Amines

The chapter has two parts sharing one element, nitrogen. Part I builds up amines: structure and the pyramidal geometry forced by nitrogen's lone pair (9.1); classification into primary, secondary and tertiary (9.2); the parallel common and IUPAC naming systems, including the N-locant used for substituents on nitrogen (9.3); six methods of preparation, from reducing a nitro group to degrading an amide (9.4); physical properties dominated by hydrogen bonding (9.5); and finally chemical reactions (9.6), which is really two chapters in one — the basicity discussion (Kb, pKb, and the inductive-versus-solvation-versus-resonance argument that explains why aliphatic and aromatic amines sit on opposite sides of ammonia) and a set of named reactions (alkylation, acylation, the carbylamine test, reaction with nitrous acid, the Hinsberg test, and electrophilic aromatic substitution) that together let you both synthesise a target amine and distinguish it from its isomers by simple test-tube chemistry. Part II is diazonium salts (9.7-9.10): how diazotisation converts a primary aromatic amine into $ArN_2^+X^-$, why that salt is so much more stable than its aliphatic cousin, its physical behaviour, and the two great reaction families — displacement of nitrogen (Sandmeyer, Gatterman, and replacement by fluoride, hydrogen, hydroxyl or nitro) and retention of the diazo group in coupling reactions that build azo dyes. The chapter's unifying thread is the nitrogen lone pair: whether it is protonated (basicity), acylated (amide formation), delocalised into a ring (aniline's weak basicity, or a diazonium salt's fleeting stability), or expelled entirely as $N_2$ gas (Sandmeyer-type reactions), everything traces back to what that one pair of electrons is doing.

Structure of amines

Like ammonia, the nitrogen atom of an amine is trivalent with one unshared electron pair, and its four electron domains (three bonds plus the lone pair) are arranged tetrahedrally around an $sp^3$-hybridised nitrogen. Because a lone pair repels bonding pairs more than bonding pairs repel each other, the actual C-N-E bond angle (E being carbon or hydrogen) is squeezed to slightly less than the ideal tetrahedral $109.5\degree$ — measured as $108\degree$ in trimethylamine. The overall molecular shape is therefore pyramidal, not flat, with the lone pair occupying the fourth corner of a distorted tetrahedron. This lone pair is the single most important feature of the whole chapter: it is what makes nitrogen a Lewis base (able to accept a proton or bond to an electrophile) and a nucleophile (able to attack an electrophilic carbon), and every reaction in this chapter is, at bottom, a story about where that lone pair goes.

Classification and nomenclature

Replacing exactly one hydrogen of ammonia by an alkyl or aryl group gives a primary (1°) amine, $R{-}NH_2$ or $Ar{-}NH_2$; replacing a second hydrogen gives a secondary (2°) amine, $R{-}NH{-}R'$; replacing the third gives a tertiary (3°) amine, $R{-}NR'{-}R''$. Secondary and tertiary amines are called simple when every substituent is identical and mixed when they differ. Two naming systems run in parallel. The common system names an aliphatic amine as one word, alkyl group plus "amine" (methylamine), with "di" or "tri" prefixed when groups repeat (trimethylamine). The IUPAC system replaces the terminal "e" of the parent alkane with "amine" (methanamine, propan-1-amine, propan-2-amine for the isomers of $C_3H_9N$... note the numeral fixes which carbon bears the group), and for a secondary or tertiary amine uses the locant "N" to show that a substituent sits on nitrogen rather than on the carbon chain — $CH_3NHC_2H_5$ is N-methylethanamine, and $(C_2H_5)_3N$ is N,N-diethylethanamine. Aromatic amines with $-NH_2$ attached directly to a benzene ring are arylamines; $C_6H_5NH_2$, common name aniline, is also an accepted IUPAC name, and the systematic alternative is benzenamine. A compound such as $C_4H_{11}N$ has eight possible isomeric amines spanning all three classes, illustrating that chain, position and (primary/secondary/tertiary) functional isomerism can all operate simultaneously on one molecular formula — always check every classification before naming an unfamiliar amine.

