An alcohol X, C4H10O, produces turbidity immediately when it is shaken with Lucas reagent (concentrated HCl and anhydrous ZnCl2) at room temperature. Which of the following is X?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 2-Methylpropan-2-ol
Alcohols dissolve in Lucas reagent, but the alkyl chlorides they form with it do not; the chloride separates and the solution turns turbid. The three classes of alcohols form their chlorides at different rates, and a tertiary alcohol forms its chloride so easily that the turbidity appears at once. So X must be the tertiary alcohol among the four isomers:
Butan-1-ol, CH3CH2CH2CH2OH — primary; no turbidity at room temperature.
Butan-2-ol, CH3CH(OH)CH2CH3 — secondary; it forms its chloride less easily than a tertiary alcohol, so the turbidity does not appear at once.
2-Methylpropan-2-ol, (CH3)3COH — tertiary; turbidity immediately.
2-Methylpropan-1-ol, (CH3)2CHCH2OH — primary; no turbidity at room temperature.
\( \displaystyle \ce{(CH3)3C-OH + HCl} \xrightarrow{\ce{ZnCl2}} \ce{(CH3)3C-Cl + H2O} \)
NCERT (2026-27 edition):
7.4.4 Chemical Reactions – “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. Primary alcohols do not produce turbidity at room temperature.”
6.4.1 From Alcohols (Unit 6) – “…The order of reactivity of alcohols with a given haloacid is 3°>2°>1°.”