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NEET · Chemistry · STD 11 - 9. Hydricarbon

નીચેની પ્રક્રિયામાં મધ્યવર્તી \((A)\) અને નીપજ \((B)\) જણાવો. \({H_3}C - C \equiv CH\,\xrightarrow[{HgS{O_4}}]{{{H_2}O,\,{H_2}S{O_4}}}\,\mathop {intermediate}\limits_{(A)} \,\)\( \to \,\mathop {\Pr oduct}\limits_{(B)} \)

  1. A \(A \,: \, \begin{array}{*{20}{c}}   {{H_3}C - C = C{H_2}} \\    | \\    {OH}  \end{array}\)     \( B\,\, :\, \begin{array}{*{20}{c}}   {{H_3}C - C = C{H_2}} \\    | \\    {S{O_4}}  \end{array}\)
  2. B \(A\,\, :\,\begin{array}{*{20}{c}}   {{H_3}C - C - C{H_3}} \\    {|\,|} \\    O  \end{array}\)    \(B\,\,:\, H_3C-C=CH\)
  3. C \(A \,: \, \begin{array}{*{20}{c}}   {{H_3}C - C = C{H_2}} \\    | \\    {OH}  \end{array}\)     \(B\,\, :\,\begin{array}{*{20}{c}}   {{H_3}C - C - C{H_3}} \\    {|\,|} \\    O  \end{array}\)
  4. D \( A\,\, :\, \begin{array}{*{20}{c}}   {{H_3}C - C = C{H_2}} \\    | \\    {S{O_4}}  \end{array}\)     \(B\,\, :\,\begin{array}{*{20}{c}}   {{H_3}C - C - C{H_3}} \\    {|\,|} \\    O  \end{array}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(A \,: \, \begin{array}{*{20}{c}}   {{H_3}C - C = C{H_2}} \\    | \\    {OH}  \end{array}\)     \(B\,\, :\,\begin{array}{*{20}{c}}   {{H_3}C - C - C{H_3}} \\    {|\,|} \\    O  \end{array}\)

Step-by-step Solution

Detailed explanation

\(C{{H}_{3}}-C\equiv CH\xrightarrow[HgS{{O}_{4}}]{{{H}_{2}}O.{{H}_{2}}S{{O}_{4}}}\begin{matrix}
   C{{H}_{3}}-C=C{{H}_{2}}  \\
   |\,\,  \\
   \,\,\,OH  \\
\end{matrix}\) \(\xrightarrow{Tautomerism}\begin{matrix}
   CH3-C-C{{H}_{3}}  \\
   ||  \\
   O  \\
\end{matrix}\)
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