The correct order of decreasing basic strength of the given amines is: (N-ethylethanamine $=$ \ce{(C2H5)2NH}; ethanamine $=$ \ce{C2H5NH2}; N-methylaniline $=$ \ce{C6H5NH(CH3)}; benzenamine/aniline $=$ \ce{C6H5NH2})

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 $=$ \ce{(C2H5)2NH} (a $2^\circ$ aliphatic amine) $\bullet$ Ethanamine $=$ \ce{C2H5NH2} (a $1^\circ$ aliphatic amine) $\bullet$ N-methylaniline $=$ \ce{C6H5NH(CH3)} (aromatic, $2^\circ$) $\bullet$ Benzenamine (aniline) $=$ \ce{C6H5NH2} (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, \ce{(C2H5)2NH} (two +I ethyl groups) $>$ \ce{C2H5NH2}. Within the aromatic pair, the \ce{N-CH3} group of N-methylaniline donates electron density (+I) to nitrogen, so N-methylaniline $>$ aniline. Since aliphatic $>$ aromatic, the overall decreasing order is: \ce{(C2H5)2NH} $>$ \ce{C2H5NH2} $>$ \ce{C6H5NH(CH3)} $>$ \ce{C6H5NH2} that is, N-ethylethanamine $>$ ethanamine $>$ N-methylaniline $>$ benzenamine, which is option (A).