CLASS 12 CHEMISTRY · CHAPTER 9 · STUDY NOTES
Amines
An exploration of amines as derivatives of ammonia, covering their classification, nomenclature, preparation methods, physical and chemical properties, and the synthetic utility of diazonium salts.
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Structure and Classification of Amines
Classify at nitrogen, then consider its shape and lone pair.Count the groups attached to nitrogen
- Amine class
- Primary, secondary and tertiary refer to one, two or three carbon groups directly bonded to neutral nitrogen.
| Nitrogen attachment | Class |
|---|---|
| one carbon group + two H | primary amine |
| two carbon groups + one H | secondary amine |
| three carbon groups, no N–H | tertiary amine |
| four carbon groups, formal N⁺ | quaternary ammonium ion |
An ordinary alkylamine N has three sigma bonds and one lone-pair region, giving a pyramidal arrangement. The chapter trimethylamine figure gives about 108°, but geometry is not identical for all amines; conjugation in arylamines matters.
RNH₂, R₂NH and R₃N are primary, secondary and tertiary. A highly branched carbon bonded to NH₂ still gives a primary amine. R₄N⁺ is a quaternary ammonium ion with four bonds and no nitrogen lone pair; it needs a counterion.
Let’s look a little closer.
Go deeper: Attachment and biological context
Aniline attaches NH₂ directly to an aromatic ring; benzylamine has a CH₂ spacer and remains an aralkyl primary amine. The spacer interrupts direct conjugation of the N lone pair with the ring, so these structures should not be treated as equivalent aromatic amines.
Amines occur in biological molecules and in synthetic materials, dyes and medicinal compounds. Identify the functional group without turning historical textbook drug examples into medical advice or present approval claims. Simple amines have identical organic groups; mixed amines have different groups.
Take a moment. What do you think?
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Question
A crowded molecule has a tert-butyl group attached to NH₂. A second has three methyl groups attached to N. Classify each and identify the rule.
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Nomenclature of Amines
Read a name as a map of which atom carries each substituent.Locate the amino group and N-substituents
- N locant
- N specifies an organic substituent attached directly to nitrogen rather than to a numbered carbon.
Name a primary alkanamine from a suitable parent chain, giving the amino group its correct locant: butan-2-amine puts NH₂ on carbon 2. A diamine needs both positions and retains the parent e, as in ethane-1,2-diamine.
For CH₃NHCH₂CH₃, ethanamine is the parent and the extra methyl is on N: N-methylethanamine. Adding another N-methyl gives N,N-dimethylethanamine. Aniline is an accepted name; benzenamine describes the same direct aromatic attachment.
Go deeper: Ring positions differ from N substitution
Number ring substituents relative to the NH₂-bearing carbon. 2-methylaniline has methyl on the ring; N-methylaniline has it on nitrogen and is a secondary amine. Ortho/meta/para refer to 2/3/4 ring relationships, not to the count of nitrogen substituents.
Unsaturation and multiple amino groups must remain in the parent name, for example prop-2-en-1-amine and hexane-1,6-diamine. Use the displayed skeleton to verify attachment before using a remembered common name.
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Question
A drawing has NH attached to an ethyl group and a methyl group. Where does the methyl locant belong, and why?
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Preparation of Amines: Reduction and Ammonolysis
Keep nitrogen and carbon ownership visible during preparation.Track bonds while keeping the carbon skeleton
- Ammonolysis
- Nucleophilic substitution of a suitable alkyl-halide C–X bond by ammonia, followed by deprotonation to obtain free amine.
Nitro reduction preserves the original carbon skeleton and nitrogen: RNO₂ + 3H₂ → RNH₂ + 2H₂O for catalytic hydrogenation. Metal/acid reduction needs subsequent base work-up to isolate the neutral amine from its protonated form.
Ammonia attacks a suitable alkyl or benzyl halide, making alkylammonium salt before deprotonation. The chapter uses ethanolic ammonia at 373 K in a sealed tube. The product amine is itself nucleophilic, so further alkylation can give secondary, tertiary and quaternary products.
