Question

Consider the acidity of the carboxylic acids.
(i) $PhCOOH$
(ii) o- $N{O}_2{C}_6{H}_{4}COOH$
(iii) p - $N{O}_2{C}_6{H}_{4}COOH$
(iv) m - $N{O}_2{C}_6{H}_{4}COOH$
which of the following order is correct ?

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

Answer 1

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

Electron withdrawing group increases the acidity of benzoic acid, o-isomer will have higher acidity than -p position have more pronounce electron-withdrawing effect than $-NO_2$ group at -m position.

Hence the order of acidity is

$o-N{O}_2{C}_6{H}_{4}COOH>p-N{O}_2{C}_6{H}_{4}COOH>m-N{O}_2{C}_6{H}_{4}COOH>Ph-COOH$

Hence, option D is correct.