Preparation of amines

Six named routes build the bridge from other functional groups to amines, and each has a signature scope and limitation worth memorising alongside the equation. Reduction of nitro compounds — passing $H_2$ over finely divided Ni, Pd or Pt, or using a metal in acid (Fe/HCl scrap iron is preferred industrially, since the $FeCl_2$ formed hydrolyses back to HCl and only a catalytic amount of acid is needed) — converts $R{-}NO_2$ or $Ar{-}NO_2$ cleanly to the primary amine. Ammonolysis of alkyl halides exploits the same $S_N$ chemistry from Class 12 Unit 6: an alkyl or benzyl halide heated with ethanolic ammonia in a sealed tube at 373 K undergoes nucleophilic substitution, halide leaving and $-NH_2$ arriving; the newly formed primary amine, itself a nucleophile, keeps attacking unreacted alkyl halide, cascading on to secondary and tertiary amine and finally a quaternary ammonium salt, from which the free amine is recovered by treating with strong base. Because this cascade is hard to stop cleanly, ammonolysis is run with a large excess of ammonia to favour the primary amine, and halide reactivity follows $RI$ greater than $RBr$ greater than $RCl$. Reduction of nitriles, with $LiAlH_4$ or catalytic hydrogenation, gives a primary amine with one carbon more than the starting nitrile — the standard method for the "ascent" of an amine series by exactly one carbon. Reduction of amides, also with $LiAlH_4$, gives an amine with the same carbon count as the amide. The Gabriel phthalimide synthesis is prized because it gives pure primary amines with no over-alkylation risk: phthalimide's ethanolic potassium salt is alkylated by an alkyl halide, then hydrolysed under alkali to liberate the primary amine and phthalate; it cannot make aromatic primary amines, because aryl halides do not undergo nucleophilic substitution with the phthalimide anion (no $S_NAr$ pathway available without strong ring activation). Finally, the Hofmann bromamide degradation treats an amide with bromine in aqueous or ethanolic NaOH; an alkyl or aryl group migrates from the carbonyl carbon to nitrogen, and the resulting amine has one carbon fewer than the starting amide — the full mechanism is derived below.

Physical properties of amines

Lower aliphatic amines are gases with a distinctive fishy odour; primary amines with three or more carbons are liquids, and still higher members are solids. Aniline and other arylamines are colourless when freshly prepared but darken on standing owing to slow atmospheric oxidation. Lower amines dissolve readily in water because nitrogen's lone pair and N-H bonds hydrogen-bond with water; solubility falls steadily as the hydrophobic alkyl portion grows, and higher amines are essentially water-insoluble, though all amines dissolve well in organic solvents such as alcohol, ether and benzene. Because nitrogen (electronegativity $3.0$) hydrogen-bonds more weakly than oxygen ($3.5$), amines are both less water-soluble and more volatile (lower-boiling) than an alcohol of comparable molar mass — between butan-1-ol and butan-1-amine, the alcohol is markedly more soluble and higher-boiling. Among amine isomers themselves, primary and secondary amines associate intermolecularly by N-H···N hydrogen bonding; a primary amine has two N-H bonds available for this, a secondary amine only one, and a tertiary amine — with no N-H bond at all — cannot hydrogen-bond to itself, though it can still accept a hydrogen bond from water or another donor. The resulting boiling-point order for isomeric amines is therefore primary greater than secondary greater than tertiary, exactly the pattern seen for four isomers of formula $C_4H_{11}N$-class molar mass in Table 9.2 of the chapter.

Basic character of amines: Kb and pKb

Amines react with acids to form ammonium salts, and those salts regenerate the free amine on treatment with a strong base such as NaOH — since the salt is water-soluble but insoluble in organic solvents like ether, this acid-base cycle is the standard method for separating an amine from non-basic organic impurities. The reaction with water itself defines the amine's base strength quantitatively.

where: $K_b$ is the base dissociation constant and $pK_b$ its negative logarithm; a larger $K_b$ (smaller $pK_b$) means a stronger base. Ammonia itself has $pK_b = 4.75$. Aliphatic amines are stronger bases than ammonia, with $pK_b$ typically between $3$ and $4.22$, because the electron-releasing ($+I$) alkyl group raises electron density on nitrogen; aromatic amines are weaker bases than ammonia — aniline's $pK_b$ is $9.38$, more than four powers of ten weaker — because the ring withdraws the lone pair by resonance. The apparent contradiction inside the aliphatic series (why does $pK_b$ not fall monotonically from primary to tertiary as the $+I$ effect alone would predict) is resolved only by adding solvation and steric effects to the inductive argument, which the trend derivation below works through in full.

