(A) CH ≡ C⁻ > CH₃ − CH₂⁻ > CH₂ = CH⁻

(B) CH₂ = CH⁻ > CH ≡ C⁻ > CH₃ − CH₂⁻

(C) CH ≡ C⁻ > CH₂ = CH⁻ > CH₃ − CH₂⁻

(D) CH₃ − CH₂⁻ > CH₂ = CH⁻ > CH ≡ C⁻

Correct Answer: (C)

Explanation

The stability of a carbanion depends mainly on the s-character of the carbon atom carrying the negative charge.

Greater the s-character, greater is the electronegativity of carbon and hence better stabilization of the negative charge.

sp (50% s-character) > sp² (33%) > sp³ (25%)

Now, analyzing each carbanion:

CH ≡ C⁻ → Carbon is sp-hybridised → maximum s-character → most stable

CH₂ = CH⁻ → Carbon is sp²-hybridised → intermediate stability

CH₃ − CH₂⁻ → Carbon is sp³-hybridised → least s-character → least stable

Therefore, correct order of stability is:

CH ≡ C⁻ > CH₂ = CH⁻ > CH₃ − CH₂⁻

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