Organic Compounds Containing Nitrogen: Amines and Diazonium Salts
OrganicWeightage: 2-3 Questions (8-12 Marks)JEE Unit 18
“Welcome back! You already know the NCERT amines chapter: classification, basic preparation and the standard tests. This guide stays on what JEE Main actually asks in this unit. Nearly every question comes down to five decisions: how basic the nitrogen is (lone pair availability), whether the amine is 1°, 2° or 3°, how many carbons a route adds or removes, whether the ring needs protecting before it meets an electrophile, and what a diazonium salt can be turned into. Settle these and the 2024–2026 papers become short multi-step arithmetic. We spend the time on the orders, exceptions and traps.”
— SCORECHEM ACADEMIC TEAM
1. Basic Strength of Aliphatic Amines
Basicity is the availability of the N lone pair to accept H+. In the gas phase it simply follows the +I effect: 3° > 2° > 1° > NH3. In water the conjugate acid must also be solvated by hydrogen bonding, and R3NH+ has only one N–H to hydrogen-bond, so the order changes.
⚠️ JEE Trap: there is no single aqueous order for all alkylamines. For methyl amines it is (CH3)2NH > CH3NH2 > (CH3)3N > NH3; for ethyl amines it is (C2H5)2NH > (C2H5)3N > C2H5NH2 > NH3. The 2° amine is first in both; the 3° amine is weaker than the 1° for methyl, stronger for ethyl. In a non-aqueous solvent the order reverts to 3° > 2° > 1°. Always read the solvent in the question.
2. Aniline and Substituent Effects on Basicity
In aniline the lone pair is delocalised into the ring, so it is far less available than in an alkylamine (pKb 9.4 against about 3.4). Anything that pulls more density from N lowers basicity further; anything that pushes density toward N raises it. The ranking among substituted anilines is a favourite one-mark question.
⚠️ JEE Trap: ortho-substituted anilines are weaker bases than aniline whatever the substituent. Even CH3, which donates electrons, makes o-toluidine weaker (pKb 9.6) than aniline and than p-toluidine (the ortho effect: after protonation the bulky NH3+ is crowded and solvates poorly). Also, for NO2 the order is m (11.5) > p (13.0) > o (14.3) as bases, because resonance withdrawal works only from ortho and para.
3. Preparing Amines: Reagent and Carbon Count
Preparation questions test two things at once: the reagent, and whether the amine has more, fewer or the same number of carbons as the starting material. Decide the carbon count first, then the reagent follows.
⚠️ JEE Trap: Gabriel gives only 1° amines, and never an aromatic amine. The phthalimide anion attacks by SN2, so Ar–X (partial double-bond character at C–X) and 3° R–X (elimination) fail. Ammonolysis of R–X gives a mixture of 1°, 2°, 3° and quaternary salts; a large excess of NH3 favours the 1° amine. Hofmann bromamide requires an unsubstituted amide, RCONH2, and the product has ONE carbon less (propanamide gives ethanamine).
4. Hinsberg, Carbylamine and Other Class Tests
Hinsberg's reagent (benzenesulphonyl chloride, or TsCl) separates all three classes by whether the sulphonamide has an N–H and so dissolves in alkali. In counting questions, the number of alkali-insoluble solids equals the number of 2° amines in the list.
⚠️ JEE Trap: classify by the groups on N, not by what the molecule looks like. N-Methylaniline, N-propylaniline and diphenylamine are 2° (insoluble sulphonamides); aniline, benzylamine and allylamine are 1° (soluble); N,N-dimethylaniline is 3° (no reaction). In the 2026 list of eleven amines, exactly four were 2°. Note also that the carbylamine test is positive for 1° amines alone, aliphatic or aromatic.
5. Nitrous Acid and Hofmann Elimination
Nitrous acid (NaNO2 + HCl, generated in situ) gives a different product for each class and for aliphatic versus aromatic substrates. The quantitative use is the N2 evolved from a 1° aliphatic amine: one mole of N2 per mole of amine.
Hofmann elimination is the amine-side elimination. Exhaustive methylation (excess CH3I, then moist Ag2O) gives a quaternary ammonium hydroxide, which on heating gives the LEAST substituted alkene because the bulky NR3 leaving group makes the base remove the most accessible β-hydrogen. This is the opposite of Saytzeff elimination of alkyl halides.
⚠️ JEE Trap: N2 evolution identifies a 1° AMINE only if it is aliphatic. A 1° aromatic amine at 273–278 K gives a stable diazonium salt with no gas; only warming above about 283 K releases N2 (as phenol forms). Tertiary aliphatic amines give a soluble nitrite salt with no gas, and 3° aromatic amines react by C-nitrosation at para instead.
