Question

The correct decreasing order of electron densities on benzene for the following compounds is:

• A. $\left(D\right)>\left(C\right)>\left(B\right)>\left(A\right)$
• B. $\left(C\right)>\left(B\right)>\left(C\right)>\left(D\right)$
• C. $\left(A\right)>\left(B\right)>\left(D\right)>\left(C\right)$
• D. $\left(C\right)>\left(D\right)>\left(A\right)>\left(B\right)$

Answer 1

The correct option is C $\left(A\right)>\left(B\right)>\left(D\right)>\left(C\right)$

When chlorine is attached to benzene group, it shows both $-I$ and $+R$ effect but $+R<-I$ , and it decreases electron density
Since, (A) has no group attached to it , so that benzene ring has the highest electron density.
More the number of halogens attached, lower is the electron density
So,
Between (C) and (D), since electronegativity decreases down the group as the size increases, so $-I$ effect also decreases. So, (D) will have more electron density in comparison to (C)

Correct order is
$\left(A\right)>\left(B\right)>\left(D\right)>\left(C\right)$