Go deeper: Excess reagent changes the mixture, not the carbon accounting
Large excess ammonia favours primary amine; it does not erase all competing alkylation. For comparable substrates the source leaving-group comparison is RI > RBr > RCl, rather than an unconditional rate ranking across all structures.
LiAlH₄ or suitable catalytic hydrogenation reduces R–C≡N to R–CH₂–NH₂. The nitrile carbon stays in the product. A preceding reaction RX + CN⁻ adds one carbon to the alkyl skeleton; nitrile reduction alone conserves the total carbon count.
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Question
A student draws CH₃–C≡N reducing to CH₃–NH₂. Which atom has been lost incorrectly?
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Preparation of Amines: Specialized Syntheses
Distinguish carbon retention, protected-N substitution and carbon shortening.Three syntheses, three different carbon outcomes
- Hofmann bromamide degradation
- Conversion of a primary amide to a primary amine with one fewer carbon using bromine and aqueous/ethanolic alkali.
LiAlH₄ reduction of an amide retains the carbonyl carbon as a CH₂ group. Hofmann degradation instead migrates the attached alkyl/aryl group from carbonyl carbon to nitrogen, removing that carbon from the amine skeleton.
Gabriel synthesis deprotonates phthalimide, alkylates the imide nitrogen with a suitable alkyl halide and hydrolyses to release a primary amine. Ordinary aryl halides do not undergo this substitution, so this route does not directly make aniline.
Go deeper: Follow the removed carbon
An isocyanate R–N=C=O is a useful skeleton to see the migration in Hofmann degradation. It is not the complete mechanism; N-bromination and base-dependent rearrangement precede it. Aqueous alkaline work-up releases amine, with the removed carbon retained in carbonate.
The net equation RCONH₂ + Br₂ + 4OH⁻ → RNH₂ + CO₃²⁻ + 2Br⁻ + 2H₂O balances carbon, oxygen, H and charge. A three-carbon amide gives a two-carbon primary amine here; its ordinary carbonyl reduction gives a three-carbon amine.
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Question
A three-carbon primary amide is split between two preparations, LiAlH₄ reduction and Hofmann degradation. Compare the amine carbon counts.
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Physical Properties of Amines
Separate donor ability, acceptor ability and hydrophobic size.See what can make a hydrogen bond
- Hydrogen-bond donor
- A group with a suitably polar H bond that can form a hydrogen-bond contact to an acceptor lone-pair region.
| Selected compound class | Source boiling point / K |
|---|---|
| primary amine | 350.8 |
| secondary amine | 329.3 |
| tertiary amine | 310.5 |
| alkane | 300.8 |
| alcohol | 390.3 |
Primary and secondary neutral amines contain N–H donors and a nitrogen lone-pair acceptor. Tertiary neutral amines lack N–H donors but still accept hydrogen bonds from water. Absence of self N–H···N bonding does not mean absence of all intermolecular attraction.
The matched source comparison gives 350.8, 329.3 and 310.5 K for selected primary, secondary and tertiary isomers, versus 300.8 K for the near-mass alkane and 390.3 K for the alcohol. These values describe particular molecules, not a universal class rule.
Go deeper: Solubility and storage
Lower amines can interact strongly with water; increasing the hydrophobic carbon region can reduce solubility despite an acceptor N. Protonation changes a neutral amine into an ionic salt, so neutral-water solubility and salt solubility are different comparisons.
The chapter describes lower aliphatic amines as gases and many higher members as liquids/solids, with dependence on structure and conditions. Aniline may darken on storage through atmospheric oxidation; a coloured sample does not by itself identify a different pure compound.
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Question
A tertiary amine has no N–H. Can a water molecule donate a hydrogen bond to it? Explain the two atoms involved.
