Question

Assign number 1 for least to 4 for most to indicate the relative basic strength of the following:
(I) $C_6{H}_{5}N{H}_2$
(II) $p-N{O}_2{C}_6{H}_{4}N{H}_2$
(III) $m-N{O}_2{C}_6{H}_{4}N{H}_2$
(IV) $p-C{H}_{3}O{C}_6{H}_{4}N{H}_2$

• A. $$\begin{array}{cccc}I& II& III& IV\\ 3& 1& 2& 4\end{array}$$
• B. $$\begin{array}{cccc}I& II& III& IV\\ 1& 2& 3& 4\end{array}$$
• C. $$\begin{array}{cccc}I& II& III& IV\\ 2& 3& 4& 1\end{array}$$
• D. $$\begin{array}{cccc}I& II& III& IV\\ 4& 1& 3& 2\end{array}$$

Answer 1

The correct option is A $$\begin{array}{cccc}I& II& III& IV\\ 3& 1& 2& 4\end{array}$$

Resonance effect will occur when the +R/-R group is present at para/ortho position.
In (IV), $-OC{H}_3$ is +R group which will increase the electron density on N by delocalistion of electron and make N more basic. Therefore, (IV) is most basic of all.
In (II), $-N{O}_2$ is -R group which will reduce the electron density on N by delocalistion of electron and make N less basic. Therefore, (II) is least basic of all.
In (I), there is no +R/-R group, it is less basic than (IV).
In (III), $-N{O}_2$ is at meta position and -I effect operates so it pulls the electron density toward itself which will reduce the electron density on N and make it less basic but not as (II) because resonance effect dominates over inductive effect.

Therefore, option (a) is correct.