The compound that will react most readily with NaOH to form methanol is
(a) $\left(\mathrm{CH}_3\right)_4 \mathrm{~N}^{+} \mathrm{I}^{-}$
(b) $\mathrm{CH}_3 \mathrm{OCH}_3$
(c) $\left(\mathrm{CH}_3\right)_3 \mathrm{~S}^{+} \mathrm{I}^{-}$
(d) $\left(\mathrm{CH}_3\right)_3 \mathrm{CCl}$