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9. Hydricarbon — Chemistry STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceChemistry9. Hydricarbon1 Mark
MCQ
$\begin{matrix}
   C{{H}_{3\,\,\,\,\,\,\,\,\,\,\,}}  \\
   |\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,  \\
   C{{H}_{3}}-C=CH-C{{H}_{3}}  \\
\end{matrix}\xrightarrow{Hg{{(OAc)}_{2}}/EtOH}\underset{(90\%)}{\mathop{(A)}}\,$

Product $(A)$ of the above reaction is

  • A
    $\begin{array}{*{20}{c}}
      {C{H_{3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}} \\ 
      {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ 
      {C{H_3} - C - CH - C{H_3}} \\ 
      {|\,\,\,\,\,\,\,\,\,\,\,|\,\,\,} \\ 
      {\,cAOgH\,\,\,\,\,\,\,OEt\,\,\,\,\,\,\,\,\,\,} 
    \end{array}$
  • ✓
    $\begin{array}{*{20}{c}}
      {C{H_{3\,\,\,\,\,\,\,\,\,\,\,\,}}} \\ 
      {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ 
      {C{H_3} - C - CH - C{H_3}} \\ 
      {|\,\,\,\,\,\,\,\,\,\,\,|\,\,\,} \\ 
      {\,\,\,\,\,\,\,\,\,\,\,\,OEt\,\,\,HgOAc\,\,\,\,\,\,} 
    \end{array}$
  • C
    $\begin{array}{*{20}{c}}
      {C{H_{3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}} \\ 
      {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ 
      {C{H_3} - C - CH - C{H_3}} \\ 
      {|\,\,\,\,\,\,\,\,\,\,\,|\,\,\,} \\ 
      {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,OH\,\,\,\,\,\,\,HgOAc\,\,\,\,\,\,\,\,\,\,} 
    \end{array}$
  • D
    $\begin{array}{*{20}{c}}
      {C{H_{3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}} \\ 
      {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ 
      {C{H_3} - C - CH - C{H_3}} \\ 
      {|\,\,\,\,\,\,\,\,\,\,\,|\,\,\,} \\ 
      {\,cAOgH\,\,\,\,\,\,\,OH\,\,\,\,\,\,\,\,\,\,\,\,\,} 
    \end{array}$
✓

Answer

Correct option: B.
$\begin{array}{*{20}{c}}
  {C{H_{3\,\,\,\,\,\,\,\,\,\,\,\,}}} \\ 
  {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ 
  {C{H_3} - C - CH - C{H_3}} \\ 
  {|\,\,\,\,\,\,\,\,\,\,\,|\,\,\,} \\ 
  {\,\,\,\,\,\,\,\,\,\,\,\,OEt\,\,\,HgOAc\,\,\,\,\,\,} 
\end{array}$
b

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