Question

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

• A. $(C{H}_3{)}_4{N}^{+}{I}^{-}$
• B. $C{H}_{3}OC{H}_3$
• C. $(C{H}_3{)}_3{S}^{+}{I}^{-}$
• D. $(C{H}_3{)}_{3}Cl$

Answer 1

The correct option is A $(C{H}_3{)}_4{N}^{+}{I}^{-}$

$EN$ of $N>S$.

Positive charge on $N$ will make (Me) group more $e^{-}$ deficient than positive charge on $S$. Therefore, (a) will undergo $S_{N}2$ reaction more rapidly than (c).
Hence option A is correct.