Organic Chemistry - Amines

SECTION 1 - Structure and Classification:

Q1. What are amines? How are they related to ammonia ?

Amines are organic compounds derived from ammonia (NH₃) by the replacement of one, two, or all three hydrogen atoms with alkyl or aryl groups. They contain nitrogen with a lone pair of electrons, which is responsible for most of their properties. Just like ammonia, amines are basic in nature. Examples are methylamine (CH₃NH₂), dimethylamine ((CH₃)₂NH), and aniline (C₆H₅NH₂).

Q2. How are amines classified? Give examples of each type?

Amines are classified based on how many hydrogen atoms of ammonia are replaced by organic groups. Primary amines (1°) have one hydrogen replaced - e.g., methylamine (CH₃NH₂) and aniline (C₆H₅NH₂). Secondary amines (2°) have two hydrogens replaced - e.g., dimethylamine ((CH₃)₂NH). Tertiary amines (3°) have all three hydrogens replaced - e.g., trimethylamine ((CH₃)₃N). Note that the classification of amines is based on the number of organic groups on nitrogen, not on the carbon bearing the nitrogen, unlike alcohols.

Q3. What is the structure of amines? What is the hybridisation of nitrogen in amines ?

In amines, the nitrogen atom is sp³ hybridised. Three of the four sp³ orbitals form bonds with hydrogen atoms or carbon atoms of organic groups, and the fourth sp³ orbital holds the lone pair of electrons. This gives the amine molecule a pyramidal shape (like ammonia). The lone pair is responsible for the basic nature and nucleophilic character of amines. The C−N−C bond angle in amines is close to 108°, slightly less than the tetrahedral angle due to the lone pair.

Q4. How are amines further classified based on the nature of the group attached to nitrogen ?

Based on the nature of groups attached, amines are classified as aliphatic amines (where alkyl groups are attached to nitrogen - e.g., methylamine, ethylamine), aromatic amines (where at least one aryl group is directly attached to nitrogen - e.g., aniline, diphenylamine), and mixed amines (where both alkyl and aryl groups are attached - e.g., N-methylaniline).

SECTION 2 - Nomenclature:

Q5. How are amines named by the common system and IUPAC system ?

In the common system, aliphatic amines are named by writing the names of alkyl groups attached to nitrogen followed by the suffix "amine." For example, CH₃NH₂ is methylamine, (CH₃)₂NH is dimethylamine, and (CH₃)₃N is trimethylamine. In the IUPAC system, primary amines are named by replacing the terminal "e" of the parent alkane with "amine." For example, CH₃NH₂ is methanamine and C₂H₅NH₂ is ethanamine. For secondary and tertiary amines, the largest group is chosen as the parent chain and other groups are named as N-substituents. For example, (CH₃)₂NH is N-methylmethanamine.

SECTION 3 - Preparation of Amines

Q6. How are amines prepared by reduction of nitro compounds ?

Nitro compounds (RNO₂) are reduced to primary amines by treating with hydrogen gas in the presence of a catalyst like Raney nickel, or by using metals like iron or tin with dilute hydrochloric acid. This is the most important method for preparing aromatic primary amines. For example, nitrobenzene (C₆H₅NO₂) is reduced to aniline (C₆H₅NH₂) using Sn/HCl or Fe/HCl. This method always gives primary amines.

Q7. How are amines prepared by ammonolysis of alkyl halides ?

When an alkyl halide is heated with alcoholic ammonia in a sealed tube, the halide undergoes nucleophilic substitution. The nitrogen of ammonia attacks the alkyl halide, first giving a primary amine salt, which loses a proton to give a primary amine. The primary amine can further react with another alkyl halide to give a secondary amine, which can react again to give a tertiary amine, and further to give a quaternary ammonium salt. The major drawback of this method is that a mixture of primary, secondary, tertiary amines, and quaternary ammonium salt is obtained.

Q8. How are amines prepared by reduction of nitriles and amides ?

Nitriles (RCN) on reduction with LiAlH₄ (lithium aluminium hydride) or hydrogen over Raney nickel give primary amines with one more carbon than the starting material - for example, CH₃CN gives CH₃CH₂NH₂ (ethylamine). This is useful for ascending the amine series. Amides (RCONH₂) on reduction with LiAlH₄ also give primary amines - RCONH₂ gives RCH₂NH₂. Secondary amides give secondary amines and tertiary amides give tertiary amines on reduction.

Q9. What is Gabriel phthalimide synthesis? What is its advantage?

