- A${K_3}[Al{({C_2}{O_4})_3}]$
- B$[Pt{(en)_2}]C{l_2}$
- C$Ag{(N{H_3})_2}Cl$
- ✓${K_2}(Ni(EDTA)]$
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$CH_3 -CH = CH -CH( OH)-Me$ is :
$(1)$ $C{H_3}C{H_2}C{H_2}Cl + $ $\begin{array}{*{20}{c}}
{\,\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,\,\,\,C{H_3}} \\
{\,\,\,|}
\end{array}} \\
{C{H_3} - C - {O^ - }} \\
{\,\,\,|} \\
{\,\,\,\,\,\,\,\,\,\,C{H_3}}
\end{array}$ $(2)$ $C{H_3}C{H_2}C{H_2}I + $$\begin{array}{*{20}{c}}
{\,\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,\,\,\,C{H_3}} \\
{\,\,\,|}
\end{array}} \\
{C{H_3} - C - {O^ - }} \\
{\,\,\,|} \\
{\,\,\,\,\,\,\,\,\,\,C{H_3}}
\end{array}$
$(3)$ $\begin{array}{*{20}{c}}
{{H_3}C - CH - C{H_3}} \\
{|\,\,\,\,} \\
{Br\,\,\,}
\end{array}$ $+CH_3O^-$ $(4)$ $\begin{array}{*{20}{c}}
{{H_3}C - CH - C{H_3}} \\
{|\,\,\,\,} \\
{Br\,\,\,}
\end{array}$ $+CH_3S^-$