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The correct order of base strength of the three isomers of methoxy aniline is

$\begin{array}{1 1}(a)\;P-CH_3OC_6H_4NH_2 > M-CH_3OC_6H_4NH_2 > O-CH_3OC_6H_4NH_2\\(b)\;P-CH_3OC_6H_4NH_2 > O-CH_3OC_6H_4NH_2 > M-CH_3OC_6H_4NH_2\\ (c)\;O-CH_3OC_6H_4NH_2 > P-CH_3OC_6H_4NH_2 > M-CH_3OC_6H_4NH_2\\(d)\;\text{None of these}\end{array}$

$-OCH_3$ group II $e^{\ominus}$-releasing group by resonance from ortho and para positions and thus increses the availability of $e^{\ominus}$ on N atom making them more basic .Ortho Bomer weaker than para isomer because of the ortho effect in which the solvation of cation is sterically hindered.
Hence (b) is the correct answer.