Question

In the preparations of ethyl chloride by Grooves’ method, $C{H}_{3}C{H}_{2}OH+HCl\stackrel{Anhyd.ZNC{l}_2}{\to }C{H}_{3}C{H}_{2}Cl+H_{2}O$ the main reason for using anhydrous $ZnC{l}_2$ as catalyst is that

• A. it absorbs water thereby pushing the equilibrium to the right
• B. it coordinates with the product CH₃CH₂Cl and
• C. it helps in producing the nucleophile Cl^– from HCl
• D. it coordinates with oxygen of ethyl alcohol and thus increases the leaving group ability of the –OH group

Answer 1

Answer: (D)

Due to the smaller size of HCl, the bond between H and Cl is stronger than that between HBr and HI, making HCl less reactive with alcohols. As a result, $ZnC{l}_2$ is required as a catalyst in its reaction with alcohols because it helps in generating a Cl- ion for substitution.

When alcohols and HCl mix, the hydroxyl group is replaced by chlorine in a substitution process. The zinc ion forms a complex with the hydroxyl group by accepting a single electron pair from O of -OH, making it a better leaving group. After that, the chloride ion attacks the last carbonium ion to create an alkyl halide.

• Grooves Process -

When an alcohol is treated with HCl in presence of Zinc chloride it leads to formation of alkyl halide and water on heating. The combination of zinc chloride and HCl is known as Lucas reagent.

Conclusion: Thus, option (B) is the correct answer.