Amines: A Comprehensive Guide for CBSE Class 12 Chemistry

What are Amines? Classification and Nomenclature

Amines are organic compounds derived from ammonia (NH₃) by replacing one or more hydrogen atoms with alkyl or aryl groups. The presence of a lone pair of electrons on the nitrogen atom makes amines basic and nucleophilic. Based on the number of hydrogen atoms replaced, amines are classified into three types:

1. Primary (1°) Amines: One hydrogen atom of ammonia is replaced by an alkyl or aryl group (R-NH₂ or Ar-NH₂). Examples include methylamine (CH₃NH₂) and aniline (C₆H₅NH₂).
2. Secondary (2°) Amines: Two hydrogen atoms of ammonia are replaced by alkyl or aryl groups (R-NH-R' or Ar-NH-R'). Examples include dimethylamine ((CH₃)₂NH) and N-methylaniline (C₆H₅NHCH₃).
3. Tertiary (3°) Amines: All three hydrogen atoms of ammonia are replaced by alkyl or aryl groups (R₃N or Ar₃N). Examples include trimethylamine ((CH₃)₃N) and N,N-dimethylaniline (C₆H₅N(CH₃)₂).

Nomenclature:

• Common Name System: Alkyl groups attached to nitrogen are named alphabetically, followed by the suffix 'amine'. For complex amines, the largest alkyl group is considered the parent, and the others are treated as N-substituted. E.g., Ethylmethylamine.
• IUPAC Name System: Primary amines are named by replacing the 'e' of the corresponding alkane with 'amine'. E.g., CH₃CH₂NH₂ is Ethanamine. For secondary and tertiary amines, the largest alkyl group is considered the parent alkane, and other groups are indicated as 'N-alkyl' or 'N,N-dialkyl' prefixes. E.g., CH₃NHCH₂CH₃ is N-Methylethanamine. Aromatic amines are named as derivatives of aniline.

Methods of Preparation of Amines

1. Reduction of Nitro Compounds — Nitro compounds (like nitrobenzene) can be reduced to amines (like aniline) using various reducing agents. Common methods include passing hydrogen gas over finely divided palladium, platinum, or nickel catalysts, or reduction with metals like tin (Sn) or iron (Fe) in the presence of concentrated hydrochloric acid. This is a common method for preparing aromatic amines. Reaction: R-NO₂ + 3H₂ (Pd/Pt/Ni) → R-NH₂ + 2H₂O Or: R-NO₂ + 6[H] (Sn/HCl or Fe/HCl) → R-NH₂ + 2H₂O
2. Ammonolysis of Alkyl Halides — Alkyl halides react with an ethanolic solution of ammonia through nucleophilic substitution to form primary amines. However, the primary amine formed is a stronger nucleophile than ammonia, and it can react further with the alkyl halide to form secondary, tertiary amines, and finally quaternary ammonium salts. This reaction is often not suitable for preparing pure primary amines due to the mixture of products. Reaction: R-X + NH₃ → R-NH₂ (1°) → R₂NH (2°) → R₃N (3°) → R₄N⁺X⁻ (Quaternary Salt)
3. Reduction of Nitriles — Nitriles (alkyl cyanides) on reduction with lithium aluminium hydride (LiAlH₄) or catalytic hydrogenation (H₂/Ni, Pt, or Pd) yield primary amines. This method is useful for ascending the amine series, meaning an amine with one more carbon atom than the starting alkyl halide (used to prepare the nitrile). Reaction: R-C≡N + 4[H] (LiAlH₄ or H₂/Ni) → R-CH₂-NH₂
4. Reduction of Amides — Amides undergo reduction with lithium aluminium hydride (LiAlH₄) to give primary amines. This is a good method for converting carboxylic acid derivatives into amines. Reaction: R-CO-NH₂ + 4[H] (LiAlH₄) → R-CH₂-NH₂ + H₂O
5. Gabriel Phthalimide Synthesis — This method is used for the preparation of pure primary aliphatic amines. It involves the reaction of phthalimide with ethanolic KOH, followed by heating with an alkyl halide, and finally hydrolysis with aqueous NaOH or acidification. Aromatic primary amines cannot be prepared by this method because aryl halides do not undergo nucleophilic substitution with the phthalimide anion. Key Steps: Phthalimide + KOH → Potassium phthalimide → N-alkylphthalimide + NaOH(aq) → Primary Amine + Sodium phthalate
6. Hoffmann Bromamide Degradation Reaction — This is a degradation reaction used to prepare primary amines by reacting an amide with bromine in an aqueous or ethanolic solution of sodium hydroxide. The amine formed contains one carbon atom less than the parent amide. This is a very important reaction as it provides a way to reduce the carbon chain length. Reaction: R-CO-NH₂ + Br₂ + 4NaOH → R-NH₂ + Na₂CO₃ + 2NaBr + 2H₂O

