Question

Which of the following binary mixture does not show the behavior of minimum boiling azeotropes?

• A. \( \text{CS}_2 + \text{CH}_3\text{COCH}_3 \)
• B. \( \text{H}_2\text{O} + \text{CH}_3\text{COC}_2\text{H}_5 \)
• C. \( \text{C}_6\text{H}_5\text{OH} + \text{C}_6\text{H}_5\text{NH}_2 \)
• D. \( \text{CH}_3\text{OH} + \text{CHCl}_3 \)

Hint:

- Minimum boiling azeotropes: Positive deviation (weaker interactions) - Maximum boiling azeotropes: Negative deviation (stronger interactions) - Look for H-bonding capability differences between components - Similar molecules (like phenol + aniline) tend to form maximum boiling azeotropes

Answer 1

The Correct Option is C

Key Concept: 
Minimum boiling azeotropes form when:
- Components show positive deviation from Raoult's law
- Molecular interactions between unlike molecules are weaker than between like molecules
- Typically occurs between molecules with different polarity or hydrogen bonding capacity

Analysis of Options:
- Option 1: CS₂ + CH₃COCH₃
- Carbon disulfide (non-polar) + acetone (polar)
- Forms minimum boiling azeotrope (shows positive deviation)
- Option 2: H₂O + CH₃COC₂H₅
- Water (strong H-bonding) + methyl ethyl ketone (weak H-bonding)
- Forms minimum boiling azeotrope
- Option 3: C₆H₅OH + C₆H₅NH₂
- Phenol + aniline (both can form strong intermolecular H-bonds)
- Shows negative deviation (forms maximum boiling azeotrope)
- Correct answer as it doesn't form minimum boiling azeotrope
- Option 4: CH₃OH + CHCl₃
- Methanol + chloroform (forms H-bonded complex)
- Shows positive deviation (minimum boiling azeotrope)