\(\underset{\text{Ester (A)}}{C_4H_8O_2} + \underset{\text{2 parts}}{CH_3MgBr} \longrightarrow \underset{\text{Alcohol (B)}}{C_4H_{10}O}\)

The alcohol (B) react the fastest with Lucas reagent. The ester (A) and alcohol (B) are respectively:

\(CH_3COOC_2H_5;\, \ (CH_3)_3COH\)

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.