Chemical Properties: Basicity and Reactions of Amines

Amines are basic in nature due to the presence of a lone pair of electrons on the nitrogen atom, which can be donated to an acid. Their basicity is influenced by the stability of the conjugate acid formed and the availability of the lone pair.

Factors Affecting Basicity:

1. Aliphatic Amines: Alkyl groups are electron-donating (+I effect), which increases the electron density on nitrogen, making the lone pair more available for donation and thus increasing basicity. The order of basicity in the gaseous phase is 3° > 2° > 1° > NH₃. However, in an aqueous solution, the order changes due to solvation effects and steric hindrance. Solvation (hydrogen bonding with water) stabilizes the conjugate acid. Primary amines form the most hydrogen bonds, followed by secondary, then tertiary. Steric hindrance also plays a role, making it difficult for water molecules to approach the nitrogen in tertiary amines. Consequently, the general order of basicity in aqueous solution for methyl substituted amines is 2° > 1° > 3° > NH₃, and for ethyl substituted amines, it's typically 2° > 3° > 1° > NH₃.
2. Aromatic Amines: Aromatic amines are generally less basic than ammonia and aliphatic amines. This is because the lone pair of electrons on the nitrogen atom is delocalised into the benzene ring through resonance, making it less available for protonation. Electron-donating groups on the benzene ring increase basicity, while electron-withdrawing groups decrease it.

Important Reactions:

• Acylation: Amines react with acid chlorides, acid anhydrides, and esters to form amides. This reaction introduces an acyl group (-COR) onto the nitrogen atom. This is also used to protect the amino group in electrophilic substitution reactions.
• Carbylamine Reaction (Isocyanide Test): Primary amines (aliphatic and aromatic) react with chloroform (CHCl₃) and alcoholic KOH to produce foul-smelling isocyanides (carbylamines). This reaction is used as a test for primary amines.

R-NH₂ + CHCl₃ + 3KOH(alc) → R-NC + 3KCl + 3H₂O

• Reaction with Nitrous Acid (NaNO₂/HCl): This reaction is crucial for distinguishing 1°, 2°, and 3° amines:
• Primary aliphatic amines: Form highly unstable alkyldiazonium salts, which immediately decompose to form alcohols with the evolution of nitrogen gas (N₂).
• Primary aromatic amines: Form stable arenediazonium salts at low temperatures (0-5°C). These are important intermediates in organic synthesis.
• Secondary amines (aliphatic & aromatic): React to form N-nitrosoamines (yellow oily compounds).
• Tertiary aliphatic amines: Form soluble salts.
• Tertiary aromatic amines: Undergo electrophilic substitution to form p-nitroso-N,N-dialkylanilines.
• Hinsberg's Test (Reaction with Benzenesulphonyl Chloride): This reagent (C₆H₅SO₂Cl) is used to distinguish between primary, secondary, and tertiary amines:
• Primary amines: React to form N-alkylbenzenesulphonamides, which are soluble in NaOH due to the acidic hydrogen attached to the nitrogen.
• Secondary amines: React to form N,N-dialkylbenzenesulphonamides, which are insoluble in NaOH because they lack an acidic hydrogen.
• Tertiary amines: Do not react with Hinsberg's reagent (C₆H₅SO₂Cl) because they do not have a replaceable hydrogen atom on the nitrogen, but instead form a salt with HCl.

