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(A) The reaction involves a nucleophilic substitution $(S_N2)$ where $OH^-$ acts as the nucleophile.In $(CH_3)_4N^+I^-$,the quaternary ammonium salt a

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$(CH_3)_4N^{+}I^{-}$

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(A) The reaction involves a nucleophilic substitution $(S_N2)$ where $OH^-$ acts as the nucleophile.In $(CH_3)_4N^+I^-$,the quaternary ammonium salt acts as a good leaving group (trimethylamine,$(CH_3)_3N$) because the positive charge on the nitrogen atom makes the attached methyl group highly electrophilic.$(CH_3)_4N^+I^- + OH^- \rightarrow CH_3OH + (CH_3)_3N + I^-$.This reaction proceeds readily due to the high reactivity of the methyl group attached to the positively charged nitrogen atom.

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

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