Preparation Methods of Amines

Complete guide to amine synthesis - Gabriel synthesis, Hoffmann bromamide, reduction methods, and reductive amination for JEE

12 min read

The Hook: Synthesizing Life-Saving Drugs

Connect: Real Life → Chemistry

Paracetamol (acetaminophen), one of the world’s most used pain relievers, is synthesized from p-nitrophenol through a series of reactions involving amine preparation!

The synthesis pathway: p-Nitrophenol → p-Aminophenol (via reduction) → Paracetamol (via acylation)

Other important amine-based drugs:

  • Sulfa drugs: First antibiotics, synthesized from aniline
  • Local anesthetics (Novocaine): Contains tertiary amine groups
  • Adrenaline: Life-saving hormone, prepared via aromatic amine synthesis

Here’s the JEE question: How do you selectively prepare a primary amine without 2° and 3° amine contamination?

The Core Concept

Why Multiple Methods?

Different methods of amine preparation are needed because:

  1. Selectivity: Some methods give only 1° amines (Gabriel synthesis)
  2. Starting materials: Different substrates require different approaches
  3. Functionality: Preserve or introduce other functional groups
  4. Scale: Industrial vs laboratory methods differ
JEE Weightage
Amine Preparation: 4-5 questions in JEE Main, 2-3 in JEE Advanced High-yield topics: Gabriel synthesis, Hoffmann bromamide degradation, reduction of nitro compounds

Method 1: Reduction of Nitro Compounds

Overview

Best for: Aromatic and aliphatic primary amines

General Reaction:

$$\boxed{\text{R-NO}_2 + 6[\text{H}] \rightarrow \text{R-NH}_2 + 2\text{H}_2\text{O}}$$

Reducing Agents

Agent 1: Sn/HCl or Fe/HCl (Acidic Medium)

For aromatic nitro compounds:

$$\text{C}_6\text{H}_5\text{-NO}_2 + 6[\text{H}] \xrightarrow{\text{Sn/HCl}} \text{C}_6\text{H}_5\text{-NH}_2 + 2\text{H}_2\text{O}$$

Mechanism:

C₆H₅-NO₂ → C₆H₅-NO → C₆H₅-NHOH → C₆H₅-NH₂
Nitrobenzene → Nitrosobenzene → Phenylhydroxylamine → Aniline

Product: Anilinium chloride (C₆H₅-NH₃⁺Cl⁻)

  • Then neutralize with NaOH to get free amine

Example - Industrial synthesis of aniline:

$$\text{C}_6\text{H}_5\text{NO}_2 \xrightarrow{\text{Fe/HCl}} [\text{C}_6\text{H}_5\text{NH}_3^+]\text{Cl}^- \xrightarrow{\text{NaOH}} \text{C}_6\text{H}_5\text{NH}_2$$

Agent 2: LiAlH₄ or H₂/Ni (Catalytic Hydrogenation)

For aliphatic and aromatic:

$$\text{CH}_3\text{-NO}_2 \xrightarrow{\text{LiAlH}_4} \text{CH}_3\text{-NH}_2$$ $$\text{C}_6\text{H}_5\text{-NO}_2 \xrightarrow{\text{H}_2/\text{Ni}, \Delta} \text{C}_6\text{H}_5\text{-NH}_2$$

Advantages:

  • Neutral conditions
  • Doesn’t affect other functional groups (selective)
Memory Trick: Choosing Reducing Agent

“Tin for Tough (aromatic), Lithium for Light (aliphatic)”

For aromatic nitro compounds:

  • Sn/HCl or Fe/HCl (cheap, industrial scale)

For aliphatic or sensitive substrates:

  • LiAlH₄ (mild, selective)
  • H₂/Ni (catalytic, clean)

JEE Strategy: If compound has acid/base sensitive groups → use LiAlH₄ or H₂/Ni

Selective Reduction of Dinitro Compounds

Problem: Reduce one NO₂ group selectively in dinitro benzene

Solution: Use (NH₄)₂S (ammonium sulfide)

$$\text{m-}\text{NO}_2\text{-C}_6\text{H}_4\text{-NO}_2 \xrightarrow{(\text{NH}_4)_2\text{S}} \text{m-}\text{NO}_2\text{-C}_6\text{H}_4\text{-NH}_2$$

Why selective?

