Step 1: Identify the starting ether. The substrate is phenetole, the aryl-alkyl ether $C_6H_5-O-CH_2CH_3$ (ethyl phenyl ether). Step 2: Apply the rule for HBr cleavage. With $HBr$, an aryl-alkyl ether always breaks at the alkyl-oxygen bond, because the aryl-oxygen bond has partial double bond character and resists cleavage. The halide goes to the alkyl group. Step 3: Write products X and Y. Cleavage gives phenol and ethyl bromide: $C_6H_5-O-CH_2CH_3+HBr \rightarrow C_6H_5OH+C_2H_5Br$. So $X=$ phenol and $Y=$ ethyl bromide. Step 4: Carry out the bromination. Bromine in acetic acid does electrophilic aromatic substitution on phenetole. The ethoxy group is strongly o,p-directing, and the para position dominates. Step 5: Identify Z. The major product is $p$-bromophenetole, so $Z=p$-bromophenetole. Step 6: Choose the option. $X=$ phenol, $Y=$ ethyl bromide, $Z=p$-bromophenetole corresponds to the figure in option (2). \[ \boxed{X=phenol,\ Y=C_2H_5Br,\ Z=p\text{-bromophenetole (Option 2)}} \]
