Correct Answer - C
`HCl` reacts with `1^(@)` and `2^(@)` alcohols to yield the corresponding chloride only in the presence of anhydrous `ZnCl_(2)`,
Thus `(I)` and `(IV)` can be used due to presence of anhydrous `ZnCl_(2)`, while `(III)` gives alkyl halide due to formation of more stable carboncation.