Chemical reactions of amines

Beyond basicity, five named reactions probe the amine's nucleophilic lone pair and, together, let a chemist both build new molecules and tell 1°, 2° and 3° amines apart in a test tube. Alkylation — an amine attacking an alkyl halide — is the same $S_N2$ chemistry used in ammonolysis, and it is why ammonolysis over-alkylates unless ammonia is in large excess. Acylation replaces an N-H hydrogen with an acyl group from an acid chloride, anhydride or ester (a base such as pyridine mops up the HCl by-product and drives the equilibrium), producing an amide; reaction specifically with benzoyl chloride is called benzoylation — methanamine plus benzoyl chloride gives N-methylbenzamide plus HCl. (Amines with carboxylic acids, by contrast, simply form ammonium carboxylate salts at room temperature — no amide forms without heat.) The carbylamine reaction is diagnostic for primary amines alone: heating with chloroform and ethanolic KOH generates a foul-smelling isocyanide (carbylamine); secondary and tertiary amines give no such odour, making this the standard qualitative test for a primary amine. Reaction with nitrous acid (generated in situ from $NaNO_2$ and a mineral acid) is class-dependent: primary aliphatic amines give unstable alkyldiazonium salts that decompose immediately, releasing $N_2$ gas quantitatively and an alcohol — a reaction exploited to estimate amino groups in amino acids and proteins — while primary aromatic amines, at the low temperature $273{-}278\ \text{K}$, give arenediazonium salts stable for a short time (Section 9.7); secondary amines give an oily N-nitrosamine, and tertiary amines (aliphatic) form unstable salts or, for aromatic tertiary amines, a p-nitroso derivative. Hinsberg's test, using benzenesulphonyl chloride ($C_6H_5SO_2Cl$), completes the diagnostic set: a primary amine gives an N-alkylbenzenesulphonamide whose remaining N-H, made strongly acidic by the adjacent electron-withdrawing sulphonyl group, dissolves in alkali; a secondary amine gives an N,N-dialkylbenzenesulphonamide with no acidic N-H, insoluble in alkali; a tertiary amine, having no N-H to substitute, does not react at all. (p-Toluenesulphonyl chloride is the modern reagent of choice, working identically.)

Electrophilic substitution in aromatic amines

The same nitrogen lone pair that makes aniline a weak Bronsted base makes its ring exceptionally electron-rich, because resonance pushes electron density specifically onto the ortho and para ring carbons — $-NH_2$ is therefore a powerful activating, ortho/para-directing group. This has a practical downside: aniline reacts with bromine water at room temperature with no catalyst at all, giving a white precipitate of 2,4,6-tribromoaniline directly, and direct nitration similarly over-reacts, producing tarry oxidation products alongside a substantial meta by-product (because strongly acidic nitrating conditions protonate the ring nitrogen to the meta-directing anilinium ion). Controlled, single-substitution reactions therefore first protect the amino group by acetylation with acetic anhydride — the acetamido nitrogen's lone pair is partly tied up in resonance with the adjacent carbonyl, so it activates the ring less strongly — carry out the desired substitution (giving, for nitration, predominantly the para product), and finally hydrolyse the amide back to the free amine. Sulphonation follows a different path entirely: aniline with concentrated $H_2SO_4$ first forms anilinium hydrogensulphate, which on heating to $453{-}473\ \text{K}$ rearranges to p-aminobenzenesulphonic acid (sulphanilic acid, a zwitterion). Aniline cannot undergo Friedel-Crafts alkylation or acylation at all, because the Lewis acid catalyst $AlCl_3$ coordinates to the basic nitrogen lone pair, converting it into a strongly ring-deactivating positively charged group before any electrophile can attack the ring.

