Predict the major product $P$ in the following sequence of reactions. The starting material is 1-methylcyclopent-1-ene: (i) \ce{HBr}, benzoyl peroxide; (ii) \ce{KCN}; (iii) \ce{Na(Hg)}/\ce{C2H5OH} $\rightarrow$ $P$ (major). The structures of the options (A)-(D) are shown in the image below:

Solution & Explanation

\textbf{Answer:} (C) A 1,2-disubstituted cyclopentane bearing \ce{-CH3} and \ce{-CH2NH2} on adjacent carbons. \textbf{Solution:} Step (i): \ce{HBr} with benzoyl peroxide adds by the anti-Markovnikov (radical) route. For 1-methylcyclopent-1-ene the double bond is between the \ce{CH3}-bearing carbon (more substituted) and the adjacent \ce{CH}; \ce{Br} adds to the less-substituted carbon, generating the more stable 3$^\circ$ radical at the methyl carbon. This gives 2-bromo-1-methylcyclopentane (\ce{Br} on the carbon adjacent to the \ce{CH3} carbon). Step (ii): \ce{KCN} effects \ce{S_N2} substitution of \ce{Br} by \ce{-CN} (cyanide is a C-nucleophile, giving a nitrile, not the isocyanide), yielding 2-methylcyclopentane-1-carbonitrile. Step (iii): \ce{Na(Hg)}/\ce{C2H5OH} reduces the nitrile \ce{-C#N} to a primary amine \ce{-CH2NH2}. The product is (2-methylcyclopentyl)methanamine: \ce{CH3} and \ce{CH2NH2} on adjacent ring carbons (option C). Options A and B retain an isocyanide (\ce{-NC}, which would need \ce{AgCN}, and would not be reduced), and option D has the wrong (geminal) regiochemistry.