Arrange the following carbanions in decreasing order of stability.

I V > I I I > I > I I

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I I I > I V > I I > I

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I > I I I > I I > I V

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I I > I > I I I > I V

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Solution

The correct option is A $I V > I I I > I > I I$
The stability order of the carbanions can be determined using the $- M$ effect of the substituents present at the para position. Higher the $- M$ power of the substituents, higher will be the stability of the carbanion.

The order of $- M$ power of the substituents is:
$- N O_{2} > - C N >$
$- C H_{3} > - O C H_{3}$ shows $+ M$ effect
Therefore, the stability order of the carbanions will be:
$I V > I I I > I > I I$

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