The order of decreasing stability of the following carbanion is:
(i) ${(C H_{3})}_{3} C^{-}$
(ii) $C {H_{3}}^{-}$
(iii) $C H_{3} C {H_{2}}^{-}$
(iv) ${(C H_{3})}_{2} C H^{-}$

i > i i > i i i > i v

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i v > i i i > i i > i

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i v > i > i i > i i i

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none of these

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Solution

The correct option is D none of these
The order of stability of carbanion is $\Rightarrow$
$m e t h y l > 1 ° > 2 ° > 3 °$

Therefore the correct order of stability of given carbanions are $\Rightarrow$

$i i > i i i > i v > i (∵ ii and iii both are primary, iv is secondary anad i is tertiary)$

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P h O^{⊖}

C H_{3} S^{⊖}

⊖ O H

M e O^{⊖}

The stability of carbanions in the following

i) $R C \equiv ⊖ C$

ii)

iii) $R_{2} C = ⊖ C H$

iv) $R_{3} C - ⊖ C H_{2}$

is in the order of

+ I

- C H_{3}

- C D_{3}

- C T_{3}

- C H D_{2}

The correct order for decreasing stability of the following free radicals is

(i) H C O O H (i i) P h - C O O H (i i i) P h - O H

(i v) C H_{3} C O O H

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