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Arrange the following carbanions in decreasing order of basicity: (A) CH3CH2−, (B) CH2=CH−, and (C) HC≡C−.

1. (B) > (A) > (C)
2. (C) > (B) > (A)
3. (A) > (C) > (B)
4. (A) > (B) > (C)

Correct answer: (A) > (B) > (C)

Solution

Carbanion basicity decreases with increasing s-character: CH3CH2- (sp3) > CH2=CH- (sp2) > HC#C- (sp), i.e. A > B > C (option 3). The stored order is reversed.

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