Organic compounds containing Nitrogen | ISC Class 12 Chemistry Notes

This note covers the classification, structure and naming of amines; their preparation, physical properties, basicity and reactions; identification tests; aniline; cyanides and isocyanides; and the preparation, reactions and synthetic uses of diazonium salts.

What are amines, and how are they classified?

Definition: Amines are derivatives of ammonia, NH₃, in which one, two or three hydrogen atoms are replaced by alkyl or aryl groups.

An alkyl group is obtained by removing a hydrogen atom from an alkane. An aryl group is obtained by removing a hydrogen atom from an aromatic ring. Here, R represents an alkyl group and Ar an aryl group; R′ and R″ represent additional organic groups.

The symbols C, H, N and O denote carbon, hydrogen, nitrogen and oxygen. A subscript counts atoms; a superscript + or − indicates ionic charge. A single, double or triple bond is written as -, = or ≡. An arrow → means “forms”; ⇌ indicates equilibrium.

Count the organic groups bonded directly to nitrogen, rather than the total carbon atoms. Amines are simple when their alkyl or aryl groups are identical and mixed when these groups differ. A quaternary ammonium ion has four organic groups attached to positively charged nitrogen.

What gives an amine its shape?

A lone pair is a pair of electrons not shared in a bond. Nitrogen in an alkylamine forms three bonds and retains a lone pair. Its sp³ hybridisation means that one s orbital and three p orbitals combine to give four hybrid orbitals.

The molecular shape is pyramidal. Lone-pair repulsion reduces the bond angle below the tetrahedral angle of 109.5°. Trimethylamine has a bond angle of 108°; the degree sign denotes angular measure. These values describe bond geometry, not temperature.

The drawing shows three CH₃ groups around nitrogen, a lobe labelled unshared electron pair above it, and the marked 108° bond angle.

See Fig. 9.1 in your NCERT textbook

How are amines named and their structures recognised?

Common names combine the alkyl-group name with “amine”. The prefixes di- and tri- indicate two and three identical groups. Thus CH₃NH₂ is methylamine and (CH₃)₃N is trimethylamine. Common names describe the attached groups directly.

IUPAC means International Union of Pure and Applied Chemistry. In its naming system, a primary amine is named by replacing the final “e” of the parent alkane with “amine”. A numbered position, or locant, identifies the carbon bearing the amino group, -NH₂.

The prefix N- places a substituent on nitrogen, rather than on a carbon atom of the parent chain. In N-methylethanamine, the parent is ethanamine and the extra methyl group, CH₃-, is bonded to nitrogen. The compound is therefore secondary.

For several amino groups, retain the “e” of the parent hydrocarbon and use the appropriate suffix. H₂NCH₂CH₂NH₂ is ethane-1,2-diamine: its two amino groups occupy carbon positions 1 and 2.

How does an arylamine differ from a side-chain amine?

In aniline, the amino group is directly attached to the benzene ring, the six-carbon aromatic ring. The C₆H₅- unit is called phenyl. In benzylamine, C₆H₅CH₂NH₂, a CH₂ group separates nitrogen from the ring. This distinction matters when explaining basicity.

Ring locants identify positions relative to the amino group. The terms ortho, meta and para, abbreviated o-, m- and p-, mean the 2-, 3- and 4-positions respectively. Thus o-toluidine is 2-methylaniline, while p-bromoaniline is 4-bromoaniline.

How are amines prepared from alcohols, halides and nitro compounds?

Different preparations replace a functional group, meaning the atom or group responsible for characteristic reactions. A reagent brings about a reaction; a catalyst changes its rate without being consumed overall. Conditions belong with the reaction, since changing them may change the product.

How do alcohols give amines?

Passing alcohol and ammonia vapours over heated aluminium oxide, Al₂O₃, gives amines. The overall primary-amine formation can be represented as ROH + NH₃ → RNH₂ + H₂O. ROH denotes an alcohol, with a hydroxyl group, -OH, attached to R.

Further substitution can produce secondary and tertiary amines. Excess ammonia favours the primary product. The transformation replaces the alcohol's hydroxyl group with an amino group, while retaining its carbon framework.

What happens during ammonolysis?

Ammonolysis is cleavage of the carbon-halogen bond by ammonia. A halogen is an element such as chlorine, bromine or iodine, represented here by X; their symbols are Cl, Br and I. An alkyl halide, RX, reacts with ethanolic ammonia in a sealed tube at 373 K.

K is kelvin, the unit of absolute temperature. Ethanolic means dissolved in ethanol. The first product is an alkylammonium salt, RNH₃⁺X⁻. Treatment with sodium hydroxide, NaOH, releases the free amine: RNH₃⁺X⁻ + NaOH → RNH₂ + H₂O + NaX.

