Question:
Consider the following anions: (i) $\text{CH}_3-\text{CH}_2^-$, (ii) $\text{CH}_2=\text{CH}^-$, (iii) $\text{HC}\equiv\text{C}^-$. Correct stability order of given anions is
Consider the following anions: (i) $\text{CH}_3-\text{CH}_2^-$, (ii) $\text{CH}_2=\text{CH}^-$, (iii) $\text{HC}\equiv\text{C}^-$. Correct stability order of given anions is
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Acidity of hydrocarbons: Alkyne>Alkene>Alkane. The stability of their conjugate bases (carbanions) follows the same order.
Updated On: Jan 28, 2026
- III>II>I
- II>III>I
- I>II>III
- I>III>II
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The Correct Option is A
Solution and Explanation
The stability of a carbanion increases with the increasing s-character of the carbon atom carrying the negative charge.
Higher s-character means the orbital is closer to the nucleus and more electronegative, thus stabilizing the negative charge.
(i) $\text{CH}_3-\text{CH}_2^-$: Carbon is $sp^3$. s-character = 25%. (Least stable).
(ii) $\text{CH}_2=\text{CH}^-$: Carbon is $sp^2$. s-character = 33.3%.
(iii) $\text{HC}\equiv\text{C}^-$: Carbon is $sp$. s-character = 50%. (Most stable).
Order: III>II>I.
Higher s-character means the orbital is closer to the nucleus and more electronegative, thus stabilizing the negative charge.
(i) $\text{CH}_3-\text{CH}_2^-$: Carbon is $sp^3$. s-character = 25%. (Least stable).
(ii) $\text{CH}_2=\text{CH}^-$: Carbon is $sp^2$. s-character = 33.3%.
(iii) $\text{HC}\equiv\text{C}^-$: Carbon is $sp$. s-character = 50%. (Most stable).
Order: III>II>I.
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