The decreasing order of acidic characters of the following is :

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 A $I I > I > I I I$
The decreasing order of acidic characters is $I I > I > I I I$ .
II is most acidic because, on deprotonation, the carbanion formed will be resonance stabilized.
III is least acidic because the carbanion formed on deprotonation will be antiaromatic and destabilized.
In the case of I, the carbanion formed by deprotonation will have negative charge on electronegative $s p$ $C$ atom.

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