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Hydrogen Bonding

The decreasin...

Question

The decreasing order of stabilities of the following carbanions is:

I. $P h_{3} ¨ ⊖ C$ II. $P h_{2} ¨ ⊖ C H$ III. $P h ¨ ⊖ C H_{2}$

I > II > III

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II > I > III

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III > II > I

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III > I > II

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Solution

The correct option is A I > II > III
$Above is the order of stability of carbanions.$
$\begin{matrix} Because if there are more benzene rings the negative charge is distributed more and have more stability. \end{matrix}$

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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 K_{a}

C H \equiv C H

C H_{3} - C H_{2}^{-}

C H_{2} = C H^{-}

C H \equiv - C

P h_{2} C = C H_{2}

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