Identify the major product of the following reaction sequence: (i) LDA (ii) CH ₃ I (iii) NaBH ₄ / MeOH

Step-by-step solution

The given starting material is cyclopentanone. Step (i): Reaction with LDA (Lithium diisopropylamide), which is a strong, bulky base. It abstracts an -proton from cyclopentanone to form a lithium enolate. Step (ii): Reaction with CH ₃ I (Methyl iodide). The enolate acts as a nucleophile and undergoes an S_N2 reaction with CH ₃ I , resulting in the -alkylation of the ketone to form 2-methylcyclopentanone. Step (iii): Reaction with NaBH ₄ in MeOH . Sodium borohydride is a reducing agent that reduces the ketone carbon