Which of the following shows the correct order of decreasing acidity?

P h C O_{2} H > P h S O_{3} H > P h C H_{2} O H > P h O H

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P h S O_{3} H > P h O H > P h C O_{2} H > P h C H_{2} O H

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P h C O_{2} H > P h O H > P h C H_{2} O H > P h S O_{3} H

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P h S O_{3} H > P h C O_{2} H > P h O H > P h C H_{2} O H

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Solution

The correct option is D $P h S O_{3} H > P h C O_{2} H > P h O H > P h C H_{2} O H$
To explain the acidity of these compounds, we need to see the stability of their conjugate bases.

The conjugate base $P h S O_{3}^{-}$ is most stable due to highest delocalization of electrons and highest number of resonating structures.

$P h C O O^{-}$ is more stable than $P h O^{-}$ .

$P h C H_{2} O^{-}$ is not stable as the negative charge is not in conjugation with ring and $+ I$ effect of alkyl group makes it less stable.

Therefore, the order of acidic strength is $P h S O_{3} H > P h C O O H > P h O H > P h C H_{2} O H$

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