According Zeisel method, when an ether reacts with cold $HI$, alkyl iodide and alcohol are formed. In case of asymmetrical ether, the alkyl halide is always formed from the smaller alkyl group provided no tertiary $\left(3^{\circ}\right)$ alkyl group is present and if any $3^{\circ}$ alkyl group is present, the halogen gets attached with it. In case of alkyl aryl ethers, the products are always phenol and an alkyl halide.
(a) $CH _{3}- O - C _{4} H _{9}+ HI \stackrel{\text { Cold}}{\longrightarrow } CH _{3} I + C _{4} H _{9} OH$