Diazonium salts: preparation and stability

A diazonium salt has the general formula $R{-}N_2^+X^-$, where the $-N_2^+$ unit is the diazonium group and $X^-$ may be $Cl^-$, $Br^-$, $HSO_4^-$, $BF_4^-$ and so on; it is named by adding "diazonium" and the anion's name to the parent hydrocarbon — $C_6H_5N_2^+Cl^-$ is benzenediazonium chloride, $C_6H_5N_2^+HSO_4^-$ is benzenediazonium hydrogensulphate. Primary aliphatic amines form alkyldiazonium salts so unstable they decompose the moment they form (already noted above); primary aromatic amines form arenediazonium salts that survive for a short time in solution at $273{-}278\ \text{K}$, stabilised by resonance delocalisation of the positive charge into the ring — the mechanism is worked through in full below. Benzenediazonium chloride, made by diazotisation — aniline treated with $NaNO_2$ and two equivalents of HCl at $273{-}278\ \text{K}$ — is a colourless, water-soluble crystalline solid, stable in the cold but reactive with water on warming, and it decomposes if allowed to dry out, so it is used immediately after preparation rather than stored. Benzenediazonium fluoroborate, by contrast, is insoluble in water and stable at room temperature, which is why it is the salt of choice whenever isolation matters.

where: the mineral acid HCl both generates nitrous acid in situ from $NaNO_2$ and supplies the chloride counter-ion of the product; the narrow temperature window matters because above about $283\ \text{K}$ the diazonium salt itself starts to hydrolyse to phenol.

Reactions of diazonium salts

Diazonium chemistry splits cleanly into two families. Reactions with displacement of nitrogen exploit $N_2$ as an outstanding leaving group: the Sandmeyer reaction uses a Cu(I) salt (CuCl, CuBr or CuCN) to replace $N_2^+$ with $-Cl$, $-Br$ or $-CN$; the closely related Gatterman reaction achieves the same halogen substitutions with copper powder and the corresponding halogen acid instead of a Cu(I) salt, though Sandmeyer generally gives the better yield. Iodide needs no catalyst at all — simply treating the diazonium salt with $KI$ gives the aryl iodide, since iodine cannot otherwise be introduced directly into a benzene ring. Fluoride is introduced by first precipitating the diazonium fluoroborate with $HBF_4$, then heating it to decompose to the aryl fluoride (the Balz-Schiemann reaction) — again, direct fluorination of benzene is not practicable, so this indirect route matters. Mild reducing agents, hypophosphorous acid ($H_3PO_2$, also called phosphinic acid) or ethanol, replace $N_2^+$ by hydrogen, reducing the diazonium salt to the parent arene while the reductant is itself oxidised (to phosphorous acid, or to ethanal). Allowing the diazonium salt solution to warm to about $283\ \text{K}$ hydrolyses it to a phenol, and heating the fluoroborate with aqueous $NaNO_2$ in the presence of copper installs a $-NO_2$ group in place of nitrogen. Reactions retaining the diazo group are dominated by coupling reactions: benzenediazonium chloride attacks the electron-rich para position of phenol to give p-hydroxyazobenzene, or of aniline to give p-aminoazobenzene, in each case forming an extended, highly conjugated $-N{=}N-$ (azo) linkage — this is an electrophilic aromatic substitution on the phenol or amine ring, and the resulting azo compounds, strongly coloured because of that extended conjugation, are the working chemistry behind the azo dye industry. Taken together, this makes the diazonium salt arguably the single most valuable synthetic intermediate in aromatic chemistry: it converts an amino group installed by nitration-then-reduction into almost any other functional group a synthesis might demand.

Board level - Ammonolysis of ethyl chloride and benzyl chloride, with over-alkylation (NCERT Example 9.1, extended)

Write the equation for the reaction of ethanolic $NH_3$ with $C_2H_5Cl$, and separately write the ammonolysis of benzyl chloride followed by reaction of the resulting amine with two moles of $CH_3Cl$.

Step 1. Ethanolic ammonia performs a nucleophilic substitution on ethyl chloride in a sealed tube at $373\ \text{K}$: the nitrogen lone pair of $NH_3$ displaces chloride, giving ethylammonium chloride, which is deprotonated by excess ammonia to the free amine. $C_2H_5Cl + NH_3 \xrightarrow{373\,\text{K, sealed tube}} C_2H_5NH_3^+Cl^- \xrightarrow{NH_3} C_2H_5NH_2 + NH_4Cl$.

