The reaction between ethanoic anhydride CH3CO 2O and water H2O to form ethanoic acid CH3COOH is a hydrolysis reaction. The mechanism for this reaction can be explained in the following steps:Step 1: Nucleophilic attackWater, which acts as a nucleophile, attacks the carbonyl carbon of one of the acyl groups in ethanoic anhydride. This leads to the formation of a tetrahedral intermediate. The oxygen atom in water donates a pair of electrons to the carbonyl carbon, while the carbonyl double bond's electrons shift to the carbonyl oxygen, creating a negative charge on the oxygen atom.Step 2: Proton transferThe negatively charged oxygen atom in the tetrahedral intermediate abstracts a proton H+ from the water molecule, which acts as a proton donor. This results in the formation of a hydroxyl group -OH attached to the carbonyl carbon.Step 3: Reforming the carbonyl groupThe negatively charged oxygen atom donates its electron pair back to form a double bond with the carbonyl carbon. At the same time, the electrons from the oxygen-carbon bond in the adjacent acyl group shift to the oxygen atom, breaking the bond between the two acyl groups.Step 4: Formation of ethanoic acidThe negatively charged oxygen atom in the leaving acyl group abstracts a proton H+ from the hydroxide ion OH- formed in step 2. This results in the formation of ethanoic acid CH3COOH .Overall, the reaction can be summarized as follows: CH3CO 2O + H2O 2 CH3COOHIn this reaction, ethanoic anhydride is hydrolyzed by water to form two molecules of ethanoic acid. The mechanism involves nucleophilic attack, proton transfer, and the breaking and reforming of bonds.