Arrange the following in the increasing order of their basic strength:
$C_{2} H_{5} N H_{2}$ , $C_{6} H_{5} N H_{2}$ , $(C_{2} H_{5})_{2} N H$

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Solution

Secondary amines have more electron density at $N$ due to electron donating effect (+I effect) of alkyl groups as compared to primary amines. Also, the lone pair of $N$ in aniline $(C_{6} H_{5}) N H_{2}$ are delocalized in the benzene ring and are therefore less available for donation. Hence makes them lesser basic than alkyl amines.
Therefore the increasing order of basic strength is:
$(C_{6} H_{5}) N H_{2}$ < $(C_{2} H_{5}) N H_{2}$ < $(C_{2} H_{5})_{2} N H$

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