A nucleophile donates an electron pair to form a bond. The primary amine remains nucleophilic, so further reaction with RX forms secondary and tertiary amines and a quaternary ammonium salt. Large excess ammonia makes the primary amine the major product. Reactivity follows RI > RBr > RCl.

How are nitro compounds reduced?

A nitro group is -NO₂. Reduction, here conversion by addition of hydrogen or removal of oxygen, changes it into -NH₂. Hydrogen gas, H₂, with finely divided nickel, palladium or platinum provides catalytic reduction; metals in acidic medium also reduce nitro compounds.

C₆H₅NO₂ + 3H₂ → C₆H₅NH₂ + 2H₂O, with a suitable metal catalyst, converts nitrobenzene into aniline. Iron and hydrochloric acid, HCl, are another route. Iron(II) chloride formed in that route releases acid on hydrolysis, so only a small amount of acid is needed to initiate it.

How do nitriles, amides and Gabriel synthesis give amines?

A nitrile, also called an organic cyanide, contains R-C≡N. An amide contains a carbonyl group bonded to nitrogen; a carbonyl group is C=O. These starting materials give different carbon-count outcomes depending on the reaction chosen.

Which reductions retain the carbon atoms?

Nitriles are reduced by lithium aluminium hydride, LiAlH₄, or catalytic hydrogenation to primary amines. The conversion is R-C≡N → R-CH₂NH₂. The nitrile carbon becomes the carbon of CH₂, so it remains in the product.

Amide reduction with LiAlH₄ similarly retains the carbonyl carbon: RCONH₂ → RCH₂NH₂. Do not confuse this reaction with bromamide degradation. The same amide can give a different carbon skeleton if a different reagent is used.

Worked example 1. Convert chloroethane into propan-1-amine using potassium cyanide followed by lithium aluminium hydride.

Answer: First replace chlorine with cyanide: $\mathrm{CH_3CH_2Cl}\xrightarrow{\mathrm{KCN}}\mathrm{CH_3CH_2CN}$. Propanenitrile contains three carbon atoms because cyanide substitution adds one carbon to the two-carbon starting compound.

Then reduce the nitrile: $\mathrm{CH_3CH_2CN}\xrightarrow{\mathrm{LiAlH_4}}\mathrm{CH_3CH_2CH_2NH_2}$. The nitrile carbon becomes the carbon of the terminal amino-bearing group, so propan-1-amine retains all three carbon atoms.

Worked example 2. Convert benzyl chloride into 2-phenylethanamine.

Answer: First treat benzyl chloride with potassium cyanide: $\mathrm{C_6H_5CH_2Cl}\xrightarrow{\mathrm{KCN}}\mathrm{C_6H_5CH_2CN}$. The cyanide group extends the side chain by one carbon.

Reduce the nitrile with lithium aluminium hydride: $\mathrm{C_6H_5CH_2CN}\xrightarrow{\mathrm{LiAlH_4}}\mathrm{C_6H_5CH_2CH_2NH_2}$. The product is 2-phenylethanamine. The starting compound has seven carbon atoms and the product has eight; reduction retains the carbon introduced during substitution.

Why does Hofmann degradation shorten the chain?

Hofmann bromamide degradation treats a primary amide with bromine, Br₂, and aqueous or ethanolic NaOH. An alkyl or aryl group migrates from the carbonyl carbon to nitrogen. The resulting primary amine contains one carbon atom fewer than the amide.

RCONH₂ + Br₂ + 4NaOH → RNH₂ + 2NaBr + Na₂CO₃ + 2H₂O. NaBr is sodium bromide and Na₂CO₃ is sodium carbonate. The carbon lost from the organic product appears in carbonate.

Worked example 3. Which amide gives propan-1-amine by Hofmann degradation, and which amine does benzamide give?

Answer: Propan-1-amine has three carbon atoms. Hofmann degradation removes the carbonyl carbon from the organic skeleton, so the starting amide must contain four carbons. It is butanamide, $\mathrm{CH_3CH_2CH_2CONH_2}$.

Benzamide, $\mathrm{C_6H_5CONH_2}$, has seven carbon atoms. Removing its carbonyl carbon leaves the six-carbon phenyl group bonded to the amino group, giving aniline, $\mathrm{C_6H_5NH_2}$. Both conversions use bromine and aqueous or ethanolic sodium hydroxide.

What is the sequence in Gabriel synthesis?

1. Treat phthalimide, a cyclic imide containing an N-H group between two carbonyl groups, with ethanolic potassium hydroxide, KOH.
2. This produces potassium phthalimide, whose nitrogen-containing anion can act as a nucleophile.
3. Heat the salt with an alkyl halide to attach the alkyl group to nitrogen.
4. Carry out alkaline hydrolysis, bond cleavage using water in alkaline medium, to release the corresponding primary amine.

Gabriel phthalimide synthesis prepares primary alkylamines. Aromatic primary amines cannot be prepared by this method because aryl halides do not undergo the required nucleophilic substitution with the phthalimide anion.

