Answer
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Hint: Lucas reagent is a mixture of HCl and \[ZnC{{l}_{2}}\]. Lucas reagent is used to test the presence of alcohol in the given compound. Alcohols in organic compounds react with Lucas reagent and form Alkyl halides as the products.
Complete step by step answer:
\[ZnC{{l}_{2}}\]is a Lewis acid because of the presence of empty d-orbitals on Zinc.
Oxygen in –OH group of alcohol forms a coordinate covalent bond with zinc and oxygen acquires a positive charge and Zinc acquires a negative charge.
Therefore oxygen becomes a good leaving group and then the reaction moves towards the formation of the product.
The formed carbonium ion reacts with chlorine in hydrochloric acid and forms alkyl halide as a product.
The above reaction can be represented as follows.
\[R-OH+HCl\xrightarrow{ZnC{{l}_{2}}}R-Cl+{{H}_{2}}O\]
In the above reaction \[ZnC{{l}_{2}}\]acts as a catalyst.
\[ZnC{{l}_{2}}\] reacts with the excess of HCl and forms the following complex.
\[ZnC{{l}_{2}}+2HCl\to {{H}_{2}}ZnC{{l}_{4}}\]
Therefore \[ZnC{{l}_{2}}\]acts a catalyst and forms \[{{H}_{2}}ZnC{{l}_{4}}\]by reacting with hydrochloric acid (HCl).
So, the correct answer is “Option A”.
Note: When primary alcohols react with Lucas reagent the solution remains colorless unless it is heated.
Secondary alcohols react with Lucas reagent and form an oily layer in 3-5 minutes.
Tertiary alcohols turn to turbid and form an oily layer immediately by the addition of Lucas reagent.
Complete step by step answer:
\[ZnC{{l}_{2}}\]is a Lewis acid because of the presence of empty d-orbitals on Zinc.
Oxygen in –OH group of alcohol forms a coordinate covalent bond with zinc and oxygen acquires a positive charge and Zinc acquires a negative charge.
Therefore oxygen becomes a good leaving group and then the reaction moves towards the formation of the product.
The formed carbonium ion reacts with chlorine in hydrochloric acid and forms alkyl halide as a product.
The above reaction can be represented as follows.
\[R-OH+HCl\xrightarrow{ZnC{{l}_{2}}}R-Cl+{{H}_{2}}O\]
In the above reaction \[ZnC{{l}_{2}}\]acts as a catalyst.
\[ZnC{{l}_{2}}\] reacts with the excess of HCl and forms the following complex.
\[ZnC{{l}_{2}}+2HCl\to {{H}_{2}}ZnC{{l}_{4}}\]
Therefore \[ZnC{{l}_{2}}\]acts a catalyst and forms \[{{H}_{2}}ZnC{{l}_{4}}\]by reacting with hydrochloric acid (HCl).
So, the correct answer is “Option A”.
Note: When primary alcohols react with Lucas reagent the solution remains colorless unless it is heated.
Secondary alcohols react with Lucas reagent and form an oily layer in 3-5 minutes.
Tertiary alcohols turn to turbid and form an oily layer immediately by the addition of Lucas reagent.
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