JEE Main 2026 — Chemistry Organic Chemistry

2026mcqmedium

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

Official previous-year question

Held on 28 Jan 2026 · Verified 6 Jul 2026.

Options

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

Did you get this right?

Solution

Carbanion stability is determined by the s-character of the hybrid orbital bearing the negative charge.

The acetylide anion $CH \equiv C^-$ has the carbanion on an sp hybridized carbon (50% s-character), providing optimal stabilization through efficient overlap with the charged orbital.

The vinyl anion $CH_2 = CH^-$ has the carbanion on an sp² hybridized carbon (33% s-character), offering moderate stabilization.

The primary alkyl anion $CH_3 - CH_2^-$ has the carbanion on an sp³ hybridized carbon (25% s-character), providing the least stabilization.

Greater s-character enables better penetration of the nucleus and enhanced stabilization of the negative charge.

Therefore, the order is $CH \equiv C^- > CH_2 = CH^- > CH_3 - CH_2^-$ .

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Options

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

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

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

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

Solution

Carbanion stability is determined by the s-character of the hybrid orbital bearing the negative charge.

The acetylide anion $CH \equiv C^-$ has the carbanion on an sp hybridized carbon (50% s-character), providing optimal stabilization through efficient overlap with the charged orbital.

The vinyl anion $CH_2 = CH^-$ has the carbanion on an sp² hybridized carbon (33% s-character), offering moderate stabilization.

The primary alkyl anion $CH_3 - CH_2^-$ has the carbanion on an sp³ hybridized carbon (25% s-character), providing the least stabilization.

Greater s-character enables better penetration of the nucleus and enhanced stabilization of the negative charge.

Therefore, the order is $CH \equiv C^- > CH_2 = CH^- > CH_3 - CH_2^-$ .

JEE Main 2026 — Chemistry Organic Chemistry

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JEE Main 2026 — Chemistry Organic Chemistry

Options

Did you get this right?

Solution

Your note

JEE Main Organic Chemistry in other years

More JEE Main Organic Chemistry questions

Other Chemistry topics for JEE Main

Physical Chemistry

Inorganic Chemistry

Keep practicing Organic Chemistry