Read the passage given and answer the question.

Phenol and alcohol have the same functional group (-OH) but phenols are stronger acids than alcohols due to the resonance stabilization of phenoxide ion than alkoxide ion \((-RO^-)\). Electron withdrawing groups tend to increase the acidic strength while electron releasing groups tend to decrease it.

Compound '\(A\)' on reaction with conc. \(HCl\) and anhydrous \(ZnCl_2\) gives turbidity immediately. Compound '\(A\)' is:

\((CH_3)_3COH\)

The correct answer is option 3. \((CH_3)_3COH\).

Let us explore the Lucas test in detail and how it helps identify the correct alcohol in this scenario.

The Lucas test is a qualitative test in organic chemistry used to differentiate between primary, secondary, and tertiary alcohols based on their reactivity with Lucas reagent (a mixture of concentrated hydrochloric acid, \(HCl\), and anhydrous zinc chloride, \(ZnCl_2\)).

The test relies on the formation of an alkyl chloride from the alcohol. The general reaction is:

\(R-OH + HCl + ZnCl_2 \longrightarrow R-Cl + H_2O\)

Here, \(R-OH\) is the alcohol, and \(R-Cl\) is the alkyl chloride formed.

The formation of an alkyl chloride makes the solution turbid (cloudy) because the alkyl chloride is insoluble in the reaction medium.

Reactivity of Alcohols

Tertiary Alcohols (3° alcohols):

These have three alkyl groups attached to the carbon that bears the hydroxyl group (-OH). The reaction with Lucas reagent is very fast because the formation of a stable carbocation is easy. Turbidity forms immediately.

Secondary Alcohols (2° alcohols):

These have two alkyl groups attached to the carbon bearing the hydroxyl group. The reaction is moderate because the carbocation is less stable than that of tertiary alcohols. Turbidity forms within a few minutes.

Primary Alcohols (1° alcohols):

These have one alkyl group attached to the carbon bearing the hydroxyl group. The reaction is very slow because the formation of a carbocation is difficult and unstable. Turbidity may take a long time to form or may not form at all under normal conditions.

Analyzing the Given Options

Option 1: \((CH_3)_2CHOH\) (Isopropanol)

\((CH_3)_2CHOH\) is a secondary alcohol. As a secondary alcohol, it would react with Lucas reagent moderately fast, forming turbidity after a few minutes.

Option 2: \(CH_3CH_2OH\) (Ethanol)

\(CH_3CH_2OH\) is a primary alcohol. As a primary alcohol, it reacts very slowly with Lucas reagent. Turbidity would take a long time to form, if at all.

Option 3: \((CH_3)_3COH\) (Tert-butyl alcohol)

\((CH_3)_3COH\) is a tertiary alcohol. As a tertiary alcohol, it reacts very rapidly with Lucas reagent, forming turbidity immediately due to the quick formation of a stable carbocation.

Option 4: \(CH_3OH\) (Methanol)

\(CH_3OH\) is a primary alcohol. Like other primary alcohols, it reacts very slowly with Lucas reagent, with turbidity taking a significant time to form.

Conclusion

The question specifies that compound \(A\) forms turbidity immediately when treated with concentrated HCl and anhydrous ZnCl₂. This behavior is characteristic of a tertiary alcohol because of the rapid formation of a stable carbocation and quick reaction with Lucas reagent.

The correct answer is option 3: \((CH_3)_3COH\) (Tert-butyl alcohol). This is the compound that forms turbidity immediately, confirming it is a tertiary alcohol.