Question

The increasing order of $p{K}_b$ for the following compounds will be

• A. (B) < (C) < (A)
• B. (C) < (A) < (B)
• C. (B) < (A) < (C)
• D. (A) < (B) < (C)

Answer 1

The correct option is C (B) < (A) < (C)

$p{K}_b\propto \frac{1}{K_b}\propto \frac{1}{BasicStrength}$

Thus, more the basic strength lesser will be the $p{K}_b$

Weaker the conjugate acid, stronger the base. (B) is the most basic as it has a guanidine like structure. It has a high tendency of accepting a proton, giving rise to a very stable conjugate acid and hence, is a very strong base.
In compound (A), the N is $s{p}^2$ hybridised and its electronegativity is higher as compared to the compound (C) which is a $2^o$ amine ($s{p}^3$ hybridised). So compound (B) is more basic compared to compound (C).

So the order of basicity is B > A > C and thus the order of $p{K}_b$ value will be B < A < C.