Amines

Written and reviewed by Akanksha Tiwari, Physics & Chemistry faculty at UniDrill (M.Sc. Physics, Banaras Hindu University · 8 years teaching Class XI–XII science). Based on NCERT Chemistry Class XII, Chapter 9.

Why does old fish smell, and why does lemon juice fix it?

The smell of stale fish comes from amines, mainly trimethylamine. Amines are volatile bases, so they drift into your nose as gases. Squeeze lemon on the fish and the citric acid turns each amine into an ammonium salt, which is ionic and does not evaporate. The smell disappears.

That one fact carries the whole chapter. Every amine has a nitrogen atom with a lone pair of electrons. The lone pair makes amines basic (it grabs H⁺), nucleophilic (it attacks electron poor carbons), and the way it is delocalised or pushed around explains almost every trend you will be asked about.

Structure and classification

Nitrogen in an amine is sp³ hybridised. Three of the four sp³ orbitals form bonds and the fourth holds the lone pair, so the shape is pyramidal. The lone pair repels the bonding pairs a little more than they repel each other, which squeezes the angle slightly below the tetrahedral 109.5°. In trimethylamine the C–N–C angle is 108°.

Amines are classified by how many hydrogens of NH₃ have been replaced by alkyl or aryl groups.

An amine is simple if all the groups on nitrogen are identical, as in (C₂H₅)₂NH, and mixed if they differ, as in CH₃NHC₂H₅. Watch the position of the NH₂: in aniline (C₆H₅NH₂) it is on the ring, but in benzylamine (C₆H₅CH₂NH₂) it is on the side chain, so benzylamine behaves like an aliphatic amine.

Beyond the textbook. For the exam: nitrogen in amines is pyramidal. In reality: the pyramid flips inside out like an umbrella in a storm, billions of times a second. Because of this fast inversion, an amine with three different groups on nitrogen cannot be separated into mirror image forms at room temperature, even though it looks chiral on paper. Why it matters: it is why carbon is the usual centre of chirality in this course and nitrogen is not.

Naming amines

In IUPAC names a primary amine drops the final "e" of the alkane and adds amine (CH₃CH₂NH₂ is ethanamine). For secondary and tertiary amines, the longest chain is the parent and the other groups are named as substituents on nitrogen with the locant N.

Worked example: naming a mixed amine. Name CH₃CH₂CH₂NHCH(CH₃)₂.

The common wrong step: treating the whole formula as "propylisopropylamine" in an IUPAC question. That is the common-style name, and CUET asks for the IUPAC one.

Making amines: six routes

C₆H₅NO₂ + 3H₂ → C₆H₅NH₂ + 2H₂O (Ni, Pd or Pt)

RCONH₂ + Br₂ + 4NaOH → RNH₂ + Na₂CO₃ + 2NaBr + 2H₂O

Why ammonolysis is messy. The first product, RNH₂, is a better nucleophile than NH₃ because the alkyl group pushes electron density onto nitrogen. It therefore attacks another R–X before the NH₃ does, giving R₂NH, then R₃N, and finally the quaternary salt R₄N⁺X⁻ (Hofmann's exhaustive alkylation). A large excess of NH₃ favours the primary amine, and an excess of R–X pushes everything to the quaternary salt.

Gabriel synthesis: how it avoids the mess.

It works only for primary alkyl halides, and aromatic primary amines cannot be made this way because aryl halides do not undergo nucleophilic substitution with the phthalimide anion.

Hofmann bromamide: what happens to the carbon.

Worked example: counting carbons. Convert (a) CH₃CH₂Br into CH₃CH₂CH₂NH₂ and (b) CH₃CONH₂ into CH₃NH₂.

The common wrong step: swapping the two. Hofmann removes a carbon, whereas nitrile reduction keeps the carbon of the CN group, so the chain gets one carbon longer than the halide it came from.

Pause and predict: can the Gabriel synthesis be used to make aniline from bromobenzene?

Check your answer: no. Bromobenzene will not undergo SN2 with the phthalimide anion because the carbon bearing Br is sp² and part of the ring. Make aniline by reducing nitrobenzene, or by Hofmann degradation of benzamide.

Physical properties

Methylamine, dimethylamine, trimethylamine and ethylamine are gases at room temperature, the higher aliphatic amines are liquids, and the lower ones smell of fish. Aniline is a colourless liquid that turns brown on standing because air slowly oxidises it. The first few amines dissolve readily in water because the nitrogen forms hydrogen bonds with water, and solubility falls as the hydrocarbon part grows.

