The compound will react most readily with NaOH of form methanol is :

Options

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

Explanation

It is understood from previous lessons that positive nitrogen is a stronger acid as compared to positive sulphur, and shall accept the electron pair more readily on an internal nucleophilic attack of the `OH^(+)` ion.
`(CH_(3))_(4)N^(+)I^(-)+NaOH rarr (CH_(3))_(4)N^(+)OH^(-)+NaI,(CH_(3))_(4)N^(+)OH^(-) overset(Delta)rarr (CH_(3))_(3)N+CH_(3)OH`