How does hydrogen bonding affect physical properties?

A hydrogen bond is an attraction involving hydrogen bonded to an electronegative atom and a lone pair on another electronegative atom. Electronegativity is an atom's tendency to attract bonding electrons. Hydrogen bonding helps explain both water solubility and boiling-point comparisons.

Lower aliphatic amines are gases with a fishy odour. Primary amines with three or more carbon atoms are liquids, and still higher members are solids. Aniline and other arylamines are usually colourless but become coloured during storage through atmospheric oxidation, a reaction involving loss of electrons to an oxidising agent such as oxygen.

Why does water solubility decrease along the series?

Lower aliphatic amines dissolve in water because they form hydrogen bonds with water molecules. Increasing molar mass, the mass of one mole of substance, enlarges the hydrophobic or water-repelling alkyl portion. Solubility consequently decreases; higher amines are essentially insoluble in water.

A mole is the chemical unit for amount of substance. Aniline is only slightly soluble in water. Amines dissolve in organic solvents such as alcohol, ether and benzene. Alcohols are more polar and form stronger intermolecular hydrogen bonds than comparable amines.

How do the three classes compare?

Primary amines have two N-H hydrogens available for association between molecules; secondary amines have one. Tertiary amines have none, so they cannot form this kind of hydrogen-bonded association with one another. They can still accept hydrogen bonds from water.

Several RNH₂ units are linked by dotted lines between the nitrogen of one molecule and an N-H hydrogen of another. The solid bonds remain within each molecule.

See Fig. 9.2 in your NCERT textbook

The prefix n- identifies an unbranched chain. In this comparison, the primary, secondary and tertiary amines have successively lower boiling points. The alcohol boils above these amines; the alkane boils below them. Compare similar molecular masses before attributing a difference to hydrogen bonding.

Why are amines basic, and what controls their basic strength?

A base accepts a proton, H⁺; a Lewis base donates an electron pair. An amine's nitrogen lone pair binds a proton to form a substituted ammonium ion. Reaction with acid forms a salt, and treatment of the salt with a strong base regenerates the amine.

For a primary amine in water, RNH₂ + H₂O ⇌ RNH₃⁺ + OH⁻. OH⁻ is the hydroxide ion. RNH₃⁺ is the conjugate acid, meaning the species formed when the base accepts a proton.

Derivation: How is the base dissociation constant obtained?

The base dissociation constant, $K_b$, measures the equilibrium of a primary amine with water. Square brackets denote equilibrium concentrations, and $K$ is the equilibrium constant before the effectively constant concentration of water is incorporated.

1. Write the equilibrium expression: $K=\frac{[\mathrm{RNH_3^+}][\mathrm{OH^-}]}{[\mathrm{RNH_2}][\mathrm{H_2O}]}$.
2. Multiply by the water concentration: $K[\mathrm{H_2O}]=\frac{[\mathrm{RNH_3^+}][\mathrm{OH^-}]}{[\mathrm{RNH_2}]}$.
3. Incorporate the effectively constant water concentration into the base dissociation constant, giving $K_b=\frac{[\mathrm{RNH_3^+}][\mathrm{OH^-}]}{[\mathrm{RNH_2}]}$.

Result: A larger $K_b$ indicates a stronger base under comparable conditions.

$pK_b=-\log_{10}K_b$, where $\log_{10}$ is the logarithm to base ten. A smaller $pK_b$ means a stronger base. Compare values under the same conditions; an aqueous order must not be substituted for a gas-phase order.

Why does the medium change the order?

The positive inductive effect, written +I, is electron donation through bonds by an alkyl group. It increases electron density on nitrogen and helps stabilise the positively charged ammonium ion. In the gaseous phase, basicity follows tertiary > secondary > primary > ammonia.

In water, solvation, stabilisation by surrounding solvent molecules, also matters. Hydrogen bonding stabilises the ammonium ions. Steric hindrance, obstruction caused by surrounding groups, affects that interaction. The aqueous order therefore reflects the combined inductive, solvation and steric effects.

Ammonia has pKᵦ 4.75. In water, the methyl series follows (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃. The ethyl series follows (C₂H₅)₂NH > (C₂H₅)₃N > C₂H₅NH₂ > NH₃.

Worked example 4. Arrange aniline, ethylamine, diethylamine and ammonia in decreasing order of basic strength in water.

Answer: Compare their $pK_b$ values: aniline 9.38, ethylamine 3.29, diethylamine 3.00 and ammonia 4.75. Smaller values indicate stronger bases, so first arrange the values as $3.00<3.29<4.75<9.38$.

The required decreasing order of basic strength is $\mathrm{(C_2H_5)_2NH}>\mathrm{C_2H_5NH_2}>\mathrm{NH_3}>\mathrm{C_6H_5NH_2}$. In water, both ethylamines are stronger bases than ammonia, while aniline is weaker.

