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B
C6HΘ5
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C
(CH3)3C−CHΘ2
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D
(CH3)2C=CHΘ
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Solution
The correct option is AHC≡CΘ
In HC≡CΘ carbanion is present on sp hybridized carbon atom which is most electronegative as s character is more so carbanion is stable.
While in C6HΘ5 and (CH3)2C=CHΘ carbonianis present on sp2 carbon atom which has less electronegative than sp. In (CH3)3C−CHΘ2 carbanion is foresent on sp3 hybridized carbon atom which is least electronegative so, CH≡CΘ carbanion is most stable.