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

Options

A. `CRCOO^(-) lt NH_(2)^(-) lt HC -= C^(-) lt R^(-)`
B. `RCO O^(-) lt HC -= C^(-) lt NH_(2)^(-) lt R^(-)`
C. `RCO O^(-) lt HC -= C^(-) lt R^(-) lt NH_(2)^(-)`
D. `R^(-) lt HC -= C^(-) lt RCO O^(-) lt NH_(2)^(-)`

Answer

B

Explanation

The conjugate acids of the given bases are `RCOOHgtCH-=CHgtNH_(3)gtRH`. Their acidic strengths are in the order : `RCOOHgtCH-=CHgtNH_(3)gtRH`. As strong acid has weak conjugate base, the basic strenths of the given bases will be in the order :
`RCOO^(-)ltHC-=C^(-)ltNH_(2)^(-)ltR^(-)` (Remember that `sp^(3)` C is less electronegative than `sp^(3)` N. Hence, alkyl carbanion `(R^(-))` is more basic than `bar(N)H_(2)`. However, sp-hybridized carbon is more electronegative than `sp^(3)` N Hence `bar(N)H_(2)` is more basic `HC-=bar(C)`.