Why acid hydrolysis of ester is slow at the beginning when dilute hydrochloric acid is used as catalyst?

Answer 1

Step 1:

An acid/base reaction. Since we only have a weak nucleophile and a poor electrophile we need to activate the ester. Protonation of the ester carbonyl makes it more electrophilic.

Step 2:

The water O functions as the nucleophile attacking the electrophilic C in the C=O, with the electrons moving towards the oxonium ion, creating the tetrahedral intermediate.

Step 3:

An acid/base reaction. Deprotonate the oxygen that came from the water molecule to neutralise the charge.

Step 4:

An acid/base reaction. Need to make the -OCH3 leave, but need to convert it into a good leaving group first by protonation.

Step 5:

Use the electrons of an adjacent oxygen to help "push out" the leaving group, a neutral methanol molecule.

Step 6:

An acid/base reaction. Deprotonation of the oxonium ion reveals the carbonyl C=O in the carboxylic acid product and regenerates the acid catalyst.

Acid produces acts as a catalyst enhances the rate of reaction