What is the order of hydrolysis of methyl acetate in the acidic medium?

Pseudo first order

The correct answer is option 2. Pseudo first order.

The hydrolysis of methyl acetate (\( \text{CH}_3\text{COOCH}_3 \)) in acidic medium typically follows a pseudo-first order reaction.

Methyl acetate hydrolyzes in acidic medium through an acid-catalyzed mechanism. In this mechanism, the acidic proton (\( \text{H}^+ \)) from the acid catalyzes the reaction by protonating the carbonyl oxygen of the methyl acetate, making it more susceptible to nucleophilic attack by water. This leads to the formation of acetic acid and methanol.

The overall reaction is as follows:

\[ \text{CH}_3\text{COOCH}_3 + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{COOH} + \text{CH}_3\text{OH} \]

The rate equation for this reaction can be written as:

\[ \text{Rate} = k [\text{CH}_3\text{COOCH}_3] [\text{H}^+] \]

In acidic medium, the concentration of \( [\text{H}^+] \) is usually kept constant (in excess), so the rate of reaction becomes dependent solely on the concentration of methyl acetate, \( [\text{CH}_3\text{COOCH}_3] \).

The reaction kinetics appear to be first-order with respect to methyl acetate because its concentration solely determines the rate. However, it's essential to note that this reaction is pseudo-first order because it depends on the constant excess concentration of \( [\text{H}^+] \) provided by the acid.

So, the correct answer is: 2. Pseudo first order