Question

Explain Grove's process?

Answer 1

Hint:Grove's process is a method used for the synthesis of haloalkanes and the reaction is carried out in by passing HCl gas and using a Lewis acid. The reaction involves bimolecular nucleophilic substitution mechanisms.

Complete step-by-step answer:First let us define which reaction is called Grove's reaction.
Grove's reaction is a reaction which is used for the synthesis of haloalkanes like chloroalkanes from primary or secondary alcohol in the presence of a Lewis acid like anhydrous zinc chloride and a hydrogen halide like HCl.
We could synthesize any haloalkane by replacing the hydrogen halide with respect to the final product required.
The order of reactivity of the hydrogen order is:$HI>HBr>HCl>HF$
The reaction mechanism involved in this reaction is bimolecular nucleophilic substitution reaction i.e. ${{S}_{N}}^{2}$ reaction.
As the mechanism involved in the Grove's method is${{S}_{N}}^{2}$, then the rate of the reaction will be second order.
Hence the Grove's method is an example of second order reaction.
Rate of the reaction will be, $rate=k[C{{l}^{-}}][alcohol]$
When a primary alcohol reacts with HCl in the presence of anhydrous$ZnC{{l}_{2}}$, primary chloroalkane is formed.
For example:\[C{{H}_{3}}C{{H}_{2}}OH+HCl\xrightarrow[heat]{Anhydrous.ZnC{{l}_{2}}}C{{H}_{3}}C{{H}_{2}}Cl+{{H}_{2}}O\]
Ethanol reacts with HCl in the presence of anhydrous $ZnC{{l}_{2}}$ and the final product formed is 1-chloroethane.
When secondary alcohol reacts with HCl in the presence of Lewis acid then the product formed is secondary alkyl halide.
For example:\[C{{H}_{3}}-\underset{OH}{\mathop{\underset{|}{\mathop{C}}\,}}\,H-C{{H}_{3}}+HCl\xrightarrow[heat]{Anhydrous\,ZnC{{l}_{2}}}C{{H}_{3}}-\underset{Cl}{\mathop{\underset{|}{\mathop{C}}\,}}\,H-C{{H}_{3}}+{{H}_{2}}O\]
Propan-2-ol reacts with HCl in the presence of anhydrous$ZnC{{l}_{2}}$ to form 2-chloropropane.

Note:Tertiary alcohol is not synthesized in this method since in tertiary alcohols two alkyl groups will be attached to the C attached to –OH functional group and due to the presence of the bulky group the C will be sterically hindered. So the reaction does not proceed since the reaction follows the ${{S}_{N}}^{2}$ mechanism.
But by using concentrated HCl we could prepare tertiary alkyl halide using tertiary alcohol.
And the anhydrous $ZnC{{l}_{2}}$ is a Lewis acid and the alcohol is a Lewis base, $ZnC{{l}_{2}}$ reacts with HCl to form a complex ion through the alcoholic –OH group and it weakens the C-OH bonds and C-Cl bond is formed.