Organic Compounds Containing Nitrogen Class 12 Chemistry NCERT Guide

Nitrogen-containing organic compounds, specifically amines and diazonium salts, are foundational to both natural biochemistry (amino acids, proteins, and alkaloids) and synthetic industrial chemistry (dyes, polymers, and pharmaceuticals). In this comprehensive CBSE Class 12 Chemistry guide on organic compounds containing nitrogen class 12 ncert, we will break down critical organic transformations, structural behavior, and basicity trends. You will learn the major synthetic pathways, such as the Hoffmann Bromamide degradation and Gabriel Phthalimide synthesis, and understand the intricate physical trends governing basicity in gas vs. aqueous phases. Let's explore these concepts with step-by-step mechanisms, exam tips, and fully worked practice questions designed to help you ace your board exams.

Structure, Classification, and Nomenclature of Amines

Amines are regarded as derivatives of ammonia (NH₃) obtained by the replacement of one, two, or all three hydrogen atoms by alkyl or aryl groups. Based on the number of hydrogen atoms replaced, they are classified as primary (1°), secondary (2°), or tertiary (3°) amines.

In terms of structure, nitrogen in amines is sp³ hybridized. The geometry is pyramidal rather than tetrahedral due to the presence of an unshared lone pair of electrons on the nitrogen atom. The C-N-H (or C-N-C) bond angle is slightly less than the ideal 109.5° tetrahedral angle (typically around 108°) because of lone pair-bonding pair repulsion.

For IUPAC naming, aliphatic amines are named as alkanamines (e.g., Methanamine, Ethanamine). For secondary and tertiary amines, the smaller alkyl groups attached to nitrogen are prefixed as 'N-alkyl' (e.g., N-Methylmethanamine).

Key Synthetic Methods for Preparing Amines

  1. Gabriel Phthalimide Synthesis — Used strictly for preparing pure aliphatic primary amines. Phthalimide is treated with ethanolic KOH to form potassium phthalimide. This salt is heated with an alkyl halide (R-X) to yield N-alkylphthalimide, which undergoes alkaline hydrolysis (with NaOH) to release a pure 1° aliphatic amine and sodium phthalate. Note: Aryl halides do not undergo nucleophilic substitution with phthalimide, so aniline cannot be prepared this way.
  2. Hoffmann Bromamide Degradation Reaction — An excellent method for preparing primary amines containing one carbon atom less than the parent amide. An amide is treated with bromine (Br₂) in the presence of an aqueous or ethanolic solution of sodium hydroxide (NaOH). The net reaction is: R-CO-NH₂ + Br₂ + 4NaOH → R-NH₂ + Na₂CO₃ + 2NaBr + 2H₂O.
  3. Reduction of Nitro Compounds, Nitriles, and Amides — Nitro compounds (R-NO₂) can be reduced using hydrogen in the presence of Pd, or with Fe/HCl or Sn/HCl. Nitriles (R-CN) yield primary amines when reduced with LiAlH₄ or via catalytic hydrogenation. Similarly, amides (R-CO-NH₂) are reduced with LiAlH₄ to primary amines containing the same number of carbon atoms.

Basicity of Amines: Gas Phase vs. Aqueous Phase

AspectDetails
Gas Phase TrendGoverned solely by the inductive (+I) effect of alkyl groups. More alkyl groups release electron density onto nitrogen, stabilizing the protonated ammonium cation.
Aqueous Phase Trend (Methyl groups)Governed by a combination of +I effect, steric hindrance, and solvation (hydration) energy of the substituted ammonium cation.
Aqueous Phase Trend (Ethyl groups)Steric hindrance of the bulkier ethyl groups reduces solvation stability of the 1° ion, placing the 3° amine ahead of the 1° amine.

Common Board Exam Traps & Distinguishing Tests

  1. The Hinsberg Test Trap: Benzenesulfonyl chloride (Hinsberg's reagent) reacts with 1° amines to form a sulfonamide soluble in alkali. With 2° amines, it forms a sulfonamide insoluble in alkali. 3° amines do not react at all. Always explain why (presence/absence of acidic H on nitrogen).
  1. Aniline vs. Aliphatic Amines: Aniline is significantly less basic than aliphatic amines. This is because the lone pair of electrons on nitrogen in aniline is conjugated with the benzene ring and participates in resonance, making it less available for protonation.
  1. Carbylamine Reaction: Only aliphatic and aromatic primary (1°) amines give this test (producing foul-smelling isocyanides when treated with CHCl₃ and KOH). Do not apply this to 2° or 3° amines!

