Question

Predict the order of electron density for the following compounds :

• A. $\left(i\right)>\left(ii\right)>\left(iii\right)>\left(iv\right)$
• B. $\left(ii\right)>\left(iii\right)>\left(i\right)>\left(iv\right)$
• C. $\left(iv\right)>\left(iii\right)>\left(ii\right)>\left(i\right)$
• D. $\left(i\right)>\left(iv\right)>\left(ii\right)>\left(iii\right)$

Answer 1

The correct option is A $\left(i\right)>\left(ii\right)>\left(iii\right)>\left(iv\right)$

All these groups show $+M$ effect.
More is the $+M$ effect, more they increase electron density at ortho/para positions.
$+M$ order :
$-N{H}_2>-NHR>-NHCOR>-OCOR$
Structure (iii) and (iv) have less $+M$ effect than (i) and (ii) because in (iii) and (iv) lone pair is also involved in conjugation with $-C=O$ group.
Comparing (i) and (ii) :

Here $-N{H}_2$ shows more $+M$ effect than $-NHR$ because the bulky alkyl group causes repulsion within the lone pair which leads to slight change in orientation.

Comparing (iii) and (iv) :
Since oxygen is more electronegative than nitrogen, lone pair in case of nitrogen is more easily involved in resonance, as compared to oxygen.
So, $+M$ effect :
$-NHCOR>-OCOR$

Correct order is :
$\left(i\right)>\left(ii\right)>\left(iii\right)>\left(iv\right)$