Animated Solution for Chemistry - Organic Chemistry: Reaction of Grignard reagent, ${\text{C}}_2{\text{H}}_5\text{MgBr}$ with ${\text{C}}_8{\text{H}}_8\text{O}$ followed by hydrolysis gives compound 'A' which reacts instantly with Lucas reagent to give compound B, ${\text{C}}_{10}{\text{H}}_{13}\text{Cl}$. The compound B is

A Journey from Ketones to Alkyl Halides

Organic synthesis problems often feel like intricate puzzles where each reagent is a clue pointing toward the final picture. In this problem, we are tasked with identifying a final compound, B, derived from a starting material with the formula ${\text{C}}_8{\text{H}}_8\text{O}$. The journey involves a Grignard addition followed by the classic Lucas test. Let's break down the logic step-by-step.

The Lucas Test

The most revealing piece of information in the problem statement is that compound A reacts instantly with the Lucas reagent (a mixture of concentrated $\text{HCl}$ and anhydrous ${\text{ZnCl}}_2$).

The Lucas test is a staple in organic chemistry for distinguishing between different classes of alcohols based on their reactivity via the ${\text{S}}_{\text{N}}1$ mechanism. Because the rate-determining step is the formation of a carbocation, the stability of that carbocation dictates the speed of the reaction: - Primary alcohols form highly unstable primary carbocations and generally do not react at room temperature. - Secondary alcohols form moderately stable secondary carbocations, leading to visible turbidity (due to the insoluble alkyl chloride) in about 5 to 10 minutes. - Tertiary alcohols form highly stable tertiary carbocations, resulting in instant turbidity.

Since compound A reacts instantly, we can definitively conclude that compound A is a tertiary alcohol.

Identifying the Starting Material

Knowing that compound A is a tertiary alcohol allows us to deduce the nature of our starting material, ${\text{C}}_8{\text{H}}_8\text{O}$. Compound A was formed by reacting this starting material with a Grignard reagent (${\text{C}}_2{\text{H}}_5\text{MgBr}$).

Recall the fundamental reactions of Grignard reagents with carbonyl compounds: - Formaldehyde + Grignard $\to$ Primary Alcohol - Other Aldehydes + Grignard $\to$ Secondary Alcohol - Ketones + Grignard $\to$ Tertiary Alcohol

Therefore, ${\text{C}}_8{\text{H}}_8\text{O}$ must be a ketone. Given the molecular formula and the presence of a benzene ring in all the multiple-choice options, the only logical structure is acetophenone ($\text{Ph}-\text{CO}-{\text{CH}}_3$).

Now that we have our starting material, let's trace the forward reactions to find compound B.

Step 1: The Grignard Addition When acetophenone reacts with ethyl magnesium bromide, the nucleophilic ethyl group attacks the electrophilic carbonyl carbon. Subsequent acidic hydrolysis yields the tertiary alcohol, compound A.

Compound A is 2-phenylbutan-2-ol. Notice that the central carbon is now bonded to a phenyl ring, a methyl group, an ethyl group, and a hydroxyl group.

Step 2: The Lucas Reaction Treating 2-phenylbutan-2-ol with the Lucas reagent initiates an ${\text{S}}_{\text{N}}1$ reaction. The hydroxyl group is protonated and leaves as water, generating a carbocation.

This intermediate is not just a tertiary carbocation; it is also benzylic, meaning it is exceptionally stabilized by resonance delocalization of the positive charge into the adjacent phenyl ring. This immense stability is exactly why the reaction is instantaneous. Finally, the chloride ion attacks the carbocation to form compound B.

Compound B is 2-chloro-2-phenylbutane. Its structure features a central carbon atom bonded to a phenyl ring, a chlorine atom, a methyl group, and an ethyl group.

Comparing this to the given options, Option (c) perfectly matches our derived structure, making it the correct answer. Understanding the mechanistic nuances of these classic reactions not only helps in solving problems but also deepens your appreciation for the elegant logic of organic chemistry.