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Inductive Effect

The stability...

Question

The stability of carbanions follows the order.
I. $C H_{3} C H_{2} C H_{2} ⊝ C H_{2}$
II. $C H_{3} ⊝ C H C H_{2} C H_{3}$
III. $(C H_{3})_{3} C^{⊝}$
IV. $C H_{3} ⊝ C (P h) C H_{2} C H_{3}$

I I I < I V < I < I I I

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I < I I < I V < I I I

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I I I < I I < I < I V

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I V < I I I < I I < I

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Solution

The correct option is D $I I I < I I < I < I V$
Stability of carbanions decreases in order: $1^{o} > 2^{o} > 3^{o}$ .
$I V$ is most stable due to resonance between negative charge and double bond of benzene ring.
$I$ is more stable than $I I$ and $I I I$ as $C H_{3} C H_{2} C H_{2} C H_{2}^{-}$ is $1^{o}$ .II is more stable than $I I I$ as $C H_{3} C H^{-} C H_{2} C H_{3}$ is $2^{o}$ . $I I I$ is least stable as it is tertiary.
So stability of carbanions follows the order $I V > I > I I > I I I$ .

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Similar questions

(C H_{3})_{3} C^{\oplus}

(C H_{3})_{2} \oplus C (O C H_{3})

C H_{3} C H_{2} C H_{2} \oplus C H_{2}

C H_{3} \oplus C H C H_{2} C H_{3}

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^{⊖}

C H_{3} C H_{2} C H_{2} C H_{2} C l

C H_{3} C H_{2} C H C l - C H_{3}

(C H_{3})_{2} C H C H_{2} C l

(C H_{3})_{3} C - C l

S_{N} 2

+ I

- C H_{3}

- C D_{3}

- C T_{3}

- C H D_{2}

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