Practice Questions with Solutions

  • Q: Arrange the following compounds in decreasing order of their basic strength in aqueous solution: Ethylamine, Diethylamine, Triethylamine, Ammonia. A: Step 1: Analyze the alkyl group attached. Here, we have ethyl groups (-C₂H₅). Step 2: Recall the combined experimental trend for ethyl-substituted amines in water, which balances the inductive effect, steric hindrance, and solvation. Step 3: The order of basicity for ethyl-substituted amines in aqueous solution is: Secondary > Tertiary > Primary > Ammonia. Final answer: (C₂H₅)₂NH > (C₂H₅)₃N > C₂H₅NH₂ > NH₃
  • Q: Why cannot aromatic primary amines (like aniline) be prepared by Gabriel Phthalimide Synthesis? A: Step 1: Identify the key step in Gabriel Phthalimide synthesis. It involves nucleophilic substitution (S_N2) of the potassium salt of phthalimide with an organic halide. Step 2: Examine the reactivity of aryl halides. Aryl halides do not easily undergo nucleophilic substitution reactions because of partial double-bond character of the C-X bond (due to resonance) and steric repulsion from the aromatic ring. Step 3: Consequently, nucleophilic attack by the phthalimide anion on aryl halides is not possible under ordinary synthetic conditions. Final answer: Aryl halides do not undergo nucleophilic substitution reactions (S_N2) with potassium phthalimide, making it impossible to prepare aromatic primary amines by this route.
  • Q: Complete and write the step-by-step chemical equations for the conversion of Aniline to Fluorobenzene (Schiemann Reaction). A: Step 1: Convert aniline to benzene diazonium chloride. Treat Aniline (C₆H₅NH₂) with nitrous acid (NaNO₂ + HCl) at cold temperatures (273-278 K) to get Benzene diazonium chloride (C₆H₅N₂⁺Cl⁻). Step 2: Treat the diazonium salt with fluoroboric acid (HBF₄) to precipitate out benzene diazonium fluoroborate (C₆H₅N₂⁺BF₄⁻). Step 3: Heat the dry benzene diazonium fluoroborate to decompose it thermally into fluorobenzene, nitrogen gas, and boron trifluoride. Final answer: C₆H₅NH₂ + NaNO₂ + 2HCl (at 273-278K) → C₆H₅N₂⁺Cl⁻ + NaCl + 2H₂O. Then, C₆H₅N₂⁺Cl⁻ + HBF₄ → C₆H₅N₂⁺BF₄⁻ + HCl. Heating C₆H₅N₂⁺BF₄⁻ yields C₆H₅F + N₂ + BF₃.
  • Q: Account for the following: Primary amines have higher boiling points than tertiary amines of comparable molecular mass. A: Step 1: Examine the intermolecular forces in primary (1°) amines. Primary amines (R-NH₂) have two hydrogen atoms bonded directly to nitrogen, allowing them to form extensive intermolecular hydrogen bonds. Step 2: Examine the structure of tertiary (3°) amines. Tertiary amines (R₃N) have no hydrogen atoms directly bonded to nitrogen, so they cannot form intermolecular hydrogen bonds with themselves. Step 3: More energy is required to break the hydrogen bonds in primary amines than the weaker dipole-dipole interactions in tertiary amines. Final answer: Due to the presence of polar N-H bonds, primary amines associate through intermolecular hydrogen bonding, whereas tertiary amines lack N-H bonds and cannot form hydrogen bonds, resulting in lower boiling points for 3° amines.

Frequently Asked Questions

What is the Hoffmann Bromamide degradation reaction?

It is a reaction where an amide is treated with bromine and sodium hydroxide to produce a primary amine containing one carbon atom less than the parent amide. This reaction is widely used in step-down organic synthetic pathways.

Why is aniline less basic than cyclohexylamine?

In aniline, the lone pair on the nitrogen atom is delocalized into the benzene ring via resonance, making it less available for protonation. In cyclohexylamine, there is no resonance, and the alkyl ring exerts a +I effect that increases electron density on nitrogen, making it more basic.

What is Hinsberg's reagent and what is its use?

Hinsberg's reagent is benzenesulfonyl chloride (C₆H₅SO₂Cl). It is used to distinguish between primary, secondary, and tertiary amines based on their distinct reactivity and the solubility of the resulting sulfonamide in alkaline solutions.