The correct order of stability for the follow | vedclass

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(C) The stability of a carbanion is directly proportional to the percentage of $s$-character in the hybrid orbital of the carbon atom bearing the nega

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$CH\equiv C^{-} > CH_2=CH^{-} > CH_3-CH_2^{-}$

Vedclass · Answer

(C) The stability of a carbanion is directly proportional to the percentage of $s$-character in the hybrid orbital of the carbon atom bearing the negative charge. $1$. In $CH\equiv C^{-}$, the carbon is $sp$ hybridized ($50\% \ s$-character). $2$. In $CH_2=CH^{-}$, the carbon is $sp^2$ hybridized ($33.3\% \ s$-character). $3$. In $CH_3-CH_2^{-}$, the carbon is $sp^3$ hybridized ($25\% \ s$-character). Higher $s$-character means the electrons are held closer to the nucleus, increasing stability. Therefore, the correct order of stability is: $CH\equiv C^{-} > CH_2=CH^{-} > CH_3-CH_2^{-}$.

The correct order of stability for the following carbanions is:
$CH_2=CH^{-}, CH_3-CH_2^{-}, CH\equiv C^{-}$

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The correct order of stability for the following carbanions is: $CH_2=CH^{-}, CH_3-CH_2^{-}, CH\equiv C^{-}$