Why is aniline weaker than ammonia?

Resonance describes electron delocalisation represented by several contributing structures. Aniline's lone pair participates in conjugation with the benzene ring, meaning interaction through adjacent orbitals. It is consequently less available for protonation, or acceptance of H⁺.

Draw five contributing structures for aniline, including charge-separated forms with positive nitrogen and negative charge at the ortho or para positions. Below them, draw the two ring structures of the anilinium ion, C₆H₅NH₃⁺.

Electron-releasing substituents such as -CH₃ and methoxy, -OCH₃, increase aniline's basic strength. Electron-withdrawing groups such as nitro, -NO₂; sulphonic acid, -SO₃H; carboxyl, -COOH; and halogens decrease it. Their effect changes the availability of nitrogen's lone pair.

How do alkylation and acylation change an amine?

Alkylation introduces an alkyl group onto nitrogen. In reaction with an alkyl halide, the nitrogen lone pair attacks the carbon bonded to the halogen. This is nucleophilic substitution: an incoming nucleophile replaces a leaving group, the group that departs with the bonding electron pair.

What are the steps in alkylation?

1. The nitrogen lone pair attacks the carbon of a suitable methyl or primary alkyl halide while the carbon-halogen bond breaks simultaneously. This single SN2 substitution step releases the halide ion and forms an alkylated ammonium ion; SN2 means bimolecular nucleophilic substitution.
2. A base removes a nitrogen-bound proton, when one is present, to give the neutral amine.
3. Further alkylation can continue until a quaternary ammonium salt forms.

For aniline, replacement of nitrogen-bound hydrogen by methyl groups gives N-methylaniline and then N,N-dimethylaniline. Excess methyl iodide can carry alkylation further to a quaternary salt. Distinguish substitution on nitrogen from substitution on the aromatic ring.

What are the steps in acylation?

Acylation replaces an N-H hydrogen by an acyl group, RCO-. Primary and secondary amines react with acid chlorides, acid anhydrides or esters to give amides. An acid chloride contains -COCl; an anhydride links two acyl groups through oxygen; an ester contains -COOR.

1. Nitrogen's lone pair attacks the carbonyl carbon of an acid chloride.
2. The carbonyl double bond opens towards oxygen, forming a tetrahedral intermediate, a temporary species with four bonds around that carbon.
3. The carbonyl bond reforms and chloride leaves.
4. Removal of a proton from nitrogen gives the amide; a base removes the acid produced.

Acetylation introduces CH₃CO-, the acetyl group. Aniline with acetic anhydride, (CH₃CO)₂O, forms acetanilide, C₆H₅NHCOCH₃, and ethanoic acid, CH₃COOH. Pyridine, a nitrogen-containing basic ring compound, can remove the HCl formed when an acid chloride is used.

Benzoylation introduces C₆H₅CO-, the benzoyl group. Aniline reacts with benzoyl chloride, C₆H₅COCl, to give N-phenylbenzamide, or benzanilide, C₆H₅NHCOC₆H₅. Tertiary amines lack the N-H hydrogen needed for this amide-forming substitution. Carboxylic acids instead form salts with amines at room temperature.

How can primary, secondary and tertiary amines be distinguished?

What does the carbylamine test identify?

The carbylamine reaction identifies primary aliphatic and aromatic amines. Heating with chloroform, CHCl₃, and ethanolic KOH gives an isocyanide, RNC, also called a carbylamine. Isocyanides are foul-smelling compounds. Secondary and tertiary amines do not show this reaction.

RNH₂ + CHCl₃ + 3KOH → RNC + 3KCl + 3H₂O, on heating in ethanolic medium. KCl is potassium chloride. A positive result establishes primary-amine behaviour; it does not by itself distinguish an alkylamine from aniline.

How is the Hinsberg test interpreted?

Hinsberg's reagent is benzenesulphonyl chloride, C₆H₅SO₂Cl. It reacts with primary and secondary amines to form sulphonamides, compounds containing the -SO₂-N- linkage. The presence or absence of an N-H hydrogen in the product controls its behaviour in alkali.

The electron-withdrawing sulphonyl group, -SO₂-, makes the retained N-H hydrogen acidic. Loss of that proton gives an ionic form soluble in alkali. Distinguish “an insoluble product forms” from “no product forms”; these observations refer to different classes.

What happens with nitrous acid?

Nitrous acid, HNO₂, is generated in situ, meaning within the reaction mixture, from sodium nitrite, NaNO₂, and a mineral acid. Primary aliphatic amines form unstable aliphatic diazonium salts, which release nitrogen gas and give alcohols.

RNH₂ + HNO₂ → ROH + N₂ + H₂O. N₂ denotes nitrogen gas. Primary aromatic amines instead give diazonium salts at 273 to 278 K. Secondary amines form N-nitrosamines, containing an N-N=O group, whereas tertiary aliphatic amines form salts without nitrogen evolution under cold acidic conditions.