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Basic Character of Amines
Compare measured aqueous data before choosing a basicity rule.Availability of a lone pair is only part of basicity
- pKb
- The negative base-10 logarithm of Kb; a lower pKb means a larger base-dissociation constant under the same conditions.
For B + H₂O ⇌ BH⁺ + OH⁻, Kb=[BH⁺][OH⁻]/[B] with solvent water omitted. Lower pKb means stronger base in the stated solvent and conditions. It does not give the pH without concentration and equilibrium information.
The source methyl series has pKb 3.38, 3.27 and 4.22 for primary, secondary and tertiary, with NH₃ 4.75. The ethyl values are 3.29, 3.00 and 3.25. Therefore aqueous methyl order is 2° > 1° > 3° > NH₃, while ethyl is 2° > 3° > 1° > NH₃.
Go deeper: Competing stabilisations and aromatic attachment
Alkyl +I effects favour protonation, while hydration and steric access affect the relative stability of amine and ammonium ion in water. The source gas-phase trend differs from the aqueous series. A hydration sketch of N–H donors is one contribution, not the final total ranking.
Aniline lone-pair conjugation stabilises neutral amine relative to its protonated form; benzylamine has a CH₂ spacer. Do not rank basicity by simply counting resonance drawings. Electron-releasing/withdrawing ring substituents and position matter. Acid/base conversion can move an amine between aqueous salt and neutral organic phases.
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Question
Two aqueous bases have pKb 3.27 and 4.22. Which has the larger Kb, and by what ratio?
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Chemical Reactions: Acylation and Carbylamine Test
Compare covalent acylation with proton-transfer salt formation.N–H count changes the reaction outcome
- Acylation of an amine
- Replacement of an N–H by an acyl group under suitable acyl-substitution conditions, producing an amide.
Primary and secondary amines can form amides with suitable acid chloride, anhydride or ester conditions. Base handles generated acid. Benzoylation uses a benzoyl group, retaining its carbonyl carbon; tertiary amines have no N–H for this stable-amide substitution.
A carboxylic acid and amine ordinarily give an ammonium carboxylate salt at room temperature, not spontaneous amide formation. Primary amines alone give the carbylamine test with CHCl₃/ethanolic KOH on heating, producing R–N≡C isocyanide rather than R–C≡N nitrile.
Go deeper: Nitrous acid and connectivity
Primary aromatic amines in cold nitrous-acid conditions give aromatic diazonium salts. Primary aliphatic diazonium species are unstable and evolve N₂; the chapter describes net alcohol formation, though rearrangement/other products can complicate real aliphatic cases.
Do not assign the primary-amine nitrogen-evolution route to every secondary/tertiary amine. The chapter states these classes react differently without deriving all products. Formal N⁺/C⁻ in isocyanide helps distinguish connectivity; the test is conceptual, not an instruction to smell reaction products.
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Question
A proposed acylation product has lost its carbonyl oxygen and is labelled as an amide. What bond must the amide retain?
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Hinsberg Test and Electrophilic Substitution
Use N–H availability to predict the test and protect an activated ring.Separate amine classes and control ring substitution
- Hinsberg test
- Distinction of amine classes by sulphonamide formation and the products’ response to alkali and acid.
Primary amine gives a sulphonamide retaining N–H; alkali can deprotonate it into a soluble salt and acidification restores insoluble neutral derivative. Secondary amine gives a derivative without N–H, insoluble in alkali. Tertiary amine gives no sulphonamide but dissolves in acid as an ammonium salt.
Aniline strongly activates ortho/para positions. Br₂ water gives 2,4,6-tribromoaniline. Acetyl protection reduces activation; ring substitution gives mainly para product, and hydrolysis restores the amine. Major product does not imply a single isomer.
Go deeper: Acidic conditions change the directing group
Direct strongly acidic nitration protonates NH₂ to meta-directing NH₃⁺ and may give oxidation products along with multiple nitro isomers. Protection permits more controlled substitution. Benzene sulphonyl chloride is the traditional reagent; the chapter also describes p-toluene sulphonyl chloride.
