activated by –OH or –NH₂) is the nucleophile. The diazonium attacks at para because that is the most electron-rich position.