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:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (CH3)3COH
Conc. HCl with anhydrous \( \ce{ZnCl2} \) is the Lucas reagent. The alcohol dissolves in it, but the alkyl chloride formed does not, so the solution turns turbid. The speed of this reaction follows the stability of the carbocation formed: 3° > 2° > 1°.
Turbidity immediately means a tertiary alcohol. \( \ce{(CH3)3COH} \), 2-methylpropan-2-ol, is the only tertiary alcohol among the options:
\( \ce{(CH3)3C-OH + HCl} \xrightarrow{\ce{ZnCl2}} \ce{(CH3)3C-Cl + H2O} \)
Why the other options are incorrect
Option (1) (CH3)2CHOH — Incorrect. Propan-2-ol is a secondary alcohol; it gives turbidity only after about five minutes.
Option (2) CH3CH2OH — Incorrect. Ethanol is a primary alcohol; it gives no turbidity at room temperature.
Option (4) CH3OH — Incorrect. Methanol is also a primary alcohol and does not give turbidity at room temperature.
NCERT (2026-27 edition):
Reaction with Hydrogen Halides – “… The difference in reactivity of three classes of alcohols with HCl distinguishes them from one another (Lucas test). Alcohols are soluble in Lucas reagent (conc. HCl and ZnCl2) while their halides are immiscible and produce turbidity in solution. In case of tertiary alcohols, turbidity is produced immediately as they form the halides easily.”