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Question

Arrange the carbanions, (CH3)3¯¯¯¯C,¯¯¯¯CCl3,(CH3)2¯¯¯¯CH, C6H5¯¯¯¯CH2 , in order of their decreasing stability:

A
C6H5¯¯¯¯CH2>¯¯¯¯CCl3>(CH3)3¯¯¯¯C>(CH3)2¯¯¯¯CH
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B
(CH3)2¯¯¯¯CH>¯¯¯¯CCl3>C6H5¯¯¯¯CH2>(CH3)3¯¯¯¯C
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C
¯¯¯¯CCl3>C6H5¯¯¯¯CH2>(CH3)2¯¯¯¯CH>(CH3)3¯¯¯¯C
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D
(CH3)3¯¯¯¯C>(CH3)2¯¯¯¯CH>C6H5¯¯¯¯CH2>¯¯¯¯CCl3
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Solution

The correct option is A ¯¯¯¯CCl3>C6H5¯¯¯¯CH2>(CH3)2¯¯¯¯CH>(CH3)3¯¯¯¯C
Due to the I effect of three chlorine atoms and due to pπdπ bonding CCl3 is extra stable.
Carbanion follow stability order.
¯¯¯¯CCl3>C6H5¯¯¯¯CH2>(CH3)2¯¯¯¯CH>(CH3)3¯¯¯¯C

In C6H5CH2 the M effect delocalizes the -ve charge on carbon. The CH3 group in 1st and 3rd destabilizes the -ve charge on carbon.

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