Question

Propan-2-ol on reaction with anhydrous ZnCl₂ and concentrated HCl gives:

• A. 1-chloropropane
• B. Ethyl chloride
• C. 2-chloro-2-methyl propane
• D. 2-chloropropane

Hint:

The speed of the Lucas test tells you the alcohol type: - Tertiary (3^): Immediate turbidity. - Secondary (2^): Turbidity in 5-10 minutes. - Primary (1^): No turbidity at room temperature.

Answer 1

The Correct Option is D

Concept: A mixture of anhydrous ZnCl₂ and concentrated HCl is known as Lucas Reagent. It is used to distinguish between primary, secondary, and tertiary alcohols by converting them into their corresponding alkyl chlorides.
Propan-2-ol (CH₃-CH(OH)-CH₃) is a secondary (2^) alcohol because the hydroxyl group is attached to a carbon atom that is further bonded to two other carbon atoms.
When a secondary alcohol reacts with Lucas reagent, a nucleophilic substitution (SₑₓtN1) reaction occurs. The -OH group is replaced by a -Cl atom at the same position. CH₃-CH(OH)-CH₃ conc. HCl + anh. ZnCl₂ CH₃-CH(Cl)-CH₃ + H₂O The product is 2-chloropropane (also known as isopropyl chloride). In this reaction, turbidity (cloudiness) usually appears within 5 to 10 minutes.