Quick Recap — Amines & Diazonium Salts
- Amines (N derivatives of ): 1 (), 2 (), 3 (); nitrogen is with a lone pair, so amines are basic.
- Basicity (aqueous): 2 1 3 ; aromatic amines (aniline) are weaker than (lone pair delocalised into the ring).
- Preparation: reduction of nitro/nitrile/amide, Hoffmann bromamide, Gabriel synthesis.
- Diazonium salts (, from aniline at 0-5C): Sandmeyer haloarene; coupling azo dyes.
Beyond-NCERT JEE Essentials
1. Basicity: the gas-phase vs aqueous clash. Amine basicity is a tug-of-war between three effects: the +I (electron-releasing) push of alkyl groups that enriches the nitrogen lone pair, the solvation (hydrogen bonding) of the PROTONATED cation by water, and steric crowding around nitrogen. In the GAS PHASE (no solvent) only induction survives, so basicity rises cleanly with the number of alkyl groups: (CH3)3N > (CH3)2NH > CH3NH2 > NH3, that is tertiary > secondary > primary > ammonia. In WATER the solvation term intervenes: the more N-H bonds the ammonium cation keeps, the more hydrogen bonds stabilise it, so solvation favours primary > secondary > tertiary, exactly opposing induction, while steric bulk further penalises the crowded tertiary amine. The net compromise makes the SECONDARY amine the strongest base in water and drags the tertiary down. The exact aqueous order is therefore alkyl-dependent: methyl series (CH3)2NH > CH3NH2 > (CH3)3N > NH3 (2 > 1 > 3), but ethyl series (C2H5)2NH > (C2H5)3N > C2H5NH2 > NH3 (2 > 3 > 1). [JEE Tip] Never memorise one universal 'aqueous order'. The only guaranteed facts are (i) the secondary amine is always strongest in water and (ii) the gas-phase order is the clean tertiary > secondary > primary. The 1-vs-3 swap between methyl and ethyl is a favourite trap: bulkier ethyl groups shift the solvation-plus-steric balance, lifting triethylamine ABOVE ethylamine even though trimethylamine sits BELOW methylamine.
2. Aromatic amines and ring substituents. Aniline (C6H5NH2) is far weaker than any aliphatic amine and even weaker than NH3, because the nitrogen lone pair is delocalised into the ring; donating it to a proton also cancels that resonance stabilisation. Overall in water: aliphatic amines > NH3 > aniline. Ring substituents shift aniline predictably: electron-donating groups (-CH3, -OCH3, -NH2) RAISE basicity, while electron-withdrawing groups (-NO2, -CN, -X, -COOH) LOWER it, so base strength runs p-toluidine > aniline > p-chloroaniline > p-nitroaniline. [JEE Tip] A substituent acts through RESONANCE only from the ortho or para position and through weaker induction from meta, so p-nitroaniline is a weaker base than m-nitroaniline. Beware the 'ortho effect' as well: almost any ortho substituent, even a donor, makes aniline a weaker base than its para isomer through steric and hydrogen-bonding disruption of the -NH2 group.
3. Preparation: mind the carbon count. Gabriel phthalimide synthesis gives PURE primary amines only, but FAILS for aromatic amines because aryl halides will not undergo the required SN2 on the phthalimide anion. Hofmann bromamide degradation (R-CONH2 + Br2 + NaOH) gives a primary amine with ONE FEWER carbon, the carbonyl carbon leaving as carbonate: CH3CONH2 (2 C) -> CH3NH2 (1 C); C6H5CONH2 -> C6H5NH2. Nitrile reduction (H2/Ni, LiAlH4, or Na/ethanol) gives a primary amine with ONE MORE carbon: R-CN -> R-CH2-NH2. Amide reduction with LiAlH4 keeps the SAME carbon count: R-CONH2 -> R-CH2-NH2, the carbonyl carbon surviving as CH2. Nitro reduction (H2/Ni, Sn/HCl, Fe/HCl) keeps the skeleton unchanged: R-NO2 -> R-NH2. Reductive amination (an aldehyde or ketone plus ammonia or an amine, then H2/Ni or NaBH3CN) builds primary, secondary or tertiary amines depending on the amine chosen. [JEE Tip] The classic whiplash is amide -> amine, so always ask WHICH reagent: LiAlH4 keeps the carbon (R-CONH2 -> R-CH2-NH2) while Br2/NaOH (Hofmann) DROPS one (R-CONH2 -> R-NH2). Pair this with 'nitrile ADDS one carbon' and you can walk up or down a carbon series at will.
