Arrange the carbanions, $\left(\mathrm{CH}_3\right)_3 \overline{\mathrm{C}}, \overline{\mathrm{C}}…

Arrange the carbanions, $\left(\mathrm{CH}_3\right)_3 \overline{\mathrm{C}}, \overline{\mathrm{C}} \mathrm{Cl}_3,\left(\mathrm{CH}_3\right)_2 \overline{\mathrm{C}} \mathrm{H}, \mathrm{C}_6 \mathrm{H}_5 \overline{\mathrm{C}} \mathrm{H}_2$, in order of their decreasing stability :
  1. $\mathrm{C}_8 \mathrm{H}_5 \overline{\mathrm{C}} \mathrm{H}_2>\overline{\mathrm{C}} \mathrm{Cl}_3>\left(\mathrm{CH}_3\right)_3 \overline{\mathrm{C}}>\left(\mathrm{CH}_3\right)_2 \overline{\mathrm{C}} \mathrm{H}$
  2. $\left(\mathrm{CH}_3\right)_2 \overline{\mathrm{C}} \mathrm{H}>\overline{\mathrm{C}} \mathrm{Cl}_3>\mathrm{C}_6 \mathrm{H}_5 \overline{\mathrm{C}} \mathrm{H}_2>\left(\mathrm{CH}_3\right)_3 \overline{\mathrm{C}}$
  3. $\overline{\mathrm{C}} \mathrm{Cl}_3>\mathrm{C}_8 \mathrm{H}_5 \overline{\mathrm{C}} \mathrm{H}_2>\left(\mathrm{CH}_3\right)_2 \overline{\mathrm{C}} \mathrm{H}>\left(\mathrm{CH}_3\right)_3 \overline{\mathrm{C}}$
  4. $\left(\mathrm{CH}_3\right)_3 \overline{\mathrm{C}}>\left(\mathrm{CH}_3\right)_2 \overline{\mathrm{C}} \mathrm{H}>\mathrm{C}_6 \mathrm{H}_5 \overline{\mathrm{C}} \mathrm{H}_2>\overline{\mathrm{C}} \mathrm{Cl}_3$

Solution

$2^{\circ}$ carbanion is more stable than $3^{\circ}$ and $\mathrm{Cl}$ is $-1$ effect group.

Asked in: JEE Main 2009

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