JEE Main 2026 — Jan 28, Shift 1 — Chemistry solution

JEE Main2026Jan 28, Shift 1Chemistry

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
$CH \equiv C ⊖ > CH_{2} = CH ⊖ > CH_{3} - CH ⊖_{2}$
D
$CH \equiv C ⊖ > CH_{3} - CH ⊖_{2} > CH_{2} = CH ⊖$

Solution

Carbanion stability depends on the hybridisation of the carbon holding the lone pair. Greater the s-character, the closer the lone pair sits to the nucleus, and the more stable the carbanion.

Hybridisation and s-character:

$CH \equiv C ⊖$ : sp (50% s)

$CH_{2} = CH ⊖$ : sp $^{2}$ (33% s)

$CH_{3} - CH ⊖_{2}$ : sp $^{3}$ (25% s)

$∴$ Stability order: $CH \equiv C ⊖ > CH_{2} = CH ⊖ > CH_{3} - CH ⊖_{2}$ , which matches option (3).

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