4. Distinguishing primary, secondary and tertiary. Hinsberg test (C6H5SO2Cl, then KOH): a primary amine gives an N-H sulphonamide whose remaining N-H is acidified by the -SO2- group and DISSOLVES in KOH; a secondary amine gives a sulphonamide with no N-H, INSOLUBLE in KOH; a tertiary amine does not react and dissolves only on adding acid. Carbylamine test (CHCl3 + alcoholic KOH): only PRIMARY amines, aliphatic or aromatic, give the foul-smelling isocyanide R-NC, while secondary and tertiary give nothing. Nitrous acid (NaNO2/HCl, cold): a primary ALIPHATIC amine forms an unstable diazonium that at once loses N2, so brisk gas effervescence appears and the alcohol is the main product; a primary AROMATIC amine forms a stable diazonium salt at 273-278 K; secondary amines of either type give a yellow oily N-nitrosamine; a tertiary aromatic amine such as N,N-dimethylaniline undergoes ring nitrosation to a green p-nitroso solid. [JEE Tip] The nitrous-acid test alone can label every class: N2 bubbles in the cold = primary aliphatic; a cold-stable solution that couples with 2-naphthol to a dye = primary aromatic; a yellow oil = secondary; a green solid = tertiary aromatic.
5. Diazonium salts: the aryl reaction hub. Diazotisation: C6H5NH2 + NaNO2 + 2 HCl -> C6H5N2+Cl- + NaCl + 2 H2O at 273-278 K (0-5 C). This salt is the single most useful branch point in aromatic synthesis, splitting into two families. REPLACEMENT of -N2+ (nitrogen leaves as gas): Sandmeyer, with CuCl/HCl -> ArCl, CuBr/HBr -> ArBr, CuCN/KCN -> ArCN; Gattermann, with Cu powder + HX -> ArCl or ArBr (cheaper, lower yield); Balz-Schiemann, add HBF4 to precipitate ArN2+BF4- then heat -> ArF + N2 + BF3 (the standard route to Ar-F); iodide, warm with KI -> ArI (no catalyst needed, iodine is the odd one out); hydrolysis, boil in water or dilute acid -> ArOH; and deamination, with H3PO2 (hypophosphorous acid) -> Ar-H, replacing -N2+ by -H. RETENTION of both nitrogens: azo coupling, in which the WEAK electrophile ArN2+ attacks a strongly activated ring at the PARA position, with phenol in mild alkali (pH about 9-10, via the more reactive phenoxide) -> p-hydroxyazobenzene, and with aniline in mild acid (pH about 4-5) -> p-aminoazobenzene, the -N=N- bridge making a coloured dye. [JEE Tip] Diazonium chemistry installs groups you CANNOT get by direct electrophilic substitution: -F (Balz-Schiemann) and -I (KI) have no clean direct route, and deamination (H3PO2) lets you use -NH2 purely as a positional director and then delete it. Temperature is the guard rail: above about 5 C the salt hydrolyses to phenol, so every replacement except hydrolysis is run ice-cold.
Worked Examples — Beyond-NCERT Essentials
Example 1 — Basicity in gas phase vs water. Arrange CH3NH2, (CH3)2NH, (CH3)3N and NH3 in order of decreasing basicity (a) in the gas phase and (b) in aqueous solution, and justify the difference.
In the gas phase only the +I effect of the methyl groups operates: each methyl feeds electron density onto nitrogen, so more methyls mean a stronger base and (CH3)3N > (CH3)2NH > CH3NH2 > NH3. In water the protonated cation must also be solvated, and the more N-H bonds it keeps the more hydrogen bonds stabilise it, so solvation favours the reverse trend (primary > secondary > tertiary), while steric bulk further hurts the tertiary amine. The two effects compromise, lifting the secondary amine to the top and pushing the tertiary down: (CH3)2NH > CH3NH2 > (CH3)3N > NH3. The demotion of the tertiary amine on going from gas to water is the fingerprint of solvation.
Example 2 — The ethyl-series trap. A student writes the aqueous basicity of the ethylamines as (C2H5)2NH > C2H5NH2 > (C2H5)3N, simply copying the methylamine pattern. Is this correct?
No. For methylamines the aqueous order is indeed secondary > primary > tertiary, but for the bulkier ethyl groups the measured order is secondary > tertiary > primary: (C2H5)2NH > (C2H5)3N > C2H5NH2 > NH3. The larger ethyl groups tilt the solvation-and-steric balance so that triethylamine ends up ABOVE ethylamine. The only features shared by both series are that the secondary amine is the strongest base and ammonia the weakest. Correct order: (C2H5)2NH > (C2H5)3N > C2H5NH2 > NH3.
