Predict the major product $P$ in the following sequence of reactions. The starting material is 1-methylcyclopent-1-ene: (i) $\mathrm{HBr}$, benzoyl peroxide; (ii) $\mathrm{KCN}$; (iii) $\mathrm{Na}(\mathrm{Hg})$/$\mathrm{C}{}_2\mathrm{H}{}_5\mathrm{OH}$ $\to$ $P$ (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 $-\mathrm{CH}{}_3$ and $-\mathrm{CH}{}_2\mathrm{NH}{}_2$ on adjacent carbons. \textbf{Solution:} Step (i): $\mathrm{HBr}$ with benzoyl peroxide adds by the anti-Markovnikov (radical) route. For 1-methylcyclopent-1-ene the double bond is between the $\mathrm{CH}{}_3$-bearing carbon (more substituted) and the adjacent $\mathrm{CH}$; $\mathrm{Br}$ adds to the less-substituted carbon, generating the more stable 3${}^{\circ }$ radical at the methyl carbon. This gives 2-bromo-1-methylcyclopentane ($\mathrm{Br}$ on the carbon adjacent to the $\mathrm{CH}{}_3$ carbon). Step (ii): $\mathrm{KCN}$ effects $\mathrm{S}{}_{\mathrm{N}}2$ substitution of $\mathrm{Br}$ by $-\mathrm{CN}$ (cyanide is a C-nucleophile, giving a nitrile, not the isocyanide), yielding 2-methylcyclopentane-1-carbonitrile. Step (iii): $\mathrm{Na}(\mathrm{Hg})$/$\mathrm{C}{}_2\mathrm{H}{}_5\mathrm{OH}$ reduces the nitrile $-\mathrm{C}\equiv \mathrm{N}$ to a primary amine $-\mathrm{CH}{}_2\mathrm{NH}{}_2$. The product is (2-methylcyclopentyl)methanamine: $\mathrm{CH}{}_3$ and $\mathrm{CH}{}_2\mathrm{NH}{}_2$ on adjacent ring carbons (option C). Options A and B retain an isocyanide ($-\mathrm{NC}$, which would need $\mathrm{AgCN}$, and would not be reduced), and option D has the wrong (geminal) regiochemistry.