Q. Arrange the carbanions, $\left( CH_{3} \right)_{3}\overline{C},\overline{C}Cl_{3},\ \left( CH_{3} \right)_{2}\overline{C}H,\ C_{6}H_{5}\overline{C}H_{2}$, in order of their decreasing stability
A) $C_{6}H_{5}\overline{C}H_{2} > \overline{C}Cl_{3} > \left( CH_{3} \right)_{2}\overline{C} > \left( CH_{3} \right)_{2}\overline{C}H$
B) $\left( CH_{3} \right)_{2}\overline{C}H > \overline{C}Cl_{3} > C_{6}H_{5}CH_{2} > \left( CH_{3} \right)_{3}\overline{C}$
C) $\overline{C}Cl_{3} > C_{6}H_{5}\overline{C}H_{2} > \left( CH_{3} \right)_{2}\overline{C}H > \left( CH_{3} \right)_{3}\overline{C}$
D) $\left( CH_{3} \right)_{3}\overline{C} > \left( CH_{3} \right)_{2}\overline{C}H > \overline{C}H_{2} > \overline{C}Cl_{3}$
Explanation
The correct answer is Option C: $\overline{C}Cl_{3} > C_{6}H_{5}\overline{C}H_{2} > \left( CH_{3} \right)_{2}\overline{C}H > \left( CH_{3} \right)_{3}\overline{C}$.
A detailed explanation for this question is being prepared.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.