Q.Propan-2-ol on reaction with anhydrous $ZnCl_2$ and concentrated HCl gives ________.

The Lucas reagent (anhydrous $ZnCl_2$ + concentrated HCl) converts an alcohol to the corresponding alkyl chloride via an $S_N1$ mechanism through a carbocation intermediate; propan-2-ol, a secondary alcohol, forms a moderately stable secondary carbocation and reacts within a few minutes to give the secondary chloride, retaining its original carbon skeleton.

The Lucas test/reaction distinguishes primary, secondary, and tertiary alcohols by their differing rates of reaction (via $S_N1$) with the Lucas reagent, based on carbocation stability. Propan-2-ol, $(CH_3)_2CHOH$, being a secondary alcohol, is protonated by the HCl, loses water to form a relatively stable secondary carbocation, $(CH_3)_2CH^+$, which is then quickly captured by chloride ion: $(CH_3)_2CHOH + HCl \xrightarrow{anhyd.\ ZnCl_2} (CH_3)_2CHCl\ (\text{2-chloropropane}) + H_2O$ Since the mechanism proceeds through the same carbon skeleton (via the carbocation at C-2), the product retains the same connectivity as the starting alcohol — 2-chlorop …