Arrange the following carbanions in decreasing order of stability.

I V > I I I > I > I I

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I V > I I > I I I > I

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I V > I I I > I I > I

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I I I > I V > I I > I

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Solution

The correct option is C $I V > I I I > I I > I$
$\begin{matrix} In IV, 3 withdrawing group, most unstable In III, II 2 withdrawing group but in II two {C H}_{3} (electron releasing group) present :. III more reactive or unstable then II \end{matrix}$
In I, only one electron-withdrawing group more stable than other
$∴ I V > I I I > I I > I$
Hence, (C) is correct

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