Gabriel phthalimide synthesis is used to prepare pure primary amines free from secondary and tertiary amines. Phthalimide is treated with KOH to form potassium phthalimide (a nucleophile). This reacts with an alkyl halide (SN2) to give N-alkylphthalimide. Alkaline hydrolysis (or hydrazinolysis) then cleaves the N−C bond to give the primary amine along with phthalic acid (or phthalyl hydrazide). The advantage is that it gives only primary amines and no secondary or tertiary amine contamination.

Q10. What is Hoffmann bromamide degradation? What type of product does it give ?

Hoffmann bromamide degradation involves treating an amide (RCONH₂) with bromine and aqueous potassium hydroxide. The product is a primary amine with one carbon less than the original amide. For example, benzamide (C₆H₅CONH₂) gives aniline (C₆H₅NH₂). The reaction goes through an isocyanate intermediate. This reaction is useful for descending the amine series and always gives a primary amine.

SECTION 4 - Physical Properties

Q11. Why do amines have lower boiling points than alcohols of comparable molecular mass?

Amines form intermolecular hydrogen bonds through N−H···N interactions, but the N−H bond is less polar than the O−H bond (nitrogen is less electronegative than oxygen). So hydrogen bonds in amines are weaker than in alcohols. Weaker hydrogen bonding means less energy is needed to overcome intermolecular forces, resulting in lower boiling points than alcohols of similar molecular mass. Tertiary amines have no N−H bond and cannot form hydrogen bonds with each other, so they have the lowest boiling points among the three types of amines.

Q12. Why are lower amines soluble in water ?

Lower aliphatic amines can form hydrogen bonds with water molecules through the lone pair on nitrogen and the N−H bonds. This makes them miscible with water. As the size of the alkyl group increases, the hydrophobic carbon chain dominates and the solubility in water decreases. Higher amines are practically insoluble in water. All amines are soluble in organic solvents.

Q13. Why do amines have a characteristic fishy smell ?

The fishy or ammonia-like smell of amines is due to the presence of the nitrogen atom with its lone pair, which makes amines volatile and easily detected by smell. Lower aliphatic amines (methylamine, dimethylamine, trimethylamine) smell like fish, which is why decaying fish produces this odour - bacterial decomposition of fish proteins releases trimethylamine. Aromatic amines have a milder characteristic smell.

SECTION 5 -Basic Nature and Chemical Properties

Q14. Why are amines basic in nature?

Amines are basic because the nitrogen atom has a lone pair of electrons that can be donated to a proton (H⁺) or to Lewis acids. When an amine accepts a proton, it forms an ammonium ion (RNH₃⁺). The reaction is: R−NH₂ + H₂O ⇌ RNH₃⁺ + OH⁻. The strength of the base depends on the availability of the lone pair on nitrogen - the more available the lone pair, the stronger the base.

Q15. Compare the basic strength of primary aliphatic amines, ammonia, and aniline. Explain the order.

The order of basic strength is: aliphatic primary amine > ammonia > aniline. Aliphatic primary amines are more basic than ammonia because the alkyl group has a positive inductive effect (+I effect), which pushes electron density towards nitrogen, making the lone pair more available for protonation. Aniline is less basic than ammonia because the lone pair on nitrogen is delocalised into the benzene ring through resonance -it becomes part of the aromatic pi system and is less available for protonation. This is why aniline is a very weak base (Kb ≈ 10⁻⁹) compared to methylamine (Kb ≈ 10⁻⁴).

Q16. Compare the basic strength among aliphatic amines - primary, secondary, and tertiary.

In the gas phase, the order is: tertiary > secondary > primary > ammonia, because more alkyl groups push more electron density onto nitrogen. However, in aqueous solution the order is: secondary > primary > tertiary > ammonia. This is because basic strength in water also depends on the stability of the protonated (ammonium) ion through solvation (hydrogen bonding with water). Tertiary ammonium ions have no N−H bonds and cannot form hydrogen bonds with water, making them less stable and thus making tertiary amines less basic than expected.

Q17. Why is aniline a weaker base than aliphatic amines ?

In aniline, the lone pair on nitrogen is involved in resonance with the benzene ring - it delocalises into the aromatic ring, making it less available for accepting a proton. This is shown by the resonance structures of aniline where electron density shifts from nitrogen into the ring. In aliphatic amines, there is no such resonance and the lone pair is fully available for protonation. Therefore aniline is a much weaker base than aliphatic amines.

