Rearrange the following in an increasing order of their basic strengths :
$C_{6} H_{5} N H_{2}$ , $C_{6} H_{5} N (C H_{3})_{2}$ , $(C_{6} H_{5})_{2} N H$ and $C H_{3} N H_{2}$ .

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Solution

$(C_{6} H_{5})_{2} N H < C_{6} H_{5} {N H}_{2} < C_{6} H_{5} N ({C H}_{3})_{2} < {C H}_{3} {N H}_{2}$
Due to the $- I$ effect of $C_{6} H_{5}$ group, the electron density on N-atom $C_{6} H_{5} - {N H}_{2}$ is lower than that on the N-atom $(C_{6} H_{5})_{2} N H$ . But in case of $C_{6} H_{5} N ({C H}_{3})_{2}$ , two ${C H}_{3}$ group attached with N atom which increases the electron density on N-atom and in ${C H}_{3} - {N H}_{3}$ only $+ I$ effect of methyl group increases the basicity of the compound

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