Step 2. Benzyl chloride, $C_6H_5CH_2Cl$, ammonolyses by the identical mechanism (the $-CH_2-$ carbon, not the aromatic ring, is attacked, so this works even though direct ammonolysis of a genuine aryl halide would not): $C_6H_5CH_2Cl + NH_3 \rightarrow C_6H_5CH_2NH_2 + HCl$, giving benzylamine (phenylmethanamine).

Step 3. Benzylamine's nitrogen still carries a lone pair and two N-H bonds, so it is itself a nucleophile and reacts further with the first mole of $CH_3Cl$: $C_6H_5CH_2NH_2 + CH_3Cl \rightarrow C_6H_5CH_2NHCH_3 + HCl$, a secondary amine.

Step 4. The secondary amine reacts with the second mole of $CH_3Cl$ in exactly the same way, consuming its one remaining N-H. Answer: $C_6H_5CH_2NHCH_3 + CH_3Cl \rightarrow C_6H_5CH_2N(CH_3)_2 + HCl$ — N,N-dimethyl-1-phenylmethanamine, a tertiary amine — illustrating exactly why ammonolysis, run without a large excess of the halide-limiting nucleophile, cascades past the primary amine unless deliberately stopped.

Board level - Identifying the amide for the Hofmann bromamide route to propanamine, and the product from benzamide (NCERT Examples 9.2 and 9.3)

(i) Give the structure and IUPAC name of the amide that yields propan-1-amine by the Hofmann bromamide reaction. (ii) Identify the amine produced by Hofmann degradation of benzamide.

Step 1. The Hofmann bromamide degradation removes exactly one carbon (the carbonyl carbon, expelled as $CO_2$, as derived above), so an amide yielding a three-carbon amine must itself contain four carbons: the amide is $CH_3CH_2CH_2CONH_2$.

Step 2. Name it by standard amide nomenclature — the parent acid is butanoic acid, so the amide is butanamide. Running the degradation strips the carbonyl carbon and delivers $CH_3CH_2CH_2NH_2$, propan-1-amine, confirming the count.

Step 3. For part (ii), benzamide is $C_6H_5CONH_2$, an aromatic amide with seven carbons total (six in the ring, one in the carbonyl); the reaction always removes the carbonyl carbon regardless of whether the migrating group is alkyl or aryl, so the amine formed retains only the six ring carbons.

Step 4. The migrating group here is the phenyl ring itself, moving intact from carbonyl carbon to nitrogen exactly as in Step 2 of the mechanism derivation. Answer: (i) the amide is butanamide, $CH_3CH_2CH_2CONH_2$; (ii) benzamide degrades to aniline (benzenamine), $C_6H_5NH_2$ — an aromatic primary amine with one carbon fewer than benzamide, obtained without ever passing through a free phenyl cation.

Competitive level (JEE / NEET) - pH of a weak-base amine solution from its NCERT pKb value

Using the chapter's own value $pK_b = 3.38$ for methanamine ($CH_3NH_2$), calculate the $[OH^-]$, pOH and pH of a $0.10\ \text{M}$ aqueous methanamine solution at 298 K, assuming the usual weak-base approximation.

Step 1. Convert $pK_b$ to $K_b$: $K_b = 10^{-3.38} = 10^{-4} \times 10^{0.62} = 4.17\times10^{-4}$. This is the equilibrium constant for $CH_3NH_2 + H_2O \rightleftharpoons CH_3NH_3^+ + OH^-$ from equation (9.1), taken directly from Table 9.3 of the chapter rather than re-measured.

Step 2. Apply the standard weak-base approximation, valid because $K_b \ll C$ (here $4.17\times10^{-4} \ll 0.10$, so less than a few per cent of the amine ionises and $C$ is not significantly depleted): $[OH^-] = \sqrt{K_b \times C} = \sqrt{4.17\times10^{-4} \times 0.10} = \sqrt{4.17\times10^{-5}} = 6.46\times10^{-3}\ \text{M}$.

