Chemistry - GOC Question with Solution | TestHub

(A) Each phenyl ring delocalises the positive charge over three more carbons. Two rings are better than one, and a benzyl cation (resonance) is more stable than a secondary alkyl cation (hyperconjugation only), which is more stable than a primary one. Correct.

(B) The order drawn is reversed. An electron-withdrawing $-\mathrm{N}{\mathrm{O}}_2$ stabilises the phenoxide ion and an electron-donating $-\mathrm{C}{\mathrm{H}}_3$ destabilises it: 4-nitrophenol $>$ phenol $>$ 4-methylphenol. Wrong.

(C) Piperidine has a localised $\mathrm{s}{\mathrm{p}}^3$ nitrogen lone pair ($\mathrm{p}{K}_{\mathrm{a}}\mathrm{H}$ about 11), pyridine has an $\mathrm{s}{\mathrm{p}}^2$ lone pair (about 5) and in pyrrole the lone pair is part of the aromatic sextet (it is hardly basic). Correct order: piperidine $>$ pyridine $>$ pyrrole. Wrong.

(D) In $\mathrm{C}{\mathrm{H}}_2=\mathrm{CH}-\mathrm{N}{\mathrm{H}}_2$ the nitrogen lone pair is delocalised into the $\mathrm{C}=\mathrm{C}$, giving the $\mathrm{C}-\mathrm{N}$ bond partial double-bond character and an $\mathrm{s}{\mathrm{p}}^2$ carbon, so it is shorter than the $\mathrm{C}-\mathrm{N}$ bond of $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2-\mathrm{N}{\mathrm{H}}_2$. Correct.

Answer: (A), (D).