6. Aniline: Protection and Ring Reactions
The NH2 group is a powerful activator, which is exactly the problem: it over-reacts, and in strong acid it is converted to a meta-directing deactivator. Acetylation is the standard fix because the amide N lone pair is shared with C=O.
⚠️ JEE Trap: three different ring outcomes from the same ring. Br2/H2O gives 2,4,6-tribromoaniline at once (no catalyst). Direct HNO3/H2SO4 gives tar plus about half m-nitroaniline (via the anilinium ion). Heating aniline with conc. H2SO4 gives sulphanilic acid (p-H2N–C6H4–SO3H), which exists as a zwitterion, with a high melting point and poor solubility in organic solvents. Friedel–Crafts fails on aniline and on acetanilide.
7. Diazonium Salts: Every Replacement
Aromatic diazonium salts are stable only at 273–278 K (resonance delocalises the charge into the ring), while aliphatic ones decompose instantly. They are the gateway from an amine to nearly any substituent on the ring. The 2026 stability order of para-substituted salts was OCH3 > H > CN > NO2: donors stabilise the cation and acceptors destabilise it.
⚠️ JEE Trap: F and I do not use the Sandmeyer route. Fluorine comes from the isolated ArN2+BF4− by heating (Balz–Schiemann); iodine comes from simple KI with no copper. H3PO2 or ethanol replaces the diazonium group by H, which is how an NH2 directing group is removed after it has done its job. Benzoic acid from aniline is two steps: CuCN (nitrile), then hydrolysis.
8. Azo Coupling and the Griess Test
The diazonium ion is a weak electrophile and attacks only strongly activated rings, at the para position (ortho if para is blocked). The pH must match the partner: mildly basic for phenol (phenoxide is the nucleophile), mildly acidic for aniline (enough acid to keep the diazonium ion, not enough to protonate the amine).
⚠️ JEE Trap: the Griess–Ilosvay test detects NITRITE, and it works by diazotising sulphanilic acid. Acidified NO2− diazotises sulphanilic acid, which then couples with α-naphthylamine to give a red azo dye. Mass questions use the dye's molar mass: aniline yellow (p-aminoazobenzene) is 197 g mol−1 and p-hydroxyazobenzene is 198 g mol−1, formed 1 : 1 from aniline.
9. Multi-Step Synthesis Planning
Most of the 2024–2026 sequence questions are the same three moves: protect or direct with the amine, do the ring chemistry, then remove or replace the amine through the diazonium salt. Use the amine as a temporary director where the target has a substitution pattern that direct halogenation cannot give.
⚠️ JEE Trap: count what each reagent does to the ring, not just the name. m-Bromoaniline cannot be made by brominating aniline (you get the tribromo compound); brominate nitrobenzene first (meta), then reduce. In the 2026 sequence nitrobenzene to m-bromoaniline to Br2/H2O to Sandmeyer gave pentabromobenzene, five Br atoms in total.
10. Cyanides, Isocyanides and Nitro Compounds
Ambident nucleophiles give two different products depending on whether the metal is ionic or covalent: KCN attacks through C (nitrile), AgCN through N (isocyanide); KNO2 attacks through O (nitrite ester), AgNO2 through N (nitroalkane).
⚠️ JEE Trap: reduction of RNC gives a 2° N-methyl amine, not a 1° amine. RNC reduced with Na/EtOH or H2/Ni gives R–NH–CH3, whereas RCN gives RCH2NH2 with one more carbon. Acid hydrolysis of RNC gives RNH2 and HCOOH; RNC resists alkali. Also remember R–CN (about 4 D) has a larger dipole moment than R–NC (about 3.5 D).
11. How JEE Frames These Questions
Sequence to count atoms: Br atoms in a final product, or % N in an intermediate (convert to moles, remember p-nitroaniline has two N).
Yield arithmetic: acetanilide from aniline (93 to 135), benzanilide (93 to 197), ethylamine to N2 by moles (1 : 1).
Hinsberg counting: how many alkali-insoluble solids = how many 2° amines, or a pair of statements about TsCl.
Back-solve from an isocyanide or alcohol: work backwards through carbylamine, Hofmann or HNO2 to the starting acid, using carbon counts.
Basicity or diazonium stability ranking: donors raise, acceptors lower; ortho is the exception.
Match-and-identify: P, Q, R identification from molecular formula (C6H7N means aniline) and a chiral clue (2-chlorobutane).