Reaction of m-methylanisole with lithium in liquid ammonia and t-butyl alcohol at 33°C generates compound X as the major product. Treatment of the compound X with dilute sulphuric acid produces compound Y as the major product. The compounds X and Y respectively are A. B. C. D.

nan
nan
nan
nan

The correct answer is:

X = 2-(t-butyl)-4-methoxyanisole
Y = 2-(t-butyl)-4-methoxyphenol

The reaction of m-methylanisole with lithium in liquid ammonia and t-butyl alcohol at 33°C is a Grignard reaction. In a Grignard reaction, an alkyl halide is reacted with magnesium metal in the presence of an anhydrous solvent, such as liquid ammonia, to form a Grignard reagent. The Grignard reagent is then reacted with a compound that contains a carbon-carbon double bond to form a new C-C bond. In this case, the Grignard reagent reacts with the carbonyl group in m-methylanisole to form compound X.

Compound X is a tertiary alcohol. Tertiary alcohols are relatively stable and do not undergo dehydration easily. Therefore, treatment of compound X with dilute sulfuric acid does not produce any significant amount of dehydration product. Instead, the reaction proceeds via an electrophilic aromatic substitution reaction, in which the proton from the sulfuric acid attacks the aromatic ring to form a carbocation intermediate. The carbocation intermediate then reacts with the methoxy group in compound X to form compound Y.

Compound Y is a phenol. Phenols are relatively acidic and can be deprotonated by dilute sulfuric acid to form the phenolate anion. The phenolate anion is a good nucleophile and can react with electrophiles, such as the t-butyl group in compound Y, to form a new C-O bond. This reaction is called an electrophilic aromatic substitution reaction.

The following is a diagram of the reaction mechanism:

In the first step, the Grignard reagent reacts with the carbonyl group in m-methylanisole to form a tetrahedral intermediate. In the second step, the tetrahedral intermediate collapses to form compound X. In the third step, the proton from the sulfuric acid attacks the aromatic ring to form a carbocation intermediate. In the fourth step, the carbocation intermediate reacts with the methoxy group in compound X to form compound Y. In the fifth step, the phenolate anion reacts with the t-butyl group in compound Y to form a new C-O bond.

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