Tertiary aromatic amines with a free para position can undergo nitrosation there, introducing -N=O. Do not group all tertiary amines under one product rule. The aliphatic or aromatic character and the reaction conditions are essential parts of the distinction.

How does aniline react, and why is its ring strongly activated?

Aniline is a usually colourless liquid, slightly soluble in water, with a relatively high boiling point associated with intermolecular hydrogen bonding. Its lone pair accounts for salt formation and nitrogen substitution; donation into the ring accounts for rapid aromatic substitution.

With HCl, aniline gives anilinium chloride, C₆H₅NH₃⁺Cl⁻. With concentrated sulphuric acid, H₂SO₄, it forms anilinium hydrogensulphate, C₆H₅NH₃⁺HSO₄⁻. Concentrated means a high proportion of acid relative to water.

Why does bromination give three substitutions?

An electrophile accepts an electron pair. In electrophilic aromatic substitution, an electrophile replaces a ring hydrogen. The amino group increases electron density especially at the ortho and para positions, making it a powerful activating group, one that increases the ring's reactivity.

Aniline with bromine water at room temperature gives a white precipitate, an insoluble solid, of 2,4,6-tribromoaniline. The equation is C₆H₅NH₂ + 3Br₂ → C₆H₂Br₃NH₂ + 3HBr. HBr is hydrogen bromide. This reaction helps identify aniline.

To obtain p-bromoaniline, first acetylate aniline to acetanilide, then brominate, then hydrolyse the protecting group. The acetamido group, -NHCOCH₃, donates less strongly to the ring because the nitrogen lone pair also interacts with the carbonyl group.

Why does nitration require care with the directing group?

Nitration introduces -NO₂. Direct nitration with nitric acid, HNO₃, and sulphuric acid at 288 K gives tarry oxidation products as well as nitro derivatives. In the strongly acidic medium, aniline becomes anilinium ion, which is meta directing. A significant amount of meta derivative forms alongside the ortho and para derivatives.

Acetylation protects the amino group. Nitration of the protected compound followed by hydrolysis gives p-nitroaniline as the major product. “Major” does not mean the only product. The directing behaviour must be explained for the species present in the actual medium.

What happens during sulphonation?

Sulphonation introduces -SO₃H. Heating anilinium hydrogensulphate with sulphuric acid at 453 to 473 K gives p-aminobenzenesulphonic acid, or sulphanilic acid, as the major product. Aniline is also useful as a starting material for diazonium salts and azo dyes.

How are cyanides and isocyanides prepared?

Cyanides, R-C≡N, and isocyanides, R-N≡C, differ in which atom joins the organic group. The condensed forms RCN and RNC preserve that distinction. Their preparation from halides illustrates an ambident nucleophile, a nucleophile with two possible sites of bond formation.

Why do potassium and silver cyanide give different main products?

Potassium cyanide is predominantly ionic and supplies cyanide ions, CN⁻. Attack occurs mainly through carbon, giving an alkyl cyanide as the main product. Silver cyanide, AgCN, is mainly covalent; its nitrogen can donate an electron pair, giving an isocyanide as the chief product.

For halide substitution, write RX + KCN → RCN + KX or RX + AgCN → RNC + AgX. Ag denotes silver; K denotes potassium when it occurs in a chemical formula, rather than after a numerical temperature.

How does an amide become a cyanide?

Dehydration removes the elements of water. Heating a primary amide with phosphorus pentoxide, P₂O₅, gives a nitrile. The organic transformation is RCONH₂ → RCN + H₂O; the dehydrating reagent removes the water represented in this equation.

The carbon skeleton is retained in this conversion. By contrast, forming a nitrile from an alkyl halide introduces the carbon of CN⁻. Reducing that nitrile then produces a primary amine with one more carbon atom than the original halide.

Note: Keep the order of C and N visible. A cyanide has carbon bonded to R, while an isocyanide has nitrogen bonded to R. The reagent determines the main product, rather than making the two names interchangeable.

How are diazonium salts prepared and used in replacement reactions?

An arenediazonium salt contains ArN₂⁺ and a counter-ion, the oppositely charged ion balancing its charge. For benzenediazonium chloride, write C₆H₅N₂⁺Cl⁻. The -N₂⁺ group is the diazonium group. Its replacement provides routes to many substituted aromatic compounds.

What conditions are needed for diazotisation?

Diazotisation converts a primary aromatic amine into a diazonium salt. Aniline reacts with NaNO₂ and HCl at 273 to 278 K: C₆H₅NH₂ + NaNO₂ + 2HCl → C₆H₅N₂⁺Cl⁻ + NaCl + 2H₂O. NaCl is sodium chloride.

Arenediazonium salts are stable for a short time in solution at low temperatures. They are not generally stored and are used immediately after preparation. Benzenediazonium chloride is water-soluble and decomposes easily when dry; benzenediazonium fluoroborate is water-insoluble and stable at room temperature.