Worked Examples

• Example 1: Nomenclature Name the following compound: CH₃CH₂CH(CH₃)NH₂ Step 1: Identify the longest carbon chain attached to the amino group. In this case, it's a three-carbon chain (propane). Step 2: Number the carbon chain starting from the end closest to the amino group. The amino group is on the second carbon. Step 3: Identify substituents. There is a methyl group on the second carbon as well. Step 4: Combine these to form the IUPAC name. The parent alkane is propane, replace 'e' with 'amine', and indicate positions. So, 2-aminopropane or propan-2-amine. Final Answer: 2-Methylpropan-1-amine (if NH₂ is on 1st C and methyl on 2nd C, but for CH₃CH₂CH(CH₃)NH₂ the correct interpretation is a primary amine where the NH2 is on the 2nd carbon of a 3-carbon chain with a methyl group on the same carbon, which would be 2-aminopropane. Re-examining the formula: CH₃CH₂CH(CH₃)NH₂. This is 2-aminobutane with an incorrect carbon count. Let's correct the example to 2-methylpropan-1-amine for clarity as the structure implies 2-amino-3-methylpropane. Let's simplify to: CH₃-CH(CH₃)-CH₂-NH₂. This is a 3-carbon chain with NH₂ on C1 and methyl on C2. Correct Example 1: Nomenclature Name the following compound: CH₃-CH(CH₃)-CH₂-NH₂ Step 1: Identify the longest carbon chain containing the amino group. This is a 3-carbon chain, so 'propane'. Step 2: Number the carbon chain so the amino group gets the lowest possible number. Here, the amino group is on C-1. The methyl group is on C-2. Step 3: Form the IUPAC name. The parent is propan-1-amine. The substituent is 2-methyl. Final Answer: 2-Methylpropan-1-amine.
• Example 2: Reaction Product Predict the major product when methylamine reacts with acetyl chloride. Step 1: Identify the reactants: Methylamine (CH₃NH₂) is a primary amine, and acetyl chloride (CH₃COCl) is an acid chloride. Step 2: Recognize the type of reaction. Amines react with acid chlorides via nucleophilic acyl substitution (acylation). Step 3: The lone pair on the nitrogen of methylamine attacks the carbonyl carbon of acetyl chloride. The chloride ion leaves. Step 4: An amide is formed, along with HCl (which reacts with another molecule of amine or base). Reaction: CH₃NH₂ + CH₃COCl → CH₃NHCOCH₃ + HCl Final Answer: N-Methylacetamide.
• Example 3: Basicity Comparison Arrange the following in increasing order of basicity in aqueous solution: Methylamine, Dimethylamine, Trimethylamine, Ammonia. Step 1: Recall the general trend for basicity in aqueous solution for methyl-substituted amines: 2° > 1° > 3° > NH₃. Step 2: Identify the given amines: Methylamine (1°), Dimethylamine (2°), Trimethylamine (3°), Ammonia (NH₃). Step 3: Apply the trend. Final Answer: Ammonia < Trimethylamine < Methylamine < Dimethylamine (NH₃ < (CH₃)₃N < CH₃NH₂ < (CH₃)₂NH).

Exam Tips & Common Pitfalls

When studying amines for your CBSE Class 12 exams, pay special attention to the following:

1. Basicity in Aqueous vs. Gaseous Phase: Understand why the order of basicity differs. In the gas phase, it's purely inductive effect (3° > 2° > 1°), but in aqueous solution, salvation and steric hindrance become significant, leading to the order (2° > 1° > 3° for methyl amines, 2° > 3° > 1° for ethyl amines).
2. Distinguishing Tests: Master the Carbylamine reaction (only for 1° amines), Hinsberg's Test (distinguishes 1°, 2°, 3° amines based on solubility in alkali), and reactions with Nitrous Acid (gives distinct products/observations for 1°, 2°, and 3° amines).
3. Gabriel Phthalimide Synthesis: Remember it's exclusively for primary aliphatic amines. Aromatic amines cannot be prepared this way due to the lack of reactivity of aryl halides in SN₂ reactions.
4. Hoffmann Bromamide Degradation: This reaction is unique because it results in an amine with one carbon atom less than the starting amide. It's often used for chain shortening.
5. Aniline's Reactivity: The amino group in aniline is a strong activating and ortho-para directing group. Be careful with direct nitration and bromination; protection of the amino group (e.g., by acetylation) is often required to control the reaction and prevent poly-substitution.
6. Nomenclature: Practice both common and IUPAC names thoroughly, especially for branched and secondary/tertiary amines, including N-substitutions.

