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Question

The order of decreasing stability of the following carbanion is:
(i) (CH3)3C
(ii) CH3
(iii) CH3CH2
(iv) (CH3)2CH

A
i>ii>iii>iv
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B
iv>iii>ii>i
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C
iv>i>ii>iii
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D
none of these
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Solution

The correct option is D none of these
The order of stability of carbanion is
methyl>1°>2°>3°

Therefore the correct order of stability of given carbanions are

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

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