Question

Explain how solubility tests could be used to differentiate between:

a. a phenol and a carboxylic acid.

b. a phenol that contains electron-withdrawing groups and a phenol that contains electron-donating groups.

c. a carboxylic acid and an amine.

d. 1-bromooctane and dibutyl ether. e. 2-pentyne and pentane.

f. 2-nitropentane and 2-aminopentane.

Answer 1

a) a phenol and carboxylic acid

Addition sodium bicarbonate solution

only carboxylic acid dissolves with effervescece and phenol does not.. Phenol is weaker acid than carbonic acid and carboxyllic acid is stronger hence replaces CO2 from bicarbonate solution.

b) A phenol that contains electron -withdrawing groups is more acidic than phenols that contain electron donating groups. Thus again using NaHCO3 solution the former which is more acidic becomes soluble with effervesence(releases of CO2).

c) A carboxylic acid dissolves in basic solution like NaHCO3 while an amine is soluble in acid solution like dil. HCl.

d) 1-bromooctane is a halo alkane which is insoluble in common polar solvents and in acids.

Dibutyl ether soluble in Conc. H2SO4 due to its electron donating Lewis behaviour , forming an oxonium salt.

e) Alkynes are soluble in acetone as they are slightly poar and acetone is also less polar solvent. However alkanes are insoluble even in acetone.

Thus 2-pentyne is soluble in acetone but pentane does not.

f) 2- nitropentane is not soluble in dil. Hcl but 2- aminopentane would be soluble in dil. HCl.