Boiling point depends on N–H bonds. A primary amine has two N–H bonds and can hydrogen bond strongly with its neighbours, a secondary amine has one, and a tertiary amine has none. For isomers: 1° > 2° > 3°. Alcohols boil higher than amines of similar mass because O is more electronegative than N, so O–H···O bonds are stronger than N–H···N bonds.

The tertiary amine boils almost where the alkane of similar mass does, because it has no N–H and so no hydrogen bonding with its own kind.

Basicity: the idea behind half the questions

An amine accepts H⁺ using its lone pair. In water, the reaction is R–NH₂ + H₂O ⇌ R–NH₃⁺ + OH⁻, and the strength is measured by $K_b$ or by the pKb (smaller pKb, stronger base):

$$\text{p}K_b = -\log K_b$$

The base is only as strong as its lone pair is available, and the conjugate acid R–NH₃⁺ must also be stable.

Aliphatic amines beat ammonia. An alkyl group pushes electrons towards nitrogen (+I effect), so the lone pair is richer and the ammonium ion that forms is stabilised. Aromatic amines lose to ammonia, because in aniline the lone pair is delocalised into the ring and is less available.

Why gas phase and water give different orders. In the gas phase only the +I effect acts, so more alkyl groups mean a stronger base: 3° > 2° > 1° > NH₃, for example (CH₃)₃N > (CH₃)₂NH > CH₃NH₂ > NH₃. In water, two more factors enter. The ammonium ion is stabilised by hydrogen bonding with water, and the more N–H bonds it has the more it is stabilised (1° has three, 2° has two, 3° has one). Crowding from large alkyl groups also hinders that solvation. The three effects compromise, so the aqueous order is irregular.

In both series the secondary amine is the strongest, and the tertiary amine is hurt by poor solvation.

Why aniline is a weak base. The lone pair on nitrogen overlaps with the ring π system. Aniline has five resonance structures (two Kekulé forms and three with a negative charge on the ortho and para carbons), while the anilinium ion, formed when aniline takes H⁺, has only two (the Kekulé forms), because the lone pair is now tied up in the N–H bond. A species with more resonance structures is more stable, so protonation of aniline is uphill, and aniline is a much weaker base than ammonia.

So p-toluidine > aniline > p-nitroaniline.

Worked example: ranking basic strength. Arrange in increasing order of basic strength: C₂H₅NH₂, (C₂H₅)₂NH, (C₂H₅)₃N, C₆H₅NH₂.

The common wrong step: ranking by the number of alkyl groups alone, which gives the gas phase order. In water the tertiary amine is not the strongest.

Worked example: how much stronger? Compare methylamine (pKb 3.38) with aniline (pKb 9.38).

Reactions of amines

Alkylation and acylation. Amines act as nucleophiles. With alkyl halides they are alkylated step by step to the quaternary salt. With acid chlorides, anhydrides and esters, 1° and 2° amines (those with an N–H) are acylated to give amides, in the presence of a base such as pyridine that removes the HCl and drives the reaction on.

R–NH₂ + CH₃COCl → R–NHCOCH₃ + HCl

Tertiary amines cannot be acylated because they have no N–H to lose. Benzoylation with C₆H₅COCl in the presence of NaOH is called the Schotten–Baumann reaction.

Tests that sort 1°, 2° and 3° amines.

R–NH₂ + CHCl₃ + 3KOH → R–NC + 3KCl + 3H₂O

R–NH₂ + HNO₂ → R–OH + N₂ + H₂O (aliphatic 1°)

R₂NH + HNO₂ → R₂N–NO + H₂O (2°)

Carbylamine works because only a primary amine has two N–H bonds to lose, and the dichlorocarbene formed from CHCl₃ and base removes both. Hinsberg works because the sulphonamide of a primary amine, C₆H₅SO₂NHR, still has an N–H made acidic by the strongly electron withdrawing SO₂, so alkali dissolves it as a sodium salt. The sulphonamide of a secondary amine, C₆H₅SO₂NR₂, has no N–H and stays undissolved.

Pause and predict: you add NaNO₂ and dilute HCl to ethylamine and to aniline, both at 0 to 5 °C. In which case do you see gas bubbling?

Check your answer: ethylamine. The aliphatic diazonium ion loses N₂ at once, giving ethanol. The aromatic diazonium ion from aniline is stabilised by the ring and stays dissolved.

Aniline: powerful ring activation, and how to tame it

The nitrogen lone pair floods electron density onto the ortho and para positions of the ring, so –NH₂ is a strong activating, ortho and para directing group.