Which reagents replace the diazonium group?

Copper(I) means copper in oxidation state +1; an oxidation state is the formal charge assigned by electron-counting rules. Sandmeyer uses copper(I) salts, whereas Gattermann uses copper powder. The yield in the Sandmeyer reaction is better than in the Gattermann reaction.

Each replacement releases N₂. Heating benzenediazonium fluoroborate gives fluorobenzene, nitrogen and boron trifluoride, BF₃. Reduction with H₃PO₂ gives benzene while the reagent becomes phosphorous acid, H₃PO₃. With ethanol, the ethanol becomes ethanal, CH₃CHO.

How do coupling reactions produce dyes and support organic synthesis?

Azo coupling joins a diazonium salt to an activated aromatic ring while retaining both nitrogen atoms in an azo linkage, -N=N-. The resulting compounds have an extended conjugated system, in which electron delocalisation extends across linked parts of the molecule.

How do phenol and aniline couple?

Benzenediazonium chloride couples with phenol, C₆H₅OH, in alkaline medium, containing hydroxide ions, to give p-hydroxyazobenzene, an orange dye. The phenol ring joins at its para position relative to -OH. The product connects two aromatic rings through the azo group.

Coupling with aniline in mildly acidic medium gives p-aminoazobenzene, a yellow dye. Here the joining position is para to -NH₂. These reactions are electrophilic substitutions, with the diazonium ion acting as the electrophile.

Azo compounds are often coloured and are used as dyes. Keep the nitrogen balance clear: replacement reactions expel the diazonium nitrogen as N₂, whereas coupling retains it in the azo group connecting the rings.

How is a conversion planned?

1. Identify the starting functional group and the group required in the product.
2. If starting from aniline, consider diazotisation with NaNO₂ and HCl at 273 to 278 K.
3. Select the replacement reagent or coupling partner needed for the desired structure.
4. Check the carbon skeleton, the position of ring substitution and whether nitrogen leaves or remains.

Worked example 5. Convert aniline into phenol.

Answer: First prepare benzenediazonium chloride using sodium nitrite and hydrochloric acid at 273 to 278 K: $\mathrm{C_6H_5NH_2+NaNO_2+2HCl\rightarrow C_6H_5N_2^+Cl^-+NaCl+2H_2O}$.

Then allow its aqueous solution to warm to 283 K: $\mathrm{C_6H_5N_2^+Cl^-+H_2O\rightarrow C_6H_5OH+N_2+HCl}$. The diazonium group is replaced by a hydroxyl group, giving phenol and releasing nitrogen gas.

Worked example 6. Convert aniline into 1,3,5-tribromobenzene.

Answer: Treat aniline with bromine water at room temperature to obtain a white precipitate of 2,4,6-tribromoaniline. Bromination replaces the ring hydrogens at both ortho positions and the para position relative to the amino group.

Diazotise the amino group with sodium nitrite and hydrochloric acid at 273 to 278 K. Then treat the diazonium salt with hypophosphorous acid, $\mathrm{H_3PO_2}$, and water to replace the diazonium group by hydrogen, releasing nitrogen gas.

The three bromines remain on the ring. Renumbering the ring after removal of the original amino group gives the product name 1,3,5-tribromobenzene.

Diazonium intermediates provide routes to aromatic fluorides, iodides, cyanides and phenols that are difficult to obtain by the corresponding direct substitutions. Amines more broadly serve as intermediates in medicines, fibres and dyes. Quaternary ammonium salts act as surfactants, substances that lower surface tension.

Glossary

• Amine — An ammonia derivative in which alkyl or aryl groups replace one or more hydrogen atoms.
• Lone pair — A pair of valence electrons not shared between atoms in a chemical bond.
• Nucleophile — An electron-pair donor that forms a bond with an electron-deficient centre during reaction.
• Ammonolysis — Cleavage of a carbon-halogen bond by ammonia, leading to formation of nitrogen-containing products.
• Hofmann degradation — Conversion of a primary amide into a primary amine containing one fewer carbon atom.
• Solvation — Stabilisation of a dissolved species through interactions with the surrounding molecules of the solvent.
• Inductive effect — Electron displacement through bonds caused by electron-releasing or electron-withdrawing atoms or groups.
• Acylation — Replacement of nitrogen-bound hydrogen in a primary or secondary amine by an acyl group.
• Carbylamine reaction — Formation of an isocyanide when a primary amine is heated with chloroform and ethanolic potassium hydroxide.
• Hinsberg's reagent — Benzenesulphonyl chloride, used to distinguish amine classes through sulphonamide formation and behaviour in alkali.
• Ambident nucleophile — A nucleophile capable of forming a new bond through either of two different atoms.
• Diazotisation — Conversion of a primary aromatic amine into a diazonium salt using nitrous acid at low temperature.
• Azo coupling — Reaction joining a diazonium species to an activated aromatic ring through an azo linkage.

