Question

2. 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. 3. What chemical tests could be used to differentiate between: a. cyclohexanone, cyclohexanol, cyclohexylamine, and cyclohexanecarboxylic acid. b. 4-methyl-1-pentanol, 4-methyl-2-pentanol, and 2-methyl-2-pentanol.

Answer 1

Dear candidate as per regulations one question with not more than four parts could be answered one time. You have asked two questions with a total of 8 parts, here we attempt first question with 6 parts. please send other one separately

2.

a) Carboxylic acids are more acidic than phenol andreact with NaHCO₃ and produce effervescence, while as phenolsdo not react with NaHCO_3.

b) Phenols with electron withdrawing groups are more basic than phenols with electron donating groups. Hence Phenols with electron withdrawing groups react ith weak bases like NaHCO3 while as phenols with electron donating groups do not.

c) carboxylic acids are acidic in nature and amines are baic. Amines form salts if few drops of acid such as HCl is added. These salts will be insoluble in orgaic solvents such as hexane . Carboxylic acids form salts with basis.

d) Ethers are miscible with water due to H-bonding while as alkanes are not. 1-bromooctane will be immiscible with water.

e) 2-pentyne when mixed with bromine water decolorize the bromine water while as pentane will not decolorize bromine water.

f) 2-aminopentane will react with acids such as HCl to form a salt insoluble in solvents such as ether heane etc. while as 2-nitropentane will not