AminesNEET Chemistry · Class 12 · NCERT Chapter 9

The key factor is availability of the lone pair on nitrogen. In aliphatic amines (like methylamine, CH₃NH₂), the alkyl group donates electrons by the +I (inductive) effect, increasing electron density on nitrogen. This makes the lone pair more available to accept a proton, so aliphatic amines are stronger bases (pKb 3–4). In aniline (C₆H₅NH₂), the lone pair on nitrogen is delocalised into the benzene ring by conjugation (resonance). The lone pair is spread over the ring and is less available to accept a proton. This lowers the basicity significantly (pKb 9.4). Quantitatively: Kb of methylamine = 5 × 10⁻⁴, Kb of aniline = 4 × 10⁻¹⁰. The resonance withdrawal in aniline reduces basicity by six orders of magnitude.

Purely by inductive effect, you would expect 3° > 2° > 1° (more alkyl groups donating more electrons). But the actual order in aqueous solution is 2° > 1° > 3°. Two opposing factors are at work: (1) Inductive effect: more alkyl groups push more electrons onto N, increasing lone pair availability. This predicts 3° > 2° > 1°. (2) Solvation (hydration): the ammonium ion (conjugate acid) is stabilised by hydrogen bonding with water. Primary ammonium R-NH₃⁺ has 3 N-H bonds and is heavily solvated. Secondary R₂NH₂⁺ has 2 N-H bonds — moderate solvation. Tertiary R₃NH⁺ has only 1 N-H bond — least solvated. Less solvation means the conjugate acid is less stable, which reduces apparent basicity of the tertiary amine. The balance of these two opposing effects gives 2° (best balance) > 1° > 3°.

Hoffmann bromamide degradation converts a primary amide (RCONH₂) to a primary amine (RNH₂) with one fewer carbon. Reagents: Br₂ + NaOH (aqueous). Mechanism: (1) Br₂ and NaOH form NaOBr (or OBr⁻). (2) OBr⁻ bromaminates the amide nitrogen: RCONH₂ → RCONHBr. (3) A strongly basic N-bromo intermediate forms and the acyl-N bond rearranges in a Curtius-type 1,2-shift: R migrates from C to N with the bonding pair. An isocyanate (RNCO) is the intermediate. (4) RNCO is hydrolysed by NaOH to give RNH₂ + CO₂. Product: a primary amine (RNH₂) with one carbon fewer than the starting amide. This is important in NEET because it distinguishes primary amides and yields primary amines specifically. Example: CH₃CONH₂ + Br₂/NaOH → CH₃NH₂.

Reagent: benzenesulfonyl chloride (C₆H₅SO₂Cl) in aqueous KOH. Primary amine (RNH₂): reacts to give a sulfonamide (RNH-SO₂C₆H₅). This sulfonamide has one acidic N-H and dissolves in excess KOH (alkali soluble). On acidification, it precipitates. Secondary amine (R₂NH): reacts to give a disubstituted sulfonamide (R₂N-SO₂C₆H₅). This has NO acidic N-H and is insoluble in KOH — precipitate forms in alkaline solution and remains on acidification. Tertiary amine (R₃N): does NOT react with benzenesulfonyl chloride because the nitrogen has no N-H. No reaction. Summary: 1° gives alkali-soluble sulfonamide; 2° gives alkali-insoluble precipitate; 3° gives no reaction.

Carbylamine reaction (also called isocyanide test): a primary amine reacts with chloroform (CHCl₃) and alcoholic KOH to give a foul-smelling isocyanide (RNC, also called carbylamine). Reagents: CHCl₃ + alc. KOH + primary amine. Observation: characteristic intensely unpleasant smell. Primary aliphatic and aromatic amines both give a positive carbylamine test. Secondary and tertiary amines do NOT react. This makes it a specific test for primary amines (both aliphatic and aromatic). In NEET questions, the carbylamine test is used to confirm whether a compound is a primary amine. The isocyanide is toxic, which is why the characteristic smell is diagnostic.

Diazonium salts (ArN₂⁺ X⁻) are formed when aromatic primary amines react with NaNO₂ + HCl at 0–5°C (diazotisation). The benzene diazonium chloride (C₆H₅N₂⁺Cl⁻) is relatively stable at low temperature and undergoes many useful reactions: (1) Sandmeyer reactions: replacement of N₂⁺ by Cl, Br (with CuCl/CuBr + HCl/HBr), CN (with CuCN), or I (with KI). (2) Balz-Schiemann reaction: N₂⁺ replaced by F via ArN₂⁺BF₄⁻ → ArF + N₂ + BF₃. (3) Reduction: N₂⁺ → ArH (deamination) using H₃PO₂. (4) Coupling reactions: the diazonium ion acts as an electrophile and couples with activated aromatic rings (phenol in alkaline medium, aniline) at the para position to give azo compounds (ArN=NAr'). These orange/red azo dyes are used in NEET synthesis questions. Diazonium salts are only stable at 0–5°C; above this temperature they decompose.

Gabriel phthalimide synthesis makes pure primary aliphatic amines without contamination by secondary or tertiary amines. Steps: (1) Phthalimide (a cyclic imide) is treated with KOH to form potassium phthalimide (the N-K salt). (2) Potassium phthalimide reacts with a primary alkyl halide (RX, usually 1°) by SN2 displacement to give N-alkylphthalimide. (3) N-alkylphthalimide is hydrolysed with NaOH (aq.) OR hydrazinolysis (NH₂NH₂) to give the primary amine (RNH₂) + phthalic acid (or phthalhydrazide with hydrazine). Why it is useful: since phthalimide has only one N-H, only one alkyl group can be introduced, guaranteeing a primary amine product. Phenyl halides (ArX) do not work because SN2 on Ar is not possible. NEET often asks which synthesis gives a pure primary amine — answer: Gabriel phthalimide.

Amines contributes 3–5 NEET questions per year from Class 12 Chemistry. The highest-frequency topics are: (1) Basicity order — aliphatic vs aromatic, 2° > 1° > 3° in water. Know the reason (inductive vs solvation). (2) Hoffmann bromamide degradation — identify that the product has one fewer carbon and is a primary amine. (3) Distinction tests — Hinsberg test (1°: alkali-soluble; 2°: alkali-insoluble; 3°: no reaction), carbylamine test (only 1°). (4) Diazonium chemistry — Sandmeyer reaction sequence (Cl, Br, CN, I replacement), Balz-Schiemann (F), coupling reactions (azo dyes). (5) Gabriel synthesis — makes pure 1° aliphatic amine. Learn the reagent–product pairs precisely, as NEET frequently tests "which reagent gives which product" in the form of match-the-following or single correct MCQs.