A) \[{{C}_{6}}{{H}_{5}}\overline{C}{{H}_{2}}>\overline{C}C{{l}_{3}}>{{(C{{H}_{3}})}_{3}}\overline{C}>{{(C{{H}_{3}})}_{2}}\overline{C}H\]
B) \[{{(C{{H}_{3}})}_{2}}\overline{C}H>\overline{C}C{{l}_{3}}>{{C}_{6}}{{H}_{5}}\overline{C}{{H}_{2}}>{{(C{{H}_{3}})}_{3}}\overline{C}\]
C) \[\overline{C}C{{l}_{3}}>{{C}_{6}}{{H}_{5}}\overline{C}{{H}_{2}}>{{(C{{H}_{3}})}_{2}}\overline{C}H>{{(C{{H}_{3}})}_{3}}\overline{C}\]
D) \[{{(C{{H}_{3}})}_{3}}\overline{C}>{{(C{{H}_{3}})}_{2}}\overline{C}H>{{C}_{6}}{{H}_{5}}\overline{C}{{H}_{2}}>\overline{C}C{{l}_{3}}\]
Correct Answer: C
Solution :
Due to the\[I\]effect of three chlorine atoms and due to\[p\pi -d\pi \]bonding\[CCl_{3}^{-}\]is extra stable. Carbanion follow stability order. \[CCl_{3}^{-}>{{C}_{6}}{{H}_{5}}\,\overline{C}{{H}_{2}}>{{(C{{H}_{3}})}_{2}}\overline{C}H>{{(C{{H}_{3}})}_{3}}\overline{C}\]
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