Arrange the following carbanions in order of their decreasing stability.
(I) $H_{3} C - C \equiv C^{-}$
(II) $H - C \equiv C^{-}$
(III) $H_{3} C - C H_{2}^{-}$

I > II > III

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II > I > III

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III > II > I

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III > I > II

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Solution

The correct option is B II > I > III
sp-hybridised carbon atom is more electronegative than $s p^{3}$ -hybridised carbon atom and hence, can accommodate the negative charge more effectively. $- C H_{3}$ group has +I effect, therefore, it intensifies the negative charge and hence, destabilises the carbanion ${C H_{3} - C \equiv C}^{-}$ .
Hence the correct order of stability is $\to$
${H - C \equiv C I I}^{-} > {H_{3} C - C \equiv C I}^{-} > {H_{3} C - C H_{2}}^{-} I I I$

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