1.

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

Answer»

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

Solution :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^(-)`


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