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Alcohols - Lucas Test

Easychemistry

When three unlabelled bottles containing 1-butanol, 2-butanol and 2-methyl-2-propanol are each treated with Lucas reagent at room temperature, which sample produces immediate turbidity?

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About This Question

Subject
chemistry
Chapter
organic compounds containing oxygen
Topic
alcohols - lucas test
Difficulty
Easy
Year
2025
Tags
Lucas testalcohol classificationSN1 mechanismcarbocation stabilitytertiary alcohol

Solution

Correct Answer:

2-methyl-2-propanol

The Lucas test distinguishes primary, secondary and tertiary alcohols by the rate at which they form an insoluble chloroalkane with concentrated HCl and anhydrous ZnCl2, a reaction that proceeds through an SN1 pathway. Because the rate-determining step is carbocation formation, the alcohol that generates the most stable cation reacts fastest. 2-methyl-2-propanol is tertiary, so it forms a tertiary carbocation stabilised by three alkyl groups through hyperconjugation and induction, and therefore gives immediate turbidity. 2-butanol is secondary and reacts in about five minutes, while 1-butanol is primary and shows essentially no reaction at room temperature because a primary carbocation is too unstable. Hence the statement that all three react at the same rate is wrong, and 1-butanol and 2-butanol are eliminated by their slower SN1 rates. This is the standard NCERT classification reaction for alcohols and a recurring JEE identification problem. As a plausibility check, the reactivity order tertiary > secondary > primary mirrors carbocation stability, confirming the chosen answer.

This easy difficulty chemistry question is from the chapter organic compounds containing oxygen, covering the topic of alcohols - lucas test. It appeared in the 2025 exam.

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