\(\underset{\text{Ester (A)}}{C_4H_8O_2} + \underset{\text{2 parts}}{CH_3MgBr} \longrightarrow \underset{\text{Alcohol (B)}}{C_4H_{10}O}\)
The alcohol (B) reacts the fastest with Lucas reagent. The ester (A) and alcohol (B) are respectively:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(CH_3COOC_2H_5;\, \ (CH_3)_3COH\)
With the Lucas reagent (conc. HCl + \( \ce{ZnCl2} \)), a tertiary alcohol gives turbidity immediately, a secondary one after a few minutes and a primary one not at room temperature. So B, \( \ce{C4H10O} \), is the tertiary alcohol 2-methylpropan-2-ol, (CH3)3COH.
An ester reacts with two moles of a Grignard reagent: the first converts it into a ketone, the second adds to the ketone to give a tertiary alcohol. Two of the three methyl groups on the carbinol carbon of \( \ce{(CH3)3COH} \) come from \( \ce{CH3MgBr} \), so the acyl part of the ester is \( \ce{CH3CO} \)–. With the formula \( \ce{C4H8O2} \), the ester is ethyl ethanoate:
\( \displaystyle \ce{CH3COOC2H5 + CH3MgBr -> CH3COCH3 + C2H5OMgBr} \)
\( \displaystyle \ce{CH3COCH3 + CH3MgBr -> (CH3)3C-OMgBr} \)
\( \displaystyle \ce{(CH3)3C-OMgBr + H2O} \xrightarrow{\ce{H+}} \ce{(CH3)3C-OH + Mg(OH)Br} \)
Why the other options are incorrect
Option (2) \(C_3H_7COOH;\, \ (CH_3)_2CHOH\) — Incorrect. \( \ce{C3H7COOH} \) is a carboxylic acid, not an ester, and its acidic H would destroy the Grignard reagent; propan-2-ol is \( \ce{C3H8O} \), not \( \ce{C4H10O} \).
Option (3) \(CH_3COOC_2H_5;\, \ (CH_3)_2CHOH\) — Incorrect. The ester is right, but it gives a tertiary alcohol; propan-2-ol is a secondary alcohol with only three carbons.
Option (4) \(C_3H_7COOH;\, \ (CH_3)_3COH\) — Incorrect. The alcohol is right, but A is an acid, not an ester; an acid only gives methane with \( \ce{CH3MgBr} \).
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. Primary alcohols do not produce turbidity at room temperature.”
Note: The same/similar question has been asked in different years by NTA as per the details given below:
| 2026 | 2025 | 2024 | 2023 | 2022 |
|---|---|---|---|---|
| – | – | – | 12-06 S3 08-06 S3 | – |