Practice Questions with Solutions

• Q: How will you convert methyl bromide into ethylamine? A: Step 1: Convert methyl bromide (CH₃Br) to methyl cyanide (CH₃CN) using potassium cyanide (KCN) in alcoholic solution. CH₃Br + KCN (alc) → CH₃CN + KBr Step 2: Reduce methyl cyanide (CH₃CN) to ethylamine (CH₃CH₂NH₂) using lithium aluminium hydride (LiAlH₄) or catalytic hydrogenation. CH₃CN + 4[H] (LiAlH₄/ether) → CH₃CH₂NH₂ Final answer: Methyl bromide can be converted to ethylamine in two steps: first by reaction with KCN to form methyl cyanide, then by reduction of methyl cyanide.
• Q: Write the structures of the products formed when ethanamine reacts with nitrous acid (NaNO₂ + HCl). A: Step 1: Ethanamine (a primary aliphatic amine) reacts with nitrous acid to form an unstable alkyldiazonium salt. CH₃CH₂NH₂ + HNO₂ (NaNO₂ + HCl, 0-5°C) → [CH₃CH₂N₂⁺Cl⁻] Step 2: The unstable alkyldiazonium salt immediately decomposes by losing nitrogen gas to form a carbocation, which then reacts with water (from the aqueous medium) to form an alcohol. [CH₃CH₂N₂⁺Cl⁻] → CH₃CH₂⁺ + N₂ (gas) → CH₃CH₂OH + H⁺ Final answer: Ethanamine reacts with nitrous acid to form ethanol and releases nitrogen gas.
• Q: Give a chemical test to distinguish between primary, secondary, and tertiary amines. A: Step 1: Use Hinsberg's reagent (benzenesulphonyl chloride, C₆H₅SO₂Cl). Step 2: Add Hinsberg's reagent to each amine sample. Primary amine (R-NH₂): Reacts to form N-alkylbenzenesulphonamide, which is soluble in aqueous KOH (due to acidic H on N). Secondary amine (R₂NH): Reacts to form N,N-dialkylbenzenesulphonamide, which is insoluble in aqueous KOH (no acidic H on N). Tertiary amine (R₃N): Does not react with Hinsberg's reagent, but dissolves in HCl to form a salt. Final answer: Hinsberg's test using benzenesulphonyl chloride followed by treatment with aqueous KOH can distinguish between primary, secondary, and tertiary amines based on the solubility of the product.
• Q: An aromatic compound 'A' on treatment with aqueous ammonia and heating forms compound 'B' which on heating with Br₂ and KOH forms a compound 'C' of molecular formula C₆H₇N. Write the structures and IUPAC names of compounds A, B, and C. A: Step 1: Identify compound C. C₆H₇N corresponds to C₆H₅NH₂ (aniline), which is a primary aromatic amine. The reaction of B with Br₂ and KOH forming C is Hoffmann bromamide degradation. This means B must be an amide with one more carbon atom than C. Step 2: Determine compound B. Since C is aniline (C₆H₅NH₂), B must be benzamide (C₆H₅CONH₂). Step 3: Determine compound A. Compound A on treatment with aqueous ammonia and heating forms benzamide. This means A must be benzoic acid (C₆H₅COOH). Reaction sequence: (A) C₆H₅COOH (Benzoic Acid) + NH₃ (aq) → (B) C₆H₅CONH₂ (Benzamide) + H₂O (B) C₆H₅CONH₂ + Br₂ + 4KOH → (C) C₆H₅NH₂ (Aniline) + K₂CO₃ + 2KBr + 2H₂O Final answer: A is Benzoic acid (C₆H₅COOH), B is Benzamide (C₆H₅CONH₂), and C is Aniline (C₆H₅NH₂).

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