MCQ
Which one of the following complexes will have four isomer 
  • $[Co(en)(NH_3)_2Cl_2]Cl$
  • B
    $[Co(PPh_3)_2(NH_3)_2Cl_2]Cl$
  • C
    $[Co(en)_3]Cl_3$
  • D
    $[Co(en)_2Cl_2]Br$

Answer

Correct option: A.
$[Co(en)(NH_3)_2Cl_2]Cl$
a
The complex $\left[ Co ( en )\left( NH _3\right)_2 Cl _2\right] Cl$ has four isomers. In one isomer, two ammonia ligands are trans to each other. In the other isomer, the chlorine ligands are trans to each other. In the third isomer the ammonia ligands are cis to each other and chlorine ligands are also cis to each other. This is optically active and exists in $d$ and $l$ forms. Thus, total four isomers are possible.

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When carboxylic acid reacts with organolithium reagents to give ketones side  reaction sometimes occur. For example,

$\begin{matrix}
   \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{{H}_{3}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O  \\
   \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,||  \\
   HOC{{H}_{2}}C{{H}_{2}}CHC{{H}_{2}}C{{H}_{2}}COH  \\
\end{matrix}$ $\xrightarrow[tetrahydro\,\,futan ]{(x)\,\,C{{H}_{3}}Li}$ $\xrightarrow[{{H}_{2}}]{N{{H}_{4}}Cl}$ $\underset{Compound\,\,\,A\,\,63\,\%}{\mathop{\begin{matrix}
   \,\,\,\,\,\,\,\,\,\,\,\,C{{H}_{3}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O\,  \\
   \,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,||  \\
   HOC{{H}_{2}}C{{H}_{2}}CHC{{H}_{2}}C{{H}_{2}}-C-C{{H}_{3}}  \\
\end{matrix}}}\,$ $\underset{37\,\%}{\mathop{+\,\,Compound\,\,(B)}}\,$

Value of $(x)$ in above reaction is

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