The order of decreasing stability of the carbanions is:
1) $(C H_{3})_{3} C^{-}$
2) $(C H_{3})_{2} {C H}^{-}$
3) $C H_{3} {C H_{2}}^{-}$
4) $C_{6} H_{5} {C H_{2}}^{-}$

1 > 2 > 3 > 4

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4 > 3 > 2 > 1

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4 > 1 > 2 > 3

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1 > 2 > 4 > 3

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Solution

The correct option is C $4 > 3 > 2 > 1$
More the $+ I$ effect, lesser is the stability of carbanions. So, stability of carbanion order is: $1^{0} > 2^{0} > 3^{0}$ . Benzyl carbanion is most stable because of resonance stablisation.

Stability order is: $4 > 3 > 2 > 1$

$C_{6} H_{5} {C H_{2}}^{-}$ > $C H_{3} {C H_{2}}^{-}$ > $(C H_{3})_{2} {C H}^{-}$ > $(C H_{3})_{3} C^{-}$

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Consider the following carbocations

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Q. Given below are isomers of

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. Which of the following statements is correct?

1.

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C H_{2} = C H - ⨁ C H_{2}

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molecule, the hybridization of carbon 1, 2, 3 and 4 respectively are

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$1) C H_{3} - ˙ C H_{2} 2) C H_{3} - C ˙ H - C H_{3} 3) C H_{3} - ˙ C - C H_{3} | C H_{3} 4) C_{3}$

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The order of decreasing stability of the carbanions is:1) (CH3)3C−2) (CH3)2CH−3) CH3CH2−4) C6H5CH2−

The correct option is C 4>3>2>1More the +I effect, lesser is the stability of carbanions. So, stability of carbanion order is: 10>20>30. Benzyl carbanion is most stable because of resonance stablisation.Stability order is: 4>3>2>1C6H5CH2− >CH3CH2−>(CH3)2CH−>(CH3)3C−

The order of decreasing stability of the carbanions is:
1) $(C H_{3})_{3} C^{-}$
2) $(C H_{3})_{2} {C H}^{-}$
3) $C H_{3} {C H_{2}}^{-}$
4) $C_{6} H_{5} {C H_{2}}^{-}$