The stability of carbanions in the following is in the order of:
A
a>c>b>d
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B
a>b>c>d
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C
b>c>d>a
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D
d>b>c>a
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Solution
The correct option is Ba>b>c>d
Among a,b,c and d q will be most stable and d will be least stable because in negative charge is kept on Sp2 hybridised carbon but in d negative charge is kept on Sp3 hybridized carbon atom. Again b is more stable than c because in b electron Π cloud is delocalized between 6 atom. So, correct order of stability : a>b>c>d.