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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:
- A \(Cl - C{H_2} - C{H_2} - \begin{array}{*{20}{c}}
{C{H_3}} \\
{|\,\,\,\,\,} \\
{CH} \\
{|\,\,\,\,\,} \\
{C{H_3}}
\end{array}\) - B \(\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_2}Cl} \\
{\,\,\,\,\,\,\,\,\,\,\,|} \\
{C{H_3} - C{H_2} - CH - C{H_3}}
\end{array}\) - C \(C{H_3} - C{H_2} - \begin{array}{*{20}{c}}
{C{H_3}\,\,\,\,\,\,\,\,} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C - C{H_3}} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{Cl\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}\) - D \(\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{H_3}C} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|} \\
{C{H_3} - CH - CH} \\
{\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\;\;|} \\
{\,\,\,\,\,\,\,Cl\,\,\,\;{H_3}C}
\end{array}\)
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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