Consider the following carbanions and arrange the decreasing order of their stability

II>III>IV>I

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: III>IV>I>II

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IV>I>II>III

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I>II>III>IV

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Solution

The correct option is
A

II>III>IV>I
Any group which decreases the magnitude of negative charge increases the stability, $- N O_{2}$ group is $(- M)$ group .so it increases the stability of benzyl carbanion as the magnitude of negative charge decreases. While $+ M$ groups decreases the stability, the carbanion with $C H_{3} O$ - group is least stable as it increases the magnitude of -ve charge. Stability of ion ∝ $\frac{1}{m a n g n i t u d e o f c h a r g e}$

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Similar questions

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

P h O^{⊖}

C H_{3} S^{⊖}

⊖ O H

M e O^{⊖}

Order of decreasing acidity of the following is

(I) HCOOH

(II) ${C H}_{3} C O O H$

(III) ${C l}_{2} C H C O O H$

(IV) ${C F}_{3} C O O H$

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