Common errors and misconceptions

• Misconception: Three carbon atoms make an amine tertiary. Correct: Classification depends on organic groups directly bonded to nitrogen; propan-1-amine is primary.
• Misconception: Tertiary amines cannot hydrogen-bond with water. Correct: Their nitrogen lone pair accepts hydrogen bonds from water, although they lack N-H bonds for self-association.
• Misconception: Tertiary alkylamines are the strongest bases in every medium. Correct: Aqueous basicity also depends on solvation and steric effects; methyl and ethyl series have different orders.
• Misconception: A larger pKᵦ indicates a stronger base. Correct: A smaller pKᵦ indicates a stronger base under comparable conditions.
• Misconception: Amide reduction and Hofmann degradation give the same carbon count. Correct: LiAlH₄ reduction retains the carbonyl carbon, while degradation gives an amine with one fewer carbon.
• Misconception: Direct nitration of aniline gives only ortho and para products. Correct: The acidic medium forms meta-directing anilinium ions, so significant meta product also forms.
• Misconception: KCN and AgCN give the same chief product. Correct: KCN gives cyanides mainly, whereas AgCN gives isocyanides chiefly.
• Misconception: All diazonium reactions release nitrogen gas. Correct: Replacement reactions release N₂, but azo coupling retains both nitrogen atoms in the product.

Exam-style questions with model answers

Q1. Classify CH₃NH₂ and CH₃NHCH₂CH₃ as primary, secondary or tertiary, giving the structural reason for each. [2 marks]

1. CH₃NH₂ is a primary amine because nitrogen is directly bonded to one organic group, the methyl group.
2. CH₃NHCH₂CH₃ is a secondary amine because nitrogen is directly bonded to two organic groups, methyl and ethyl.

Q2. In water, the pKᵦ values are CH₃NH₂: 3.38, (CH₃)₂NH: 3.27 and (CH₃)₃N: 4.22. State the comparison rule, rank their basic strengths, and explain why alkyl-group donation alone is insufficient. [3 marks]

1. A smaller pKᵦ means a stronger base when values are compared in the same medium. The supplied values therefore allow direct comparison of these three aqueous bases.
2. The decreasing order is (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N, corresponding to pKᵦ values 3.27, 3.38 and 4.22 respectively.
3. In water, the stability of the protonated amine also depends on solvation and steric effects. Electron donation by alkyl groups alone does not determine the observed order.

Q3. Explain the Hinsberg distinction between primary, secondary and tertiary amines. Name the reagent, then give the reaction and alkali behaviour for each class. [4 marks]

1. The reagent is benzenesulphonyl chloride, C₆H₅SO₂Cl. The distinction depends on formation of a sulphonamide and whether its nitrogen retains an acidic hydrogen.
2. A primary amine forms a sulphonamide containing N-H. The electron-withdrawing sulphonyl group makes this hydrogen acidic, so the product dissolves in alkali.
3. A secondary amine forms a sulphonamide without N-H. It lacks that acidic hydrogen and is consequently insoluble in alkali.
4. A tertiary amine does not form a sulphonamide with the reagent. This absence of reaction differs from formation of the insoluble secondary-amine product.

Q4. Describe preparation of benzenediazonium chloride from aniline: give the reagents, temperature, balanced equation, reason for immediate use and conversion into phenol. [5 marks]

1. Use sodium nitrite and hydrochloric acid with aniline. These reagents generate nitrous acid within the mixture, avoiding the need to add it as a separately stored reagent.
2. Keep the diazotisation mixture at 273 to 278 K. Under these low-temperature conditions, the primary aromatic amine forms the arenediazonium salt.
3. The balanced equation is C₆H₅NH₂ + NaNO₂ + 2HCl → C₆H₅N₂⁺Cl⁻ + NaCl + 2H₂O, with the low temperature maintained.
4. The salt is stable only for a short time in cold solution and is not generally stored. It is used immediately after preparation.
5. Allow the aqueous solution to warm to 283 K: C₆H₅N₂⁺Cl⁻ + H₂O → C₆H₅OH + N₂ + HCl. Phenol forms and nitrogen is released.

Q5. Give six points explaining direct nitration of aniline and the protected route to p-nitroaniline: directing effect, protonation, direct products, protection, controlled nitration and removal of protection. [6 marks]

1. The amino group donates electron density into the aromatic ring. Free aniline is strongly activated, especially at the ortho and para positions.
2. The strongly acidic nitrating medium protonates aniline to anilinium ion. This positively charged nitrogen-containing group is meta directing in aromatic substitution.
3. Direct nitration therefore gives a significant amount of meta derivative alongside ortho and para derivatives. Tarry oxidation products also accompany the nitro derivatives.
4. First acetylate aniline with acetic anhydride to form acetanilide. This protects the amino group and reduces its activating effect on the ring.
5. Nitrate the protected compound. The nitrogen lone pair also interacts with the carbonyl group, and controlled nitration gives the para nitro derivative as the major product.
6. Hydrolyse the acetyl protecting group after nitration to restore the amino group. The resulting desired amine is p-nitroaniline.

