The correct order of decreasing basic strength of the given amines is: (N-ethylethanamine $=$ $(\mathrm{C}{}_2\mathrm{H}{}_5){}_2\mathrm{NH}$; ethanamine $=$ $\mathrm{C}{}_2\mathrm{H}{}_5\mathrm{NH}{}_2$; N-methylaniline $=$ $\mathrm{C}{}_6\mathrm{H}{}_5\mathrm{NH}(\mathrm{CH}{}_3)$; benzenamine/aniline $=$ $\mathrm{C}{}_6\mathrm{H}{}_5\mathrm{NH}{}_2$)

Solution & Explanation

\textbf{Answer:} (A) Aliphatic amines are more basic than aromatic amines; among each set, alkyl substitution (+I) raises basicity. \textbf{Solution:} Identify the four amines: $\bullet$ N-ethylethanamine $=$ $(\mathrm{C}{}_2\mathrm{H}{}_5){}_2\mathrm{NH}$ (a $2^{\circ }$ aliphatic amine) $\bullet$ Ethanamine $=$ $\mathrm{C}{}_2\mathrm{H}{}_5\mathrm{NH}{}_2$ (a $1^{\circ }$ aliphatic amine) $\bullet$ N-methylaniline $=$ $\mathrm{C}{}_6\mathrm{H}{}_5\mathrm{NH}(\mathrm{CH}{}_3)$ (aromatic, $2^{\circ }$) $\bullet$ Benzenamine (aniline) $=$ $\mathrm{C}{}_6\mathrm{H}{}_5\mathrm{NH}{}_2$ (aromatic, $1^{\circ }$) Key principle: in aliphatic amines the lone pair is fully available and reinforced by the +I effect of alkyl groups, whereas in aromatic amines the lone pair is delocalised into the ring, lowering basicity. Within the aliphatic pair, $(\mathrm{C}{}_2\mathrm{H}{}_5){}_2\mathrm{NH}$ (two +I ethyl groups) $>$ $\mathrm{C}{}_2\mathrm{H}{}_5\mathrm{NH}{}_2$. Within the aromatic pair, the $\mathrm{N}-\mathrm{CH}{}_3$ group of N-methylaniline donates electron density (+I) to nitrogen, so N-methylaniline $>$ aniline. Since aliphatic $>$ aromatic, the overall decreasing order is: $(\mathrm{C}{}_2\mathrm{H}{}_5){}_2\mathrm{NH}$ $>$ $\mathrm{C}{}_2\mathrm{H}{}_5\mathrm{NH}{}_2$ $>$ $\mathrm{C}{}_6\mathrm{H}{}_5\mathrm{NH}(\mathrm{CH}{}_3)$ $>$ $\mathrm{C}{}_6\mathrm{H}{}_5\mathrm{NH}{}_2$ that is, N-ethylethanamine $>$ ethanamine $>$ N-methylaniline $>$ benzenamine, which is option (A).