Arrange the following carbanions in order of their decreasing stabilit...
‘sp’ hybridized, more ‘s character’ more stable ‘-’ charge.
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Arrange the following carbanions in order of their decreasing stabilit...
To determine the stability of carbanions, we need to consider the factors that influence stability. These factors include:
1. Inductive effect: The presence of electron-withdrawing groups adjacent to the carbanion stabilizes it by withdrawing electron density through the sigma bond. This effect helps to disperse the negative charge and stabilize the carbanion.
2. Hyperconjugation: The presence of adjacent C-C sigma bonds or C-H sigma bonds allows for the delocalization of the negative charge through the overlap of orbitals. This delocalization stabilizes the carbanion.
Now, let's analyze the given carbanions and determine their stability:
(A) H3C-CC
- This carbanion does not have any adjacent electron-withdrawing groups. However, it does have adjacent C-C sigma bonds, which allow for hyperconjugation and stabilization.
(B) H-C-C
- This carbanion has an adjacent electron-withdrawing group, the methyl group (CH3). The inductive effect of the methyl group stabilizes the carbanion by withdrawing electron density and dispersing the negative charge. Additionally, it has an adjacent C-H sigma bond, which allows for hyperconjugation and further stabilization.
(C) H3C-CH2
- This carbanion has an adjacent electron-withdrawing group, the methyl group (CH3). The inductive effect of the methyl group stabilizes the carbanion by withdrawing electron density and dispersing the negative charge. However, it does not have any adjacent C-C or C-H sigma bonds for hyperconjugation.
Based on the analysis above, we can arrange the carbanions in order of decreasing stability as follows:
B > A > C
Therefore, the correct answer is option 'B', which states that carbanion B is the most stable, followed by carbanion A, and then carbanion C.