Question

The sequence for the stability of carbanions $H-C\equiv C^{-},H_{2}C=\stackrel{-}{C}H$ and ${}^{-}C{H}_3$ is:

• A. ${}^{-}C{H}_3>H-C\equiv C^{-}>H_{2}C=\stackrel{-}{C}H$
• B. $H-C\equiv C^{-}>H_{2}C=\stackrel{-}{C}H{>}^{-}C{H}_3$
• C. $H_{2}C=\stackrel{-}{C}H>H-C\equiv C^{-}{>}^{-}C{H}_3$
• D. ${}^{-}C{H}_3>H_{2}C=\stackrel{-}{C}H>H-C\equiv C^{-}$

Answer 1

The correct option is B $H-C\equiv C^{-}>H_{2}C=\stackrel{-}{C}H{>}^{-}C{H}_3$

The carbon atoms containing negative charges are $sp,s{p}^2$ and $s{p}^3$ hybridised in $H-C\equiv C^{-},H_{2}C=\stackrel{-}{C}H$ and ${}^{-}C{H}_3$ respectively.
The stability of carbanion depends upon the electronegativity of the respective carbon bearing negative atom. more is the s-character more will be electronegativity thus more electronegative carbon will hold the negative charge more strongly, therefore the stability of carbanion will be more.
the correct correct order of s-character and electronegativity of C-atom is sp>$s{p}^2$>$s{p}^3$
Hence the correct order of stability of carbanion will be:
$H-C\equiv C^{-}>H_{2}C=\stackrel{-}{C}H{>}^{-}C{H}_3$