Question

In a reaction of $C_6{H}_{5}Y,$ the major product (>60%) is m-isomer, so the group Y is

• A. $$-COOH$$
• B. $$-N{H}_2$$
• C. $$-OH$$
• D. $$-Cl$$

Answer 1

The correct option is A $$-COOH$$

$$-COOH$$ because it is meta directing group while $-N{H}_2,-OH$ and $-Cl$ are ortho and para directing groups.

$-N{H}_2$ and $-OH$ are one of the strongest and powerful ortho/para directing group due to the electron donating ability of $-N{H}_2$. It shows +R effect. Hence they more activating group than others.

$-COOH$ is one of the strongest deactivating group, it reduces the electron density at ortho and para position by -R effect. Hence, it is the strongest meta-directing group than others.

$-Cl$ weak meta directing groups due to -I effect and strong ortho and para directing groups due to +R effect.