1.

A 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 overset(+)(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 dificient than + charge on S. HENCE `(CH_(3))_(4)NI^(+)` will UNDERGO nucleophilic substitution reaction more readily than `(CH_(3))_(3)S^(+)I^(-)`


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