The correct order of increasing basicity of the given conjugate bases `(R=CH_(3))` is

Options

A. `RCOO^(-) lt NH_(2)^(-)ltHC-=bar(C)lt R^(-)`
B. `RCOO^(-) ltHC-=bar(C) ltNH_(2)^(-) ltR^(-)`
C. `RCOO^(-) ltHC-=bar(C) lt R^(-) ltNH_(2)^(-)`
D. `R^(-) ltHC -=bar(C)ltRCOO^(-)ltNH_(2)^(-)`

Answer

B

Explanation

The conjugate acids of the given bases are RCOOH,`NH_(3),HC-=CH` and RH . Their acidic strength will be in the order.
`RCHOO gtHC -= CH gt NH_(3) gt RH`. As strong acid has a weak conjugate base, the basic strength will be in the order `RCOO^(-) ltHC -= C.CNH_(2)^(-) ltR'` Note that `sp^(3)C` is less electro negative than `sp^(3)N`. Hence alkyl carbanion `(R^(-))` is more basic than `bar(N) H_(2)`. However, sp hybridized C is more electronegative than `HC-= bar(C)`