Chemistry Organic - How to Solve: Amines (Basicity, Hofmann vs Curtius, Carbylamine Test, Hinsberg Reagent) – NEET UG Guide

How to Solve: Amines (Basicity, Hofmann vs Curtius, Carbylamine Test, Hinsberg Reagent) – NEET UG Guide

Introduction Mastering amines unlocks 5-7 marks in NEET Chemistry—enough to push you from 600 to 650+. These reactions appear in mechanism-based questions, organic conversions, and qualitative analysis, making them a high-yield topic for both theory and numerical problems.

WHAT YOU NEED TO KNOW FIRST

1. Nucleophilicity vs Basicity – How lone pairs on nitrogen behave.
2. Rearrangement Reactions – Migration of alkyl/aryl groups in organic chemistry.
3. Qualitative Tests – How to distinguish primary, secondary, and tertiary amines.

KEY TERMS & FORMULAS

1. Basicity of Amines

Formula: - pKb = -log(Kb) - pKb = Basicity constant (lower pKb = stronger base) - Kb = Base dissociation constant

MEMORISE THIS: - Aliphatic amines (R-NH₂) > Ammonia (NH₃) > Aromatic amines (Ar-NH₂) in basicity. - Electron-donating groups (EDGs) increase basicity (e.g., -CH₃, -OCH₃). - Electron-withdrawing groups (EWGs) decrease basicity (e.g., -NO₂, -CN).

2. Hofmann vs Curtius Rearrangement

MEMORISE THIS: - Hofmann: Uses Br₂ + NaOH (basic conditions). - Curtius: Uses NaN₃ + heat (neutral conditions). - Both involve isocyanate (R-N=C=O) intermediate.

3. Carbylamine Test (Isocyanide Test)

Reaction: Primary amine + CHCl₃ + 3KOH → R-NC (isocyanide) + 3KCl + 3H₂O - Positive test: Foul-smelling isocyanide (R-NC) formed. - Negative test: No reaction (secondary/tertiary amines).

MEMORISE THIS: - Only primary amines give this test. - Secondary & tertiary amines do NOT react.

4. Hinsberg Reagent Test

Reagent: Benzene sulfonyl chloride (C₆H₅SO₂Cl) Reactions:
1. Primary amine (R-NH₂) → Soluble sulfonamide (R-NH-SO₂C₆H₅) - Dissolves in alkali (NaOH).
2. Secondary amine (R₂NH) → Insoluble sulfonamide (R₂N-SO₂C₆H₅) - Does NOT dissolve in alkali.
3. Tertiary amine (R₃N) → No reaction.

MEMORISE THIS: - Primary amine → Soluble in NaOH. - Secondary amine → Insoluble in NaOH. - Tertiary amine → No reaction.

STEP-BY-STEP METHOD

Step 1: Identify the Type of Amine

• Primary (1°): R-NH₂ (two H atoms on N).
• Secondary (2°): R₂NH (one H atom on N).
• Tertiary (3°): R₃N (no H atom on N).

Step 2: Predict Basicity

1. Compare pKb values (lower pKb = stronger base).
2. Check for EDGs/EWGs (EDGs increase basicity, EWGs decrease it).
3. Aliphatic > Ammonia > Aromatic in basicity.

Step 3: Distinguish Between Hofmann & Curtius

1. Hofmann: Uses Br₂ + NaOH (basic conditions).
2. Curtius: Uses NaN₃ + heat (neutral conditions).
3. Both form isocyanate (R-N=C=O) intermediate.

Step 4: Apply Carbylamine Test

1. Add CHCl₃ + KOH to the amine.
2. Foul smell → Primary amine.
3. No smell → Secondary/Tertiary amine.

Step 5: Apply Hinsberg Test

1. Add C₆H₅SO₂Cl (Hinsberg reagent).
2. Dissolves in NaOH → Primary amine.
3. Insoluble in NaOH → Secondary amine.
4. No reaction → Tertiary amine.

WORKED EXAMPLES

Example 1 – Basic (Basicity Comparison)

Question: Which is more basic: CH₃NH₂ or C₆H₅NH₂? Solution:
1. CH₃NH₂ (Methylamine) is aliphatic → Stronger base.
2. C₆H₅NH₂ (Aniline) is aromatic → Weaker base (lone pair delocalized into benzene ring). Answer: CH₃NH₂ > C₆H₅NH₂ in basicity.

What we did and why: - Compared aliphatic vs aromatic amines (aliphatic are stronger bases). - Remember: Aromatic amines are less basic due to resonance.

Example 2 – Medium (Hofmann vs Curtius)

Question: Which reaction converts CH₃COCl → CH₃NH₂? Options: A) Br₂ + NaOH B) NaN₃ + Heat C) Both A & B D) Neither

Solution:
1. Hofmann (Br₂ + NaOH) → Converts amides to amines.
2. Curtius (NaN₃ + Heat) → Converts acyl azides to amines.
3. CH₃COCl (acid chloride) can be converted to amide (CH₃CONH₂) first, then Hofmann.
4. CH₃COCl can also form acyl azide (CH₃CON₃), then Curtius. Answer: C) Both A & B

What we did and why: - Recognized that both reactions can produce primary amines from different precursors. - Hofmann needs an amide, Curtius needs an acyl azide.

Example 3 – Exam-Style (Carbylamine & Hinsberg Test)

Question: An organic compound X gives a foul smell with CHCl₃ + KOH and dissolves in NaOH after treatment with C₆H₅SO₂Cl. Identify X. Solution:
1. Foul smell with CHCl₃ + KOH → Primary amine (Carbylamine test).
2. Dissolves in NaOH after Hinsberg test → Primary amine. Answer: X is a primary amine (e.g., CH₃NH₂).

What we did and why: - Carbylamine test confirms primary amine. - Hinsberg test confirms primary amine (soluble in NaOH).

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP (Night Before Exam)

"Listen up! Amines are high-yield for NEET. Here’s the crash course:

1. Basicity: Aliphatic > Ammonia > Aromatic. EDGs increase, EWGs decrease basicity.
2. Hofmann vs Curtius: Both make primary amines, but Hofmann uses Br₂ + NaOH, Curtius uses NaN₃ + heat.
3. Carbylamine test: Foul smell = primary amine only.
4. Hinsberg test:
5. Dissolves in NaOH → Primary amine.
6. Insoluble in NaOH → Secondary amine.
7. No reaction → Tertiary amine.

Memorise these 4 points, and you’ll nail every amine question in NEET!