Question

Arrange the following carbanions in order of their decreasing stability.

(I) $H_{3}C - C \equiv C$ (II) H – C $\equiv$C

(III)$H_{3}C - CH_{2}$

• A. I > II > III
• B. II > I > III
• C. III > II > I
• D. III > I >II

Answer 1

Answer: (B)

II > I > III

Answer 2

: The order of decreasing stability of carbanions is:

$H - C \equiv C^{-} > CH_{3} - C \equiv C^{-} > CH_{3} - CH_{2}^{-}$

(II) (I) (III)

Sp- hybridized carbon atom is more electronegative than $sp^{3}$hybridized carbon atom and hence, can accommodate the negative charge more effectively. $- CH_{3}$group has +l effect, therefore, it intensifies the negative charge and hence, destabilizes the carbanion $CH_{3} \rightarrow C \equiv C^{-}$