MCQ
The compound that will react most readily with $ NaOH $ to form methanol is
- ✓${(C{H_3})_4}{N^ + }{I^ - }$
- B$C{H_3}OC{H_3}$
- C${(C{H_3})_3}{S^ + }{I^ - }$
- D${(C{H_3})_3}Cl$
The $CH _3$ group with positive $N$ atom is more electron deficient and to more reactive towards the nucleophile.
$\left( CH _3\right)_4 N ^{+} I ^{-}+ OH ^{-} \stackrel{ SN ^2}{\longrightarrow} CH _3 OH$
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$2$ $[image]$ + $\begin{array}{*{20}{c}}
{O\,\,\,\,\,\,} \\
{||\,\,\,\,\,\,} \\
{H - C - CC{l_3}}
\end{array}$ $\xrightarrow{{{H_2}S{O_4}}}$
The major product formed is