$MeOCH _{2} Cl$ $\xrightarrow{Slow}$$\underset{(i)}{\mathop{Me\underset{\,\,\centerdot \,\,\centerdot }{\overset{\centerdot \,\,\centerdot }{\mathop{O}}}\,\overset{+}{\mathop{C{{H}_{2}}}}\,+C{{l}^{-}}}}\,$ $\leftrightarrow $ $\underset{(ii)}{\mathop{Me-\underset{\,\,\centerdot \,\,\centerdot }{\overset{+}{\mathop{O}}}\,=C{{H}_{2}}}}\,$
Though $(ii)$ contains positive charge on oxygen, octet around each atom in $(ii)$ is complete, structure $(ii)$ is more stable than $(i)$.
$(a)$ $\begin{array}{*{20}{c}}
{C{H_3} - CH - Br} \\
{|\,} \\
{\,\,\,\,\,\,{C_2}{H_5}}
\end{array}$
$(b)$ $\begin{array}{*{20}{c}}
{\begin{array}{*{20}{c}}
{\,\,Br} \\
|
\end{array}} \\
{C{H_3} - C - C{H_3}} \\
| \\
{\,\,\,\,\,\,\,\,\,{C_2}{H_5}}
\end{array}$
$(c)$ $\begin{array}{*{20}{c}}
{C{H_3}{\mkern 1mu} - CH{\mkern 1mu} - {\mkern 1mu} {\mkern 1mu} C{H_2}Br} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{{C_2}{H_5}\,\,\,}
\end{array}$
$(i)\,\,(CH_3)_2CH - CH_2Br \xrightarrow{{{C_2}{H_5}OH}}$ $ (CH_3)_2CH - CH_2OC_2H_5 + HBr$
$(ii)\,\,(CH_3)_2CH - CH_2Br \xrightarrow{{{C_2}{H_5}O^-}} $ $(CH_3)_2CH - CH_2OC_2H_5 + Br^-$
પ્રકિયા ની પદ્ધતિ $(i)$ અને $(ii)$ અનુક્રમે શું હશે ?
$RCH _2 Br + I ^{-} \stackrel{\text { Acetone }}{\longrightarrow} \underset{\text { }}{ RCH _2 I + Br ^{-}}$
સાચું વિધાન શોધો.