How does ethanoic acid react with alcohols?

Ethanoic acid reacts with alcohols in the presence of a catalyst to form an ester and water.

In more detail, this reaction is known as esterification. When ethanoic acid (CH3COOH) reacts with an alcohol (R-OH), an ester (R-COOH) and water (H2O) are formed. The general equation for this reaction is: CH3COOH + R-OH → R-COOH + H2O. The 'R' in the equation represents the alkyl group from the alcohol.

The reaction requires a strong acid catalyst, usually concentrated sulphuric acid, to proceed. The catalyst is needed to donate a proton (H+) to the alcohol, which allows the reaction to occur. The reaction is also reversible, meaning it can go in both directions. This is indicated by the arrow pointing both ways in the equation.

The ester that is formed depends on the alcohol that is used. For example, if the alcohol is ethanol (C2H5OH), the ester formed is ethyl ethanoate (CH3COOC2H5). This ester has a sweet, fruity smell and is often used in perfumes and flavourings.

The reaction between ethanoic acid and alcohols is an important one in organic chemistry. It is used in the production of a wide range of products, from plastics to pharmaceuticals. Understanding this reaction can help you to understand the wider principles of organic chemistry, including the role of catalysts and the concept of reversible reactions.