eithyl 4-aminobenzoate reaction with 1.0M NaOH formula. the add HCl 1.0HCl formula

2 formulas and explain what happends plz and thank you :)

ethyl 4-aminobenzoate is an ester. The NaOH cleaves the ester and forms CH3CH2OH and the sodium salt of 4-aminobenzoic acid. Then the addition of HCl replaces the Na^+ with a H^+ to form the 4-aminobenzoic acid

The reaction of ethyl 4-aminobenzoate (also known as ethyl p-aminobenzoate or benzocaine) with 1.0 M NaOH and 1.0 M HCl can be described by the chemical equations below:

1) Reaction with 1.0 M NaOH:
C9H11NO2 + NaOH → C9H11NO2Na + H2O

The ethyl 4-aminobenzoate reacts with sodium hydroxide (NaOH) to form sodium ethyl 4-aminobenzoate (a salt) and water. The carboxyl group of the p-aminobenzoate reacts with sodium hydroxide to produce the salt form of the compound.

2) Reaction with 1.0 M HCl:
C9H11NO2Na + HCl → C9H11NO2 + NaCl

The sodium ethyl 4-aminobenzoate reacts with hydrochloric acid (HCl) to regenerate the original ethyl 4-aminobenzoate compound (benzocaine) and form sodium chloride (NaCl). This reaction is a neutralization reaction, where the salt is converted back to its original form.

To perform these reactions, you would mix the appropriate amounts of ethyl 4-aminobenzoate, NaOH, and HCl according to the given concentrations (1.0 M). The reaction with NaOH would result in the formation of the sodium salt, while the reaction with HCl would convert the salt back to the original compound. It is important to note that these reactions would occur in a suitable solvent or medium, typically an aqueous solution.