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Arrange the following carbanions in decreasing order of stability: (I) RC\equiv C^- (II) C_6H_5^- (III) R_2C=CH^- (IV) R_3C–CH_2^-

1. (I) > (II) > (III) > (IV)
2. (II) > (III) > (IV) > (I)
3. (IV) > (II) > (III) > (I)
4. (IV) > (III) > (II) > (I)

Correct answer: (IV) > (III) > (II) > (I)

Solution

Carbanions are stabilized by higher s-character: sp > sp2 > sp3. Here, the alkyl carbanion is least stable, the vinylic/aryl types are more stable than simple alkyl carbanions, and the acetylide carbanion is most stable due to sp hybridization. Thus the decreasing order is (IV) > (III) > (II) > (I).

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