  • Mild reducing agent
  • Reduces only ONE nitro group
  • Used for preparing nitro-amines
JEE Trap: Complete vs Partial Reduction

Question type: “What product forms when m-dinitrobenzene is treated with (NH₄)₂S?”

Wrong answer: m-Phenylenediamine (both NO₂ reduced)

Correct answer: m-Nitroaniline (only one NO₂ reduced)

Key: (NH₄)₂S is mild, Sn/HCl would reduce both groups!

Interactive Demo: Visualize Reduction Mechanisms

See how nitro groups are reduced to amines step-by-step.

Method 2: Reduction of Nitriles (Cyanides)

Overview

Best for: Primary amines with one extra carbon

General Reaction:

$$\boxed{\text{R-CN} + 4[\text{H}] \rightarrow \text{R-CH}_2\text{-NH}_2}$$

Reducing Agents

1. LiAlH₄ (Lithium Aluminium Hydride)

$$\text{CH}_3\text{-CN} \xrightarrow{\text{LiAlH}_4} \text{CH}_3\text{-CH}_2\text{-NH}_2$$

2. H₂/Ni (Catalytic Hydrogenation)

$$\text{C}_6\text{H}_5\text{-CN} \xrightarrow{\text{H}_2/\text{Ni}} \text{C}_6\text{H}_5\text{-CH}_2\text{-NH}_2$$

3. Na/C₂H₅OH (Sodium in ethanol)

$$\text{R-CN} \xrightarrow{\text{Na/C}_2\text{H}_5\text{OH}} \text{R-CH}_2\text{-NH}_2$$

Application: Chain Extension

Key advantage: Adds one carbon while forming amine!

Example: Prepare propanamine from ethyl bromide

Step 1: Form nitrile

$$\text{CH}_3\text{CH}_2\text{Br} + \text{KCN} \rightarrow \text{CH}_3\text{CH}_2\text{-CN}$$

Step 2: Reduce nitrile

$$\text{CH}_3\text{CH}_2\text{-CN} \xrightarrow{\text{LiAlH}_4} \text{CH}_3\text{CH}_2\text{-CH}_2\text{-NH}_2$$

Result: Propanamine (3 carbons from 2 carbons!)

JEE Application: Retrosynthesis

Q: How to prepare butanamine from propanol?

Retrosynthetic analysis:

CH₃-CH₂-CH₂-CH₂-NH₂ (target)
         ↑
   (reduce nitrile)
         ↑
CH₃-CH₂-CH₂-CN
         ↑
   (from halide)
         ↑
CH₃-CH₂-CH₂-Br
         ↑
   (from alcohol)
         ↑
CH₃-CH₂-CH₂-OH (starting)

Solution:

  1. CH₃CH₂CH₂OH → CH₃CH₂CH₂Br (PBr₃)
  2. CH₃CH₂CH₂Br → CH₃CH₂CH₂CN (KCN)
  3. CH₃CH₂CH₂CN → CH₃CH₂CH₂CH₂NH₂ (LiAlH₄)

Related: Organic synthesis strategies

Method 3: Reduction of Amides

Overview

Best for: Primary amines (same carbon count)

General Reaction:

$$\boxed{\text{R-CO-NH}_2 \xrightarrow{\text{LiAlH}_4} \text{R-CH}_2\text{-NH}_2}$$

Mechanism

Reagent: LiAlH₄ (powerful reducing agent)

Example:

$$\text{CH}_3\text{-CO-NH}_2 \xrightarrow{\text{LiAlH}_4} \text{CH}_3\text{-CH}_2\text{-NH}_2$$

(Acetamide → Ethylamine)

Key point: C=O is reduced to CH₂

For N-substituted amides:

$$\text{R-CO-NH-R'} \xrightarrow{\text{LiAlH}_4} \text{R-CH}_2\text{-NH-R'}$$

(Forms secondary amine)

$$\text{R-CO-NR'}_2 \xrightarrow{\text{LiAlH}_4} \text{R-CH}_2\text{-NR'}_2$$

(Forms tertiary amine)

Amide Structure Determines Amine Type

Crucial JEE concept:

  • 1° Amide (R-CO-NH₂) → 1° Amine (R-CH₂-NH₂)
  • 2° Amide (R-CO-NH-R’) → 2° Amine (R-CH₂-NH-R')
  • 3° Amide (R-CO-NR’₂) → 3° Amine (R-CH₂-NR’₂)

The amide degree = amine degree!

