Rearrange the following in the increasing order of acidic strength.
(i) benzoic acid
(ii) p-methoxybenzoic
(iii) o-methyoxybenzoic

i < i i < i i i

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i i i < i < i i

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i i < i < i i i

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i i i < i i < i

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Solution

The correct option is D $i i i < i i < i$
The order of acidity will be-
O-methoxybenzoic acid > Benzoic acid> P -methoxybenzoic acid.
Reason-
[1] As the paramethoxybenzoic acid has an electron donating group $(- O C H_{3})$ at the para position so this group will increase the electron density in the ring that will unstable the negative charge that is left after the removal of $H^{+}$ .

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Rearrange the following in the increasing order of acidic strength.(i) benzoic acid (ii) p-methoxybenzoic (iii) o-methyoxybenzoic

Rearrange the following in the increasing order of acidic strength.
(i) benzoic acid
(ii) p-methoxybenzoic
(iii) o-methyoxybenzoic