Alkyl halides are prepared by which of the following reaction(s)? I. $CH_3CH_2OH + HCl \xrightarrow{anh. ZnCl_2}$ II. $CH_3CH_2OH + HCl \rightarrow$ III. $(CH_3)_3COH + HCl \rightarrow$ IV. $(CH_3)_2CHOH + HCl \xrightarrow{anh. ZnCl_2}$

The reaction of alcohols with HCl to form alkyl halides involves the cleavage of the C-O bond. The reactivity order of alcohols towards this reaction is Tertiary ($3^\circ$) > Secondary ($2^\circ$) > Primary ($1^\circ$).

1. Reactions I & II (Primary Alcohol): Primary alcohols are the least reactive. The reaction with HCl requires a catalyst, anhydrous zinc chloride ($ZnCl_2$), to facilitate the cleavage of the strong C-O bond (Groves' process). Therefore, Reaction I proceeds, but Reaction II (without catalyst) does not occur efficiently at room temperature.
2. Reaction III (Tertiary Alcohol): Tertiary alcohols are highly reactive due to the stability of the intermediate carbocation. They react readily with concentrated HCl even in the absence of a catalyst (simply by shaking) at room temperature. Thus, Reaction III proceeds.
3. Reaction IV (Secondary Alcohol): Secondary alcohols have intermediate reactivity and, like primary alcohols, generally require the presence of anhydrous $ZnCl_2$ as a catalyst to generate the alkyl halide. Thus, Reaction IV proceeds.