This is different from Hoffmann bromamide (see next section)

Method 4: Hoffmann Bromamide Degradation

Overview

Best for: Primary amines with one less carbon than amide

General Reaction:

$$\boxed{\text{R-CO-NH}_2 + \text{Br}_2 + 4\text{NaOH} \rightarrow \text{R-NH}_2 + \text{Na}_2\text{CO}_3 + 2\text{NaBr} + 2\text{H}_2\text{O}}$$

Key feature: Carbon chain decreases by one (decarboxylation)

Mechanism

Step 1: Formation of N-bromoamide

$$\text{R-CO-NH}_2 + \text{Br}_2 + 2\text{NaOH} \rightarrow \text{R-CO-NH-Br} + \text{NaBr} + 2\text{H}_2\text{O}$$

Step 2: Formation of isocyanate (key intermediate)

$$\text{R-CO-NH-Br} + \text{NaOH} \rightarrow \text{R-N=C=O} + \text{NaBr} + \text{H}_2\text{O}$$

(Isocyanate)

Step 3: Hydrolysis to amine

$$\text{R-N=C=O} + 2\text{NaOH} \rightarrow \text{R-NH}_2 + \text{Na}_2\text{CO}_3$$

Examples

Example 1: Acetamide to Methylamine

$$\text{CH}_3\text{-CO-NH}_2 \xrightarrow{\text{Br}_2/\text{NaOH}} \text{CH}_3\text{-NH}_2 + \text{CO}_2$$

Carbon count: 2 → 1 (lost as CO₂)

Example 2: Benzamide to Aniline

$$\text{C}_6\text{H}_5\text{-CO-NH}_2 \xrightarrow{\text{Br}_2/\text{NaOH}} \text{C}_6\text{H}_5\text{-NH}_2$$

Example 3: Propionamide to Ethylamine

$$\text{CH}_3\text{CH}_2\text{-CO-NH}_2 \xrightarrow{\text{Br}_2/\text{NaOH}} \text{CH}_3\text{CH}_2\text{-NH}_2$$
Memory Trick: Hoffmann Degradation

“Hoffmann Hates Carbon”

  • Hoffmann degradation degrades (loses) one carbon
  • Amide loses CO₂
  • Forms amine with n-1 carbons

Contrast:

  • LiAlH₄ reduction of amide → same carbon count
  • Hoffmann degradation → one less carbon

JEE Memory aid: “LiAlH₄ is Loyal (keeps carbons), Hoffmann is Harsh (loses carbon)”

JEE Trap: Hoffmann vs LiAlH₄

Question: “Benzamide is treated with (i) Br₂/NaOH and (ii) LiAlH₄. Identify products.”

Solution:

(i) Br₂/NaOH (Hoffmann):

$$\text{C}_6\text{H}_5\text{-CO-NH}_2 \rightarrow \text{C}_6\text{H}_5\text{-NH}_2$$

Product: Aniline (lost CO₂)

(ii) LiAlH₄ (Reduction):

$$\text{C}_6\text{H}_5\text{-CO-NH}_2 \rightarrow \text{C}_6\text{H}_5\text{-CH}_2\text{-NH}_2$$

Product: Benzylamine (kept all carbons)

Key difference: Carbon count!

Method 5: Gabriel Phthalimide Synthesis

Overview

Best for: Pure primary aliphatic amines (no 2° or 3° contamination)

Limitation: Only for 1° aliphatic amines (not aromatic)

Why important for JEE: Only method giving exclusively primary amines!

Mechanism

Step 1: Formation of phthalimide anion

    O                          O
    ‖                          ‖
  C-NH      + KOH    →      C-N⁻K⁺
    |                          |
  C-NH                       C=O
    ‖
    O
(Phthalimide)          (Phthalimide salt)

Step 2: N-alkylation (SN2 reaction)

$$\text{Phthalimide-N}^-\text{K}^+ + \text{R-X} \rightarrow \text{Phthalimide-N-R} + \text{KX}$$

Step 3: Hydrolysis

$$\text{Phthalimide-N-R} \xrightarrow{\text{H}^+/\text{H}_2\text{O or NaOH}} \text{R-NH}_2 + \text{Phthalic acid}$$

Complete Example

Synthesis of ethylamine:

