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STD 11 - 9. Hydricarbon — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 9. Hydricarbon4 Marks
MCQ
The compound $\begin{array}{*{20}{c}}
  {C{H_{3\,\,}}\,\,\,\,\,\,\,} \\ 
  {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ 
  {C{H_3} - C = CH - C{H_3}} 
\end{array}$ on reaction with $NaI{O_4}$ in the presence of $KMn{O_4}$ gives
  • A
    $C{H_3}CHO + C{O_2}$
  • B
    $C{H_3}COC{H_3}$
  • C
    $C{H_3}COC{H_3} + C{H_3}CHO$
  • ✓
    $C{H_3}COC{H_3} + C{H_3}COOH$
✓

Answer

Correct option: D.
$C{H_3}COC{H_3} + C{H_3}COOH$
d
(d)$\begin{array}{*{20}{c}}
  {C{H_{3\,\,}}\,\,\,\,\,\,\,} \\ 
  {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ 
  {C{H_3} - C = CH - C{H_3}} 
\end{array}$ $\mathop {\xrightarrow{{NaI{O_4}}}}\limits_{KMn{O_4}} C{H_3}COC{H_3}$

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