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Arrange the carbanions (CH3)3C(-), CCl3(-), (CH3)2CH(-) and C6H5CH2(-) in order of decreasing stability.

1. CCl3(-) > C6H5CH2(-) > (CH3)2CH(-) > (CH3)3C(-)
2. (CH3)3C(-) > (CH3)2CH(-) > C6H5CH2(-) > CCl3(-)
3. C6H5CH2(-) > CCl3(-) > (CH3)3C(-) > (CH3)2CH(-)
4. (CH3)2CH(-) > CCl3(-) > C6H5CH2(-) > (CH3)3C(-)

Correct answer: CCl3(-) > C6H5CH2(-) > (CH3)2CH(-) > (CH3)3C(-)

Solution

CCl3(-) is most stable due to strong -I withdrawal by three Cl atoms. C6H5CH2(-) is next, stabilized by delocalization into the ring. Among the alkyl carbanions, less substituted ones are more stable, so (CH3)2CH(-) (secondary) beats (CH3)3C(-) (tertiary, most destabilized by +I of three methyls).

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