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NEET · Chemistry · STD 11 - 9. Hydricarbon

An alkene \(A\) on reaction with \(\mathrm{O}_{3}\) and \(\mathrm{Zn}-\mathrm{H}_{2} \mathrm{O}\) gives propanone and ethanal in equimolar ratio. Addition of \(HCl\) to alkene \(A\) gives \(B\) as the major product. The structure of product \(B\) is:

  1. A \(Cl - C{H_2} - C{H_2} - \begin{array}{*{20}{c}}
      {C{H_3}} \\ 
      {|\,\,\,\,\,} \\ 
      {CH} \\ 
      {|\,\,\,\,\,} \\ 
      {C{H_3}} 
    \end{array}\)
  2. B \(\begin{array}{*{20}{c}}
      {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_2}Cl} \\ 
      {\,\,\,\,\,\,\,\,\,\,\,|} \\ 
      {C{H_3} - C{H_2} - CH - C{H_3}} 
    \end{array}\)
  3. C \(C{H_3} - C{H_2} - \begin{array}{*{20}{c}}
      {C{H_3}\,\,\,\,\,\,\,\,} \\ 
      {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ 
      {C - C{H_3}} \\ 
      {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ 
      {Cl\,\,\,\,\,\,\,\,\,\,\,\,\,\,} 
    \end{array}\)
  4. D \(\begin{array}{*{20}{c}}
      {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{H_3}C} \\ 
      {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|} \\ 
      {C{H_3} - CH - CH} \\ 
      {\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\;\;|} \\ 
      {\,\,\,\,\,\,\,Cl\,\,\,\;{H_3}C} 
    \end{array}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(C{H_3} - C{H_2} - \begin{array}{*{20}{c}}
  {C{H_3}\,\,\,\,\,\,\,\,} \\ 
  {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ 
  {C - C{H_3}} \\ 
  {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ 
  {Cl\,\,\,\,\,\,\,\,\,\,\,\,\,\,} 
\end{array}\)

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