Question

Explain the results for the tube in which 1.0 M HCl was added to ethyl 4-aminobenzoate. Write a balanced equation for the reaction giving complete structures for all reactants and products. Now explain what happened when 6.0 M NaOH was added to this same tube, and provide a balanced equation. Explain why the different forms of the molecules display such dramatically different solubility behavior.

Answer 1

When 1.0 M HCl is added to ethyl 4-aminobenzoate, a reaction occurs resulting in the formation of carboxylic acid. When 6.0 M NaOH is added to the same tube, another reaction occurs resulting in the formation of a water-soluble salt. The different solubility behavior of these molecules is due to the presence of different functional groups.

Step-by-step explanation:

When 1.0 M HCl is added to ethyl 4-aminobenzoate, a reaction occurs resulting in the formation of a new compound. The balanced equation for this reaction is:

C6H5CO2C2H5 + HCl -> C6H5CO2H + C2H5Cl

When 6.0 M NaOH is then added to the same tube, another reaction occurs. The balanced equation for this reaction is:

C6H5CO2H + NaOH -> C6H5CO2Na + H2O

The different solubility behavior of these molecules is due to the presence of different functional groups. In the first reaction, the ester group (C6H5CO2) is converted to a carboxylic acid group (C6H5CO2H), which increases the solubility. In the second reaction, the carboxylic acid group reacts with the strong base NaOH to form a water-soluble salt (C6H5CO2Na).