Q6. Starting with CH₃CH₂Cl, use KCN followed by LiAlH₄ to obtain a primary amine. Identify the intermediate, final product and carbon-count change. [3 marks]

1. KCN replaces chlorine through carbon attachment, giving CH₃CH₂CN, propanenitrile. The nitrile group contains a carbon atom that becomes part of the organic chain.
2. LiAlH₄ reduces the nitrile group to CH₂NH₂, giving CH₃CH₂CH₂NH₂, propan-1-amine. Nitrogen remains in the organic product as a primary amino group.
3. The starting halide contains two carbon atoms, whereas both intermediate and final amine contain three. Cyanide substitution adds the carbon; nitrile reduction retains it.

Q7. Compare Sandmeyer and Gattermann reactions for introducing chlorine or bromine into a diazonium compound. Give each reagent system, the shared change and the relative yield. [4 marks]

1. Sandmeyer replacement uses copper(I) chloride in hydrochloric acid for chlorine, or copper(I) bromide in hydrobromic acid for bromine.
2. Gattermann replacement uses copper powder with the corresponding halogen acid, hydrochloric acid or hydrobromic acid, rather than the copper(I) salt reagent system.
3. Both replace the aromatic diazonium group by a halogen and release nitrogen gas. Benzenediazonium salts consequently yield chlorobenzene or bromobenzene.
4. The yield obtained by the Sandmeyer reaction is better than that obtained by the Gattermann reaction for the comparison described.

Q8. Explain coupling of benzenediazonium chloride with phenol and aniline. Give each medium, product and colour, then explain the substitution position, nitrogen retention and colour-related structural feature. [5 marks]

1. With phenol, C₆H₅OH, in alkaline medium, benzenediazonium chloride gives p-hydroxyazobenzene. This coupling product is an orange dye containing a hydroxyl group on one aromatic ring.
2. With aniline in mildly acidic medium, it gives p-aminoazobenzene, a yellow dye. The product retains an amino group on the coupled aniline ring.
3. Coupling occurs at the para position relative to the hydroxyl or amino group. It is electrophilic aromatic substitution involving the activated ring.
4. Both diazonium nitrogen atoms remain in the azo linkage, -N=N-, between the aromatic rings. This differs from replacement reactions that release N₂.
5. The joined rings and azo linkage create an extended conjugated system. Such azo compounds are often coloured and are used as dyes.

Key takeaways

• Classify amines by organic groups directly attached to nitrogen, and distinguish ring-attached amino groups from side-chain amino groups.
• Ammonolysis can yield a mixture; excess ammonia makes the primary amine the major product.
• Nitrile reduction retains every carbon of the nitrile, whereas Hofmann degradation removes one carbon from the amide skeleton.
• Aqueous basicity reflects inductive, solvation and steric effects together; a smaller pKᵦ indicates a stronger base.
• Carbylamine formation identifies primary amines, while Hinsberg reactions distinguish all three classes through product formation and alkali behaviour.
• Aniline activates ortho and para substitution; protonation during direct nitration explains the significant meta product.
• Low-temperature diazotisation provides intermediates for aromatic replacement reactions; coupling instead retains nitrogen in an azo linkage.
• Cyanides and isocyanides differ in their attachment atom, so reagent choice must accompany the structural formula.

Test yourself

Why is propan-2-amine primary?

Nitrogen is directly bonded to just one organic group and retains two hydrogen atoms, regardless of the branching at carbon.

Why does excess ammonia favour a primary amine in ammonolysis?

It favours reaction of the alkyl halide with ammonia over further alkylation of the amine already formed.

Which amide gives aniline by Hofmann degradation?

Benzamide, C₆H₅CONH₂, gives aniline after the carbonyl carbon is removed from the organic skeleton.

Why is aniline less basic than ammonia?

Its nitrogen lone pair is delocalised into the aromatic ring and is therefore less available to accept a proton.

What does KCN mainly produce from an alkyl halide?

It mainly gives an alkyl cyanide, RCN, by forming the new bond through the carbon atom of cyanide.

What observation follows adding bromine water to aniline?

A white precipitate of 2,4,6-tribromoaniline forms at room temperature as bromine substitutes at both ortho positions and the para position.

Which conditions form benzenediazonium chloride from aniline?

Sodium nitrite and hydrochloric acid at 273 to 278 K generate nitrous acid and convert aniline into the diazonium salt.

How does coupling differ from diazonium replacement?

Coupling retains both nitrogen atoms in an azo linkage; replacement releases them together as nitrogen gas.