Single Choice

Arrange the carbanions, $$(\mathrm{C}\mathrm{H}_3)_{3}\overline{\mathrm{C}},\overline{\mathrm{C}}\mathrm{C}l_{3},(\mathrm{C}\mathrm{H}_3)_{2}\overline{\mathrm{C}}\mathrm{H},\ \mathrm{C}{}_{6}\mathrm{H}_{5}\overline{\mathrm{C}}\mathrm{H}_{2}$$ , in order of their decreasing stability:

A$$\mathrm{C}{}_{6}\mathrm{H}_{5}\overline{\mathrm{C}}\mathrm{H}_{2}>\overline{\mathrm{C}}\mathrm{C}l_{3}>(\mathrm{C}\mathrm{H}_{3})_{3}\overline{\mathrm{C}}>(\mathrm{C}\mathrm{H}_{3})_{2}\overline{\mathrm{C}}\mathrm{H}$$
B$$(\mathrm{C}\mathrm{H}3)_{2}\overline{\mathrm{C}}\mathrm{H}>\overline{\mathrm{C}}\mathrm{C}l_{3}>\mathrm{C}_{6}\mathrm{H}_{5}\overline{\mathrm{C}}\mathrm{H}_{2}>(\mathrm{C}\mathrm{H}_{3})_{3}\overline{\mathrm{C}}$$
C$$\overline{\mathrm{C}}\mathrm{C}l_{3}>\mathrm{C}_{6}\mathrm{H}_{5}\overline{\mathrm{C}}\mathrm{H}_{2}>(\mathrm{C}\mathrm{H}_{3})_{2}\overline{\mathrm{C}}\mathrm{H}>( \mathrm{C}\mathrm{H}_{3})_{3}\overline{\mathrm{C}}$$
Correct Answer
D$$(\mathrm{C}\mathrm{H}_{3})_{3}\overline{\mathrm{C}}>( \mathrm{C}\mathrm{H}_{3})_{2}\overline{\mathrm{C}}\mathrm{H}>\mathrm{C}_{6}\mathrm{H}_{5}\overline{\mathrm{C}}\mathrm{H}_{2}>\overline{\mathrm{C}}\mathrm{C}l_{3}$$

Solution

Due to the $$-I$$ effect of three chlorine atoms and due to $$\mathrm{p}\pi-\mathrm{d}\pi$$ bonding $$\mathrm{C}\mathrm{C}l_{3}^{-}$$ is extra stable.
Carbanion follow stability order.
$$\overline{\mathrm{C}}\mathrm{C}l_{3}>\mathrm{C}_{6}\mathrm{H}_{5}\overline{\mathrm{C}}\mathrm{H}_{2}>(\mathrm{C}\mathrm{H}_{3})_{2}\overline{\mathrm{C}}\mathrm{H}>( \mathrm{C}\mathrm{H}_{3})_{3}\overline{\mathrm{C}}$$

In $$C_6H_5CH_2^-$$ the $$-M$$ effect delocalizes the -ve charge on carbon. The $$CH_3$$ group in $$ 1^{st}$$ and $$3^{rd}$$ destabilizes the -ve charge on carbon.

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