The major product formed in the following reaction is:
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In reactions involving enolate formation, the nucleophilic enolate reacts with alkyl halides to form carbon-carbon bonds. The reaction is followed by hydrolysis to yield the final product.
This reaction involves a sequence of three steps:
1. The first reagent used is KNH\(_2\) (potassium amide), which will abstract a proton from the carbonyl group to form an enolate ion. This enolate ion is highly nucleophilic.
2. In the second step, n-BuBr (n-butyl bromide) is used as the electrophile, which reacts with the enolate ion to form a C-C bond, resulting in the substitution of the hydrogen atom on the alpha-carbon with a n-butyl group. This creates an alkylated intermediate.
3. Finally, in the third step, the reaction is treated with dilute aqueous NaOH, which hydrolyzes the enolate back to the carbonyl group, forming the final product. The major product is the ester, where the alkylation of the alpha-carbon and the ester functionality are retained. The reaction sequence effectively introduces a butyl group at the alpha position of the ester.
Thus, the major product formed is the one where the butyl group (from n-BuBr) is attached to the ester at the alpha-carbon, resulting in the structure \(\text{Me}-\text{O}-\text{CH}_2\text{CH}_2\text{Me}\), which corresponds to option (B).