Step 1:

$$\text{Phthalimide} + \text{KOH} \rightarrow \text{Potassium phthalimide}$$

Step 2:

$$\text{Potassium phthalimide} + \text{CH}_3\text{CH}_2\text{Br} \rightarrow \text{N-Ethylphthalimide}$$

Step 3:

$$\text{N-Ethylphthalimide} \xrightarrow{\text{H}_3\text{O}^+} \text{CH}_3\text{CH}_2\text{NH}_2 + \text{Phthalic acid}$$

Advantages and Limitations

Advantages:

  1. ✓ Pure primary amine (no 2° or 3° byproducts)
  2. ✓ Clean reaction
  3. ✓ Easy separation (phthalic acid is insoluble)

Limitations:

  1. ✗ Only for aliphatic primary amines
  2. ✗ Cannot prepare aromatic amines (C₆H₅-NH₂)
  3. ✗ Requires alkyl halide (RX)
Why Gabriel Synthesis Fails for Aniline

Question: Can you prepare aniline using Gabriel synthesis?

Answer: NO!

Reason:

  • Requires C₆H₅-X (aryl halide)
  • Aryl halides don’t undergo SN2 reactions
  • C-X bond in aromatic ring is too strong
  • No backside attack possible (sp² carbon)

For aniline: Use nitrobenzene reduction instead!

$$\text{C}_6\text{H}_5\text{NO}_2 \xrightarrow{\text{Sn/HCl}} \text{C}_6\text{H}_5\text{NH}_2$$

Related: Haloalkanes and haloarenes

JEE Problem: Gabriel Synthesis

Q: How will you prepare 1-aminopropane using Gabriel synthesis?

Solution:

Required: CH₃-CH₂-CH₂-NH₂

Step 1: Prepare potassium phthalimide

$$\text{Phthalimide} + \text{KOH} \rightarrow \text{Potassium phthalimide} + \text{H}_2\text{O}$$

Step 2: React with 1-bromopropane

$$\text{Potassium phthalimide} + \text{CH}_3\text{CH}_2\text{CH}_2\text{Br} \rightarrow \text{N-Propylphthalimide} + \text{KBr}$$

Step 3: Hydrolyze

$$\text{N-Propylphthalimide} \xrightarrow{\text{HCl/}\text{H}_2\text{O}} \text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2 + \text{Phthalic acid}$$

Advantage: Pure 1° amine, no contamination with 2° or 3° amines!

Method 6: Reduction of Imines and Oximes

Reduction of Imines (Reductive Amination)

Best for: Secondary and tertiary amines

General Reaction:

$$\text{R-CHO} + \text{R'-NH}_2 \rightarrow \text{R-CH=N-R'} \xrightarrow{[\text{H}]} \text{R-CH}_2\text{-NH-R'}$$

Mechanism:

Step 1: Imine formation

$$\text{R-CHO} + \text{R'-NH}_2 \rightarrow \text{R-CH=N-R'} + \text{H}_2\text{O}$$

(Carbonyl + Amine → Imine)

Step 2: Reduction

$$\text{R-CH=N-R'} \xrightarrow{\text{H}_2/\text{Ni or NaBH}_4} \text{R-CH}_2\text{-NH-R'}$$

(Secondary amine)

Example:

$$\text{CH}_3\text{CHO} + \text{CH}_3\text{NH}_2 \rightarrow \text{CH}_3\text{CH=N-CH}_3 \xrightarrow{[\text{H}]} \text{CH}_3\text{CH}_2\text{-NH-CH}_3$$

Reduction of Oximes

For primary amines:

$$\text{R}_2\text{C=N-OH} \xrightarrow{\text{LiAlH}_4} \text{R}_2\text{CH-NH}_2$$

Example:

$$\text{CH}_3\text{-CO-CH}_3 \xrightarrow{\text{NH}_2\text{OH}} \text{CH}_3\text{-C(=N-OH)-CH}_3 \xrightarrow{\text{LiAlH}_4} (\text{CH}_3)_2\text{CH-NH}_2$$

Method 7: Alkylation of Ammonia

Overview

Produces: Mixture of 1°, 2°, 3° amines and quaternary salt

Reaction:

$$\text{NH}_3 + \text{R-X} \rightarrow \text{RNH}_2 + \text{R}_2\text{NH} + \text{R}_3\text{N} + \text{R}_4\text{N}^+\text{X}^-$$

Why Mixture Forms?