Q18. What happens when amines react with acids ?

Amines being basic react with acids to form salts. With hydrochloric acid: RNH₂ + HCl → RNH₃⁺Cl⁻ (alkylammonium chloride). With sulphuric acid: 2RNH₂ + H₂SO₄ → (RNH₃⁺)₂SO₄²⁻. These salts are ionic and soluble in water but insoluble in non-polar solvents. The free amine can be regenerated by treating the salt with a strong base like NaOH.

Q19. What is alkylation of amines ?

When amines react with alkyl halides (nucleophilic substitution), the nitrogen acts as a nucleophile and attacks the alkyl halide. This is called alkylation. A primary amine reacts with an alkyl halide to give a secondary amine, which can react further to give a tertiary amine, and then a quaternary ammonium salt. This is why ammonolysis of alkyl halides gives a mixture - successive alkylation keeps occurring unless controlled.

Q20. What is acylation of amines? What is the product ?

Acylation is the reaction of an amine with an acid chloride (RCOCl) or an acid anhydride ((RCO)₂O) to form an amide (RCONHR'). For example, aniline reacts with acetic anhydride or acetyl chloride to form acetanilide (C₆H₅NHCOCH₃). The product amide has a much less basic nitrogen because the lone pair is delocalised into the carbonyl group. Acylation is used to protect the amino group in synthesis.

Q21. What is the carbylamine reaction? What is its importance?

The carbylamine reaction (also called isocyanide test) is used to detect primary amines. When a primary amine is heated with chloroform (CHCl₃) and alcoholic potassium hydroxide, a foul-smelling isocyanide (carbylamine, RNC) is formed. Secondary and tertiary amines do not give this reaction. So the carbylamine reaction is a test to identify primary amines. The equation is: RNH₂ + CHCl₃ + 3KOH → RNC + 3KCl + 3H₂O.

Q22. How do primary, secondary, and tertiary amines react differently with nitrous acid (HNO₂)?

Primary aliphatic amines react with nitrous acid (NaNO₂ + HCl at 0°C) to give an unstable diazonium salt that immediately decomposes to give alcohol, nitrogen gas, and other products - N₂ gas is evolved. Primary aromatic amines give stable diazonium salts at 0–5°C (used in synthesis). Secondary amines (aliphatic or aromatic) react with nitrous acid to give yellow oily N-nitrosamines (R₂N−N=O). Tertiary aliphatic amines do not give a stable product with nitrous acid they form soluble salts that decompose. Tertiary aromatic amines undergo ring nitrosation (C-nitroso compound is formed on the ring). This difference in reaction with HNO₂ is used to distinguish primary, secondary, and tertiary amines.

Q23. What is the Hinsberg test? How does it distinguish primary, secondary, and tertiary amines ?

The Hinsberg test uses benzenesulphonyl chloride (C₆H₅SO₂Cl) -also called Hinsberg's reagent. Primary amines react to form a sulphonamide that is soluble in alkali (because it has an acidic N−H on the sulphonamide which can be removed by base, giving a soluble sodium salt). Secondary amines react to form a sulphonamide that is insoluble in alkali (no N−H on the sulphonamide nitrogen). Tertiary amines do not react with Hinsberg's reagent at all (no N−H to react). This test clearly distinguishes the three classes of amines.

Q24. What are the electrophilic substitution reactions of aniline ?

The amino group (−NH₂) in aniline is a powerful activating and ortho/para-directing group. Due to electron donation from nitrogen into the ring, aniline undergoes electrophilic substitution very readily. Three important reactions are:

Bromination - aniline reacts with bromine water immediately to give 2,4,6-tribromoaniline (white precipitate), without needing a catalyst. This is used as a test for aniline.

Nitration - direct nitration of aniline with HNO₃/H₂SO₄ is not done directly because acid protonates aniline to anilinium ion (C₆H₅NH₃⁺), which is a deactivating meta-director. To get ortho/para products, aniline is first acetylated to protect the amino group and then nitrated.

Sulphonation - aniline reacts with concentrated H₂SO₄ at 180°C to give sulphanilic acid (4-aminobenzenesulphonic acid), a major product of para-sulphonation.

SECTION 6 - Diazonium Salts

Q25. What are diazonium salts? How are they prepared ?

