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

An azeotrope boils at a constant temperature because the vapour phase has the same composition as the liquid phase.
  • If the total vapour pressure of the mixture is higher than that predicted by Raoult’s law (positive deviation), the boiling point of the mixture becomes lower than that of either pure component. Such an azeotrope is called a minimum-boiling azeotrope.
  • If the total vapour pressure is lower than Raoult’s law (negative deviation), the boiling point is higher than that of both components, giving a maximum-boiling azeotrope.

Whether a binary liquid shows positive or negative deviation depends on the relative strength of intermolecular forces:
  • Interactions in the mixture weaker than those within the pure liquids ⇒ molecules escape more easily ⇒ higher vapour pressure ⇒ positive deviation ⇒ minimum-boiling azeotrope.
  • Interactions in the mixture stronger than in the pure liquids (e.g. new hydrogen bonds) ⇒ molecules escape with difficulty ⇒ lower vapour pressure ⇒ negative deviation ⇒ maximum-boiling azeotrope.

Now examine each option.

Case A: $$H_2O + CH_3COOC_2H_5$$ (water + ethyl acetate)
Water is highly polar; ethyl acetate is much less polar and cannot hydrogen-bond effectively with water. Hence, the A-B (unlike) interactions are weaker than A-A and B-B interactions. The mixture shows a large positive deviation. Therefore it forms a minimum-boiling azeotrope (bp ≈ $$70^{\circ}C$$).

Case B: $$C_6H_5OH + C_6H_5NH_2$$ (phenol + aniline)
Both phenol and aniline possess -OH/-NH2 groups that can form strong intermolecular hydrogen bonds with each other. In the mixture the A-B attractions are stronger than those in the pure liquids, so the vapour pressure drops (negative deviation). Consequently, if an azeotrope is formed at all, it will be a maximum-boiling one; it can never be a minimum-boiling azeotrope. Hence this binary system does not show minimum-boiling behaviour.

Case C: $$CS_2 + CH_3COCH_3$$ (carbon disulphide + acetone)
CS2 is non-polar while acetone is polar but cannot hydrogen-bond strongly with CS2. Thus A-B interactions are weaker ⇒ positive deviation ⇒ the pair forms a minimum-boiling azeotrope (bp ≈ $$39^{\circ}C$$).

Case D: $$CH_3OH + CHCl_3$$ (methanol + chloroform)
Though each component can engage in hydrogen bonding, the two form a 1 : 1 complex in the liquid phase which is less stable than the separate hydrogen-bond networks in the pure liquids. Hence the net A-B interaction is weaker, giving a positive deviation and a minimum-boiling azeotrope (bp ≈ $$55.9^{\circ}C$$ at 1 atm).

Only the mixture in Option B fails to exhibit minimum-boiling behaviour.

Correct answer: Option B ($$C_6H_5OH + C_6H_5NH_2$$)