####27. The compound that will react most readily with $NaOH$ to form methanol is
(2001, 1M)
(a) $\left(CH _3\right) _4 N^{+} I^{-}$
(b) $CH _3 OCH _3$
(c) $\left(CH _3\right) _3 S^{+} I^{-}$
(d) $\left(CH _3\right) _3 Cl$
Lack of $\beta-H$ on quaternary ammonium iodide leads to $S _N 2$ reaction otherwise E2 elimination usually takes place.
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