Diazonium salts have the general formula Ar−N₂⁺X⁻, where Ar is an aryl group and X⁻ is an anion like Cl⁻ or HSO₄⁻. They are prepared by the process called diazotisation - treating a primary aromatic amine (like aniline) with sodium nitrite (NaNO₂) and hydrochloric acid at 0–5°C. The reaction is: C₆H₅NH₂ + NaNO₂ + 2HCl → C₆H₅N₂⁺Cl⁻ + NaCl + 2H₂O. The low temperature (0–5°C) is essential to prevent decomposition of the diazonium salt.

Q26. Why must diazotisation be carried out at 0–5°C ?

Diazonium salts are unstable and decompose rapidly at higher temperatures, releasing nitrogen gas and forming phenol or other by-products. At 0–5°C, the diazonium salt is stable enough to be used in further reactions. Above 5°C, decomposition is too fast to control the reaction.

Q27. What are the reactions of diazonium salts involving replacement of the diazonium group ?

The diazonium group (−N₂⁺) is an excellent leaving group and can be replaced by various groups - these are called Sandmeyer reactions and related reactions.

Replacement by Cl - treatment with CuCl/HCl gives chlorobenzene (Sandmeyer reaction).

Replacement by Br - treatment with CuBr/HBr gives bromobenzene (Sandmeyer reaction).

Replacement by CN - treatment with CuCN gives benzonitrile (Sandmeyer reaction).

Replacement by F - treatment with HBF₄ gives fluorobenzene (Balz-Schiemann reaction - the diazonium tetrafluoroborate is heated).

Replacement by I - treatment with KI gives iodobenzene (no copper catalyst needed).

Replacement by OH — boiling with water gives phenol.

Replacement by H - treatment with hypophosphorous acid (H₃PO₂) replaces the diazonium group with hydrogen.

Replacement by NO₂ - treatment with NaNO₂/CuCl gives nitrobenzene.

Q28. What are the reactions of diazonium salts involving retention of the nitrogen (coupling reactions)?

Coupling reactions involve the diazonium ion acting as an electrophile and attacking an activated aromatic ring (phenol or amine) at the para position. The diazonium ion couples with phenol (in alkaline medium) to give an orange/yellow azo compound -for example, benzenediazonium chloride + phenol → p-hydroxyazobenzene. Coupling with aniline gives p-aminoazobenzene. Azo compounds contain the −N=N− (azo) group and are intensely coloured - they are used as dyes (azo dyes). These are electrophilic aromatic substitution reactions.

Q29. What is the importance of diazonium salts in synthesis ?

Diazonium salts are extremely important in synthetic organic chemistry because the diazonium group can be replaced by a wide variety of groups — F, Cl, Br, I, CN, OH, NO₂, and H - making it possible to introduce groups that are otherwise difficult to attach directly to a benzene ring. This makes diazonium salts a key intermediate for synthesising a wide range of aromatic compounds starting from aniline. They are also used in the manufacture of azo dyes, which are the largest class of synthetic dyes used in textiles and printing.

Q30. How can you distinguish between primary, secondary, and tertiary amines using chemical tests ?

Three tests are commonly used. The carbylamine test - only primary amines give a foul-smelling isocyanide with CHCl₃/KOH; secondary and tertiary do not react. The Hinsberg test - primary amines give an alkali-soluble sulphonamide; secondary amines give an alkali-insoluble sulphonamide; tertiary amines do not react. Reaction with nitrous acid - primary amines evolve N₂ gas (aliphatic) or give a stable diazonium salt (aromatic); secondary amines give yellow N-nitrosamines; tertiary aliphatic amines give no stable characteristic product.

Q31. What is acetanilide? How is it formed and what is its use in synthesis ?

Acetanilide is formed by the acylation of aniline with acetic anhydride or acetyl chloride - aniline reacts to give C₆H₅NHCOCH₃ (acetanilide) with the loss of acetic acid. The purpose of converting aniline to acetanilide is to protect the amino group during nitration. The −NHCOCH₃ group is less activating than −NH₂ and prevents over-bromination or over-nitration. After the desired substitution, the acetyl group is removed by hydrolysis with acid or base to regenerate the amino group. This strategy of protection-deprotection is widely used in synthesis.

Q32. Why does aniline not undergo Friedel-Crafts reaction ?

Friedel-Crafts reactions require a Lewis acid catalyst like AlCl₃. However, aniline is a base and it reacts with AlCl₃ to form a complex (C₆H₅NH₂·AlCl₃) - the amine donates its lone pair to AlCl₃. This complex makes the amino group electron-withdrawing rather than electron-donating, deactivating the ring. As a result, Friedel-Crafts alkylation and acylation do not proceed with aniline.