Question

The increasing order of $PK_{b}$ for the following compounds will be
$\left(\right.1\left.\right)$ $NH_{2}-CH=NH,$
$\left(\right.2\left.\right)$
$\left(\right.3\left.\right)$ $CH_{3}NHCH_{3}$

• A. $\left(\right.2\left.\right) < \left(\right.1\left.\right) < \left(\right.3\left.\right)$
• B. $\left(\right.3\left.\right) < \left(\right.1\left.\right) < \left(\right.2\left.\right)$
• C. $\left(\right.2\left.\right) < \left(\right.3\left.\right) < \left(\right.1\left.\right)$
• D. $\left(\right.1\left.\right) < \left(\right.2\left.\right) < \left(\right.3\left.\right)$

Answer 1

Answer: (A)

$pK_{b} \propto \frac{1}{K_{b}} \propto \frac{1}{B a s i c \, s t r e n g t h}$
As basic nature decreases in the order (2) > (1) > (3). So $pK_{b}$ order is as follows
$\left(\right.2\left.\right) < \left(\right.1\left.\right) < \left(\right.3\left.\right)$