The decreasing order for acidic strength is:

N O_{2} C H_{2} C O O H > N C C H_{2} C O O H > F C H_{2} C O O H > C l C H_{2} C O O H

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F C H_{2} C O O H > N C C H_{2} C O O H > N O_{2} C H C O O H > C l C H_{2} C O O H

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N O_{2} C H_{2} C O O H > F C H_{2} C O O H > C N C H_{2} C O O H > C l C H_{2} C O O H

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C N C H_{2} C O O H > O_{2} N C H_{2} C O O H > F C H_{2} C O O H > C l C H_{2} C O O H

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Solution

The correct option is A $N O_{2} C H_{2} C O O H > N C C H_{2} C O O H > F C H_{2} C O O H > C l C H_{2} C O O H$
The electron-withdrawing group increases the acidic strength.

Here the strength of electron-withdrawing group is in the following order:

$N O_{2} > C N > F > C l$

Thus, acidic strength is given as:

$N O_{2} C H_{2} C O O H > N C C H_{2} C O O H > F C H_{2} C O O H > C l C H_{2} C O O H$

Hence, the correct option is A.

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Similar questions

p K_{a}

C F_{3} C O O H, C C l_{3} C O O H, H C C l_{2} C O O H, N O_{2} C H_{2} C O O H, F C H_{2} C O O H, C l C H_{2} C O O H, B r C H_{2} C O O H, H C O O H, C l C H_{2} C H_{2} C O O H, C_{6} H_{5} C O O H, C_{6} H_{5} C H_{2} C O O H, C H_{3} C O O H, C H_{3} C H_{2} C O O H

{C l}_{3} C C O O H > {C l}_{2} C H C O O H > C l {C H}_{2} C O O H

{C H}_{3} C H_{2} C O O H > {(C H}_{3})_{2} C H C O O H > {(C H}_{3})_{3} C C O O H

F_{2} C H_{2} C O O H > F C H_{2} C O O H > C l {C H}_{2} C O O H

$C H_{3} C O C H_{2} C l O H^{-}, C l_{2} - ---- \to$ Product P is

C l_{3} C C O O H > C l_{2} C H C O O H > C l C H_{2} C O O H

C l_{3} C C O O H > C l_{2} C H C O O H > C l C H_{2} C O O H

C H_{3} C H_{2} C O O H > (C H_{3})_{2} C H C O O H > (C H_{3})_{3} C . C O O H

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