Question:

Arrange the carbanions, $\left(CH_{3}\right)_{3}\, \bar{C}, \bar{C}Cl_{3}, \left(CH_{3}\right)_{2}\, \bar{C}H, C_{6}H_{5}\, \bar{C}H_{2}$, in order of their decreasing stability :

Updated On: Jun 23, 2023
  • $C_{6}H_{5}\,\bar{C}H_{2} > \bar{C}Cl_{3} > \left(CH_{3}\right)_{3}\,\bar{C} > \left(CH_{3}\right)_{2}\,\bar{C}H$
  • $\left(CH_{3}\right)_{2}\,\bar{C}H > \bar{C}Cl_{3} > C_{6}H_{5}\,\bar{C}H_{2} > \left(CH_{3}\right)_{3}\,\bar{C}$
  • $\bar{C}Cl_{3} > C_{6}H_{5}\,\bar{C}H_{2} > \left(CH_{3}\right)_{2}\,\bar{C}H > \left(CH_{3}\right)_{3}\,\bar{C}$
  • $\left(CH_{3}\right)_{3}\,\bar{C }> \left(CH_{3}\right)_{2}\,\bar{C}H > C_{6}H_{5}\,\bar{C}H_{2} > \bar{C}Cl_{3}$
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The Correct Option is C

Solution and Explanation

$2^o$ carbanion is more stable than $3^o$ and $Cl$ is $-I$ effect group.
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