C₆H₅NH₂ + 3Br₂ → C₆H₂Br₃NH₂ + 3HBr

Taming aniline: the acetanilide route. Acetylation turns –NH₂ into –NHCOCH₃. The lone pair is now partly delocalised towards the C=O, so it activates the ring less, and the bulky group blocks the ortho positions. Substitution then gives mainly the para product, and hydrolysis restores the amine.

The route in one line: aniline → (CH₃CO)₂O, pyridine → acetanilide → Br₂ in CH₃COOH → p-bromoacetanilide → H₃O⁺ → p-bromoaniline

The same protection trick converts aniline cleanly into p-nitroaniline, avoiding the tar and meta product of direct nitration.

Pause and predict: what do you get when aniline is shaken with bromine water, and what do you get if you want only p-bromoaniline?

Check your answer: bromine water gives 2,4,6-tribromoaniline at once. For p-bromoaniline, protect the amine as acetanilide, brominate, then hydrolyse.

Diazonium salts: the master key

Arenediazonium salts have the form Ar–N₂⁺X⁻. They are made by diazotisation: a primary aromatic amine reacts with nitrous acid (made in the flask from NaNO₂ and HCl) at 273–278 K.

C₆H₅NH₂ + NaNO₂ + 2HCl → C₆H₅N₂⁺Cl⁻ + NaCl + 2H₂O

The cold is essential. Above about 5 °C the salt decomposes. Benzenediazonium chloride is a colourless, water soluble salt that is stable in cold solution, but dry diazonium salts can explode, so they are always used immediately in solution. The fluoroborate (BF₄⁻) salt is insoluble in water and stable at room temperature. An aromatic diazonium ion survives because its positive charge is spread over the ring by resonance, whereas an aliphatic one has no such help and loses N₂ immediately.

Replacement reactions (N₂ is lost, a new group takes its place).

C₆H₅N₂⁺Cl⁻ + KI → C₆H₅I + KCl + N₂

C₆H₅N₂⁺BF₄⁻ → C₆H₅F + BF₃ + N₂ (heat)

C₆H₅N₂⁺Cl⁻ + H₃PO₂ + H₂O → C₆H₆ + N₂ + H₃PO₃ + HCl

Why this matters. Direct halogenation cannot put F or I on a ring cleanly, and chlorobenzene will not undergo nucleophilic substitution with CN⁻. The diazonium route introduces F, Cl, Br, I, CN, OH and H easily, so aniline becomes the gateway to a huge range of aromatic compounds.

Coupling reactions: the diazonium ion keeps its N₂. Here the diazonium ion acts as a weak electrophile and attacks the para position of a strongly activated ring, giving an azo compound (Ar–N=N–Ar′). The extended conjugation across –N=N– makes these compounds coloured, so they are azo dyes.

The pH is chosen for a reason. Phenol needs base so that it becomes the far more reactive phenoxide ion. Aniline needs a little acid so that the diazonium ion survives, but not too much, or the amine would be protonated and deactivated. Nitrobenzene and benzene do not couple, because their rings are not activated enough to attack the weakly electrophilic diazonium ion.

Beyond the textbook. For the exam: coupling with phenol and aniline gives orange and yellow azo dyes. In reality: most synthetic dyes are azo compounds, and so is the indicator methyl orange, which is made by coupling diazotised sulphanilic acid (the product of aniline sulphonation above) with N,N-dimethylaniline. Why it matters: the sulphonic acid group makes the dye water soluble, and the colour changes with pH because protonation alters the conjugated azo system.

Worked example: identify the amine. An amine C₃H₉N gives a foul smell with CHCl₃ and alcoholic KOH. With NaNO₂ and HCl it evolves N₂ and forms an alcohol that gives a yellow precipitate with I₂ and NaOH. What is it?

The common wrong step: concluding propan-1-amine. Its alcohol, propan-1-ol, does not give the iodoform test.

Reagent to product: quick recognition

Exam traps

Reaction sheet

60-second revision

Where this chapter connects

Ammonolysis and Gabriel are SN2 reactions of alkyl halides, so Haloalkanes and Haloarenes explains why aryl halides refuse. The amides used in Hofmann degradation and the carbonyl compounds that form imines and hydrazones come from Aldehydes, Ketones and Carboxylic Acids. Aliphatic amines turn into alcohols and aryldiazonium salts into phenols, which links to Alcohols, Phenols and Ethers. The inductive and resonance arguments behind every basicity order are introduced in Organic Chemistry: Some Basic Principles and Techniques, and amino acids, which carry an amine group, appear next in Biomolecules.