Consider the following carbanions. I)CH3−CH−2 II)CH2=CH− III)CH≡−C Correct order of stability of these carbanions in decreasing order is:
A
I > II > III
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B
III > II > I
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C
I > III > II
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D
III > I > II
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Solution
The correct option is B III > II > I As s-character of orbital increases, the stability of carbanion increases and corresponding conjugate acid character increases. So, stability of anions is - III (sp) > II (sp2) > I (sp3)