Step 1: Primary amine formation

$$\text{NH}_3 + \text{R-X} \rightarrow \text{R-NH}_2 + \text{HX}$$

Problem: Primary amine is more nucleophilic than NH₃!

Step 2: Secondary amine formation

$$\text{R-NH}_2 + \text{R-X} \rightarrow \text{R}_2\text{NH} + \text{HX}$$

Step 3: Tertiary amine formation

$$\text{R}_2\text{NH} + \text{R-X} \rightarrow \text{R}_3\text{N} + \text{HX}$$

Step 4: Quaternary salt formation

$$\text{R}_3\text{N} + \text{R-X} \rightarrow \text{R}_4\text{N}^+\text{X}^-$$

Controlling Product Distribution

Use excess ammonia to favor primary amine:

$$\text{NH}_3 + \text{CH}_3\text{I} \xrightarrow{\text{excess NH}_3} \text{CH}_3\text{NH}_2 \text{ (major)}$$
JEE Concept: Why This Method is Problematic

Disadvantages of direct alkylation:

  1. ✗ Mixture of products (difficult separation)
  2. ✗ 1° amine is more nucleophilic than NH₃ (continues reacting)
  3. ✗ Low yield of desired product

JEE Strategy:

  • If question asks for pure 1° amine → Use Gabriel synthesis
  • If mixture acceptable → Direct alkylation with excess NH₃

Better alternatives:

  • Gabriel synthesis (for pure 1° aliphatic)
  • Hoffmann degradation (for pure 1°)
  • Reduction of nitro/nitrile (for pure 1°)

Comparison Table: Preparation Methods

MethodProduct TypeCarbon ChangeSelectivityBest For
Nitro reduction1°No changeHighAromatic & aliphatic
Nitrile reduction1°+1 carbonHighChain extension
Amide + LiAlH₄1°, 2°, 3°No changeHighSame carbon count
Hoffmann1° only-1 carbonHighFrom amides
Gabriel1° aliphaticNo changeHighestPure 1° aliphatic
Alkylation of NH₃MixtureNo changeLowAvoid for pure products
Reductive amination2° or 3°VariesModerateSecondary/tertiary

Common Mistakes to Avoid

Mistake #1: Wrong Method for Aromatic Amines

Wrong: “Prepare aniline using Gabriel synthesis”

Why wrong: Gabriel synthesis requires SN2 reaction with RX. Aryl halides (Ar-X) don’t undergo SN2!

Correct method for aniline:

$$\text{C}_6\text{H}_5\text{NO}_2 \xrightarrow{\text{Sn/HCl}} \text{C}_6\text{H}_5\text{NH}_2$$

Rule: Gabriel = Aliphatic 1° amines ONLY

Mistake #2: Confusing LiAlH₄ with Hoffmann

Question: Benzamide → Product?

Wrong reasoning: “Both give amines, so same product”

Correct analysis:

With LiAlH₄:

$$\text{C}_6\text{H}_5\text{CO-NH}_2 \rightarrow \text{C}_6\text{H}_5\text{CH}_2\text{NH}_2$$

(Benzylamine - 7 carbons)

With Br₂/NaOH (Hoffmann):

$$\text{C}_6\text{H}_5\text{CO-NH}_2 \rightarrow \text{C}_6\text{H}_5\text{NH}_2$$

(Aniline - 6 carbons)

Key: Count carbons!

Mistake #3: Expecting Pure Product from NH₃ Alkylation

Wrong: “CH₃I + NH₃ → only CH₃NH₂”

Correct: Mixture of CH₃NH₂, (CH₃)₂NH, (CH₃)₃N, and (CH₃)₄N⁺I⁻

Why: Primary amine is more nucleophilic than NH₃, keeps reacting

Solution for pure 1°: Use Gabriel synthesis or reduce RNO₂/RCN

Practice Problems

Level 1: Foundation (NCERT)

Problem 1: Method Selection

Q: How will you convert: (a) Nitrobenzene to aniline (b) Ethanamide to methanamine

Solution:

(a) Nitrobenzene to Aniline:

$$\text{C}_6\text{H}_5\text{NO}_2 \xrightarrow{\text{Sn + HCl}} [\text{C}_6\text{H}_5\text{NH}_3^+]\text{Cl}^- \xrightarrow{\text{NaOH}} \text{C}_6\text{H}_5\text{NH}_2$$

Method: Reduction of nitro compound

(b) Ethanamide to Methanamine:

$$\text{CH}_3\text{-CO-NH}_2 \xrightarrow{\text{Br}_2 + 4\text{NaOH}} \text{CH}_3\text{NH}_2 + \text{Na}_2\text{CO}_3$$

Method: Hoffmann bromamide degradation Note: Carbon count decreases (2 → 1)

Problem 2: Gabriel Synthesis

Q: Write the steps involved in Gabriel synthesis of butylamine.

