The correct order of decreasing basic character of the three aliphatic primary amines is
The correct order of decreasing basic character of the three aliphatic primary amines is
$\mathrm{I}>\mathrm{II}>\mathrm{III}$
$\mathrm{III}>\mathrm{II}>\mathrm{I}$
$\mathrm{I}>\mathrm{II} \approx \mathrm{III}$
$\mathrm{I}=\mathrm{II} \equiv \mathrm{III}$
Solution
Note the point of difference in the given compounds which here lies at $\beta$ -carbon. In I, II, III, the $\beta$ -carbon atoms are $s p^{3}, s p^{2}$ and $s p$ hybridised respectively which in turn cause the difference in their $s$ -character. We know that more is the $s$ character of an atom, greater will be its electron-withdrawing nature. Thus $s p\left(50 \% sight.$ character) hybridised carbon is most electron-withdrawing, while $s p^{3}(25 \%$ $s$ -character) is least electron-withdrawing. Further, we know that presence of an electron-withdrawing group decreases basicity of an amine. Thus
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