Question

Correct order of electron density in aromatic ring of following compounds is:

• A. $\text{IV > III > II > I}$
• B. $\text{I > II > III > IV}$
• C. $\text{IV > II > I > III}$
• D. $\text{IV > II > III > I}$

Answer 1

The correct option is D $\text{IV > II > III > I}$


In $I$, the group attached to benzene ring is electron withdrawing in nature and hence it withdraws electrons from the benzene ring as a result electron density on benzene ring is decreased.

In $II$, due to lone pair of electrons present on oxygen are delocalized and there will come in resonance with the benzene ring, it will lead to increase in the electron density of the ring.

In $III$, $-C{H}_3$ group is electron donating in nature and it generates $+I$ effect in the ring and will increase the density of the benzene ring but as we know that the resonance effect is stronger than the $+I$ effect therefore in this case the electron density of $II$ will be more than the density of third ring.

In $IV$, lone pair present on nitrogen are in resonance with the $C=O$ group and also with the benzene ring so the electron density will increase here. As nitrogen is less electronegative than oxygen. Hence, electron density will be highest in $IV$.
The correct order will be: $\text{IV > II > III > I}$