The acid-catalyzed cleavage of an ether involves the protonation of the ether oxygen, followed by the nucleophilic attack of a halide ion, and finally the deprotonation of the intermediate to give the products. Here's the stepwise mechanism for the reaction between ethyl ether CH3CH2-O-CH2CH3 and hydrochloric acid HCl :Step 1: Protonation of the ether oxygenThe lone pair of electrons on the oxygen atom of the ether interacts with the hydrogen atom of the hydrochloric acid, forming a bond and resulting in the protonation of the ether. This generates a positively charged oxonium ion CH3CH2-O + -CH2CH3 and a chloride ion Cl- .Step 2: Nucleophilic attack by the halide ionThe chloride ion Cl- acts as a nucleophile and attacks one of the carbon atoms bonded to the oxygen in the oxonium ion. This leads to the formation of a new carbon-chlorine bond and the breaking of the carbon-oxygen bond. In this case, the chloride ion can attack either of the carbon atoms, resulting in the formation of two different intermediates: CH3CH2-O-CH2CH2-Cl and CH3CH2-Cl .Step 3: Deprotonation of the intermediateThe oxygen atom in the intermediate CH3CH2-O-CH2CH2-Cl donates its lone pair of electrons to form a new bond with the hydrogen atom of another hydrochloric acid molecule. This results in the breaking of the hydrogen-chlorine bond, generating water H2O and another chloride ion Cl- . The chloride ion then attacks the other carbon atom, forming a new carbon-chlorine bond and breaking the carbon-oxygen bond, resulting in the formation of the second product CH3CH2-Cl .Overall, the reaction between ethyl ether and hydrochloric acid results in the formation of two products: ethyl chloride CH3CH2-Cl and chloroethane CH3CH2-Cl .