Concentrated sulphuric acid gives anilinium hydrogensulphate before heated conversion at 453–473 K to mainly sulphanilic acid, represented as the zwitterion. Aniline coordination/protonation in Lewis-acid conditions suppresses ordinary Friedel–Crafts alkylation/acylation; it is not failure of every electrophilic ring reaction.
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Question
An amine gives no sulphonamide in a Hinsberg scheme but its unreacted material dissolves after acid addition. Which class fits and what causes dissolution?
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Diazonium Salts: Preparation and Properties
Identify both nitrogen atoms and keep the counterion separate.Keep aromatic diazotisation cold
- Diazotisation
- Conversion of a primary aromatic amine into an aromatic diazonium salt using nitrous acid generated in cold acidic solution.
NaNO₂ and HCl generate nitrous acid in situ. At 273–278 K, primary aromatic amine gives ArN₂⁺Cl⁻. One nitrogen comes from the amine and the second from nitrous acid; the ring–N≡N unit is covalent while Cl⁻ is a separate counterion.
Aromatic diazonium salts persist briefly in cold solution and are normally used promptly. Warming aqueous solution promotes hydrolysis to phenol with N₂ release. Temperature labels should not be read as a universal sharp threshold or an exact kinetic lifetime.
Go deeper: Counterion and stability are conditional
The chapter contrasts water-soluble benzenediazonium chloride with less-water-soluble fluoroborate and its relative room-temperature stability. These descriptions do not mean all diazonium salts have equal handling/storage behaviour; dry diazonium salts can be unstable.
Aromatic resonance supports relative stability compared with alkyl diazonium species. A diagram must not show a free nitrogen lone pair on a positively charged triple-bonded diazonium N, nor make a covalent N–Cl bond simply because chloride appears in the salt name.
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Question
A salt diagram draws ring–N≡N–Cl with every connection covalent. Repair the chloride attachment and identify its charge.
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Chemical Reactions of Diazonium Salts
Decide whether a route releases N₂ or retains both N atoms.Release N₂ or retain both nitrogen atoms
- Azo linkage
- A nitrogen–nitrogen double bond joining two organic groups; aromatic azo products can have extended conjugation.
| Selected product | Source route |
|---|---|
| ArCl / ArBr / ArCN | CuCl / CuBr / CuCN |
| ArI | KI |
| ArF | fluoroborate, then heat |
| ArOH | warm aqueous hydrolysis |
| ArH | H₃PO₂/H₂O or ethanol |
| ArNO₂ | fluoroborate, NaNO₂/Cu, heat |
| Ar–N=N–Ar′ | azo coupling; both N atoms retained |
Diazonium replacement releases N₂. CuCl/CuBr/CuCN support Sandmeyer routes; corresponding halogen acid/Cu powder gives the source Gattermann alternative. KI introduces I, fluoroborate formation then heating introduces F, warm water gives OH, and suitable reducing agents replace the group with H.
For NO₂, the chapter uses diazonium fluoroborate with aqueous NaNO₂/Cu and heat. Azo coupling instead retains both nitrogens in Ar–N=N–Ar′: phenol in alkaline medium or aniline under suitable controlled acidic conditions gives mainly para-linked products.
Go deeper: Conjugation and synthesis
Two joined aromatic systems can extend conjugation, supporting coloured azo compounds. A coloured swatch is an illustrative product cue; without data it is not an absorption spectrum, exact wavelength or proof of a specific dye shade under all conditions.
The route nitrobenzene → aniline → diazonium → benzonitrile → benzoic acid uses different reagents at each operation. Cyanide supplies the new carboxyl carbon. The textbook’s broad claim that aryl F/I cannot be made by direct halogenation should be read as the ordinary routes considered here, not a prohibition of every possible synthetic method.
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Question
Compare the nitrogen count retained in the aromatic product after hydroxyl replacement and after azo coupling.
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