- A[Ni(CN)4]2–
- B[Ni(NH3)2Cl2]
- C[Fe(NH3)4Cl2]+
- D[Co(NH3)4Cl2]+
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Kaushal's method : $\begin{array}{*{20}{c}}
{C{H_3} - CH - C{H_3}\xrightarrow{{KCl}}} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{OH\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}$
Preeti's method : $C{H_3} - CH = C{H_2}\xrightarrow{{C{l_2} + {H_2}O}}$
Raghav's method : $\begin{array}{*{20}{c}}
{C{H_3} - CH - C{H_3}\xrightarrow[{Pyridine}]{{SOC{l_2}}}} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{OH\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}$
[Given : atomic mass of $Cr =52$ $amu$ and $Cl =35\, amu ]$
$2$ $[image]$ + $\begin{array}{*{20}{c}}
{O\,\,\,\,\,\,} \\
{||\,\,\,\,\,\,} \\
{H - C - CC{l_3}}
\end{array}$ $\xrightarrow{{{H_2}S{O_4}}}$
The major product formed is