Question

Lucas reagent is
(A) conc. HCl and anhydrous $ZnC{{l}_{2}}$
(B) conc. HCl and hydrous $ZnC{{l}_{2}}$
(C) conc. $HN{{O}_{3}}$ and hydrous $ZnC{{l}_{2}}$
(D) conc. $HN{{O}_{3}}$ and anhydrous $ZnC{{l}_{2}}$

Answer 1

The Lucas test in alcohols is a test to differentiate between primary, secondary, and tertiary alcohols. It is based on the difference in reactivity of the three classes of alcohols with hydrogen halides via an ${{S}_{N}}1$ reaction:
$ROH+HCl\to RCl+{{H}_{2}}O$

Complete step by step solution:
We have been asked about Lucas reagent,
Lucas' reagent is a solution of anhydrous zinc chloride in concentrated hydrochloric acid. This solution is used to classify alcohols of low molecular weight. The reaction is a substitution in which the chloride replaces a hydroxyl group.
So, we can say that: Lucas reagent is a mixture of conc. HCl and anhydrous $ZnC{{l}_{2}}$.
It is used to classify the low molecular weight alcohols into primary secondary and tertiary alcohols.
A primary alcohol is alcohol in which the hydroxy group is bonded to a primary carbon atom. It can also be defined as a molecule containing a $C{{H}_{2}}OH$ group. In contrast, secondary alcohol has a formula $CHROH$ and tertiary alcohol has a formula $C{{R}_{2}}OH$ where “R” indicates a carbon-containing group.
$ROH+HCl\to RCl+{{H}_{2}}O$
In this substitution reaction, hydroxide ion is replaced by chloride ion.

Therefore, we can conclude that option (A) is correct.

Note: Lucas’ reagent is a solution of zinc chloride in concentrated hydrochloric acid, used to classify alcohols of low molecular weight. Zinc chloride forms a complex with the oxygen of alcohol and converts the (-OH) to a much better-leaving group thereby giving way to a rapid formation of chloroalkane.