Question

1) 2-Fluorobenzoic acid has a pKa of 3.27. 4-Nitrobenzoic acid has a pKa of 3.41. p-Cresol has a
pKa of 10.26. Acetanilide is neutral. You will need the following information: The pKa of HCl is
–7; pKa of carbonic acid is 6.52 and the pKa of water is 15.
Mixture A is a mixture of 2-fluorobenzoic acid and p-cresol
Mixture B is a mixture of 4-nitrobenzoic acid and p-cresol
Mixture C is a mixture of 2-fluorobenzoic acid and 4-nitrobenzoic acid.
All mixtures are dissolved in methylene chloride.

a) What substance (among HCl, NaOH and NaHCO3) could you use to separate the components
of Mixture A? What substance would exist in the aqueous layer AFTER the extraction?

b) Why would reacting Mixture B with an aqueous solution of NaOH NOT separate one
substance from the other?

c) Why would reacting Mixture C with an aqueous solution of NaHCO3 NOT separate one
substance from the other?

2) The distribution coefficient between methylene chloride and water for solute Y is 12. An
amount of 95.0 g of Y is dissolved in 120 mL of water.

a) What weight of Y would be removed from water with a single extraction with 120-mL of
methylene chloride? Report to 1 decimal place.

b) What weight of Y would be removed from water (the original solution) with two successive
extractions with 60-mL portions each of methylene chloride? Report to 1 decimal place.

In what case would column chromatography be used to purify a desired compound because techniques like recrystallization, extraction and distillation are not adequate to do so?

2) You have 122.0 g of solute Q dissolved in 150-mL of water. The distribution coefficient between ether and water for solute Q is 9.

Answer 1

1 a) What substance (among HCl, NaOH and NaHCO3) could you use to separate the components of Mixture A? What substance would exist in the aqueous layer AFTER the extraction?

Ans: Mixture A contains 2-fluorobenzoic acid and p-cresol. Here the pKa of 2-fluorobenzoic acid is 3.27 and the pKa of p-cresol is 10.26. Both of these are dissolve in methylene dichloride.

If we use aqeous solution of HCl which has a pKa of -7, the 2-fluorobenzoic acid will exist in undissociated form. Since the pKa of 2-fluorobenzoic acid is 3.27, it would exist in disscociated form as benzoic acid anion and H⁺ ion. Under acidic conditions ie in HCl this benzoic acid mostly exists as COOH rather than COO^- and H⁺ ion.So it is not possible to extract 2-fluorobenzoic acid into aqeous layer.

If we use aqeous solution of NaOH solution it is strongly basic in nature, so it has a capacity to abstract protons form both the benzoic acid and p-cresol. This leads to movement of both these compounds in the Mixture A in methylene chloride into aqeous layer.

If we use aqeous solution of NaHCO₃ - is has a pKa of 6.52. It is basic enough to form a sodium salt of carboxylic acid which will be soluble in aqueous layer. NaHCO3 is not basic enough to form a sodium phenoxide ion and as a result the phenol exists in undissociated form and cannot move into aqueous layer.

We should use aqeous solution of NaHCO3 to seperate the mixture of 2-fluoro benzoic acid and p-cresol. The sodium salt of 2-fluorobenzoic acid moves to aqeous layer.

b) Why would reacting Mixture B with an aqueous solution of NaOH NOT separate one
substance from the other?

Ans: Mixture B is a mixture of 4-nitrobenzoic acid and p-cresol

an aqueous solution of NaOH NOT separate one substance from the other because the aqeous solution of NaOH is strongly basic in nature and would form sodium salt of benzoic acid and sodium salt of p-cresol where both would go into aqeous layer. So we cannot seperate them by using aqueous solution of NaOH. It is required that only one should be soluble in aqeous layer and other should stay in methylene dichloride to separate this mixture. But it is not possible with aqueous solution of NaOH.

c) Why would reacting Mixture C with an aqueous solution of NaHCO3 NOT separate one
substance from the other?

Ans: Mixture C is a mixture of 2-fluorobenzoic acid and 4-nitrobenzoic acid.

Both of these would form sodium salt of benzoic acid and would move into aqeous layer. So it is impossible to separate this mixture by using aqueous solution of NaHCO3

2) The distribution coefficient between methylene chloride and water for solute Y is 12. An
amount of 95.0 g of Y is dissolved in 120 mL of water.

a) What weight of Y would be removed from water with a single extraction with 120-mL of
methylene chloride? Report to 1 decimal place.

Ans: P =12 ;

P = [Y]_{organic layer} / [Y ]_Water

12 = x/ 120/(95-x)/120

12 = x/(95-x)

12 (95-x) = x

12x95 - 12x = x

13x = 12 x 95

x = 87.6 gm of Y

87.6 gm weight of Y would be removed from water with a single extraction with 120-mL of
methylene chloride.

b) What weight of Y would be removed from water (the original solution) with two successive
extractions with 60-mL portions each of methylene chloride? Report to 1 decimal place.

Ans: First extraction:

12 = x/60/(95-x)/120

12(95-x) = 2x

12 x 95 - 12x = 2x

14x = 12 x 95

x= 81.4 gm

So the amount of Y removed from water after first extraction is 81.4 gm.

Now the amount of Y remaining in water is = 95 - 81.4 = 13.6 gm (Remained in water)

Second extraction:

12 = x /60/ (13.6-x)/ 120

12 = 2x / (13.6-x)

12 (13.6-x) =2x

12 x 13.6 - 12x = 2x

14x = 12 x 13.6

x = 11.5 gm

11.5 gm Y is removed from water after second extraction with methylene dichloride.