Example 3 — Positive carbylamine test. Among aniline, N-methylaniline, N,N-dimethylaniline and acetamide, which give a positive carbylamine test?
The carbylamine (isocyanide) test needs a PRIMARY amine, that is two N-H bonds on an amino nitrogen, which CHCl3 with alcoholic KOH converts into the foul-smelling isocyanide R-NC. Aniline (C6H5NH2) is a primary amine, so it is positive. N-methylaniline (C6H5NHCH3) is secondary and N,N-dimethylaniline is tertiary, so both are negative. Acetamide (CH3CONH2) does carry N-H bonds, but its nitrogen belongs to an amide with the lone pair tied into the C=O; it is not an amine and gives no carbylamine reaction. Answer: only aniline. Trap: amide N-H bonds do not qualify.
Example 4 — Hofmann degradation and the carbon count. Propanamide, CH3CH2CONH2, is treated with Br2 and aqueous NaOH. Give the product, and contrast it with LiAlH4 reduction of the same amide.
Hofmann bromamide degradation expels the carbonyl carbon as carbonate and yields a primary amine with ONE FEWER carbon. Propanamide has 3 carbons, so the product is ethylamine, CH3CH2NH2 (2 carbons). LiAlH4, by contrast, reduces the amide WITHOUT losing carbon, since the carbonyl carbon survives as a CH2: CH3CH2CONH2 -> CH3CH2CH2NH2, propylamine (3 carbons). Answer: Br2/NaOH gives ethylamine (2 C) while LiAlH4 gives propylamine (3 C). One amide, two carbon counts, decided entirely by the reagent.
Example 5 — Distinguish three amines by Hinsberg. Three unlabelled bottles hold ethylamine, diethylamine and triethylamine. Using only Hinsberg's reagent followed by aqueous KOH, how do you tell them apart?
Add C6H5SO2Cl to each and then shake with KOH. Ethylamine (primary) gives C6H5SO2NHC2H5, whose lone N-H is acidified by the -SO2- group, so the product DISSOLVES in KOH to a clear solution. Diethylamine (secondary) gives C6H5SO2N(C2H5)2 with no N-H, an INSOLUBLE solid that stays as a precipitate in KOH. Triethylamine (tertiary) has no N-H and does NOT react; it remains as an insoluble oil that dissolves only when the mixture is acidified. Three distinct observations pin down the three classes.
Example 6 — Predict the diazonium products. Benzenediazonium chloride, C6H5N2+Cl-, is treated separately with (a) CuBr/HBr, (b) KI with warming, (c) HBF4 then heat, (d) H3PO2, and (e) boiling water. Name each product.
Every route replaces -N2+ and releases N2 gas. (a) The Sandmeyer reaction gives bromobenzene, C6H5Br. (b) KI needs no catalyst and gives iodobenzene, C6H5I. (c) Balz-Schiemann: HBF4 precipitates C6H5N2+BF4-, which on heating gives fluorobenzene, C6H5F, plus N2 and BF3. (d) Hypophosphorous acid replaces -N2+ by -H (deamination), giving benzene, C6H6. (e) Boiling water hydrolyses the salt to phenol, C6H5OH. Note that -F and -I have no clean DIRECT route on benzene, which is exactly why the diazonium detour is used.
Example 7 — Azo coupling: product, position and medium. Benzenediazonium chloride is added to phenol in mild alkali. Give the product, the position of attack, and explain the choice of medium.
This is azo coupling, an electrophilic aromatic substitution in which ArN2+ is a WEAK electrophile and so needs a strongly activated ring. Mild alkali converts phenol into phenoxide, and -O- activates the ring more strongly than -OH, so coupling proceeds readily; attack falls on the PARA position (most activated and least hindered), giving p-hydroxyazobenzene, C6H5-N=N-C6H4-OH, an orange dye. A strongly acidic medium would leave only the less reactive neutral phenol, while a strongly alkaline medium would convert the diazonium into an unreactive diazotate; hence the pH is kept only mildly alkaline, about 9-10.
Example 8 — Rank aniline basicity with substituents. Arrange in decreasing base strength in water: aniline, p-toluidine, p-nitroaniline and cyclohexylamine.
Cyclohexylamine is an ALIPHATIC amine with a fully localised lone pair, so it is far more basic than any aniline. Among the anilines the lone pair is delocalised into the ring, which lowers basicity; a para electron-donating -CH3 (p-toluidine) partly restores electron density and lifts basicity above aniline, whereas a para electron-withdrawing -NO2 (p-nitroaniline) drains the lone pair by resonance and sharply lowers it. Order: cyclohexylamine > p-toluidine > aniline > p-nitroaniline.