Which of the following orders is not correct for the decreasing order of acidic character?

I C H_{2} C O O H < B r C H_{2} C O O H < C l C H_{2} C O O H < F C H_{2} C O O H

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I C H_{2} C O O H > B r C H_{2} C O O H > C l C H_{2} C O O H > F C H_{2} C O O H

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

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H C O O H > C H_{3} C O O H > C_{2} H_{5} C O O H > (C H_{3})_{2} C H C O O H

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Solution

The correct option is B $I C H_{2} C O O H > B r C H_{2} C O O H > C l C H_{2} C O O H > F C H_{2} C O O H$
The correct order of acidity is:

$I C H_{2} C O O H < B r C H_{2} C O O H < C l C H_{2} C O O H < F C H_{2} C O O H$

As the $- I$ effect increases the acidity of carboxylic acid increases.

While the electron-donating group decreases the acidity.
So option D is correct order of acidity.

In halogen group down the group, the electronegativity decreases i.e. from fluorine to iodine.

Option B is correct.

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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 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

C H_{3} C O O H

C H_{3} C H_{2} C O O H

C_{6} H_{5} C O O H

C H_{3} C O O H

C_{2} H_{5} C O O H

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