I was watching a video. The instructor said he was skipping a "cleanup step" where I circled in red, and went directly to the product of the reaction. Can someone draw the missing mechanism so I can better understand? **Carboxylic Acids - Addition-Elimination Reaction**This diagram illustrates the mechanism of an addition-elimination reaction involving carboxylic acids.1. **Starting Materials:** - A hydroxyl group (\(-OH\)) on a carboxylic acid, represented by the first structure with a carbonyl group (\(C=O\)). - Methanol (\(CH_3OH\)) and a proton (\(H\)) participate as reactants.2. **Reaction Mechanism:** - **First Step:** The hydroxyl group acts as a nucleophile, attacking the carbon of the carbonyl group. This is indicated by a curved arrow pointing from the oxygen of the hydroxyl group to the carbon of the carbonyl group. - **Intermediate Formation:** The intermediate with a positively charged oxygen is represented. A proton (\(+\)H) is lost in this step. - **Methanol Attack:** Simultaneously, another curly arrow represents attack by methanol (\(CH_3OH\)) forming an ester linkage. Water (\(H_2O\)) is a byproduct. - **Product Formation:** The final product is shown in a separate box, depicting an ester with the formula \(\text{OCH}_3\).3. **Red Encircled Section:** - This section details more about the attachment of \(CH_3OH\). The curly arrow shows proton transfer to form an ester while removing the water molecule.This illustration explains how carboxylic acids undergo nucleophilic substitution to form esters, highlighting critical structural transformations during the reaction.

In the given reaction mechanism for the formation of ester from acid(CH3COOH) and alcohol(CH3OH), H+(or H3O+) is used as a catalyst to speed up the reaction. Since H+(or H3O+) is a catalyst, it is not used up in the reaction. Hence the catalyst, H+(or H3O+) that is used in the 1st step of the mechanism, is regenerated in the cleanup step. The missing cleanup step releasing the catalyst, H+(or H3O+) is shown below: