CORRECT order of stability for the following is $\mathrm{CH}_{2}=\mathrm{CH}^{-}, \mathrm{CH}_{3}-\mathrm{CH}_{2}^{-}, \mathrm{CH} \equiv \mathrm{C}^{-}$

(C) $\mathrm{CH} \equiv \mathrm{C}^{-}>\mathrm{CH}_{2}=\mathrm{CH}^{-}>\mathrm{CH}_{3}-\mathrm{CH}_{2}^{-}$

(A) $\mathrm{CH}_{3}-\mathrm{CH}_{2}^{-}>\mathrm{CH}_{2}=\mathrm{CH}^{-}>\mathrm{CH} \equiv \mathrm{C}^{-}$

(B) $\mathrm{CH}_{2}=\mathrm{CH}^{-}>\mathrm{CH} \equiv \mathrm{C}^{-}>\mathrm{CH}_{3}-\mathrm{CH}_{2}^{-}$

(D) $\mathrm{CH} \equiv \mathrm{C}^{-}>\mathrm{CH}_{3}-\mathrm{CH}_{2}^{-}>\mathrm{CH}_{2}=\mathrm{CH}^{-}$

JEE Main · 2026 · 28 Jan 2026 · Shift 1 · General Organic Chemistry

CORRECT order of stability for the following is

CH_{2} = CH^{-}, CH_{3} - CH_{2}^{-}, CH \equiv C^{-}

(A) $CH_{3} - CH_{2}^{-} > CH_{2} = CH^{-} > CH \equiv C^{-}$
(B) $CH_{2} = CH^{-} > CH \equiv C^{-} > CH_{3} - CH_{2}^{-}$
(C) $CH \equiv C^{-} > CH_{2} = CH^{-} > CH_{3} - CH_{2}^{-}$ ✓ Correct
(D) $CH \equiv C^{-} > CH_{3} - CH_{2}^{-} > CH_{2} = CH^{-}$

Answer: (C)

Solution

The stability of a carbanion grows as the s-character of the orbital holding the lone pair grows, because higher s-character keeps the electron pair closer to the nucleus and lowers its energy. The carbanion carbon is

s p

-hybridised (50% s) in

CH \equiv C^{-}

s p^{2}

(33% s) in

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

and

s p^{3}

(25% s) in

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

. Hence the order of stability is

CH \equiv C^{-} > C H_{2} = C H^{-} > C H_{3} - C H_{2}^{-}

JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Single correct option

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JEE Main · 2026 · 28 Jan 2026 · Shift 1 · General Organic Chemistry

Solution

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