\(\underset{\text{Ester (A)}}{C_4H_8O_2} + \underset{\text{2 parts}}{CH_3MgBr} \longrightarrow \underset{\text{Alcohol (B)}}{C_4H_{10}O}\)
The alcohol (B) react 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. \(CH_3COOC_2H_5;\, \ (CH_3)_3COH\).
Let us analyze the reaction step by step to identify the correct ester (A) and alcohol (B).
Given,
Ester (A): \(C_4H_8O_2\)
Alcohol (B): \(C_4H_{10}O\)
The ester reacts with 2 equivalents of \(CH_3MgBr\) (methyl magnesium bromide) to give the alcohol \(B\), which reacts fastest with Lucas reagent.
When an ester reacts with 2 equivalents of a Grignard reagent (\(RMgX\)), it forms a tertiary alcohol.
General Reaction:
Given that the product alcohol (B) is \(C_4H_{10}O\), which suggests that alcohol (B) is a tertiary alcohol because tertiary alcohols react fastest with Lucas reagent.
Molecular Formula: \(C_4H_{10}O\)
The possible structure for alcohol \(B\) is tert-butyl alcohol, \((CH_3)_3COH\).
tert-Butyl alcohol is a tertiary alcohol and is known to react fastest with Lucas reagent due to the formation of a stable carbocation during the reaction.
Since the product is \(C_4H_{10}O\) (tert-butyl alcohol), the ester must be such that when it reacts with 2 equivalents of \(CH_3MgBr\), it produces tert-butyl alcohol.
For this to happen, the ester must be ethyl acetate \(CH_3COOC_2H_5\).
Reaction:
Ester (A): \(CH_3COOC_2H_5\) (ethyl acetate)
Alcohol (B): \((CH_3)_3COH\) (tert-butyl alcohol)
The correct ester (A) and alcohol (B) are: Option 1: \(CH_3COOC_2H_5\); \ (CH_3)_3COH\)
This matches the reaction and the conditions given in the problem, where alcohol (B) reacts fastest with Lucas reagent, confirming it as a tertiary alcohol.