Solution:

Target: CH₃CH₂CH₂CH₂NH₂

Step 1: Phthalimide + KOH → Potassium phthalimide

Step 2:

$$\text{Potassium phthalimide} + \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Br} \rightarrow \text{N-Butylphthalimide}$$

Step 3:

$$\text{N-Butylphthalimide} \xrightarrow{\text{H}_3\text{O}^+ \text{ or } \text{NaOH}} \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{NH}_2$$

Byproduct: Phthalic acid (removed by washing)

Level 2: JEE Main

Problem 3: Product Identification

Q: Benzamide is treated with Br₂ and KOH. The product formed is: (A) Benzylamine (B) Aniline (C) Benzyl alcohol (D) Benzoic acid

Solution: (B) Aniline

Reaction: Hoffmann bromamide degradation

$$\text{C}_6\text{H}_5\text{-CO-NH}_2 \xrightarrow{\text{Br}_2/\text{KOH}} \text{C}_6\text{H}_5\text{-NH}_2 + \text{CO}_2$$

Mechanism:

  1. Formation of N-bromobenzamide
  2. Rearrangement to phenyl isocyanate (C₆H₅-N=C=O)
  3. Hydrolysis to aniline

Key: One carbon lost as CO₂!

(A) is wrong: Benzylamine would be product with LiAlH₄ reduction

Problem 4: Multi-step Synthesis

Q: How will you prepare propylamine from ethanol?

Solution:

Retrosynthetic analysis: CH₃CH₂CH₂NH₂ ← CH₃CH₂CN ← CH₃CH₂Br ← CH₃CH₂OH

Forward synthesis:

Step 1: Alcohol → Alkyl bromide

$$\text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{PBr}_3} \text{CH}_3\text{CH}_2\text{Br}$$

Step 2: Alkyl bromide → Nitrile (SN2)

$$\text{CH}_3\text{CH}_2\text{Br} + \text{KCN} \rightarrow \text{CH}_3\text{CH}_2\text{CN} + \text{KBr}$$

Step 3: Nitrile → Amine (reduction)

$$\text{CH}_3\text{CH}_2\text{CN} \xrightarrow{\text{LiAlH}_4} \text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2$$

Advantage: Chain extended by one carbon via nitrile!

Related: Haloalkanes

Level 3: JEE Advanced

Problem 5: Selective Reduction

Q: Suggest a method to convert:

$$\text{O}_2\text{N-C}_6\text{H}_4\text{-CO-NH}_2 \rightarrow \text{H}_2\text{N-C}_6\text{H}_4\text{-CO-NH}_2$$

(Reduce only the nitro group, not the amide)

Solution:

Challenge: Selective reduction of NO₂ in presence of CONH₂

Reagent: Fe/HCl or Sn/HCl (selective for aromatic NO₂)

$$\text{p-}\text{NO}_2\text{-C}_6\text{H}_4\text{-CO-NH}_2 \xrightarrow{\text{Fe/HCl}} \text{p-}\text{H}_2\text{N-C}_6\text{H}_4\text{-CO-NH}_2$$

Why selective:

  • Aromatic NO₂ is easily reduced by Sn/HCl
  • Amide group is stable under these conditions
  • LiAlH₄ would reduce BOTH (avoid!)

Product: p-Aminobenzamide

JEE Concept: Choice of reducing agent determines selectivity!

Problem 6: Mechanism-Based Problem

Q: An organic compound A (C₃H₇NO₂) on reduction with Sn/HCl followed by treatment with NaOH gives compound B. B on treatment with CHCl₃ and KOH gives foul smell. Identify A and B.

