[Milind_95]

How do I separate a mixture of N-Butanol and water which is forming due to the decomposition of a Tin catalyst in esterification reactions?

[chenbeier]

Probably destillation under reduced pressure. I dont know if an azeoptrop is built.

[pgk]

n-Butanol (b.p. 118oC) and water (b.p. 100oC) form an azeotrope at 92.5oC.
So, you distill at 92-93oC under normal pressure, in order to remove water and when the distillation temperature exceeds 93oC, you cool  below 100oC and apply reduced pressure, in order to remove the rest of n-butanol.
Hint 1: If butanol exists in very low amount, you may not observe the azeotrope point due to fast distillation.
Hint 2: Do not apply reduced pressure from the beginning because you will destroy the azeotrope and risk an incomplete removing of water.
Hint 2: In presence of other solvents (e.g. toluene) you may form ternary azeotropes below 92.5oC.

[Babcock_Hall]

I am confused by the question.  Water and n-butanol are immiscible, a property that happens to be useful when pouring polyacrylamide gels.  Obviously they are still slightly soluble in each other.

[pgk]

n-Butanol has log P = 0.9 and aqueous solubility = 7.3% at 25 °C. Consequently, n-butanol and water cannot be considered as immiscible, neither as “slightly” soluble in each other. One the other hand, the low solubility of n-butanol and water in each other, prevents sedimentation when pouring polyacrylamide gels.

[Babcock_Hall]

It is my understanding that immiscible solvents form two layers, which is what n-butanol and water do.  What is your definition?

[pgk]

The definition is that indeed, immiscible solvents form two separate layers, but not only. Solvents with low miscibility to each other, can also form two separate layers. However, the content of each solvent in each separate layer depends on their logP value that reflects on their solubility to each other.
Thus, when mixing n-butanol and water, the aqueous layer contains n-butanol up to 7.3%. Similarly, when extracting n-butanol with equal volumes of an organic solvent and water, ≈12% of n-butanol is passing into the aqueous phase. 

[pgk]

In contrast, when mixing water and toluene that has aqueous solubility = 0.05% at 20 °C, the aqueous layer contains toluene up to 0.05%. Besides, when extracting toluene that has logP = 2.7 with equal volumes of an organic solvent and water, ≈ 0.2% of toluene is passing into the aqueous phase.

[hollytara]

There is a desire to create a dichotomy: soluble or miscible vs. insoluble or immiscible.  But real world solubility doesn't work that way - there are some compounds that are "infinitely miscible" and make solutions in any proportions (like water and methanol), but most of the time there is a solubility limit.  In order to make the dichotomy, you have to define an arbitrary cut off.  I have seen different cutoff values, usually around 5-10 g in 100 mL, but sometimes higher or lower.