The compound that will react most readily with `NaOH` to from methanol is

Options

A. `(CH_(3))_(4)N^(+)I^(-)`
B. `CH_(3)OCH_(3)`
C. `(CH_(3))_(3)S^(+)I^(-)`
D. `(CH_(3))_(3)C Cl`.

Answer

A

Explanation

Due to greater electronegativity of N over S, +ve charge on N will make `CH_(3)` group more electron deficient than +ve charge on S. therefore, `(CH_(3))_(4)N^(+)I^(-)` will undergo nucleophilic substitution reaction more readily than `(CH_(3))_(3)S^(+)I^(-)`