Solution:

Analysis of clues:

Clue 1: Reduction with Sn/HCl → suggests nitro compound Clue 2: Foul smell with CHCl₃ + KOH → Carbylamine test (positive for 1° amine)

Molecular formula: C₃H₇NO₂

Possible structures:

  • CH₃-CH₂-CH₂-NO₂ (1-Nitropropane)
  • (CH₃)₂CH-NO₂ (2-Nitropropane)
  • CH₃CH(NO₂)CH₃ would give 2° amine (no carbylamine test)

Since carbylamine test is positive → 1° amine → A must be 1-nitropropane

Compound A: CH₃CH₂CH₂NO₂ (1-Nitropropane)

Reduction:

$$\text{CH}_3\text{CH}_2\text{CH}_2\text{NO}_2 \xrightarrow{\text{Sn/HCl, NaOH}} \text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2$$

Compound B: CH₃CH₂CH₂NH₂ (1-Propylamine)

Carbylamine test:

$$\text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2 + \text{CHCl}_3 + 3\text{KOH} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{-NC} + 3\text{KCl} + 3\text{H}_2\text{O}$$

Foul smell: Propyl isocyanide (carbylamine)

Related: Classification of amines

Quick Revision Box

Starting MaterialReagentProduct TypeCarbon ChangeKey Point
R-NO₂Sn/HCl or LiAlH₄1° amineNoneBest for aromatic
R-CNLiAlH₄ or H₂/Ni1° amine+1Chain extension
R-CO-NH₂LiAlH₄1° amineNoneKeeps carbons
R-CO-NH₂Br₂/NaOH1° amine-1Hoffmann degradation
Phthalimide(1) KOH (2) RX (3) H₃O⁺1° aliphaticNonePure 1° only
NH₃R-XMixtureNoneAvoid (impure)
R-CHO + R’-NH₂H₂/Ni2° amineVariesReductive amination

Decision Tree: Choosing Preparation Method

Need to prepare amine?
│
├─ Primary amine?
│  ├─ Pure 1° aliphatic needed?
│  │  └─ Gabriel synthesis (phthalimide route)
│  │
│  ├─ From nitro compound?
│  │  └─ Reduce with Sn/HCl (aromatic) or LiAlH₄
│  │
│  ├─ Need one more carbon?
│  │  └─ RX → RCN → RCH₂NH₂ (nitrile route)
│  │
│  ├─ From amide (same carbons)?
│  │  └─ LiAlH₄ reduction
│  │
│  └─ From amide (one less carbon)?
│     └─ Hoffmann bromamide degradation
│
├─ Secondary amine?
│  ├─ R-CHO + R'-NH₂ → Reductive amination
│  └─ R-CO-NHR' + LiAlH₄ → Reduction
│
└─ Tertiary amine?
   └─ R-CO-NR'₂ + LiAlH₄ → Reduction

Connection to Other Topics

Prerequisites:

Related Topics:

Applications:

Summary

Key Takeaways

1. High-Selectivity Methods (JEE Favorites):

Gabriel Synthesis:

  • Only method for pure primary aliphatic amines
  • Cannot prepare aromatic amines
  • Phthalimide → N-alkyl phthalimide → 1° amine

Hoffmann Degradation:

  • Primary amines with one less carbon
  • R-CO-NH₂ + Br₂/NaOH → R-NH₂ + CO₂
  • Loses carbon as CO₂

2. Reduction Methods:

Nitro → Amine: (No carbon change)

  • Sn/HCl for aromatic (industrial)
  • LiAlH₄ for sensitive substrates

Nitrile → Amine: (+1 carbon)

  • R-CN → R-CH₂-NH₂ (chain extension)
  • Useful for retrosynthesis

Amide → Amine:

  • LiAlH₄: Keeps carbon count
  • Hoffmann: Loses one carbon

3. Low-Selectivity Method:

Direct Alkylation (NH₃ + RX):

  • Gives mixture (1°, 2°, 3°, quaternary)
  • Avoid for pure products
  • Use excess NH₃ if necessary

4. JEE Strategy:

RequirementMethod
Pure 1° aliphaticGabriel synthesis
Aromatic 1°Nitro reduction
Chain extensionNitrile reduction
From amide (same C)LiAlH₄
From amide (C-1)Hoffmann
2° or 3° amineReductive amination

“For amines, choose your method wisely - Gabriel for purity, Hoffmann for degradation, reduction for retention!”

Next, study amine properties and reactions to understand how these prepared amines react!