Question

$(C{H}_3{)}_{3}C-OH+HCl\stackrel{ZnC{l}_2}{\to }(C{H}_3{)}_{3}C-Cl+H_{2}O$
In the above reaction, the leaving group is:

• A. $H{O}^{\ominus }$
• B. $H_{2}O$
• C. $HOZ\stackrel{\ominus }{n}C{l}_2$
• D. $H_3{O}^{\oplus }$

Answer 1

The correct option is C $HOZ\stackrel{\ominus }{n}C{l}_2$

Lucas reagent reacts with alcohols to form precipitates of alkyl halides.
$ZnC{l}_2$ is a Lewis acid so oxygen of -OH forms coordinate bond with Zn or in other words complexes with zinc.
Oxygen acquires a positive charge and Zn acquires a negative charge. Thus, oxygen becomes a good leaving group and the reaction progresses in the forward reaction. At last chloride ion as a nucleophile attack the carbocation to form alkylhalide.
First step in Lucas test reaction​ is acid-base reaction with $ZnC{l}_2$