Step 3. Convert to pOH: $pOH = -\log(6.46\times10^{-3}) = 3 - \log(6.46) = 3 - 0.810 = 2.19$.

Step 4. Use $pH + pOH = 14$ at 298 K: $pH = 14 - 2.19 = 11.81$. Answer: $[OH^-] = 6.46\times10^{-3}\ \text{M}$, $pOH = 2.19$, $pH = 11.81$ — comfortably basic, consistent with methanamine being a somewhat stronger base than ammonia (whose own $pK_b = 4.75$ would give a lower $[OH^-]$ and a lower pH at the same concentration). Timing note: under two minutes. The whole calculation is a single weak-base $\sqrt{K_bC}$ chain — $pK_b \to K_b \to [OH^-] \to pOH \to pH$ — identical in structure to any weak-acid pH problem with $K_a$ replaced by $K_b$ and pH replaced by pOH at the midpoint. The one trap is forgetting the final $14 - pOH$ step and reporting pOH as if it were pH; always ask "did the question want pH or pOH" before circling a final answer, and sanity-check that a base gives $pH$ greater than $7$.

Students very often invert the basicity comparison, assuming "more alkyl groups always means more basic" and applying the gas-phase order ($3\degree$ greater than $2\degree$ greater than $1\degree$) directly to an aqueous exam question — the correct aqueous order depends on solvation and sterics too, and for methyl-substituted amines the tertiary amine is actually the weakest base of the three. A second common slip is forgetting that aniline is a weaker base than ammonia, not a stronger one — the intuition "amines are always stronger bases than ammonia" holds only for aliphatic amines; resonance delocalisation of the lone pair into an aromatic ring always pulls basicity the other way. Students frequently misapply the carbylamine test to secondary or tertiary amines expecting a positive result — it answers only for primary amines, aliphatic or aromatic. In the Hofmann bromamide reaction, a very common arithmetic error is forgetting that the amide needs one carbon more than the target amine, not the same number or one fewer. On Hinsberg's test, students often confuse which sulphonamide is alkali-soluble: it is the one from the primary amine (an acidic N-H survives), not the secondary. Finally, many answers state that the diazonium salt is "stable" without qualification — it is only stable for a short time at low temperature ($273{-}278\ \text{K}$); warming it, or letting it dry out, decomposes it, which is why it is always used immediately after preparation rather than isolated and stored.

• Classify any amine as primary, secondary or tertiary from its structure, and name it correctly by both the common and IUPAC systems, including the N-locant for substituents on nitrogen.
• Explain why amine nitrogen is pyramidal rather than planar, and state the approximate C-N-C bond angle and its cause.
• Select the correct preparation method (nitro reduction, ammonolysis, nitrile or amide reduction, Gabriel synthesis, or Hofmann degradation) for a given target amine, and state each method's key limitation.
• Predict and explain the relative boiling points of isomeric primary, secondary and tertiary amines, and compare an amine's boiling point and solubility with those of an alcohol of similar molar mass.
• Compare the basicity of two given amines (aliphatic versus aliphatic, aliphatic versus aromatic, or substituted aniline versus aniline), citing inductive, resonance, solvation and steric arguments as appropriate, in both gas phase and aqueous solution.
• Use the carbylamine reaction, reaction with nitrous acid, and Hinsberg's test to distinguish primary, secondary and tertiary amines, and write the products of each test for a given amine.
• Explain why the -NH2 group is a powerful ortho, para-directing activator, and describe how acetylation is used to control mono-substitution on aniline.
• Write the diazotisation reaction for a primary aromatic amine and explain why the arenediazonium ion is more stable than an alkyldiazonium ion.
• Predict the product of a diazonium salt reacting with CuCl, CuBr, CuCN, KI, HBF4-then-heat, H3PO2, warm water, or a phenol or aromatic amine (coupling), and design a short synthetic route from benzene to a target substituted aromatic compound using a diazonium intermediate.

Ask the tutor: Want to work through predicting the major product of a diazonium-salt reaction step by step, or practice ranking a mixed set of aliphatic and aromatic amines by aqueous basicity?