MCQ
Dilute aqueous $KMn{O_4}$, at room temperature reacts with ${\rm{ }}R - CH = CH - R$ to give
- A$R -CHO$
- B$R -COOH$
- ✓$RCHOH -CHOHR$
- D$C{O_2} + {H_2}O$
✓
Answer
Correct option: C.
$RCHOH -CHOHR$
c
(c) $R-CH=CH-R\underset{\text{room temp}\text{.}}{\mathop{\xrightarrow{\text{dil}\text{. aqueous }KMn{{O}_{4}}}}}\,\underset{(\text{Alcohol)}}{\mathop{R-\underset{\begin{smallmatrix}
| \\
OH
\end{smallmatrix}}{\mathop{CH}}\,-\underset{\begin{smallmatrix}
| \\
OH
\end{smallmatrix}}{\mathop{CH}}\,\,-R}}\,$
(c) $R-CH=CH-R\underset{\text{room temp}\text{.}}{\mathop{\xrightarrow{\text{dil}\text{. aqueous }KMn{{O}_{4}}}}}\,\underset{(\text{Alcohol)}}{\mathop{R-\underset{\begin{smallmatrix}
| \\
OH
\end{smallmatrix}}{\mathop{CH}}\,-\underset{\begin{smallmatrix}
| \\
OH
\end{smallmatrix}}{\mathop{CH}}\,\,-R}}\,$
$R-CH=CH-R\underset{\text{heat}}{\mathop{\xrightarrow{\text{Conc}\text{. }KMn{{O}_{4}}}}}\,R-COOH+R-COOH$
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