What ester is formed when methanol reacts with acetic acid?

Answer 1

The reaction of acetic acid and methanol affords water and an ester, namely methyl methanoate. The mechanism of this reaction can involves two pathways. In one, the hydroxyl hydride is displaced by a strong base, leaving a methanoate anion, which nucleophilically attacks the methyl group of the alcohol. Also, the carbonyl double bond of the acid can form a bond with the methyl group. The hydroxyl group of what used to be methanol leaves. Immediately following, the O-H bond of the acetic acid collapses to form a carbonyl, a proton is abstracted, which reacts with the hydroxide left from the loss of hydrogen from the water to form methyl methanoate and water. See esterfication and Fischer esterfication

Answer 2

Acetic acid: CH3COOH

2-methyl-1-propanol: CH3CH2CH(CH3)OH

Final product: CH3COOCH3CHCH2CH3

The reaction between the two is a condensation reaction, generating an ester and losing an H2O. Here's a simplified version of it:

Acid: R1CO-OH

Alcohol: R2O-H

Final product: R1CO-OR2 (dashed line is to show where each part came from)

An easy to to remember what the product will be is to take away an -OH from the acid, take an -H from the alcohol, and forge a link between the two to create the ester.

Some people may create questions to help remember this. For example, if the alcohol has a labeled O18 and whether it will be on the ester or in H20, the answer would be the alcohol that retained the labeled oxygen.

Answer 3

methyl ethanoate + water