The correct increasing order of basic strengths in,
$C H_{3} C H_{2} C N, C H_{3} C H_{2} N H_{2}, C H_{3} N = C H C H_{3}$ is :

C H_{3} N = C H C H_{3} > C H_{3} C H_{2} N H_{2} > C H_{3} C H_{2} C N

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C H_{3} C H_{2} N H_{2} > C H_{3} N = C H C H_{3} > C H_{3} C H_{2} C N

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C H_{3} C H_{2} C N > C H_{3} N = C H C H_{3} > C H_{3} C H_{2} N H_{2}

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C H_{3} C H_{2} C N > C H_{3} C H_{2} N H_{2} > C H_{3} N = C H C H_{3}

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Solution

The correct option is B $C H_{3} C H_{2} N H_{2} > C H_{3} N = C H C H_{3} > C H_{3} C H_{2} C N$
$k e y p o i n t :$
More the availability of $N -$ atom to donate its lone pair, more will be the basicity of amine.
Presence of electron releasing groups like $(- C H_{3}, - C_{2} H_{5})$ increases the basicity
The presence of an electron-withdrawing group (Like $(- C N)$ decreases the basicity.
$s p^{2}$ hybridized N is less electronegative than $s p$ hybridized.

Thus, the order of basic strength of the given compounds is $C H_{3} C H_{2} N H_{2} > C H_{3} N = C H C H_{3}, > C H_{3} C H_{2} C N$

Option B is correct.

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