Question

The increasing order of $p{K}_b$ values of the following compounds is :

• A. $II<I<III<IV$
• B. $II<IV<III<I$
• C. $I<II<IV<III$
• D. $I<II<III<IV$

Answer 1

The correct option is C $I<II<IV<III$

$p{K}_b\propto \frac{1}{\text{Basic strength}\left(K_b\right)}$

$K_b\propto +I,$ $\text{+M effect}$

$K_b\propto \frac{1}{-I,-M\text{effect}}$

Due to +R effect of $-OC{H}_3$ group, the electron density on N of compound (I) increases and it is the most basic among the following.
Compound (III) and (IV) are less basic compared to I and II because they have meta-directing group which reduce the electron density at meta position. Comparing III and IV, compound III is less basic because $CN$ is strong electron withdrawing group so by -I effect , the electron density on nitrogen decrease and it becomes the least basic.
Thus, correct order of basicity is
$I>II>IV>III$

Hence, correct order of $p{K}_b$ value is
$I<II<IV<III$