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MCQEasyJEE 2026Electronic Effects (Inductive, Resonance, Hyperconjugation)

JEE Chemistry 2026 Question with Solution

The correct order of stability for the following carbanions is: CH2=CH−,CH3−CH2−,CH\equivC−CH_2=CH^- ,\quad CH_3-CH_2^- ,\quad CH\equivC^-

  • A

    CH≡C−>CH2=CH−>CH3−CH2−CH\equiv C^- > CH_2=CH^- > CH_3-CH_2^-

  • B

    CH3−CH2−>CH2=CH−>CH≡C−CH_3-CH_2^- > CH_2=CH^- > CH\equiv C^-

  • C

    CH2=CH−>CH≡C−>CH3−CH2−CH_2=CH^- > CH\equiv C^- > CH_3-CH_2^-

  • D

    CH≡C−>CH3−CH2−>CH2=CH−CH\equiv C^- > CH_3-CH_2^- > CH_2=CH^-

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The carbanions are CH2=CH−CH_2=CH^-, CH3−CH2−CH_3-CH_2^-, and CH≡C−CH\equiv C^-.

Find: The correct order of stability.

Concept: The stability of a carbanion depends mainly on the hybridization of the negatively charged carbon. Greater the ss-character of the orbital holding the negative charge, greater is the stability.

Step 1: Analyze the hybridization of each carbanion.

  • CH3−CH2−CH_3-CH_2^-: negatively charged carbon is sp3sp^3-hybridized, so it has 25%25\% ss-character.
  • CH2=CH−CH_2=CH^-: negatively charged carbon is sp2sp^2-hybridized, so it has 33%33\% ss-character.
  • CH≡C−CH\equiv C^-: negatively charged carbon is spsp-hybridized, so it has 50%50\% ss-character.

Step 2: Apply the stability rule.

sp>sp2>sp3sp > sp^2 > sp^3

Higher ss-character means greater electronegativity and better stabilization of the negative charge.

Step 3: Write the correct order.

CH≡C−>CH2=CH−>CH3−CH2−CH\equiv C^- > CH_2=CH^- > CH_3-CH_2^-

Therefore, the correct option is A.

Common mistakes

  • Assuming all carbanions have similar stability because they all carry one negative charge is incorrect. The hybridization of the negatively charged carbon matters. Compare the ss-character first.

  • Reversing the stability order by using the alkyl carbocation trend is wrong. For carbanions, greater ss-character stabilizes the negative charge, so spsp is more stable than sp2sp^2 and sp3sp^3.

  • Ignoring which carbon actually carries the negative charge leads to a wrong conclusion. Identify the hybridization of the charged carbon, not the molecule as a whole.

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